Question 15 Marks
Evaluate the following integrals:$\int_{\frac{\pi}{6}}^\limits{\frac{\pi}{3}}(\tan\text{x}+\cot\text{x})^2\text{dx}$
Answer$\int_{\frac{\pi}{6}}^\limits{\frac{\pi}{3}}(\tan\text{x}+\cot\text{x})^2\text{dx}$$=\int_{\frac{\pi}{6}}^\limits{\frac{\pi}{3}}\big(\tan^2\text{x}+\cot^2\text{x}+2\tan\text{x }\cot\text{x})\text{dx}$
$=\int_{\frac{\pi}{6}}^\limits{\frac{\pi}{3}}\big(\sec^2\text{x}-1+\text{cosec}^2\text{x}-1+2\big)\text{dx}$
$=\int_{\frac{\pi}{6}}^\limits{\frac{\pi}{3}}\sec^2\text{x dx}+\int_{\frac{\pi}{6}}^\limits{\frac{\pi}{3}}\text{cosec}^2\text{x dx}$
$=\big[\tan\text{x}+(-\cot\text{x})\big]^{\frac{\pi}{3}}_\frac{\pi}{6}$
$=\Big(\tan\frac{\pi}{3}-\tan\frac{\pi}{6}\Big)-\Big(\cot\frac{\pi}{3}-\cot\frac{\pi}{6}\Big)$
$=\Big(\sqrt{3}-\frac{1}{\sqrt{3}}\Big)-\Big(\frac{1}{\sqrt{3}}-\sqrt{3}\Big)$
$=2\sqrt{3}-\frac{2}{3}$
$=\frac{4}{\sqrt{3}}$
View full question & answer→Question 25 Marks
Evaluate the following integrals:$\int^\limits{1}_0\big(\cos^{-1}\text{x}\big)^2\text{dx}$
AnswerWe have,$\int^\limits{1}_0\big(\cos^{-1}\text{x}\big)^2\text{dx}=(\cos^{-1}\text{x}\big)^2\int\limits^1_0\text{dx}-\int\limits^1_0\big(\int\text{dx}\big)\frac{\text{d}\big(\cos^{-1}\text{x}\big)^2}{\text{dx}}\text{ dx}$
$=\Big[\text{x}\big(\cos^{-1}\text{x}\big)^2\Big]^1_0+\int^\limits{1}_0\frac{\text{x}\cdot2\cos^{-1}}{\sqrt{1-\text{x}^2}}\text{ dx}$
Now, Let $\cos^{-1}\text{x}=\text{t}\Rightarrow-\frac{1}{\sqrt{1-\text{x}^2}}\text{ dx}=\text{dt}$ When $\text{x}=0\Rightarrow\text{t}=\frac{\pi}{2}$$\text{x}=1\Rightarrow\text{t}=0$
$\therefore\ \int^\limits{1}_0\frac{2\text{ x}\cos^{-1}}{\sqrt{1-\text{x}^2}}\text{ dx}=-2\int^0_\limits{\frac{\pi}{2}}\text{t}\cos\text{t dt}=2\int^\limits{\frac{\pi}{2}}_0\text{t}\cos\text{t dt}$
$=2\Big[\text{t}\int\cos\text{t dt}-\int\big(\cos\text{t dt}\big)\frac{\text{dt}}{\text{dt}}\text{ dt}\Big]^{\frac{\pi}{2}}_0$
$=2\Big[\text{t}\sin\text{t}-\int\sin\text{t dt}\Big]^{\frac{\pi}{2}}_0$
$=2\Big[\text{t}\sin\text{t}+\cos\text{t}\Big]^{\frac{\pi}{2}}_0$
$=2\Big[\frac{\pi}{2}-1\Big]$
$\int^\limits{1}_0\big(\cos^{-1}\text{x}\big)^2\text{dx}=\Big[\text{x}\big(\cos^{-1}\text{x}\big)^2\Big]^1_0+\int^\limits{1}_0\frac{\text{x}\cdot2\cos^{-1}}{\sqrt{1-\text{x}^2}}\text{ dx}\\=\Big[\text{x}\big(\cos^{-1}\text{x}\big)^2\Big]^1_0+2\Big(\frac{\pi}{2}-1\Big)$
$=0-0+2\Big(\frac{\pi}{2}-1\Big)$
$=(\pi-2)$
View full question & answer→Question 35 Marks
Evaluate the following integrals as limit of sum:$\int\limits^\text{b}_{\text{a}}\cos\text{x dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=\text{a},\text{ b}=\text{b},\text{ f(x)}=\cos\text{x},\text{ h}=\frac{\text{b}-\text{a}}{\text{n}}$
Therefore, $\text{I}=\int\limits^\text{b}_{\text{a}}\cos\text{x dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\ ....\ +\text{f}\big\{\text{a}+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\cos(\text{a})+\cos(\text{a}+\text{h})+\ ....+\ \cos\big\{\text{a}+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Bigg[\frac{\cos\big\{\text{a}+(\text{n}-1)\frac{\text{h}}{2}\big\}\sin\frac{\text{nh}}{2}}{\sin\frac{\text{h}}{2}}\Bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\Bigg[\frac{\frac{\text{h}}{2}}{\sin\frac{\text{h}}{2}}2\cos\Big(\text{a}+\frac{\text{b}-\text{a}}{2}-\frac{\text{h}}{2}\Big)\sin\Big(\frac{\text{b}-\text{a}}{2}\Big)\Bigg]$ (Using nh = b - a)
$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{\text{h}}{2}}{\sin\frac{\text{h}}{2}}\times\lim\limits_{\text{h}\rightarrow0}2\cos\Big(\frac{\text{a}+\text{b}}{2}-\frac{\text{h}}{2}\Big)\sin\Big(\frac{\text{b}-\text{a}}{2}\Big)$
$=2\cos\Big(\frac{\text{a}+\text{b}}{2}\Big)\sin\Big(\frac{\text{b}-\text{a}}{2}\Big)$
$=\sin\text{b}-\sin\text{a}$ $\Big[\text{Since},2\cos\text{A}\sin\text{B}=\sin(\text{A}+\text{B})-\sin(\text{A}-\text{B})\Big]$
View full question & answer→Question 45 Marks
Evaluate the following integrals:$\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{1}{1+\sqrt{\tan\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{1}{1+\sqrt{\tan\text{x}}}\text{ dx}\ ....(\text{i})$$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{1}{1+\sqrt{\tan\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{1}{1+\sqrt{\cot\text{x}}}\text{ dx}\ ....(\text{ii})$
Adding (i) and (ii)$2\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\Big(\frac{1}{1+\sqrt{\tan\text{x}}}+\frac{1}{1+\sqrt{\cot\text{x}}}\Big)\text{dx}$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\bigg(\frac{1+\sqrt{\cot\text{x}}+1+\sqrt{\tan\text{x}}}{1+\sqrt{\cot\text{x}}+1+\sqrt{\tan\text{x}}+\sqrt{\tan\text{x}\cot\text{x}}}\bigg)\text{dx}$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{2+\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}{2+\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\text{dx}=\Big[\text{x}\Big]^{\frac{\pi}{3}}_\frac{\pi}{6}$
$=\frac{\pi}{3}-\frac{\pi}{6}$
$\therefore\ 2\text{I}=\frac{\pi}{6}$
Hence, $\text{I}=\frac{\pi}{12}$
View full question & answer→Question 55 Marks
Evaluate the following integrals as limit of sum:$\int\limits^2_{1}\text{x}^2\text{ dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=1,\text{ b}=2,\text{ f(x)}=\text{x}^2,\text{ h}=\frac{2-1}{\text{n}}=\frac{1}{\text{n}}$
Therefore, $\text{I}=\int\limits^2_{1}\text{x}^2\text{ dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(1)+\text{f}(1+\text{h})\ ....\ +\text{f}\big(1+(\text{n}-1)\text{h}\big)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(1)+(\text{h}+1)^2+\ ....\ +\big((\text{n}-1)\text{h}+1\big)^2\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+\text{h}^2\big\{1^2+2^2+3^2+\ ....\ +(\text{n}-1)^2\big\}\\+2\text{h}\big\{1+2+3+\ ....\ +(\text{n}-1)\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[\text{n}+\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}+2\text{h}\frac{\text{n}(\text{n}-1)}{2}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{1}{\text{n}}\bigg[\text{n}+\frac{(\text{n}-1)(2\text{n}-1)}{6\text{n}}+\text{n}-1\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\bigg\{2+\frac{1}{6}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)-\frac{1}{\text{n}}\bigg\}$
$=2+\frac{1}{3}$
$=\frac{7}{3}$
View full question & answer→Question 65 Marks
Evaluate the following integrals:$\int\limits^{\frac{\pi}{2}}_0\big(2\log\cos\text{x}-\log\sin2\text{x}\big)\text{dx}$
AnswerWe have,$\text{I}=\int\limits^{\frac{\pi}{2}}_0\big(2\log\cos\text{x}-\log\sin2\text{x}\big)\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\big(\log\cos^2\text{x}-\log\sin2\text{x}\big)\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\log\frac{\cos^2\text{x}}{\sin\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\log\frac{\cos^2\text{x}}{2\sin\text{x}\cdot\cos\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\log\frac{\cos\text{x}}{2\sin\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\big(\log\cos\text{x}-\log\sin\text{x}-\log2\big)\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\log\cos\text{x dx}-\int\limits^{\frac{\pi}{2}}_0\log\sin\text{x dx}-\int\limits^{\frac{\pi}{2}}_0\log2$
We know that$\int\limits^{\frac{\pi}{2}}_0\log\cos\text{x dx}=\int\limits^{\frac{\pi}{2}}_0\log\sin\text{x dx}\ ...(\text{i})$
Hence from equation (i)$\text{I}=-\int\limits^{\frac{\pi}{2}}_0\log2=-\frac{\pi}{2}\log2$
View full question & answer→Question 75 Marks
Evaluate the following definite integrals:$\int_{1}^\limits{4}\frac{\text{x}^2+\text{x}}{\sqrt{2\text{x}+1}}\text{ dx}$
AnswerLet $\text{I}=\int_{1}^\limits{4}\frac{\text{x}^2+\text{x}}{\sqrt{2\text{x}+1}}\text{ dx}$ Let $2\text{x}+1=\text{u}$$\Rightarrow\text{x}=\frac{\text{u}-1}{2}$
$\Rightarrow\text{dx}=\frac{\text{du}}{2}$
$\therefore\ \text{I}=\int\frac{\big(\frac{\text{u}-1}{2}\big)^2+\frac{\text{u}-1}{2}}{\sqrt{\text{u}}}\frac{\text{du}}{2}$
$\Rightarrow\text{I}=\frac{1}{8}\int\frac{\text{u}^2+1-2\text{u}+2\text{u}-2}{\sqrt{\text{u}}}\text{ du}$
$\Rightarrow\text{I}=\frac{1}{8}\int\frac{(\text{u}^2-1)}{\sqrt{\text{u}}}\text{ du}$
$\Rightarrow\text{I}=\frac{1}{8}\int\Big(\text{u}^{\frac{3}{2}}-\text{u}^{-\frac{1}{2}}\Big)\text{du}$
$\Rightarrow\text{I}=\frac{1}{8}\bigg[\frac{2\text{u}^{\frac{5}{2}}}{5}-\frac{2\text{u}^{\frac{1}{2}}}{1}\bigg]$
$\Rightarrow\text{I}=\frac{1}{8}\Big[\frac{2}{5}\times243-6-\frac{2}{5}\times9\sqrt{3}+2\sqrt{3}\Big]$
$\Rightarrow\text{I}=\frac{1}{8}\Big[\frac{456}{5}-\frac{8\sqrt{3}}{5}\Big]$
$\Rightarrow\text{I}=\frac{57-\sqrt{3}}{5}$
View full question & answer→Question 85 Marks
Evaluate the following integrals:$\int_{0}^\limits{\frac{\pi}{2}}\frac{\text{x}+\sin\text{x}}{1+\cos\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\text{x}+\sin\text{x}}{1+\cos\text{x}}\text{ dx}$$=\int_{0}^\limits{\frac{\pi}{2}}\frac{\text{x}+\sin\text{x}}{2\cos^2\frac{\text{x}}{2}}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\bigg[\frac{\text{x}}{2\cos^2\frac{\text{x}}{2}}+\frac{\sin\text{x}}{2\cos^2\frac{\text{x}}{2}}\bigg]\text{dx}$
$=\frac{1}{2}\int_{0}^\limits{\frac{\pi}{2}}\text{x}\sec^2\frac{\text{x}}{2}\text{ dx}+\int_{0}^\limits{\frac{\pi}{2}}\frac{2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}\text{ dx}$
$=\frac{1}{2}\bigg[\text{x}\frac{\tan\frac{\text{x}}{2}}{\frac{1}{2}}\bigg]^{\frac{\pi}{2}}_0-\frac{1}{2}\int_{0}^\limits{\frac{\pi}{2}}\frac{\tan\frac{\text{x}}{2}}{\frac{1}{2}}\text{ dx}+\int_{0}^\limits{\frac{\pi}{2}}\tan\frac{\text{x}}{2}\text{ dx}$
$=\Big[\text{x}\tan\frac{\text{x}}{2}\Big]^{\frac{\pi}{2}}_0-\int_{0}^\limits{\frac{\pi}{2}}\tan\frac{\text{x}}{2}\text{ dx}+\int_{0}^\limits{\frac{\pi}{2}}\tan\frac{\text{x}}{2}\text{ dx}$
$=\Big[\frac{\pi}{2}\tan\frac{\pi}{4}\Big]$
$=\frac{\pi}{2}\times1$
$=\frac{\pi}{2}$
View full question & answer→Question 95 Marks
Evaluate the following definite integrals:$\int_{0}^\limits{\frac{\pi}{4}}\sec\text{x}\text{ dx}$
AnswerLet $\int_{0}^\limits{\frac{\pi}{4}}\sec\text{x}\text{ dx}$ Then,$\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\sec\text{x}\frac{\sec\text{x}+\tan\text{x}}{\sec\text{x}+\tan\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\frac{\sec^2\text{x}+\sec\text{x}\tan\text{x}}{\sec\text{x}+\tan\text{x}}\text{ dx}$
Put $\text{u}=\sec\text{x}+\tan\text{x}$$\Rightarrow\text{du}=\sec^2\text{x}+\sec\text{x}\tan\text{x dx}$
$\therefore\ \int_{0}^\limits{\frac{\pi}{4}}\frac{\sec^2\text{x}+\sec\text{x}\tan\text{x}}{\sec\text{x}+\tan\text{x}}\text{ dx}=\int\frac{\text{du}}{\text{u}}$
$\Rightarrow\text{I}=\big[\log\text{u}\big]$
$\Rightarrow\text{I}=\big[\log(\sec\text{x}+\tan\text{x})\big]^{\frac{\pi}{4}}_0$
$\Rightarrow\text{I}=\log\Big(\sec\frac{\pi}{4}+\tan\frac{\pi}{4}\Big)-\log(\sec0+\tan0)$
$\Rightarrow\text{I}=\log\big(\sqrt{2}+1\big)-\log1$
$\Rightarrow\text{I}=\log\big(\sqrt{2}+1\big)$
View full question & answer→Question 105 Marks
Evaluate the following integrals as limit of sum:$\int\limits^{\frac{\pi}{2}}_{0}\sin\text{x dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=\frac{\pi}{2},\text{ f(x)}=\sin\text{x},\text{ h}=\frac{\frac{\pi}{2}-0}{\text{n}}=\frac{\pi}{2\text{n}}$
Therefore, $\text{I}=\int\limits^{\frac{\pi}{2}}_{0}\sin\text{x dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big(0+(\text{n}-1)\text{h}\big)\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\sin0+\sin\text{h}+\sin2\text{h}+\ ....+\ \sin(\text{n}-1)\text{h}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Bigg[\frac{\sin\big((\text{n}-1)\frac{\text{h}}{2}\big)\sin\frac{\text{nh}}{2}}{\sin\frac{\text{h}}{2}}\Bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\Bigg[\frac{\frac{\text{h}}{2}}{\sin\frac{\text{h}}{2}}\times2\sin\Big(\frac{\pi}{2}-\frac{\text{h}}{2}\Big)\sin\frac{\pi}{4}\Bigg]$ (Using nh = b - a)
$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{\text{h}}{2}}{\sin\frac{\text{h}}{2}}\times\lim\limits_{\text{h}\rightarrow0}2\sin\Big(\frac{\pi}{2}-\frac{\text{h}}{2}\Big)\sin\frac{\pi}{4}$
$=2\sin\frac{\pi}{4}\sin\frac{\pi}{4}$
$=2\times\frac{1}{\sqrt{2}}\times\frac{1}{\sqrt{2}}$
$=1$
View full question & answer→Question 115 Marks
Evaluate the following integrals:$\int\limits^\pi_0\Big(\frac{\text{x}}{1+\sin^2\text{x}}+\cos^7\text{x}\Big)\text{dx}$
AnswerLet $\text{I}=\int\limits^\pi_0\Big(\frac{\text{x}}{1+\sin^2\text{x}}+\cos^7\text{x}\Big)\text{dx}\ ....(\text{i})$ Then,$\text{I}=\int\limits^\pi_0\Big(\frac{\pi-\text{x}}{1+\sin^2(\pi-\text{x})}+\cos^7(\pi-\text{x})\Big)\text{dx}$
$=\int\limits^\pi_0\Big(\frac{\pi-\text{x}}{1+\sin^2\text{x}}+\cos^7\text{x}\Big)\text{dx}\ ...(\text{ii})$
Adding (i) and (ii) we get$2\text{I}=\int\limits^\pi_0\Big(\frac{\text{x}}{1+\sin^2\text{x}}+\cos^7\text{x}+\frac{\pi-\text{x}}{1+\sin^2\text{x}}-\cos^7\text{x}\Big)\text{dx}$
$\Rightarrow2\text{I}=\pi\int\limits^\pi_0\frac{1}{1+\sin^2\text{x}}\text{ dx}$
Dividing the numerator and denominator by $\cos^2\text{x},$ we get$2\text{I}=\pi\int\limits^\pi_0\frac{\sec^2\text{x}}{\sec^2\text{x}+\tan^2\text{x}}\text{ dx}$
$\Rightarrow2\text{I}=\pi\int\limits^\pi_0\frac{\sec^2\text{x}}{1+2\tan^2\text{x}}\text{ dx}$
$\Rightarrow2\text{I}=2\pi\int\limits^\pi_0\frac{\sec^2\text{x}}{1+2\tan^2\text{x}}\text{ dx}$ $\begin{bmatrix}\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=\begin{cases}\int\limits^{2\text{a}}_0\text{f(x)}\text{dx},&\text{if }\text{f}(2\text{a}-\text{x})=\text{f(x)}\\0,&\text{if }\text{f}(2\text{a}-\text{x})=\text{f(x)}\end{cases}\end{bmatrix}$
Put $\tan\text{x}=\text{z}$ Then $\sec^2\text{x dx}=\text{dz}$ When $\text{x}\rightarrow0,\text{ z}\rightarrow0$ When $\text{x}\rightarrow\frac{\pi}{2},\text{ z}\rightarrow\infty$$\therefore\ 2\text{I}=2\pi\int\limits^{\infty}_0\frac{\text{dz}}{1+\big(\sqrt{2}\text{z}\big)^2}$
$\Rightarrow2\text{I}=2\pi\times\bigg[\frac{\tan^{-1}\sqrt{2}\text{z}}{\sqrt{2}}\bigg]^{\infty}_0$
$\Rightarrow\text{I}=\frac{\pi}{\sqrt{2}}\Big(\tan^{-1}\infty-\tan^{-1}0\Big)$
$\Rightarrow\text{I}=\frac{\pi}{\sqrt{2}}\times\Big(\frac{\pi}{2}-0\Big)$
$\Rightarrow\text{I}=\frac{\pi^2}{2\sqrt{2}}$
View full question & answer→Question 125 Marks
Evaluate the following integrals:$\int\limits^1_{-1}|\text{x}\cos\pi\text{x}|\text{dx}$
AnswerLet $\text{I}=\int\limits^1_{-1}|\text{x}\cos\pi\text{x}|\text{dx}$ Consider $\text{f(x)}=|\text{x}\cos\pi\text{x}|$$\text{f}(-\text{x})=\big|(\text{x})\cos\pi(-\text{x})\big|=|-\text{x}\cos\pi\text{x}|=|\text{x}\cos\pi\text{x}|\text{f(x)}$
$\therefore\ \text{I}=\int\limits^1_{-1}|\text{x}\cos\pi\text{x}|\text{dx}$
$=2\int\limits^1_{-1}|\text{x}\cos\pi\text{x}|\text{dx}$ $\begin{bmatrix}\int\limits^{\text{a}}_{-\text{a}}\text{f(x)}\text{dx}=\begin{cases}2\int\limits^{\text{a}}_{0}\text{f(x)}\text{dx},&\text{ if }\text{f}(-\text{x})=\text{f(x)}\\0,&\text{ if }\text{f}(-\text{x})=-\text{f}(\text{x})\end{cases}\end{bmatrix}$
Now,$|\text{x}\cos\pi\text{x}|=\begin{cases}\text{x}\cos\pi\text{x},&\text{if }0\leq\text{x}\leq\frac{1}{2}\\-\text{x}\cos\pi\text{x},&\text{if }\frac{1}{2}\text{x}\leq1\end{cases}$
$\therefore\ \text{I}=2\Bigg[\int\limits^{{1}/{2}}_0\text{x}\cos\pi\text{x dx}+\int\limits^1_{1/2}(-\text{x}\cos\text{dx})\text{dx}\Bigg]$
$=2\Big(\frac{1}{2\pi}\sin\frac{\pi}{2}-0\Big)-\frac{2}{\pi}\times\Big[\Big(-\frac{\cos\pi\text{x}}{\pi}\Big)\Big]^{\frac{1}{2}}_0\\-2\Big(\frac{1}{\pi}\sin\pi-\frac{1}{2\pi}\sin\frac{\pi}{2}\Big)+\frac{2}{\pi}\times\Big(-\frac{\cos\pi\text{x}}{\pi}\Big)\Big]^1_{\frac{1}{2}}$
$=\frac{1}{\pi}+\frac{2}{\pi^2}\Big(\cos\frac{\pi}{2}-\cos0\Big)+\frac{1}{\pi}-\frac{2}{\pi}\Big(\cos\pi-\cos\frac{\pi}{2}\Big)$
$=\frac{1}{\pi}-\frac{2}{\pi^2}+\frac{1}{\pi}+\frac{2}{\pi^2}$
$=\frac{2}{\pi}$
View full question & answer→Question 135 Marks
Evaluate the following integrals:$\int\limits^{\text{a}}_{-\text{a}}\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)\text{d}\theta$
AnswerLet $\text{I}=\int\limits^{\text{a}}_{-\text{a}}\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)\text{d}\theta$ Consider $\text{f}(\theta)=\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)$$\text{f}(-\theta)=\log\Big(\frac{\text{a}-\sin(-\theta)}{\text{a}+\sin(-\theta)}\Big)$
$=\log\Big(\frac{\text{a}+\sin\theta}{\text{a}-\sin\theta}\Big)$ $[\sin(-\text{x})=-\sin\text{x}\big]$
$=\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)^{-1}$
$=-\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)$ $\Big[\log\text{a}^{\text{b}}=\text{b}\log\text{a}\Big]$
$=-\text{f}(\theta)$
$\therefore\ \text{f}(-\theta)=-\text{f}(\theta)$
$\Rightarrow\text{I}=\int\limits^{\text{a}}_{-\text{a}}\log\Big(\frac{\text{a}-\sin\theta}{\text{a}+\sin\theta}\Big)\text{d}\theta=0$ $\begin{bmatrix}\int\limits^{\text{a}}_{-\text{a}}\text{f(x)}\text{dx}=\begin{cases}2\int\limits^{\text{a}}_{0}\text{f(x)}\text{dx},&\text{ if }\text{f}(-\text{x})=\text{f(x)}\\0,&\text{ if }\text{f}(-\text{x})=-\text{f}(\text{x})\end{cases}\end{bmatrix}$
View full question & answer→Question 145 Marks
Evaluate the following integrals:$\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\cos^{2}\text{x}}{1+\text{e}^{\text{x}}}\text{ dx}$
AnswerWe know,$\int\limits_{\text{a}}^{\text{b}}\text{f}\text{(x)}\text{dx}=\int\limits_{\text{a}}^{\text{b}}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}$
Hence,
$\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{x}}{1+\text{e}^\text{x}}\text{dx}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{(-x)}}{1+\text{e}^\text{-x}}\text{dx}$
$\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{x}}{1+\text{e}^\text{x}}\text{dx}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{x}}{1+\text{e}^\text{-x}}\text{dx}$
If,
$\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{x}}{1+\text{e}^\text{x}}\text{dx}$
Then
$\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{x}}{1+\text{e}^\text{x}}\text{dx}$
So,
$2\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{x}}{1+\text{e}^\text{x}}+\frac{\cos^2\text{x}}{1+\text{e}^\text{-x}}\text{dx}$
$2\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{x}}{1+\text{e}^\text{x}}+\frac{\text{e}^\text{x}\cos^2\text{x}}{1+\text{e}^\text{-x}}\text{dx}$
$2\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{(1+\text{e}^\text{x})\cos^2\text{x}}{1+\text{e}^\text{x}}$
$2\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2\text{x}\text{dx}$
$2\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{1+\cos2\text{x}}{2}\text{dx}$
$\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{1}{4}\bigg\{\text{x}+\frac{\sin2\text{x}}{2}\bigg\}^\frac{\pi}{2}_{-\frac{\pi}{2}}$
$\text{I}=\frac{1}{4}\bigg\{\bigg(\frac{\pi}{2}\bigg)-\bigg(-\frac{\pi}{2}\bigg)\bigg\} $
$\text{I}=\frac{\pi}{4}$
View full question & answer→Question 155 Marks
Evaluate the following integrals as limit of sum:$\int\limits^2_{1}\big(\text{x}^2-1\big)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=1,\text{ b}=2,\text{ f(x)}=\text{x}^2-1,\text{ h}=\frac{2-1}{\text{n}}=\frac{1}{\text{n}}$
Therefore, $\text{I}=\int\limits^2_{1}\big(\text{x}^2-1\big)\text{ dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(1)+\text{f}(1+\text{h})+\ ....\ +\text{f}\big\{1+(\text{n}-1)\text{h}\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[(1-1)+(\text{h}^2-1)+\ ....\ +\big\{(\text{n}-1)^2\text{h}^2-1\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{n}-1+\text{h}^2\big\{1^2+2^2+3^2\ ....\ +(\text{n}-1)^2\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[\text{n}-1+\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{1}{\text{n}}\bigg[\text{n}-1+\frac{(\text{n}-1)(2\text{n}-1)}{6\text{n}}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\bigg\{1-\frac{1}{\text{n}}+\frac{1}{6}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)\bigg\}$
$=1+\frac{1}{3}$
$=\frac{4}{3}$
View full question & answer→Question 165 Marks
Evaluate the following integrals as limit of sum:$\int\limits^{4}_{0}\big(\text{x}+\text{e}^{2\text{x}}\big)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=4,\text{ f(x)}=\text{x}+\text{e}^{2\text{x}},\text{ h}=\frac{4-0}{\text{n}}=\frac{4}{\text{n}}$
Therefore, $\text{I}=\int\limits^{4}_{0}\big(\text{x}+\text{e}^{2\text{x}}\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big\{0+(\text{n}-1)\text{h}\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\Big[(0+\text{e}^0)+(\text{h}+\text{e}^{2\text{h}})+\ .....\ +\Big\{(\text{n}-1)\text{h}+\text{e}^{2(\text{n}-1)\text{h}}\Big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{h}\big\{1+2+\ ...+ (\text{n}-1)\text{h}\big\}+\text{e}^0+\text{e}^{2\text{h}}+\text{e}^{4\text{h}}+\ ....+\ \text{e}^{2(\text{n}-1)\text{h}}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[\text{h}\frac{\text{n}(\text{n}-1)}{2}+\frac{(\text{e}^{2\text{h}})^\text{n}-1}{\text{e}^{2\text{h}}-1}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{16}{\text{n}^2}\times\frac{\text{n}(\text{n}-1)}{2}+\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^8-1}{\frac{\text{e}^{2\text{h}}-1}{\text{h}}}$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{16}{\text{n}^2}\times\frac{\text{n}(\text{n}-1)}{2}+\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^8-1}{\frac{2(\text{e}^{2\text{h}}-1)}{\text{h}}}$
$=8+\frac{\text{e}^8-1}{2}$
$=\frac{15+\text{e}^8}{2}$
View full question & answer→Question 175 Marks
Evaluate the following integrals:$\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\tan^{2}\text{x}}{1+\text{e}^{\text{x}}}\text{ dx}$
AnswerWe now $\int_\limits{a}^{b}\text{f}\text{(x)}\text{dx}=\int_\limits{a}^{b} \text{f}(\text{a}+\text{b}-\text{x}) \text{dx}$
Hence,
$\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{{1}+\text{e}^\text{x}}\text{dx}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{(-x)}}{1-\text{e}^\text{-x}}\text{dx}$
$\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{{1}+\text{e}^\text{x}}\text{dx}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{1-\text{e}^\text{-x}}\text{dx}$
If,
$\text{I}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{{1}+\text{e}^\text{x}}\text{dx}$
Then
$\text{I}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{{1}+\text{e}^\text{-x}}\text{dx}$
So,
$2\text{I}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{{1}+\text{e}^\text{x}}+\frac{\tan^2\text{x}}{1+\text{e}^\text{-x}}\text{dx}$
$2\text{I}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{{1}+\text{e}^\text{x}}+\frac{\tan^2\text{x}}{1+\text{e}^\text{-x}}\text{dx}$
$2\text{I}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{{1}+\text{e}^\text{x}}+\frac{\text{e}^x\tan^2\text{x}}{1+\text{e}^\text{x}}\text{dx}$
$2\text{I}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{1+\text{e}^\text{x}}+\frac{\text{e}^\text{x}\tan^2\text{x}}{1+\text{e}^2}\text{dx}$
$2\text{I}=\int_\limits{\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}+\text{e}^\text{x}\tan^2\text{x}}{1+\text{e}^\text{x}}\text{dx}$
$2\text{I}=\int_\limits{\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{(1+\text{e}^\text{x})\tan^2\text{x}}{1+\text{e}^\text{x}}\text{dx}$
$2\text{I}=\int_\limits{\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}+\text{e}^\text{x}\tan^2\text{x}}{1+\text{e}^\text{x}}\text{dx}$
$2\text{I}=\int_\limits{\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{(1+\text{e}^\text{x})\tan^2\text{x}}{1+\text{e}^\text{x}}\text{dx}$
$2\text{I}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\tan^2\text{x}\text{dx}$.
$\text{I}=\frac{1}{2}\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\tan^2\text{x}\text{dx}$
We know
If f(x)is even
$\int_\limits{-a}^{a} \text{f}\text{(x)}\text{dx}=2\int_\limits{0}^{a}\text{f}\text{(x)}\text{dx}$
If f(x)is odd
$\int_\limits{-a}^{a} \text{f}\text{(x)}\text{dx}=0$
Here
$\text{f}\text{(x)}=\tan^2\text{x}$
f(x)is even,hence
$\text{I}=\int\limits_{0}^{\frac{\pi}{4}}\tan^2\text{x}\text{dx}$
$\text{I}=\int\limits_{0}^{\frac{\pi}{4}}\sec^2\text{x}-1\text{dx}$.
$\text{I}=\big\{\tan\text{x}-\text{x}\big\}\frac{\frac{\pi}{4}}{0 }$
$\text{I}=1-\frac{\pi}{4}$
View full question & answer→Question 185 Marks
Evaluate the following integrals as limit of sum:$\int\limits^{\frac{\pi}{2}}_{0}\cos\text{x dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=\frac{\pi}{2},\text{ f(x)}=\cos\text{x},\text{ h}=\frac{\frac{\pi}{2}-0}{\text{n}}=\frac{\pi}{2\text{n}}$
Therefore, $\text{I}=\int\limits^{\frac{\pi}{2}}_{0}\cos\text{x dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big(0+(\text{n}-1)\text{h}\big)\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\cos0+\cos\text{h}+\cos2\text{h}+\ ....+\ \cos(\text{n}-1)\text{h}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Bigg[\frac{\cos\big((\text{n}-1)\frac{\text{h}}{2}\big)\sin\frac{\text{nh}}{2}}{\sin\frac{\text{h}}{2}}\Bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\Bigg[\frac{\cos\big(\frac{\pi}{4}-\frac{\text{h}}{2}\Big)\sin\frac{\pi}{4}}{\sin\frac{\text{h}}{2}}\Bigg]$ $\Big(\text{Using, nh}=\frac{\pi}{2}\Big)$
$=\lim\limits_{\text{h}\rightarrow0}\Bigg[\frac{\frac{\text{h}}{2}}{\sin\frac{\text{h}}{2}}\times2\cos\Big(\frac{\pi}{4}-\frac{\text{h}}{2}\Big)\sin\frac{\pi}{4}\Bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{\text{h}}{2}}{\sin\frac{\text{h}}{2}}\times\lim\limits_{\text{h}\rightarrow0}2\cos\Big(\frac{\pi}{4}-\frac{\text{h}}{2}\Big)\sin\frac{\pi}{4}$
$=2\cos\frac{\pi}{4}\sin\frac{\pi}{4}$
$=2\times\frac{1}{\sqrt{2}}\times\frac{1}{\sqrt{2}}$
$=1$
View full question & answer→Question 195 Marks
Evaluate the following integrals:$\int\limits^{\pi}_0\frac{\text{x}}{1+\cos\alpha\sin\text{x}}\text{ dx},0<\alpha<\pi$
AnswerWe have,$\text{I}=\int\limits^{\pi}_0\frac{\text{x}}{1+\cos\alpha\sin\text{x}}\text{ dx}\ ...(\text{i})$
$=\int\limits^{\pi}_0\frac{\pi-\text{x}}{1+\cos\alpha\sin(\pi-\text{x})}\text{ dx}$
$=\int\limits^{\pi}_0\frac{\pi-\text{x}}{1+\cos\alpha\sin\text{x}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get,$2\text{I}=\int\limits^{\pi}_0\frac{\text{x}+\pi-\text{x}}{1+\cos\alpha\sin\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\frac{\pi}{2}\int\limits^{\pi}_0\frac{1}{1+\cos\alpha\sin\text{x}}\text{ dx}$
$=\frac{\pi}{2}\int\limits^{\pi}_0\frac{1}{1+\cos\alpha\sin\text{x}}$
$=\frac{\pi}{2}\int\limits^{\pi}_0\frac{1}{1+\cos\alpha\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}}\text{ dx}$
$=\frac{\pi}{2}\int\limits^{\pi}_0\frac{1+\tan^{2}\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}+2\cos\alpha\tan\frac{\text{x}}{2}}\text{ dx}$
$=\frac{\pi}{2}\int\limits^{\pi}_0\frac{\sec^{2}\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}+2\cos\alpha\tan\frac{\text{x}}{2}}\text{ dx}$
Putting $\tan\frac{\text{x}}{2}=\text{t}$$\frac{1}{2}\sec^2\text{x dx}=\text{dt}$
When $\text{x}\rightarrow0;\text{t}\rightarrow0$ and $\text{x}\rightarrow\pi;\text{t}\rightarrow\infty$$\therefore\ \text{I}=\frac{\pi}{2}\int\limits^{\infty}_0\frac{2}{1+\text{t}^2+2\cos\alpha\text{t}}\text{ dt}$
$=\frac{\pi}{2}\int\limits^{\infty}_0\frac{2}{(\text{t}+\cos\alpha)^2-\cos^2\alpha+1}\text{ dt}$
$={\pi}\int\limits^{\infty}_0\frac{1}{(\text{t}+\cos\alpha)^2+\sin^2\alpha}\text{ dt}$
$=\pi\bigg[\frac{1}{\sin\alpha}\tan^{-1}\Big(\frac{1+\cos\alpha}{\sin\alpha}\Big)\bigg]^1_0$
$=\frac{\pi}{\sin\alpha}\Big[\tan^{-1}(\infty)-\tan^{-1}(\cot\alpha)\Big]$
$=\frac{\pi}{\sin\alpha}\bigg[\frac{\pi}{2}-\tan^{-1}\Big(\tan\Big(\frac{\pi}{2}-\alpha\Big)\Big)\bigg]$
$=\frac{\pi\alpha}{\sin\alpha}$
View full question & answer→Question 205 Marks
Evaluate the following integrals as limit of sum:$\int\limits^{2}_{0}\big(\text{x}^2+\text{x}\big)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=2,\text{ f(x)}=\text{x}^2+\text{x},\text{ h}=\frac{2-0}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^{2}_{0}\big(\text{x}^2+\text{x}\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big\{0+(\text{n}-1)\text{h}\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[(0+0)+(\text{h}^2+\text{h})+\ ....\ +\big\{(\text{n}-1)^2\text{h}^2+\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{h}^2\big(1^2+2^2+3^2\ ....\ +(\text{n}-1)^2\big)+\\\text{h}\big\{1+2+3+\ ...\ +(\text{n}-1)\text{h}\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}+\text{h}\frac{\text{n}(\text{n}-1)}{2}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2}{\text{n}}\Big[\frac{2\text{n}(\text{n}-1)(2\text{n}-1)}{3\text{n}}+\text{n}-1\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\Big[\frac{2}{3}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)+1-\frac{1}{\text{n}}\Big]$
$=\frac{8}{3}+2$
$=\frac{14}{3}$
View full question & answer→Question 215 Marks
Evaluate the following integrals:$\int\limits^{\pi}_2\log(1-\cos\text{x})\text{dx}$
AnswerLet $\text{I}=\int\limits^{\pi}_2\log(1-\cos\text{x})\text{dx}$$=\int\limits^{\pi}_2\Big(2\sin^2\frac{\text{x}}{2}\Big)\text{dx}$
$=\int\limits^{\pi}_2\log2\text{ dx}+\int\limits^{\pi}_2\log\sin\frac{\text{x}}{2}\text{ dx}$
Let $\text{t}=\frac{\text{x}}{2}$ in these cong integral then $\text{dt}=\frac{1}{2}\text{ dx}$ When $\text{x}\rightarrow0;\text{t}\rightarrow0$ and $\text{x}\rightarrow\pi;\text{t}\rightarrow\frac{\pi}{2}$$\text{I}=\log2\big[\text{x}\big]^{\pi}_0+4\int\limits^{\frac{\pi}{2}}_0\log\sin\text{t dt}$
$=\pi\log2+4\times\Big(-\frac{\pi}{2}\log2\Big)$ $\Bigg[\text{Where,}\int\limits^{\frac{\pi}{2}}_0\log\sin\text{t dt}=-\frac{\pi}{2}\log2\Bigg]$
$=-\pi\log2$
View full question & answer→Question 225 Marks
Evaluate the following integrals:$\int^\limits{\frac{\pi}{2}}_{0}\frac{\cos\text{x}}{\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)^\text{n}}\text{ dx}$
AnswerLet $\text{I}=\int^\limits{\frac{\pi}{2}}_{0}\frac{\cos\text{x}}{\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)^\text{n}}\text{ dx}$$=\int^\limits{\frac{\pi}{2}}_{0}\frac{\cos^2\frac{\text{x}}{2}-\sin^2\frac{\text{x}}{2}}{\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)^\text{n}}\text{ dx}$
$=\int^\limits{\frac{\pi}{2}}_{0}\frac{\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)}{\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)^{\text{n}}}\text{ dx}$
$=\int^\limits{\frac{\pi}{2}}_{0}\frac{\big(\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{2}\big)}{\big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\big)^{\text{n}-1}}\text{ dx}$
Put $\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}=\text{z}$$\therefore\ \Big(-\sin\frac{\text{x}}{2}\times\frac{1}{2}+\cos\frac{\text{x}}{2}\times\frac{1}{2}\Big)\text{dx}=\text{dz}$
$\Big(\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{2}\Big)\text{dx}=2\text{dz}$
When $\text{x}\rightarrow0,\text{ z}\rightarrow1$ When $\text{x}\rightarrow\frac{\pi}{2},\text{ z}\rightarrow\sqrt{2}$ $\Big(\text{z}=\cos\frac{\pi}{4}+\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\Big)$$\therefore\ \text{I}=2\int^\limits{\sqrt{2}}_{1}\frac{\text{dz}}{\text{z}^{\text{n}-1}}$
$=2\times\Big[\frac{\text{z}^{\text{n}-1}}{2-\text{n}}\Big]^{\sqrt{2}}_1$
$=\frac{2}{(2-\text{n})}\Big[\big(\sqrt{2}\big)^{2-\text{n}}-1\Big]$
$=\frac{2}{(2-\text{n})}\bigg(2^{\frac{2}{(2-\text{n})}}-1\bigg)$
$=\frac{2}{(2-\text{n})}\Big(2^{1-\frac{\text{n}}{2}}-1\Big)$
View full question & answer→Question 235 Marks
Evaluate the following integrals:$\int\limits^{\infty}_0\frac{\text{x}}{(1+\text{x})(1+\text{x}^2)}\text{ dx}$
AnswerWe have$\text{I}=\int\limits^{\infty}_0\frac{\text{x}}{(1+\text{x})(1+\text{x}^2)}\text{ dx}$
Putting $\text{x}=\tan\theta$$\text{dx}=\sec^2\theta\text{ d}\theta$
When $\text{x}\rightarrow0;\theta\rightarrow0$ and $\text{x}\rightarrow\infty;\theta\rightarrow\frac{\pi}{2}$ Now, integral becomes,$\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\tan\theta}{(1+\tan\theta)\sec^2\theta }\sec^2\theta\text{ d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\tan\theta}{1+\tan\theta}\text{ d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\frac{\sin\theta}{\cos\theta}}{1+\frac{\sin\theta}{\cos\theta}}\text{ d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sin\theta}{\sin\theta+\cos\theta}\text{ d}\theta\ ....(\text{i})$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sin\big(\frac{\pi}{2}-\theta\big)}{\sin\big(\frac{\pi}{2}-\theta\big)+\cos\big(\frac{\pi}{2}-\theta\big)}\text{ d}\theta$ $\Bigg[\because\ \int\limits^{\text{a}}_{0}\text{f(x)}\text{dx}=\int\limits^{\text{a}}_{\text{0}}\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\cos\theta}{\cos\theta+\sin\theta}\text{ d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\cos\theta}{\sin\theta+\cos\theta}\text{ d}\theta\ ....(\text{ii})$
Adding (i) and (ii) we get$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sin\theta+\cos\theta}{\sin\theta+\cos\theta}\text{ d}\theta$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\text{d}\theta$
$\Rightarrow2\text{I}=\frac{\pi}{2}$
$\Rightarrow\text{I}=\frac{\pi}{4}$
$\therefore\ \int\limits^{\infty}_0\frac{\text{x}}{(1+\text{x})(1+\text{x}^2)}\text{ dx}=\frac{\pi}{4}$
View full question & answer→Question 245 Marks
Evaluate the following integrals:$\int^\limits{\text{a}}_0\sin{-1}{\sqrt\frac{\text{x}}{\text{a}+\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int^\limits{\text{a}}_0\sin{-1}{\sqrt\frac{\text{x}}{\text{a}+\text{x}}}\text{ dx}$ Let $\text{x}=\text{x}\tan^{2}\theta\Rightarrow\theta=\tan^{-1}\sqrt{\frac{\text{x}}{\text{a}}}$ When, $\text{x}\rightarrow\text{x};\ \theta\rightarrow0$ and $\text{x}\rightarrow\text{a};\ \theta\rightarrow\frac{\pi}{4}$ and $\text{dx}=2\text{a}\tan\theta\ \sec^2\theta\text{ d}\theta$ Then,$\text{I}=\int^\limits{\frac{\pi}{4}}_0\sin^{-1}\sqrt{\frac{\text{a}\tan^{2}\theta}{\text{a}+\text{a}\tan^{2}\theta}}2\text{a}\tan\theta\sec^2\theta\text{ d}\theta$
$\Rightarrow\text{I}=2\text{a}\int^\limits{\frac{\pi}{4}}_0\sin^{-1}(\sin\theta)\tan\theta\sec^2\theta\text{ d}\theta$
$\text{I}=\int^\limits{\frac{\pi}{4}}_0\theta\tan\theta\sec^2\theta\text{ d}\theta$
Let $\tan\theta=\text{t}\Rightarrow\theta=\tan^{-1}\text{t}$$\Rightarrow\sec^2\theta\text{ d}\theta=\text{dt}$
When, $\theta\rightarrow0;\text{ t}\rightarrow0$ and $\theta\rightarrow\frac{\pi}{4};\text{ t}\rightarrow1$ Then, $\text{I}=2\text{a}\int^\limits1_0\tan^{-1}\text{t dt}$$\text{I}=2\text{a}\int^\limits1_0\tan^{-1}\text{t dt}$
View full question & answer→Question 255 Marks
Evaluate the following integrals:$\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\text{x}^{11}-3\text{x}^9+5\text{x}^7-\text{x}^5+1}{\cos^2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\text{x}^{11}-3\text{x}^9+5\text{x}^7-\text{x}^5+1}{\cos^2\text{x}}\text{ dx}$$=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\text{x}^{11}-3\text{x}^9+5\text{x}^7-\text{x}^5}{\cos^2\text{x}}\text{ dx}+\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{1}{\cos^2\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\text{x}^{11}-3\text{x}^9+5\text{x}^7-\text{x}^5}{\cos^2\text{x}}\text{ dx}+\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\sec^2\text{x}\text{ dx}$
$=\text{I}_1+\text{I}_2$
Now,
Consider, $\text{f(x)}=\frac{\text{x}^{11}-3\text{x}^9+5\text{x}^7-\text{x}^5}{\cos^2\text{x}}\text{ dx}$
$\therefore\ \text{f}(-\text{x})=\frac{(-\text{x})^{11}-3(-\text{x})^9+5(-\text{x})^7-(-\text{x})^5}{\cos^2(-\text{x})}$
$=\frac{-\text{x}^{11}+3\text{x}^9-5\text{x}^7+\text{x}^5}{\cos^2\text{x}}$
$=-\frac{\text{x}^{11}-3\text{x}^9+5\text{x}^7-\text{x}^5}{\cos^2\text{x}}=\text{f(x)}$
$\Rightarrow\text{I}_1=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\text{x}^{11}-3\text{x}^9+5\text{x}^7-\text{x}^5}{\cos^2\text{x}}\text{ dx}=0$ $\begin{bmatrix}\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\begin{cases}\int\limits^{\text{a}}_0\text{f(x)}\text{dx},&\text{ if }\text{ f}(-\text{x})=\text{f(x)}\\0,&\text{ if }\text{ f}(-\text{x})=-\text{f(x)}\end{cases}\end{bmatrix}$
$\text{g(x)}=\sec^2\text{x}$
Let $\text{g}(-\text{x})=\sec^2(-\text{x})=\sec^2\text{x}=\text{g}(\text{x})$
$\Rightarrow\text{I}_1=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\sec^2\text{x dx}$
$=2\int\limits^{\frac{\pi}{4}}_0\sec^2\text{x dx}$ $\begin{bmatrix}\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\begin{cases}\int\limits^{\text{a}}_0\text{f(x)}\text{dx},&\text{ if }\text{ f}(-\text{x})=\text{f(x)}\\0,&\text{ if }\text{ f}(-\text{x})=-\text{f(x)}\end{cases}\end{bmatrix}$
$=2\times\Big[\tan\text{x}\Big]^{\frac{\pi}{4}}_0$
$=2\Big(\tan\frac{\pi}{4}-\tan0\Big)$
$=2\times(1-0)$
$=2$
$\therefore\ \text{I}=\text{I}_1+\text{I}_2$
$\text{I}=0+2$
$\text{I}=2$
View full question & answer→Question 265 Marks
Evaluate the following integrals:$\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{5\cos\text{x}+3\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{5\cos\text{x}+3\sin\text{x}}\text{ dx}$ Then,$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{5\Bigg(\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)+3\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)}\text{ dx}$ $\Bigg[\because\sin\text{A}=\bigg(\frac{2\tan\frac{\text{A}}{2}}{1+\tan^2\frac{\text{A}}{2}}\bigg),\cos\text{A}=\bigg(\frac{1-\tan^2\frac{\text{A}}{2}}{1+\tan^2\frac{\text{A}}{2}}\bigg)\Bigg]$
$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{1+\tan^2\frac{\text{x}}{2}}{5-5\tan^2\frac{\text{x}}{2}+6\tan\frac{\text{x}}{2}}\text{ dx}$
$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\sec^2\frac{\text{x}}{2}}{5-5\tan^2\frac{\text{x}}{2}+6\tan\frac{\text{x}}{2}}\text{ dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$ Then, $\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{ dx}=\text{dt}$ Also, $\text{x}=0,\text{t}=0$ and $\text{x}=\frac{\pi}{2}, \text{t}=1$$\therefore\ \text{I}=\int_{0}^\limits{1}\frac{2\text{ dt}}{5-5\text{t}^2+6\text{t}}$
$\text{I}=\frac{1}{5}\int_{0}^\limits{1}\frac{2\text{dt}}{1-\text{t}^2+\frac{6}{5}\text{t}+\frac{36}{100}-\frac{36}{100}}$
$=\frac{2}{5}\int_{0}^\limits{1}\frac{\text{dt}}{-\big(\text{t}-\frac{6}{10}\big)^2+\frac{136}{100}}$
$=\frac{2}{5}\times\frac{10}{\sqrt{136}}\Bigg[-\log\Bigg(\frac{\text{t}-\frac{6}{10}-\frac{\sqrt{136}}{10}}{\text{t}-\frac{6}{10}+\frac{\sqrt{136}}{10}}\Bigg)\Bigg]^1_0$
$=\frac{1}{\sqrt{34}}\bigg[-\log\bigg(\frac{4-2\sqrt{34}}{4+2\sqrt{34}}\bigg)+\log\bigg(\frac{-6-2\sqrt{34}}{-6+2\sqrt{34}}\bigg)\bigg]$
$=\frac{1}{\sqrt{34}}\log\bigg(\frac{6+2\sqrt{34}}{6-2\sqrt{34}}\times\frac{4+2\sqrt{34}}{4-2\sqrt{34}}\bigg)$
$=\frac{1}{\sqrt{34}}\log\bigg(\frac{160+20\sqrt{34}}{160-20\sqrt{34}}\bigg)$
$=\frac{1}{\sqrt{34}}\log\bigg(\frac{8+\sqrt{34}}{8-\sqrt{34}}\bigg)$
View full question & answer→Question 275 Marks
Evaluate the following integrals as limit of sum:$\int\limits^{2}_{0}\big(\text{x}^2+2\text{x}+1\big)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=2,\text{ f(x)}=\text{x}^2+2\text{x}+1,\text{ h}=\frac{2-0}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^{2}_{0}\big(\text{x}^2+2\text{x}+1\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big\{0+(\text{n}-1)\text{h}\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[(0+0+1)+(\text{h}^2+2\text{h}+1)+\\\ ....\ +\big\{(\text{n}-1)^2\text{h}^2+2(\text{n}-1)\text{h}+1\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+\text{h}^2\big(1^2+2^2+3^2\ ....\ +(\text{n}-1)^2\big)+\\2\text{h}\big\{1+2+\ ...\ +(\text{n}-1)\text{h}\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[\text{n}+\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}+2\text{h}\frac{\text{n}(\text{n}-1)}{2}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2}{\text{n}}\Big[\text{n}+\frac{2\text{n}(\text{n}-1)(2\text{n}-1)}{3\text{n}}+2\text{n}-2\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\Big\{{3}+\frac{2}{3}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)-\frac{2}{\text{n}}\Big\}$
$=6+\frac{8}{3}$
$=\frac{26}{3}$
View full question & answer→Question 285 Marks
Evaluate the following integrals:$\int\limits^2_{-1}\big(|\text{x}+1|+|\text{x}|+|\text{x}-1|\big)\text{dx}$
AnswerWe know that,$\Rightarrow|\text{x}+1|=\begin{cases}\text{x}+1,&\text{ if }\text{ x}+1\geq0\\-(\text{x}+1),&\text{ if }\text{ x}+1<0\end{cases}\\=\begin{cases}\text{x}+1,&\text{ if }\text{ x}\geq-1\\-(\text{x}+1),&\text{ if }\text{ x}<-1\end{cases}$
$\Rightarrow|\text{x}|=\begin{cases}\text{x},&\text{ if }\text{ x}\geq0\\-\text{x},&\text{ if }\text{ x}<0\end{cases}$
$\Rightarrow|\text{x}-1|=\begin{cases}\text{x}-1,&\text{ if }\text{ x}-1\geq0\\-(\text{x}+1),&\text{ if }\text{ x}-1<0\end{cases}\\=\begin{cases}\text{x}-1,&\text{ if }\text{ x}\geq1\\-(\text{x}-1),&\text{ if }\text{ x}<1\end{cases}$
When $-1\leq\text{x}\leq0,$$\Rightarrow|\text{x}+1|+|\text{x}|+|\text{x}-1|\\=\text{x}+1+(-\text{x})+\big[-(\text{x}-1)\big]\\=2-\text{x}$
When $0\leq\text{x}\leq1,$$\Rightarrow|\text{x}+1|+|\text{x}|+|\text{x}-1|\\=\text{x}+1+\text{x}+\big[-(\text{x}-1)\big]\\=\text{x}+2$
When $1\leq\text{x}\leq2,$$\Rightarrow|\text{x}+1|+|\text{x}|+|\text{x}-1|\\=\text{x}+1+\text{x}+\text{x}-1\\=3\text{x}$
$\therefore\ \int\limits^2_{-1}\big(|\text{x}+1|+|\text{x}|+|\text{x}-1|\big)\text{dx}$
$=\int\limits^0_{-1}(2-\text{x})\text{dx}+\int\limits^1_{0}(\text{x}+2)\text{dx}+\int\limits^2_{1}3\text{x dx}$
$=\bigg[\frac{(2-\text{x})^2}{2\times(-1)}\bigg]^0_{-1}+\bigg[\frac{(\text{x}+2)^2}{2}\bigg]^1_0+3\times\Big[\frac{\text{x}^2}{2}\Big]^2_1$
$=-\frac{1}{2}(4-9)+\frac{1}{2}(9-4)+\frac{3}{2}(4-1)$
$=\frac{5}{2}+\frac{5}{2}+\frac{9}{2}$
$=\frac{19}{2}$
View full question & answer→Question 295 Marks
Evaluate the following integrals:$\int\limits^2_{-2}\frac{3\text{x}^3+2|\text{x}|+1}{\text{x}^2+|\text{x}|+1}\text{ dx}$
AnswerLet $\text{I}=\int\limits^2_{-2}\frac{3\text{x}^3+2|\text{x}|+1}{\text{x}^2+|\text{x}|+1}\text{ dx}$$=\int\limits^2_{-2}\frac{3\text{x}^3}{\text{x}^2+|\text{x}|+1}\text{ dx}+\int\limits^2_{-2}\frac{2|\text{x}|+1}{\text{x}^2+|\text{x}|+1}\text{ dx}$
$=\text{I}_1+\text{I}_2$
Consider $\text{f(x)}=\int\limits^2_{-2}\frac{3\text{x}^3}{\text{x}^2+|\text{x}|+1}\text{ dx}$$\text{f}(-\text{x})=\int\limits^2_{-2}\frac{3(-\text{x}^3)}{(-\text{x}^2)+|-\text{x}|+1}\text{ dx}=\int\limits^2_{-2}\frac{-3\text{x}^3}{\text{x}^2+|\text{x}|+1}\text{ dx}=-\int\limits^2_{-2}\frac{3\text{x}^3}{\text{x}^2+|\text{x}|+1}\text{ dx}=-\text{f(x)}$
$\therefore\ \text{I}_1=\int\limits^2_{-2}\frac{3\text{x}^3}{\text{x}^2+|\text{x}|+1}\text{ dx}=0$ $\begin{bmatrix}\int\limits^{\text{a}}_{-\text{a}}\text{f(x)}\text{dx}=\begin{cases}2\int\limits^{\text{a}}_{0}\text{f(x)}\text{dx},&\text{ if }\text{f}(-\text{x})=\text{f(x)}\\0,&\text{ if }\text{f}(-\text{x})=-\text{f}(\text{x})\end{cases}\end{bmatrix}$
Now, Consider$\text{g(x)}=\int\limits^{2}_{-2}\frac{2|\text{x}|+1}{\text{x}^2+|\text{x}|+1}\text{ dx}$
$\text{g}(-\text{x})=\int\limits^{2}_{-2}\frac{2|-\text{x}|+1}{(-\text{x}^2)+|-\text{x}|+1}\text{ dx}=\text{g(x)}=\int\limits^{2}_{-2}\frac{2|\text{x}|+1}{\text{x}^2+|\text{x}|+1}\text{ dx}=\text{g}(\text{x)}$
$\therefore\ \text{I}_2=\int\limits^{2}_{-2}\frac{2|\text{x}|+1}{\text{x}^2+|\text{x}|+1}\text{ dx}$
$=2\int\limits^{2}_{0}\frac{2|\text{x}|+1}{\text{x}^2+|\text{x}|+1}\text{ dx}$ $\begin{bmatrix}\int\limits^{\text{a}}_{-\text{a}}\text{f(x)}\text{dx}=\begin{cases}2\int\limits^{\text{a}}_{0}\text{f(x)}\text{dx},&\text{ if }\text{f}(-\text{x})=\text{f(x)}\\0,&\text{ if }\text{f}(-\text{x})=-\text{f}(\text{x})\end{cases}\end{bmatrix}$
$=2\int\limits^{2}_{0}\frac{2\text{x}+1}{\text{x}^2+\text{x}+1}\text{ dx}$ $\begin{bmatrix}|\text{x}|\begin{cases}\text{x},&\text{x}\geq0\\-\text{x},&\text{x}<0\end{cases}\end{bmatrix}$
$=2\times\Big[\log(\text{x}^2+\text{x}+1)\Big]^2_0$
$=2\times\big(\log7-\log1\big)$
$=2\times\big(7-0\big)$
$=2\log7$
$\therefore\ \text{I}=\text{I}_1+\text{I}_2$
$=0+2\log7$
$=2\log7$
View full question & answer→Question 305 Marks
Evaluate the following integrals:$\int\limits^{\frac{\pi}{2}}_0\frac{\text{x}\sin\text{x}\cos\text{x}}{\sin^4\text{x}+\cos^4\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\text{x}\sin\text{x}\cos\text{x}}{\sin^4\text{x}+\cos^4\text{x}}\text{ dx}\ ...(\text{i})$$=\int\limits^{\frac{\pi}{2}}_0\frac{\big(\frac{\pi}{2}-\text{x}\big)\sin\big(\frac{\pi}{2}-\text{x}\big)\cos\big(\frac{\pi}{2}-\text{x}\big)}{\sin^4\big(\frac{\pi}{2}-\text{x}\big)+\cos^4\big(\frac{\pi}{2}-\text{x}\big)}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\big(\frac{\pi}{2}-\text{x}\big)\cos\text{x}\sin\text{x}}{\cos^4\text{x}+\sin^4\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\big(\frac{\pi}{2}-\text{x}\big)\sin\text{x}\cos\text{x}}{\sin^4\text{x}+\cos^4\text{x}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\Big(\text{x}+\frac{\pi}{2}-\text{x}\Big)\frac{\sin\text{x}\cos\text{x}}{\sin^4\text{x}+\cos^4\text{x}}\text{ dx}$
$=\frac{\pi}{2}\int\limits^{\frac{\pi}{2}}_0\frac{\sin\text{x}\cos\text{x}}{\sin^4\text{x}+\cos^4\text{x}}\text{ dx}$
Let $\sin^2\text{x}=\text{t},$ Then $2\sin\text{x}\cos\text{x dx}=\text{dt}$ When $\text{x}=0,\text{t}=0,\text{x}=\frac{\pi}{2},\text{t}=1$ Therefore,$2\text{I}=\frac{\pi}{4}\int\limits^1_0\frac{\text{dt}}{\text{t}^2+(1-\text{t}^2)}$
$=\frac{\pi}{8}\int\limits^1_0\frac{\text{dt}}{\big(\text{t}-\frac{1}{2}\big)^2+\frac{1}{4}}$
$=\frac{\pi}{8}\times2\Big[\tan^{-1}(2\text{t}-1)\Big]^1_0$
$=\frac{\pi}{4}\Big(\frac{\pi}{4}+\frac{\pi}{4}\Big)$
Hence, $\text{I}=\frac{\pi^2}{16}$
View full question & answer→Question 315 Marks
Evaluate the following integrals as limit of sum:$\int\limits^{2}_{0}\big(\text{x}^2+2\big)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=2,\text{ f(x)}=\text{x}^2+2,\text{ h}=\frac{2-0}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^{2}_{0}\big(\text{x}^2+2\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big\{0+(\text{n}-1)\text{h}\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(0-2)+(\text{h}^2-2)+\ ....\ +\{(\text{n}-1)^2\text{h}^2-2\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[2\text{n}+\text{h}^2\big\{1^2+2^2+3^2+\ ....\ +(\text{n}-1)^2\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[2\text{n}+\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2}{\text{n}}\Big[2\text{n}+\frac{2(\text{n}-1)(2\text{n}-1)}{3\text{n}}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\bigg\{2+\frac{2}{3}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)\bigg\}$
$=4+\frac{8}{3}$
$=\frac{20}{3}$
View full question & answer→Question 325 Marks
Evaluate the following integrals:$\int\limits^{\pi}_0\text{x}\log\sin\text{x}\text{ dx}$
Answer$\int\limits^{\pi}_0\text{x}\log\sin\text{x}\text{ dx}$Let $\text{I}=\int\limits^{\pi}_0\text{x}\log\sin\text{x}\text{ dx}\ ...(\text{i})$
$\text{I}=\int\limits^{\pi}_0(\pi-\text{x})\log\sin(\pi-\text{x})\text{dx}$
$\text{I}=\int\limits^{\pi}_0(\pi-\text{x})\log(\sin\text{x})\text{dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\pi\int\limits^{\pi}_0\log\sin\text{x}\text{ dx}$
$=2\pi\int\limits^{\frac{\pi}{2}}_0\log\sin\text{x}\text{ dx}$
$\text{I}=\pi\int\limits^{\frac{\pi}{2}}_0\log\sin\text{x}\text{ dx}\ ....(\text{iii})$
Let $\int\limits^{\frac{\pi}{2}}_0\log\sin\text{x}\text{ dx}=\text{I}_2$
$\text{I}_2=\int\limits^{\frac{\pi}{2}}_0\log\sin\Big(\frac{\pi}{2}-\text{x}\Big)\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\log\cos\text{x dx}$
$2\text{I}_2=\int\limits^{\frac{\pi}{2}}_0(\log\sin\text{x}+\log\cos\text{x})\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\log(\sin\text{x}\cos\text{x})\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\log(\sin2\text{x})\text{dx}-\int\limits^{\frac{\pi}{2}}_0\log2\text{ dx}$
Let $2\text{x}=\text{t}$
$2\text{dx}=\text{dt}$
When, $\text{x}=0\Rightarrow\text{t}=0$
$\text{x}=0\Rightarrow\text{t}=\pi$
$2\text{I}_2=\frac{1}{2}\int\limits^{{\pi}}_0\log(\sin\text{t})\text{dt}-\frac{\pi}{2} \log2 $
$2\text{I}_2=\frac{2}{2}\int\limits^{{\pi}}_0\log(\sin\text{t})\text{dt}-\frac{\pi}{2} \log2 $
$2\text{I}_2=\text{I}_2-\frac{\pi}{2} \log2 $
$\text{I}_2=-\frac{\pi}{2}\log2$
From (iii),
$\text{I}=\pi\int\limits^{{\pi}}_0\log\sin\text{x}\text{dx}=\pi\text{I}_2$
$\text{I}=\pi\Big(-\frac{\pi}{2}\log2\Big)$
$\text{I}=\frac{-\pi^2\log2}{2}$
View full question & answer→Question 335 Marks
Evaluate the following integrals as limit of sum:$\int\limits^{4}_{1}\big(3\text{x}^2+2\text{x}\big)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=1,\text{ b}=4,\text{ f(x)}=3\text{x}^2+2\text{x},\text{ h}=\frac{4-1}{\text{n}}=\frac{3}{\text{n}}$
Therefore, $\text{I}=\int\limits^{4}_{1}\big(3\text{x}^2+2\text{x}\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(1)+\text{f}(1+\text{h})+\ ....\ +\text{f}\big(1+(\text{n}-1)\text{h}\big)\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\big(3.1^2+2\times1)+(3+(1+\text{h})^2+2(1+\text{h})\big)+\\\ ....+\ \Big\{3\big(1+(\text{n}-1)\text{h}\big)^2+2\big(1+(\text{n}-1)\text{h}\big)\Big\}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[3\Big\{1^2+\big(1+\text{h}^2+(1+2\text{h})\big)^2+\ ....\ +\big(1+(\text{n}-1)\text{h}^2\big)\Big\}\\+2\Big\{1+(1+\text{h})+\ ...+\ \big(1+(\text{n}-1)\text{h}\big)\Big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[3\text{n}+3\text{h}^2\big\{1^2+2^2+3^2+\ ....\ +(\text{n}-1)^2\big\}+6\text{h}\big\{1+2+\ ...+\\\ (\text{n}-1)\text{h}\big\}+2\text{n}+2\text{h}\big\{1+2+\ .....\ +(\text{n}-1)\text{h}\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[5\text{n}+3\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}+8\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{3}{\text{n}}\Big[5\text{n}+\frac{9(\text{n}-1)(2\text{n}-1)}{2\text{n}}+12\text{n}-12\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}3\Big[17-\frac{12}{\text{n}}+\frac{9}{2}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)\Big]$
$=51+27$
$=78$
View full question & answer→Question 345 Marks
Evaluate the following integrals:$\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin\text{x}|\sin\text{x}|\text{dx}$
Answer$\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin\text{x}|\sin\text{x}|\text{dx}$$=\int\limits^{0}_{-\frac{\pi}{4}}\sin\text{x}|\sin\text{x}|\text{dx}+\int\limits_{0}^{\frac{\pi}{2}}\sin\text{x}|\sin\text{x}|\text{dx}$
$=\int\limits^{0}_{-\frac{\pi}{4}}\sin\text{x}(-\sin\text{x})\text{dx}+\int\limits_{0}^{\frac{\pi}{2}}\sin\text{x}(\sin\text{x})\text{dx}$ $\begin{pmatrix}|\sin\text{x}|=\begin{cases}\sin\text{x},&0\leq\text{x}\leq\frac{\pi}{2}\\-\sin\text{x},&-\frac{\pi}{4}\leq\text{x}\leq0\end{cases} \end{pmatrix} $
$=\int\limits^{0}_{-\frac{\pi}{4}}\sin^2\text{x dx}+\int\limits_{0}^{\frac{\pi}{2}}\sin^2\text{x dx}$
$=-\int\limits^{0}_{-\frac{\pi}{4}}\frac{1-\cos2\text{x}}{2}\text{ dx}+\int\limits_{0}^{\frac{\pi}{2}}\frac{1-\cos2\text{x}}{2}\text{ dx}$
$=-\frac{1}{2}\int\limits^{0}_{-\frac{\pi}{4}}\text{dx}+\frac{1}{2}\int\limits^{0}_{-\frac{\pi}{4}}\cos2\text{x dx}+\frac{1}{2}\int\limits_{0}^{\frac{\pi}{2}}\text{dx}-\frac{1}{2}\int\limits_{0}^{\frac{\pi}{2}}\cos2\text{x dx}$
$=-\frac{1}{2}\times\big[\text{x}\big]^0_\frac{\pi}{4}+\frac{1}{2}\times\Big[\frac{\sin2\text{x}}{2}\Big]^0_{\frac{\pi}{2}}+\frac{1}{2}\times\big[\text{x}\big]^{\frac{\pi}{2}}_0-\frac{1}{2}\times\Big[\frac{\sin2\text{x}}{2}\Big]^{\frac{\pi}{2}}_0$
$=-\frac{1}{2}\Big(0+\frac{\pi}{4}\Big)+\frac{1}{4}(0+1)+\frac{\pi}{4}-\frac{1}{4}(0-0)$
$=\frac{\pi}{8}+\frac{1}{4}$
View full question & answer→Question 355 Marks
Evaluate the following integrals:$\int_{0}^\limits{{\pi}}\frac{1}{5+3\cos\text{x}}\text{ dx}$
AnswerWe know that,$\cos\text{x}=\frac{1-\tan^{2}\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}}$
$\Rightarrow\ \frac{1}{5+3\cos\text{x}}=\frac{1}{5+3\Bigg(\frac{1-\tan^{2}\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}}\Bigg)}\\=\frac{1+\tan^{2}\frac{\text{x}}{2}}{5\big(1+\tan^{2}\frac{\text{x}}{2}\big)+3\big(1-\tan^{2}\frac{\text{x}}{2}\big)}=\frac{\sec^2\frac{\text{x}}{2}\text{ dx}}{8+2\tan^{2}\frac{\text{x}}{2}}$
$\therefore\ \int_{0}^\limits{{\pi}}\frac{1}{5+3\cos\text{x}}\text{ dx}=\frac{1}{2}\int_{0}^\limits{{\pi}}\frac{\sec^2\frac{\text{x}}{2}}{2^2+2\tan^{2}\frac{\text{x}}{2}}\text{ dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$ Differentiating w.r.t. x, we get$\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{ dx}=\text{dt}$
Now, $\text{x}=0\Rightarrow\text{t}=0$$\text{x}=\pi\Rightarrow\text{t}=\infty$
$\therefore\ \frac{1}{2}\int^\limits\pi_0\bigg(\frac{\sec^2\frac{\text{x}}{2}\text{ dx}}{2^2+\tan^{2}\frac{\text{x}}{2}}\bigg)\text{dx}$
$=\int^\limits\infty_0\frac{\text{dt}}{2^2+\text{t}^2}$
$=\Big[\frac{1}{2}\tan^{-1}\Big(\frac{\text{t}}{2}\Big)\Big]^{\infty}_0$
$=\frac{1}{2}\Big[\tan^{-1}(\infty)-\tan^{-1}(0)\Big]$
$=\frac{1}{2}\Big[\tan^{-1}\Big(\tan\frac{\pi}{2}\Big)-\tan^{-1}\big(\tan0\big)\Big]$
$=\frac{1}{2}\Big[\frac{\pi}{2}-0\Big]$
$=\frac{\pi}{4}$
View full question & answer→Question 365 Marks
Evaluate the following integrals:$\int^\limits1_0\frac{1-\text{x}^2}{\text{x}^4+\text{x}^2+1}\text{ dx}$
AnswerLet $\text{I}=\int^\limits1_0\frac{1-\text{x}^2}{\text{x}^4+\text{x}^2+1}\text{ dx}$$=-\int\frac{\text{x}^2-1}{\text{x}^4+\text{x}^2+1}\text{ dx}$
$=-\int\frac{1-\frac{1}{\text{x}^2}}{\text{x}^2+1+\frac{1}{\text{x}^2}}\text{ dx}$
$=-\int\frac{1-\frac{1}{\text{x}^2}}{\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-1}\text{ dx}$
Let, $\text{x}+\frac{1}{\text{x}}=\text{t}$$\Big(1-\frac{1}{\text{x}^2}\Big)\text{dx}=\text{dt}$
Then integral becames,$\text{I}=-\int\frac{1}{\text{t}^2-1}\text{dt}$
$=-\frac{1}{2}\log\Big|\frac{\text{t}-1}{\text{t}+1}\Big|$
$=\frac{1}{2}\log\Big|\frac{\text{t}+1}{\text{t}-1}\Big|$
$=\frac{1}{2}\log\Big|\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big|$
i,e., $\int\frac{1-\text{t}^2}{\text{x}^4+\text{x}^2+1}\text{ dx}=\frac{1}{2}\log\Big|\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big|$$\int^\limits1_0\frac{1-\text{x}^2}{\text{x}^4+\text{x}^2+1}\text{ dx}=\Big[\frac{1}{2}\log\Big|\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}+1}\Big|\Big]^1_0$
$=\frac{1}{2}\log3$
View full question & answer→Question 375 Marks
Evaluate the following integrals as limit of sum:$\int\limits^{3}_{0}\big(2\text{x}^2+3\text{x}+5\big)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=3,\text{ f(x)}=2\text{x}^2+3\text{x}+5,\text{ h}=\frac{3-0}{\text{n}}=\frac{3}{\text{n}}$
Therefore, $\text{I}=\int\limits^{3}_{0}\big(2\text{x}^2+3\text{x}+5\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big\{0+(\text{n}-1)\text{h}\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[(0+0+5)+(2\text{h}^2+3\text{h}+5)+\\\ ....\ +\big\{2(\text{n}-1)^2\text{h}^2+3(\text{n}-5)\text{h}+1\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[5\text{n}+2\text{h}^2\big(1^2+2^2+3^2+\ ....\ +(\text{n}-1)^2\big)+\\3\text{h}\big\{1+2+\ ...\ +(\text{n}-1)\text{h}\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[5\text{n}+2\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}+3\text{h}\frac{\text{n}(\text{n}-1)}{2}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{3}{\text{n}}\Big[\text{n}+\frac{3\text{n}(\text{n}-1)(2\text{n}-1)}{\text{n}}+\frac{9(\text{n}-1)}{2}\Big]$
$=15+18+\frac{27}{2}$
$=\frac{93}{3}$
View full question & answer→Question 385 Marks
Evaluate the following integrals:$\int\limits^{{\pi}}_{{-\pi}}\frac{2\text{x}(1+\sin\text{x})}{1+\cos^2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{{\pi}}_{{-\pi}}\frac{2\text{x}(1+\sin\text{x})}{1+\cos^2\text{x}}\text{ dx}$ Then,$\text{I}=\int\limits^{{\pi}}_{{-\pi}}\frac{2\text{x}}{1+\cos^2\text{x}}\text{ dx}+\int\limits^{{\pi}}_{{-\pi}}\frac{2\text{x}\sin\text{x}}{1+\cos^2\text{x}}\text{ dx}$
$=\text{I}_1+\text{I}_2$
Consider $\text{f(x)}=\frac{2\text{x}}{1+\cos^2-\text{x}}$ Now,$\text{f}(-\text{x})=\frac{2(-\text{x})}{1+\cos^2(\pi-\text{x})}=-\frac{2\text{x}}{1+(-\cos\text{x})^2}=-\frac{2\text{x}}{1+\cos^2\text{x}}=-\text{f(x)}$
$\therefore\ \text{I}_1=\int\limits^{{\pi}}_{{-\pi}}\frac{2\text{x}}{1+\cos^2\text{x}}\text{ dx}=0$ $\begin{bmatrix}\int\limits^{\text{a}}_{-\text{a}}\text{f(x)}\text{dx}=\begin{cases}2\int\limits^{\text{a}}_{0}\text{f(x)}\text{dx},&\text{ if }\text{f}(-\text{x})=\text{f(x)}\\0,&\text{ if }\text{f}(-\text{x})=-\text{f}(\text{x})\end{cases}\end{bmatrix}$
Again, consider $\text{g(x)}=\frac{2\text{x}\sin\text{x}}{1+\cos^2\text{x}}$$\text{g}(-\text{x})=\frac{2(-\text{x})\sin(-\text{x})}{1+\cos^2(-\text{x})}=\frac{2\text{x}\sin\text{x}}{1+\cos^2\text{x}}=\text{g(x)}$ $\big[\sin(-\text{x})=-\sin\text{x}\text{ and }\cos(-\text{x})=\cos\text{x}\big]$
$\therefore\ \text{I}_2=\int\limits^{\pi}_{-\pi}\frac{2\text{x}\sin\text{x}}{1+\cos^2\text{x}}\text{ dx}$
$=2\times2\int\limits^{\pi}_0\frac{\text{x}\sin\text{x}}{1+\cos^2\text{x}}\text{ dx}$ $\begin{bmatrix}\int\limits^{\text{a}}_{-\text{a}}\text{f(x)}\text{dx}=\begin{cases}2\int\limits^{\text{a}}_{0}\text{f(x)}\text{dx},&\text{ if }\text{f}(-\text{x})=\text{f(x)}\\0,&\text{ if }\text{f}(-\text{x})=-\text{f}(\text{x})\end{cases}\end{bmatrix}$
$=4\int\limits^{\pi}_0\frac{\text{x}\sin\text{x}}{1+\cos^2\text{x}}\text{ dx}\ ...(\text{i})$
Then,$\text{I}_2=4\int\limits^{\pi}_0\frac{(\pi-\text{x})\sin(\pi-\text{x})}{1+\cos^2(\pi-\text{x})}\text{ dx}$
$=4\int\limits^{\pi}_0\frac{(\pi-\text{x})\sin\text{x}}{1+\cos^2\text{x}}\text{ dx}\ ....(\text{ii})$
Adding (i) and (ii) we get$2\text{I}_2=4\int\limits^{\pi}_0\frac{\pi\sin\text{x}}{1+\cos^2\text{x}}\text{ dx}$
$\Rightarrow2\text{I}_2=4\int\limits^{\pi}_0\frac{\sin\text{x}}{1+\cos^2\text{x}}\text{ dx}$
Put $\cos\text{x}=\text{z}$$\Rightarrow-\sin\text{x}\text{dx}=\text{dz}$
When $\text{x}\rightarrow0,\text{ z}\rightarrow1$ When $\text{x}\rightarrow\pi,\text{ z}\rightarrow-1$$\therefore\ 2\text{I}_2=-4\pi\int\limits^{-1}_1\frac{\text{dz}}{1+\text{z}^2}$
$\Rightarrow2\text{I}_2=-4\pi\times\Big[\tan^{-1}\text{z}\Big]^{-1}_1$
$\Rightarrow2\text{I}_2=-4\pi\Big[\tan^{-1}(-1)-\tan^{-1}1\Big]$
$\Rightarrow2\text{I}_2=-4\pi\Big(-\frac{\pi}{4}-\frac{\pi}{4}\Big)=2\pi^2$
$\Rightarrow\text{I}_2=\pi^2$
$\therefore\ \text{I}=\text{I}_1+\text{I}_2$
$\Rightarrow\text{I}=0+\pi^2=\pi^2$
View full question & answer→Question 395 Marks
Evaluate the following integrals:$\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{5+4\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{5+4\sin\text{x}}\text{ dx}$ Then,$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{5+4\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{1+\tan^2\frac{\text{x}}{2}}{5\Big(1+\tan^2\frac{\text{x}}{2}\Big)+8\tan\frac{\text{x}}{2}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\sec^2\frac{\text{x}}{2}}{5\tan^2\frac{\text{x}}{2}+8\tan\frac{\text{x}}{2}+5}\text{ dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$ Then, $\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{ dx}=\text{dt}$ When $\text{x}=0,\text{t}=0$ and $\text{x}=\frac{\pi}{2},\text{t}=1$$\therefore\ \text{I}=2\int\limits_0^1\frac{1}{5\text{t}^2+8\text{t}+5}\text{ dt}$
$\Rightarrow\text{I}=2\int\limits_0^1\frac{1}{\big(\sqrt{5\text{t}}\big)^2+8\text{t}+5+\Big(\frac{4}{\sqrt{5}}\Big)^2-\Big(\frac{4}{\sqrt{5}}\Big)^2}\text{ dt}$
$\Rightarrow\text{I}=2\int\limits_0^1\frac{1}{\Big(\sqrt{5}\text{t}+\frac{4}{\sqrt{5}}\Big)^2+\frac{9}{5}}\text{ dt}$
$\Rightarrow\text{I}=\frac{2}{3}\begin{bmatrix}\tan^{-1}\begin{pmatrix}\frac{\sqrt{5}\text{t}+\frac{4}{\sqrt{5}}}{\frac{3}{\sqrt{5}}} \end{pmatrix}\end{bmatrix}^1_0$
$\Rightarrow\text{I}=\frac{2}{3}\Big[\tan^{-1}3-\tan^{-1}\frac{4}{3}\Big]$
$\Rightarrow\text{I}=\frac{2}{3}\Bigg[\tan^{-1}\Bigg(\frac{3-\frac{4}{3}}{1+3\times\frac{4}{3}}\Bigg)\Bigg]$
$\Rightarrow\text{I}=\frac{2}{3}\tan^{-1}\frac{1}{3}$
View full question & answer→Question 405 Marks
Evaluate the following integrals:$\int\limits^{\frac{\pi}{2}}_0\frac{\tan^{7}\text{x}}{\tan^{7}\text{x}+\cot^7\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\tan^{7}\text{x}}{\tan^{7}\text{x}+\cot^7\text{x}}\text{ dx}\ ...(\text{i})$ Then,$\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\tan^{7\big(\frac{\pi}{2}-\text{x}\big)}}{\tan^{7}{\big(\frac{\pi}{2}-\text{x}\big)}+\cot^7{\big(\frac{\pi}{2}-\text{x}\big)}}\text{ dx}$ $\Bigg[\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\cot^7\text{x}}{\cot^7\text{x}+\tan^{7}\text{x}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\tan^{7}\text{x}+\cot^7\text{x}}{\tan^{7}\text{x}+\cot^7\text{x}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\text{dx}$
$\Rightarrow2\text{I}=\Big[\text{x}\Big]^{\frac{\pi}{2}}_0$
$\Rightarrow2\text{I}=\frac{\pi}{2}-0=\frac{\pi}{2}$
$\Rightarrow\text{I}=\frac{\pi}{4}$
View full question & answer→Question 415 Marks
Evaluate the following integrals:$\int^\limits{(\pi)^\frac{2}{3}}_{0}\sqrt{\text{x}}\cos^2\text{x}^{\frac{3}{2}}\text{ dx}$
AnswerLet $\text{I}=\int^\limits{(\pi)^\frac{2}{3}}_{0}\sqrt{\text{x}}\cos^2\text{x}^{\frac{3}{2}}\text{ dx}$ Then, Let $\text{x}^{\frac{3}{2}}=\text{t}$ Then, $\frac{3}{2}\sqrt{\text{x}}\text{ dx}=\text{dt}$ When, $\text{x}=0,\text{t}=0$ and $\text{x}=\big(\pi\big)^{\frac{2}{3}},\text{t}=\pi$$\therefore\ \text{I}=\frac{2}{3}\int^\limits{\pi}_{0}\cos^2\text{t}\text{ dt}$
$\Rightarrow\text{I}=\frac{2}{3}\int^\limits{\pi}_{0}\frac{1+\cos2\text{x}}{2}\text{ dx}$
$\Rightarrow\text{I}=\frac{1}{3}\Big[\text{x}+\frac{\sin2\text{x}}{2}\Big]^{\pi}_0$
$\Rightarrow\text{I}=\frac{1}{3}\big(\pi+0\big)$
$\Rightarrow\text{I}=\frac{\pi}{3}$
View full question & answer→Question 425 Marks
Evaluate the following integrals:$\int^\limits{\text{a}}_{-\text{a}}\sqrt{\frac{\text{a}-\text{x}}{\text{a}+\text{x}}}\text{ dx}$
AnswerLet $\text{x}=\text{a}\cos2\theta$ Differentiating w.r.t. x, we get$\text{dx}=-2\text{a}\sin2\theta$
Now, $\text{x}=-\text{a}\Rightarrow\theta=\frac{\pi}{2}$$\text{x}=\text{a}\Rightarrow\theta=0$
$\therefore\ \int^\limits{\text{a}}_{-\text{a}}\sqrt{\frac{\text{a}-\text{x}}{\text{a}+\text{x}}}\text{ dx}=\int^\limits0_\frac{\pi}{2}\sqrt{\frac{\text{a}(1-\cos2\theta)}{\text{a}\big(1+\cos2\theta)}}(-2\sin2\theta\big)\text{d}\theta$
$=2\text{a}\int^\limits{\frac{\pi}{\text{2}}}_{0}\frac{\sin\theta}{\cos\theta}\cdot\sin2\theta\text{ d}\theta$ $\begin{bmatrix}\because1-\cos2\theta=2\sin^2\theta\\1+\cos2\theta=2\cos^2\theta\\-\int^\limits\text{b}_\text{a}\text{f(x)}\text{dx}=\int^\limits\text{a}_\text{b}\text{f}(\text{x})\text{dx} \end{bmatrix}$
$=2\text{a}\int^\limits{\frac{\pi}{\text{2}}}_{0}\frac{\sin\theta\cdot2\sin\theta\cos\theta}{\cos\theta}$
$=4\text{a}\int^\limits{\frac{\pi}{\text{2}}}_{0}\sin^{2}\theta\text{ d}\theta$
$=2\text{a}\int^\limits{\frac{\pi}{\text{2}}}_{0}\big(1-\cos2\theta\big)\text{d}\theta$
$=2\text{a}\Big[\theta-\frac{\sin2\theta}{2}\Big]^{\frac{\pi}{2}}_0$
$=2\text{a}\Big[\frac{\pi}{2}-0-0+0\Big]$
$=\pi\text{a}$
View full question & answer→Question 435 Marks
Evaluate the following definite integrals:$\int_{0}^\limits{1}\frac{1-\text{x}}{1+\text{x}}\text{ dx}$
AnswerWe have,$\int_{0}^\limits{1}\frac{1-\text{x}}{1+\text{x}}\text{ dx}$
Let $\text{x}=\cos2\theta$$\text{dx}=-2\sin2\theta\text{ d}\theta$
Now,$\int_{0}^\limits{1}\frac{1-\text{x}}{1+\text{x}}\text{ dx}$
$=\int_{\frac{\pi}{4}}^\limits{0}\frac{1-\cos2\theta}{1+\cos2\theta}\times(-2\sin2\theta)\text{d}\theta$
$\int_{\frac{\pi}{4}}^\limits{0}\frac{2\sin^2\theta}{2\cos^2\theta}\times2\sin2\theta\text{ d}\theta$ $\bigg[\because\ -\int_{\text{a}}^\limits{\text{b}}\text{f(x)}\text{dx}=\int_{\text{b}}^\limits{\text{a}}\text{f(x)}\text{dx}\bigg]$
$=\int_{\frac{\pi}{4}}^\limits{0}\frac{4\sin^3\theta}{\cos\theta}\text{ d}\theta$
Let $\cos\theta=\text{t}$$-\sin\theta\text{ d}\theta=\text{dt}$
Now,$\theta=0\Rightarrow\text{t}=1$
$\theta=\frac{\pi}{4}\Rightarrow\text{t}=\frac{1}{\sqrt{2}}$
$\therefore\ \int_{0}^\limits{\frac{\pi}{4}}\frac{4\sin^3\theta}{\cos\theta}\text{ d}\theta$
$=-4\int_{1}^\limits{\frac{1}{\sqrt{2}}}\frac{\big(1-\text{t}^2\big)}{\text{t}}\text{ dt}$
$=-4\Big[\log\text{t}-\frac{\text{t}^2}{2}\Big]^{\frac{1}{\sqrt{2}}}_1$
$=-4\Big[\log\Big(\frac{1}{\sqrt{2}}\Big)-\frac{1}{4}-0+\frac{1}{2}\Big]$
$=-4\Big[\log\sqrt{2}+\frac{1}{4}\Big]$
$\therefore\ \int_{0}^\limits{1}\frac{1-\text{x}}{1+\text{x}}\text{ dx}=2\log2-1$
View full question & answer→Question 445 Marks
Evaluate the following integrals:$\int^\limits2_1\frac{1}{\text{x}(1+\log\text{x})^2}\text{ dx}$
AnswerLet $\text{I}=\int^\limits2_1\frac{1}{\text{x}(1+\log\text{x})^2}\text{ dx}$ Then, Let $(1+\log\text{x})=\text{t}$ Then, $\frac{1}{\text{x}}\text{ dx}=\text{dt}$ When, $\text{x}=1,\text{t}=1$ and $\text{x}=2,\text{t}=1+\log2$$\therefore\ \text{I}=\int^\limits{(1+\log2)}_{1}\frac{1}{\text{t}}\text{ dt}$
$\Rightarrow\text{I}=\Big[\frac{-1}{\text{t}}\Big]^{(1+\log2)}_1$
$\Rightarrow\text{I}=-\frac{1}{1+\log2}+1$
$\Rightarrow\text{I}=\frac{\log2}{1+\log2}$
View full question & answer→Question 455 Marks
Evaluate the following definite integrals:$\int_{0}^\limits{1}\frac{1}{\sqrt{(\text{x}-1)(2-\text{x})}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\frac{1}{\sqrt{(\text{x}-1)(2-\text{x})}}\text{ dx}$ Put $\text{x}=\cos^2\theta+2\sin^2\theta$$\therefore\ \text{dx}=2\cos\theta(-\sin\theta)\text{d}\theta+4\sin\theta\cos\theta\text{ d}\theta$
$=2\sin\theta\cos\theta\text{ d}\theta$
Also, $\text{x}=\cos^2\theta+2\sin^2\theta$$\Rightarrow\text{x}=1+\sin^2\theta$
$\Rightarrow\sin\theta=\sqrt{\text{x}-1}$
When $\text{x}\rightarrow1,\sin\theta\rightarrow0$ or $\theta\rightarrow0$ When $\text{x}\rightarrow2,\sin\theta\rightarrow1$ or $\theta\rightarrow\frac{\pi}{2}$$\therefore\ \text{I}=\int\limits^2_1\frac{1}{\sqrt{(\text{x}-1)(2-\text{x})}}\text{ dx}$
$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{2\sin\theta\cos\theta\text{ d}\theta}{\sqrt{(\cos^2\theta+2\sin^2\theta-1)(2-\cos^2\theta-2\sin^2\theta)}}$
$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{2\sin\theta\cos\theta\text{ d}\theta}{\sqrt{\sin^2\theta\cos^2\theta}}$ $\big(\sin^2\theta+\cos^2\theta=1\big)$
$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{2\sin\theta\cos\theta\text{ d}\theta}{\sin\theta\cos\theta}$
$\Rightarrow\text{I}=2\int\limits^\frac{\pi}{2}_0\text{d}\theta$
$\Rightarrow\text{I}=2\big[\theta\big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=2\Big(\frac{\pi}{2}-0\Big)$
$\Rightarrow\text{I}=\pi$
View full question & answer→Question 465 Marks
Evaluate the following integrals:$\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}-\cos^3\text{x}}(\sec^3\text{x}-1)\cos^2\text{x dx}$
AnswerLet $\text{I}=\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}-\cos^3\text{x}}(\sec^3\text{x}-1)\cos^2\text{x dx}$$=\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}(1-\cos^2\text{x}\big)}(-\tan^2\text{x})\cos^2\text{x dx}$
$=-\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}(\sin^2\text{x})}\sin^2\text{x dx}$
$=-\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}}(\sin\text{x})\sin^2\text{x dx}$
$=-\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}}\big(1-\cos^2\text{x}\big)\sin\text{x dx}$ $\Big(\sin\text{x}=\sin\text{x}\text{ for }0\leq\text{x}\leq\frac{\pi}{2}\Big)$
Put $\cos\text{x}=\text{z}^2$$\therefore\ -\sin\text{x dx}=2\text{z dz}$
When $\text{x}\rightarrow0,\text{ z}\rightarrow0$ When $\text{x}\rightarrow\frac{\pi}{2},\text{ z}\rightarrow0$$\therefore\ \text{I}=-\int^\limits0_1\text{z}(1-\text{z}^4)2\text{z dz}$
$=-2\int^0\limits_1\text{z}^2\text{ dz}+2\int^\limits0_1\text{z}^6\text{ dz}$
$=-2\times\Big[\frac{\text{z}^3}{3}\Big]^0_1+2\times\Big[\frac{\text{z}^7}{7}\Big]^0_1$
$=-\frac{2}{3}\big(0-1\big)+\frac{2}{7}\big(0-1\big)$
$=\frac{2}{3}-\frac{2}{74}$
$=\frac{8}{21}$
View full question & answer→Question 475 Marks
Evaluate the following definite integrals:$\int_{0}^\limits{\frac{\pi}{2}}\sin^3\text{x}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\sin^3\text{x}\text{ dx}$ Then,$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\sin\text{x }\sin^2\text{x}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\sin\text{x}(1-\cos^2\text{x})\text{dx}$
Let $\text{u}=\cos\text{x},\text{ du}=-\sin\text{x dx}$$\therefore\ \text{I}=\int-(1-\text{u}^2)\text{du}$
$\Rightarrow\text{I}=\Big[\frac{\text{u}^3}{3}-\text{u}\Big]$
$\Rightarrow\text{I}=\Big[\frac{\cos^3\text{x}}{3}-\cos\text{x}\Big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=0-\frac{1}{3}+1$
$\Rightarrow\text{I}=\frac{2}{3}$
View full question & answer→Question 485 Marks
Evaluate the following definite integrals:$\int_{0}^\limits{1}\sqrt{\text{x}(1-\text{x})}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\sqrt{\text{x}(1-\text{x})}\text{ dx}$ Then,$\text{I}=\int_{0}^\limits{1}\sqrt{\frac{1}{4}-\big(\text{x}-\frac{1}{2}\big)^2}\text{ dx}$
$\Rightarrow\text{I}=\frac{1}{2}\int_{0}^\limits{1}\sqrt{1-\frac{\big(\text{x}-\frac{1}{2}\big)^2}{\frac{1}{4}}}\text{ dx}$
$\Rightarrow\text{I}=\frac{1}{2}\int_{0}^\limits{1}\sqrt{1-\Bigg(\frac{\text{x}-\frac{1}{2}}{\frac{1}{2}}\Bigg)^2}\text{ dx}$
Let $\Bigg(\frac{\text{x}-\frac{1}{2}}{\frac{1}{2}}\Bigg)=\sin\text{u}$$\Rightarrow2\text{dx}=\cos\text{u du}$
$\therefore\ \text{I}=\frac{1}{4}\int_{-\frac{\pi}{2}}^\limits{\frac{\pi}{2}}\sqrt{1-\sin^2\text{u}}\cos\text{u du}$
$\Rightarrow \text{I}=\frac{1}{4}\int_{-\frac{\pi}{2}}^\limits{\frac{\pi}{2}}\cos^2\text{u du}$
$\Rightarrow \text{I}=\frac{1}{4}\int_{-\frac{\pi}{2}}^\limits{\frac{\pi}{2}}\Big(\frac{\cos2\text{u}+1}{2}\Big)\text{du}$
$\Rightarrow \text{I}=\frac{1}{8}\Big[\frac{\sin2\text{u}}{2}+\text{u}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}$
$\Rightarrow \text{I}=\frac{1}{8}\Big[\frac{\pi}{2}+\frac{\pi}{2}\Big]$
$\Rightarrow\text{I}=\frac{\pi}{8}$
View full question & answer→Question 495 Marks
Evaluate the following integrals:$\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{3+\sin2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{3+\sin2\text{x}}\text{ dx}$$=\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{4-(1-\sin2\text{x})}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{4-(\sin^2\text{x}+\cos^2\text{x}-2\sin\text{x}\cos\text{x})}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{4-(\sin\text{x}-\cos\text{x})^2}\text{ dx}$
Put $\sin\text{x}-\cos\text{x}=\text{z}$$\therefore\ (\cos\text{x}+\sin\text{x})\text{dx}=\text{dz}$
When $\text{x}\rightarrow0,\text{z}\rightarrow-1$ $(\text{z}=\sin0-\cos0=0-1=-1)$ When $\text{x}\rightarrow\frac{\pi}{4},\text{z}\rightarrow0$ $\Big(\text{z}=\sin\frac{\pi}{4}-\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=0\Big)$$\therefore\ \text{I}=\int_{-1}^\limits{0}\frac{\text{dz}}{2^2-\text{z}^2}$
$=\frac{1}{2\times2}\log\Big[\log\Big(\frac{2+\text{z}}{2-\text{z}}\Big)\Big]^0_{-1}$
$=\frac{1}{4}\Big(\log1-\log\frac{1}{3}\Big)$
$=\frac{1}{4}\big[0-\big(\log1-\log3\big)\big]$
$=-\frac{1}{4}\big(0-\log3\big)$
$=\frac{1}{4}\log3$
View full question & answer→Question 505 Marks
Evaluate the following definite integrals:$\int_{1}^\limits{2}\frac{\text{x}+3}{\text{x}(\text{x}+2)}\text{ dx}$
AnswerWe have$\int_{1}^\limits{2}\frac{\text{x}+3}{\text{x}(\text{x}+2)}\text{dx}$
$=\int_{1}^\limits{2}\frac{\text{x}}{\text{x}(\text{x}+2)}\text{ dx}+\int_{1}^\limits{2}\frac{3}{\text{x}(\text{x}+2)}\text{ dx}$
$=\int_{1}^\limits{2}\frac{\text{x}}{(\text{x}+2)}\text{ dx}+\int_{1}^\limits{2}\frac{3}{\text{x}(\text{x}+2)}\text{ dx}$
$=\big[\log(\text{x}+2)\big]^2_1+\frac{3}{2}\int_{1}^\limits{2}\frac{1}{\text{x}}-\frac{1}{\text{x}+2}\text{ dx}$ [using partial fraction]
$=\big[\log(\text{x}+2)\big]^2_1+\Big[\frac{3}{2}\log\text{x}-\frac{3}{2}\log(\text{x}+2)\Big]^2_1$
$=\Big[\frac{3}{2}\log\text{x}-\frac{1}{2}\log(\text{x}+2)\Big]^2_1$
$=\frac{1}{2}\big[3\log2-\log4+\log3\big]$
$=\frac{1}{2}\big[3\log2-2\log2+\log3\big]$ $\big[\because\log4=2\log2\big]$
$=\frac{1}{2}\big[\log2+\log3\big]$
$=\frac{1}{2}\big[\log6\big]$
$=\frac{1}{2}\log6$
$\therefore\ \int_{1}^\limits{2}\frac{\text{x}+3}{\text{x}(\text{x}+2)}\text{dx}=\frac{1}{2}\log6$
View full question & answer→Question 515 Marks
Evaluate the following integrals:$\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\text{x}}}{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\text{x}}}{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}\text{ dx}\ ...(\text{i})$ Then,$\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}}{\sqrt{\sin\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}+\sqrt{\cos\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}}\text{ dx}$ $\Bigg[\int\limits^{\text{b}}_{\text{a}}\text{f(x)}\text{dx}=\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\big(\frac{\pi}{2}-\text{x}\big)}}{\sqrt{\sin\big(\frac{\pi}{2}-\text{x}\big)}+\sqrt{\cos\big(\frac{\pi}{2}-\text{x}\big)}}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\cos\text{x}}}{\sqrt{\cos\text{x}}+\sqrt{\sin\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get$2\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\text{dx}$
$\Rightarrow2\text{I}=\Big[\text{x}\Big]^{\frac{\pi}{3}}_{\frac{\pi}{6}}$
$\Rightarrow2\text{I}=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}$
$\Rightarrow\text{I}=\frac{\pi}{12}$
View full question & answer→Question 525 Marks
Evaluate the following definite integrals:$\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}(\tan\text{x}+\cot\text{x})^2\text{ dx}$
AnswerWe have,$\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}(\tan\text{x}+\cot\text{x})^2\text{ dx}$
$=\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}\Big(\frac{\sin^2\text{x}+\cot^2\text{x}}{\sin\text{x}\cos\text{x}}\Big)^2\text{dx}$
$=\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}\Big(\frac{1}{\sin\text{x}\cos\text{x}}\Big)^2\text{dx}$
Multiplying numberator and denominator by 2$=\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}\Big(\frac{2}{2\sin\text{x}\cos\text{x}}\Big)^2\text{dx}$
$=\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}\Big(\frac{2}{\sin2\text{x}}\Big)^2\text{dx}$ $\big[\because2\sin\text{x}\cos\text{x}=\sin2\text{x}\big]$
$=4\int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}\text{cosec}^2\text{x dx}$
$=4\Big[-\frac{\cot2\text{x}}{2}\Big]^{\frac{\pi}{4}}_\frac{\pi}{3}$
$=2\Big[-\cot\frac{\pi}{2}+\cot2\frac{\pi}{3}\Big]$
$=2\Big[\frac{-1}{\sqrt{3}}-0\Big]$
$=\frac{-2}{\sqrt{3}}$
$\therefore\ \int_{\frac{\pi}{3}}^\limits{\frac{\pi}{4}}(\tan\text{x}+\cot\text{x})^2\text{ dx}=\frac{-2}{\sqrt{3}}$
View full question & answer→Question 535 Marks
Evaluate the following integrals:$\int\limits^2_{-2}\text{xe}^{|\text{x}|}\text{ dx}$
Answer$\int\limits^0_{-2}\text{xe}^{-\text{x}}\text{ dx}+\int\limits^2_{0}\text{xe}^{\text{x}}\text{ dx}$For
$\int\limits^0_{-2}\text{xe}^{-\text{x}}\text{ dx}$
Using integration by parts
$\int\text{f}^\text{t}\text{g}=\text{fg}-\int\text{fg}^{\text{t}}$
$\text{f}^\text{t}=\text{e}^{-\text{x}},\text{g}=\text{x}$
$\text{f}=-\text{e}^{-\text{x}},\text{g}^\text{t}=1$
$\int\limits^0_{-2}\text{xe}^{-\text{x}}\text{ dx}=\big\{-\text{xe}^{-\text{x}}\big\}^0_{-2}+\int\limits^0_{-2}\text{e}^{-\text{x}}\text{ dx}$
$\int\limits^0_{-2}\text{xe}^{-\text{x}}\text{ dx}=\big\{-\text{xe}^{-\text{x}}-\text{e}^{-\text{x}}\big\}^0_{-2}$
$\int\limits^0_{-2}\text{xe}^{-\text{x}}\text{ dx}=\big\{(-1)-\big(2\text{e}^2-\text{e}^2\big)\big\}$
$\int\limits^0_{-2}\text{xe}^{-\text{x}}\text{ dx}=\big\{-1-\text{e}^2\big\}$
For
$\int\limits^2_{0}\text{xe}^{\text{x}}\text{ dx}$
Using integration by parts
$\int\text{f}^\text{t}\text{g}=\text{fg}-\int\text{fg}^{\text{t}}$
$\text{f}^\text{t}=\text{e}^{\text{x}},\text{g}=\text{x}$
$\text{f}=-\text{e}^{-\text{x}},\text{g}^\text{t}=1$
$\int\limits^2_{0}\text{xe}^{\text{x}}\text{ dx}=\big\{\text{xe}^{\text{x}}\big\}^2_{0}-\int\limits^2_{0}\text{e}^{\text{x}}\text{ dx}$
$\int\limits^2_{0}\text{xe}^{\text{x}}\text{ dx}=\big\{\text{xe}^{\text{x}}-\text{e}^{\text{x}}\big\}^2_{0}$
$\int\limits^2_{0}\text{xe}^{\text{x}}\text{ dx}=2\text{e}^2-\text{e}^2+1$
$\int\limits^2_{0}\text{xe}^{\text{x}}\text{ dx}=\text{e}^2+1$
Hence answer is,
$\int\limits^2_{-2}\text{xe}^{|\text{x}|}\text{ dx}=-1-\text{e}^2+\text{e}^2+1=0$
View full question & answer→Question 545 Marks
Evaluate the following integrals:$\int_\limits{\frac{1}{3}}^{1}\frac{\big(\text{x}-\text{x}^3\big)^{\frac{1}{3}}}{\text{x}^4}\text{ dx}$
AnswerLet $\text{I}=\int_\limits{\frac{1}{3}}^{1}\frac{\big(\text{x}-\text{x}^3\big)^{\frac{1}{3}}}{\text{x}^4}\text{ dx}$$=\int_\limits{\frac{1}{3}}^{1}\frac{\Bigg[\text{x}^3\Big(\frac{\text{x}}{\text{x}^3}-1\Big)\Bigg]^{\frac{1}{3}}}{\text{x}^4}\text{ dx}$
$=\int_\limits{\frac{1}{3}}^{1}\frac{\text{x}\big(\frac{1}{\text{x}^2}-1\big)^{\frac{1}{3}}}{\text{x}^4}\text{ dx}$
$=\int_\limits{\frac{1}{3}}^{1}\frac{\text{x}\big(\frac{1}{\text{x}^2}-1\big)^{\frac{1}{3}}}{\text{x}^3}\text{ dx}$
Put $\Big(\frac{1}{\text{x}^2}-1\Big)=\text{Z}$$\therefore\ -\frac{2}{\text{x}^3}\text{ dx}=\text{dz}$
$\Rightarrow\frac{\text{dx}}{\text{x}^3}=-\frac{\text{dz}}{2}$
When $\text{x}\rightarrow\frac{1}{3},\text{z}\rightarrow8$ When $\text{x}\rightarrow1,\text{z}\rightarrow0$$\therefore\ \text{I}=-\frac{1}{2}\int^\limits0_8\text{z}^{\frac{1}{3}}\text{ dz}$
$=-\frac{1}{2}\times\Bigg[\frac{\text{z}^{\frac{4}{3}}}{\frac{4}{3}}\Bigg]^0_8$
$=-\frac{3}{8}\Big[0-(8)^{\frac{4}{3}}\Big]$
$=-\frac{3}{8}\times(-16)$
$=6$
View full question & answer→Question 555 Marks
Evaluate the following integrals:$\int_{0}^\limits{\frac{\pi}{3}}\frac{\cos\text{x}}{3+4\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{3}}\frac{\cos\text{x}}{3+4\sin\text{x}}\text{ dx}$ Let $\sin\text{x}=\text{t}$ Then, $\cos\text{x}\text{dx}=\text{dt}$ When $\text{x}=0,\text{t}=0$ and $\text{x}=\frac{\pi}{3},\text{t}=\frac{\sqrt{3}}{2}$$\therefore\ \text{I}=\int_{0}^\limits{\frac{\pi}{3}}\frac{\cos\text{x}}{3+4\sin\text{x}}\text{ dx}$
$=\int_{0}^\limits{\frac{\sqrt{3}}{2}}\frac{1}{3+4\text{t}}\text{ dt}$
$=\frac{1}{4}\big[\log\big(3+-4\text{t}\big)\big]^{\frac{\sqrt{3}}{2}}_0$
$=\frac{1}{4}\big(\log\big(3+2\sqrt{3}\big)-\log3\big)$
$=\frac{1}{4}\log\Big(\frac{3+2\sqrt{3}}{3}\Big)$
View full question & answer→Question 565 Marks
Evaluate the following integrals:$\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\cot\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\cot\text{x}}\text{ dx}\ ...(\text{i})$$=\int\limits^{\frac{\pi}{2}}_0\frac{\text{dx}}{1+\cot\big(\frac{\pi}{2}-\text{x}\big)}\text{dx}$ $\Big[\text{Using},\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Big]$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\text{dx}}{1+\tan\text{x}}\ ...(\text{ii})$
Adding (i) and (ii)$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\cot\text{x}}+\frac{1}{1+\tan\text{x}}\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{1+\tan\text{x}+1+\cot\text{x}}{(1+\cot\text{x})(1+\tan\text{x})}\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{2+\tan\text{x}+\cot\text{x}}{1+\tan\text{x}+\cot\text{x}+\tan\text{x}\cot\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{2+\tan\text{x}+\cot\text{x}}{2+\tan\text{x}+\cot\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\text{dx}$
$=\Big[\text{x}\Big]^{\frac{\pi}{2}}_0=\frac{\pi}{2}$
Hence, $\text{I}=\frac{\pi}{4}$
View full question & answer→Question 575 Marks
Evaluate the following integrals as limit of sum:$\int\limits^5_{3}(2-\text{x})\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ + \\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=3,\text{ b}=5,\text{ f(x)}=2-\text{x},\text{ h}=\frac{5-3}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^5_{3}(2-\text{x})\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(2)+\text{f}(2+\text{h})+\ ....\ +\text{f}\big\{2+(\text{n}-1)\text{h}\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(2-2)+(2-\text{h}-2)+\\\ ....\ +\big(2-(\text{n}-1)\text{h}-2\big)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[-\text{h}\big(1+2+3+\ ....\ +(\text{n}-1)\big)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[-2\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2}{\text{n}}\big[-2\text{n}+2\big]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\Big({-2}+\frac{2}{\text{n}}\Big)$
$=-4$
View full question & answer→Question 585 Marks
Evaluate the following integrals:$\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\sqrt{\tan\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\sqrt{\tan\text{x}}}\text{ dx}\ ....(\text{i})$$=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\sqrt{\tan\big(\frac{\pi}{2}-\text{x}}\big)}\text{dx}$ $\Big[\text{Using},\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Big]$
$=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\sqrt{\cot\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii)$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\sqrt{\tan\text{x}}}+\frac{1}{1+\sqrt{\cot\text{x}}}\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{1+\sqrt{\cot\text{x}}+1+\sqrt{\tan\text{x}}}{\big(1+\sqrt{\tan\text{x}}\big)\big(1+\sqrt{\cot\text{x}}\big)}\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{1+\sqrt{\cot\text{x}}+1+\sqrt{\tan\text{x}}}{1+\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}+\sqrt{\tan\text{x}\cot\text{x}}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{2+\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}{2+\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\text{dx}$
$=\Big[\text{x}\Big]^{\frac{\pi}{2}}_0$
$=\frac{\pi}{2}$
Hence, $\text{I}=\frac{\pi}{4}$
View full question & answer→Question 595 Marks
Evaluate the following integrals as limit of sum:$\int\limits^3_{2}\big(2\text{x}^2+1\big)\text{ dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=2,\text{ b}=3,\text{ f(x)}=2\text{x}^2+1,\text{ h}=\frac{3-2}{\text{n}}=\frac{1}{\text{n}}$
Therefore, $\text{I}=\int\limits^3_{2}\big(2\text{x}^2+1\big)\text{ dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(2)+\text{f}(2+\text{h})+\ ....\ +\text{f}\big\{2+(\text{n}-1)\text{h}\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[2(2.2)^2+1+\big\{2(2+\text{h})^2+1\big\}+\\\ ....\ +\big\{2((2+\text{n}-1)\text{h})^2+1\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+2\Big\{2^2+(2+\text{h})^2+\ .....\big((2+\text{n}-1\big)\text{h}\big)^2\Big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+8\text{n}+2\text{h}^2\Big\{1^2+2^2+3^3+\ .....\ +(\text{n}-1)^2\Big\}\\+8\text{h}\big\{1+2+\ ....\ +(\text{n}-1)\big\}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\text{h}\bigg[9\text{n}+\text{h}^2\frac{2\text{n}(\text{n}-1)(2\text{n}-1)}{6}+8\text{h}\frac{\text{n}(\text{n}-1)}{2}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{1}{\text{n}}\bigg[9\text{n}+\frac{(\text{n}-1)(2\text{n}-1)}{3\text{n}}+4\text{n}-4\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\bigg\{13+\frac{1}{3}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)-\frac{4}{\text{n}}\bigg\}$
$=13+\frac{2}{3}$
$=\frac{41}{3}$
View full question & answer→Question 605 Marks
Evaluate the following integrals:$\int_{0}^\limits{{\frac{\pi}{4}}}(\tan\text{x}+\cot\text{x})^{-2}\text{ dx}$
Answer$\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}(\tan\text{x}+\cot\text{x})^{-2}\text{ dx}$$\Rightarrow\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}\frac{1}{(\tan\text{x}+\cot\text{x})^{2}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}\frac{1}{\Big(\frac{\sin^2\text{x}+\cos^2\text{x}}{\sin\text{x}\cos\text{x}}\Big)^2}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}(\sin\text{x}\cos\text{x})^2\text{dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}\sin^2\text{x}(1-\sin^2\text{x})\text{dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}\sin^2\text{x dx}-\int_{0}^\limits{{\frac{\pi}{4}}}\sin^4\text{x dx}$
We know that by reduction formula,
$\int\sin^{\text{n}}\text{x dx}=\frac{\text{n}-1}{\text{n}}\int\sin^{\text{n}-2}\text{x dx}-\frac{\cos\text{x}\sin^{\text{n}-1}}{\text{n}}$
For n = 2
$\Rightarrow\int\sin^2\text{x dx}=\frac{2-1}{2}\int1\text{ dx}-\frac{\cos\text{x}\sin\text{x}}{2}$
$\Rightarrow\int\sin^2\text{x dx}=\frac{1}{2}\text{x}-\frac{\cos\text{x}\sin\text{x}}{2}$
For n = 4
$\Rightarrow\int\sin^4\text{x dx}=\frac{4-1}{4}\int\sin^2\text{x dx}-\frac{\cos\text{x}\sin^3\text{x}}{4}$
$\Rightarrow\int\sin^4\text{x dx}=\frac{3}{4}\Big\{\frac{1}{2}\times-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}-\frac{\cos\text{x}\sin^3\text{x}}{4}$
Hence,
$\text{I}=\Big\{\frac{1}{2}\times-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}^{\frac{\pi}{4}}_0-\Big\{\frac{3}{4}\Big\{\frac{1}{2}\times-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}-\frac{\cos\text{x}\sin^3\text{x}}{4}\Big\}^{\frac{\pi}{4}}_0$
$\Rightarrow\text{I}=\Big\{\frac{\pi}{8}-\frac{1}{4}\Big\}-\Big\{\frac{3}{4}\Big(\frac{\pi}{8}-\frac{1}{4}\Big)-\frac{1}{16}\Big\}$
$\Rightarrow\text{I}=\frac{\pi}{32}$
$\Rightarrow\int_{0}^\limits{\frac{\pi}{2}}(\sin^{\text{x}}\cos\text{x})^2\text{dx}$
$\Rightarrow\int_{0}^\limits{\frac{\pi}{2}}\sin^2\text{x}(1-\sin^2\text{x})\text{dx}$
$\Rightarrow\int_{0}^\limits{\frac{\pi}{2}}\sin^2\text{x}-\sin^4\text{x dx}$
$\Rightarrow\int_{0}^\limits{\frac{\pi}{2}}\sin^2\text{x dx}-\int_{0}^\limits{\frac{\pi}{2}}\sin^4\text{x dx}$
We know, by reduction formula,
$\int\sin^{\text{n}}\text{x dx}=\frac{\text{n}-1}{\text{n}}\int\sin^{\text{n}-2}\text{x dx}-\frac{\cos\text{x }\sin^{\text{n}-1}\text{x}}{\text{n}}$
For n = 2
$\Rightarrow\int\sin^2\text{x dx}=\frac{2-1}{2}\int1\text{ dx}-\frac{\cos\text{x}\sin\text{x}}{2}$
$\Rightarrow\int\sin^2\text{x dx}=\frac{1}{2}\text{x}-\frac{\cos\text{x}\sin\text{x}}{2}$
For n = 4
$\Rightarrow\int\sin^4\text{x dx}=\frac{4-1}{4}\int\sin^2\text{x dx}-\frac{\cos\text{x}\sin^3\text{x}}{4}$
$\Rightarrow\int\sin^4\text{x dx}=\frac{3}{4}\Big\{\frac{1}{2}\text{x}-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}-\frac{\cos\text{x}\sin^3\text{x}}{4}$
Hence,
$\Big\{\frac{1}{2}\text{x}-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}^{\frac{\pi}{2}}_0-\Big\{\frac{3}{4}\Big\{\frac{1}{2}\text{x}-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}-\frac{\cos\text{x}\sin^3\text{x}}{4}\Big\}^{\frac{\pi}{2}}_0$
$\Rightarrow\frac{\pi}{4}-\frac{3}{4}\Big\{\frac{\pi}{4}\Big\}$
$\Rightarrow\frac{\pi}{16}$
View full question & answer→Question 615 Marks
Evaluate the following definite integrals:$\int_{1}^\limits{2}\frac{\text{x}}{(\text{x}+1)(\text{x}+2)}\text{ dx}$
AnswerLet $\text{I}=\int_{1}^\limits{2}\frac{\text{x}}{(\text{x}+1)(\text{x}+2)}\text{ dx}$ Then,$\text{I}=\int_{1}^\limits{2}\Big(\frac{-1}{(\text{x}+1)}+\frac{2}{(\text{x}+2)}\Big)\text{dx}$
$\Rightarrow\text{I}=-\int_{1}^\limits{2}\frac{1}{(\text{x}+1)}\text{ dx}+2\int_{1}^\limits{2}\frac{1}{(\text{x}+2)}\text{ dx}$
$\Rightarrow\text{I}=\big[-\log(\text{x}+1)+2\log(\text{x}+2)\big]^2_1$
$\Rightarrow\text{I}=-\log3+2\log4+\log2-2\log3$
$\Rightarrow\text{I}=5\log2-3\log3$
$\Rightarrow\text{I}=\log2^5-\log3^3$
$\Rightarrow\text{I}=\log\frac{32}{27}$
View full question & answer→Question 625 Marks
Evaluate the following integrals:$\int\limits^{\pi}_0\text{x}\sin\text{x}\cos^2\text{x dx}$
AnswerLet $\text{I}=\int\limits^{\pi}_0\text{x}\sin\text{x}\cos^2\text{x dx}\ ...(\text{i})$ Then,$\text{I}=\int\limits^{\pi}_0(\pi-\text{x})\sin(\pi-\text{x})\cos^2(\pi-\text{x})\text{dx}$ $\Bigg[\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\pi}_0(\pi-\text{x})\sin\text{x}\cos^2\text{x dx}\ ...(\text{ii})$
Adding (i) and (ii) we get$2\text{I}=\int\limits^{\pi}_0(\pi-\text{x}+\text{x})\sin\text{x}\cos^2\text{x dx}$
$\Rightarrow2\text{I}=\pi\int\limits^{\pi}_0\sin\text{x}\cos^2\text{x dx}$
$\Rightarrow2\text{I}=-\pi\int\limits^{\pi}_0\cos^2\text{x}(-\sin\text{x})\text{dx}$
$\Rightarrow2\text{I}=-\pi\Big[\frac{\cos^3\text{x}}{3}\Big]^{\pi}_0$ $\Bigg[\int\big[\text{f(x)}\big]^{\text{n}}\text{f}'(\text{x})\text{dx}=\frac{\big[\text{f(x)}\big]^{\text{n}+1}}{\text{n}+1}+\text{C}\Bigg]$
$\Rightarrow2\text{I}=-\frac{\pi}{3}\big(\cos^3\text{x}-\cos^20\big)$
$\Rightarrow2\text{I}=-\frac{\pi}{3}(-1-1)=\frac{2\pi}{3}$
$\Rightarrow\text{I}=\frac{\pi}{3}$
View full question & answer→Question 635 Marks
Evaluate the following integrals:$\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}$
AnswerWe know,$\int\limits^{2\pi}_0\text{f(x)}\text{dx}=\int\limits^{2\pi}_0\text{f}(2\pi-\text{x})\text{dx}$
Hence,$\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}=\int\limits^{2\pi}_0\log\big(\sec(2\pi-\text{x})+\tan(2\pi-\text{x})\big)\text{dx}$
$\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}=\int\limits^{2\pi}_0\log(\sec\text{x}-\tan\text{x})\text{dx}$
If$\text{I}=\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}$
$2\text{I}=\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}+\int\limits^{2\pi}_0\log(\sec\text{x}-\tan\text{x})\text{dx}$
$2\text{I}=\int\limits^{2\pi}_0\log(\sec\text{x}+\tan\text{x})\text{dx}+\log(\sec\text{x}-\tan\text{x})\text{dx}$
$2\text{I}=\int\limits^{2\pi}_0\log(\sec^2\text{x}-\tan^2\text{x})\text{dx}$
$2\text{I}=\int\limits^{2\pi}_0\log(1)\text{dx}$
$2\text{I}=0$
$\text{I}=0$
View full question & answer→Question 645 Marks
Evaluate the following integrals:$\int\limits^\pi_0\sin^{100}\text{x}\cos^{101}\text{x dx}$
AnswerLet $\text{I}=\int\limits^\pi_0\sin^{100}\text{x}\cos^{101}\text{x dx}$ Consider $\text{f(x)}=\sin^{100}\text{x}\cos^{101}\text{x}$ Now,$\text{f}(2\pi-\text{x})=\sin^{100}(2\pi-\text{x})\cos^{101}(2\pi-\text{x})$
$=(-\sin\text{x})^{100}(\cos\text{x})^{101}=\sin^{100}\text{x}\cos^{101}\text{x}=\text{f}(\text{x)}$
$\therefore\ \text{I}=\int\limits^{2\pi}_0\sin^{100}\text{x}\cos^{101}\text{x dx}=2\int\limits^{\pi}_0\sin^{100}\text{ x}\cos^{101}\text{x dx}$$\begin{bmatrix}\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=\begin{cases}\int\limits^{2\text{a}}_0\text{f(x)}\text{dx},&\text{if }\text{f}(2\text{a}-\text{x})=\text{f(x)}\\0,&\text{if }\text{f}(2\text{a}-\text{x})=\text{f(x)}\end{cases}\end{bmatrix}$
Again,$\text{f}(\pi-\text{x})=\sin^{100}(\pi-\text{x})\cos^{101}(\pi-\text{x})$
$=(\sin\text{x})^{100}(-\cos\text{x})^{101}=-\sin^{100}\text{x}\cos^{101}\text{x}=-\text{f(x)}$
$\therefore\ \text{I}=2\times0=0$ $\begin{bmatrix}\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=\begin{cases}\int\limits^{2\text{a}}_0\text{f(x)}\text{dx},&\text{if }\text{f}(2\text{a}-\text{x})=\text{f(x)}\\0,&\text{if }\text{f}(2\text{a}-\text{x})=\text{f(x)}\end{cases}\end{bmatrix}$
View full question & answer→Question 655 Marks
Evaluate the following definite integrals:$\int_{0}^\limits{\frac{\pi}{2}}\sqrt{1+\sin\text{x}}\text{ dx}$
AnswerLet $\int_{0}^\limits{\frac{\pi}{2}}\sqrt{1+\sin\text{x}}\text{ dx}$ Then,$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\sqrt{1+\sin\text{x}}\times\frac{\sqrt{1-\sin\text{x}}}{\sqrt{1-\sin\text{x}}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\sqrt{1-\sin^2\text{x}}}{\sqrt{1-\sin^2\text{x}}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\cos\text{x}}{\sqrt{1-\sin\text{x}}}\text{ dx}$
Let $1-\sin\text{x}=\text{u}$$\Rightarrow-\cos\text{x dx}=\text{du}$
$\therefore\ \text{I}=\int\frac{-\text{du}}{\sqrt{\text{u}}}$
$\Rightarrow\text{I}=\big[-2\sqrt{\text{u}}\big]$
$\Rightarrow\text{I}=\big[-2\sqrt{1-\sin\text{x}}\big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=0+2$
$\Rightarrow\text{I}=2$
View full question & answer→Question 665 Marks
Evaluate the following integrals as limit of sum:$\int\limits^5_{0}(\text{x}+1)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=5,\text{ f(x)}=\text{x}+1,\text{ h}=\frac{5-1}{\text{n}}=\frac{5}{\text{n}}$
Therefore, $\text{I}=\int\limits^5_{0}(\text{x}+1)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big\{0+(\text{n}-1)\text{h}\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(0+1)+(\text{h}+1)+\ ....\ +\{(\text{n}-1)\text{h}+1\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+\text{h}\big\{1+2+3+\ ....\ +(\text{n}-1)\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{5}{\text{n}}\Big[\text{n}+\frac{5\text{n}-5}{2}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}5\Big(\frac{7}{2}-\frac{5}{\text{n}}\Big)$
$=\frac{35}{2}$
View full question & answer→Question 675 Marks
Evaluate the following integrals as limit of sum:$\int\limits^{5}_{0}(\text{x}+1)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=5,\text{ f(x)}=\text{x}+1,\text{ h}=\frac{5-0}{\text{n}}=\frac{5}{\text{n}}$
Therefore, $\text{I}=\int\limits^{5}_{0}(\text{x}+1)\text{ dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big\{0+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[(0+1)+(\text{h}+1)+\ ....+\ \big\{(\text{n}-1)\text{h}+1\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{n}+\text{h}\big\{1+2+3+\ ....\ +(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}\frac{5}{\text{n}}\Big[\text{n}+\frac{5(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}5\Big[1+\frac{5}{2}\Big(1-\frac{1}{\text{n}}\Big)\Big]$
$=5+\frac{25}{2}$
$=\frac{35}{2}$
View full question & answer→Question 685 Marks
Evaluate the following integrals:$\int_{0}^\limits{1}\frac{\tan^{-1}\text{x}}{1+\text{x}^2}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\frac{\tan^{-1}\text{x}}{1+\text{x}^2}\text{ dx}$ Then, Let $\tan^{-1}\text{x}=\text{t}$ Then, $\frac{1}{1+\text{x}^2}\text{ dx}=\text{dt}$ When $\text{x}=0,\text{t}=0$ and $\text{x}=1,\text{t}=\frac{\pi}{4}$$\therefore\ \text{I}=\int_{0}^\limits{\frac{\pi}{4}}\text{t}\text{ dt}$
$\Rightarrow\text{I}=\Big[\frac{\text{t}^2}{2}\Big]^{\frac{\pi}{4}}_0$
$\Rightarrow\text{I}=\frac{\pi^2}{32}$
View full question & answer→Question 695 Marks
Evaluate the following integrals:$\int\limits^{7}_0\frac{\sqrt[3]{\text{x}}}{\sqrt[3]{\text{x}}+\sqrt[3]{7}-\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{7}_0\frac{\sqrt[3]{\text{x}}}{\sqrt[3]{\text{x}}+\sqrt[3]{7-\text{x}}}\text{ dx}\ ...(\text{i})$ We know that $\int\limits^{\text{a}}_0\text{f(x)}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})$ Hence,$\text{I}=\int\limits^{7}_0\frac{\sqrt[3]{7-\text{x}}}{\sqrt[3]{7-\text{x}}+\sqrt[3]{\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) & (ii)$2\text{I}=\int\limits^{7}_0\frac{\sqrt[3]{\text{x}}}{\sqrt[3]{\text{x}}+\sqrt[3]{7-\text{x}}}\text{ dx}+\frac{\sqrt[3]{7-\text{x}}}{\sqrt[3]{7-\text{x}}+\sqrt[3]{\text{x}}}\text{ dx}$
$2\text{I}=\int\limits^{7}_0\frac{\sqrt[3]{\text{x}}+\sqrt[3]{7-\text{x}}}{\sqrt[3]{\text{x}}+\sqrt[3]{7-\text{x}}}\text{ dx}$
$2\text{I}=\int\limits^{7}_0\text{dx}$
$2\text{I}=\big[\text{x}\big]^7_0$
$\text{I}=\frac{7}{2}$
View full question & answer→Question 705 Marks
Evaluate the following definite integrals:$\int_{0}^\limits{\frac{\pi}{2}}\big(\text{a}^2\cos^2\text{x}+\text{b}^2\sin^2\text{x}\big)\text{dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\big(\text{a}^2\cos^2\text{x}+\text{b}^2\sin^2\text{x}\big)\text{dx}$ Then,$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\big(\text{a}^2\cos^2\text{x}+\text{b}^2(1-\cos^2\text{x})\big)\text{dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\big(\text{b}^2+(\text{a}^2-\text{b}^2)\cos^2\text{x}\big)\text{dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\Bigg(\text{b}^2+\frac{\big(\text{a}^2-\text{b}^2\big)\big(1+\cos2\text{x}\big)}{2}\Bigg)\text{dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\bigg[\text{b}^2\text{x}+\frac{\text{a}^2-\text{b}^2}{2}\Big(\text{x}+\frac{\sin2\text{x}}{2}\Big)\bigg]_0^{\frac{\pi}{2}}$
$\Rightarrow\text{I}=\frac{\text{b}^2\pi}{2}+\frac{\text{a}^2-\text{b}^2}{2}\frac{\pi}{2}+0$
$\Rightarrow\text{I}=\frac{\pi}{4}\big(\text{a}^2+\text{b}^2\big)$
View full question & answer→Question 715 Marks
Evaluate the following integrals as limit of sum:$\int\limits^{3}_{1}\big(3\text{x}^2+1\text{x}\big)\text{dx}$
AnswerWe have,$\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx} = \lim\limits_{\text{h}\rightarrow0}\text{f}(\text{a}+\text{h})+(\text{a}+2\text{h})+\\...+\text{f}[(\text{a}+(\text{n}-1)\text{h})]\}$
Here, $\text{a}=1, \text{b}=3\text{ f}(\text{x})=3\text{x}^2+1$ and $\text{h}=\frac{3-1}{\text{n}}=\frac{2}{\text{n}}\Rightarrow \text{nh}=2$$\therefore\int\limits^3_1(3\text{x}^2+1)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\{\text{f}(1)+\text{f}(1+\text{h})+\text{f}(1+2\text{h})...+\text{f}[1+(\text{n}-1)\text{h}]\}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\{[3\times1^2+1]+[3\times(1+\text{h})^2+1]\\+[3\times(1+2\text{h})^2+1]+.....+[3\times(1+(\text{n}-1)\text{h}^2+1]$
$=\lim\limits\text{h}\{3[1+(1+2\text{h}+\text{h}^2)+(1+4\text{h}+2^2\text{h}^2)+\\...+(1+2(\text{n}-1)\text{h}+(\text{n}-1)^2\text{h}^2)]+\text{n}\}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\{3[\text{n}+2(1+2+...+(\text{n}-1))\text{h}+(1^2+2^2+\\.....+(\text{n}-1)^2)\text{h}^2]+\text{n}\}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[4\text{n}+6\times\frac{\text{n}(\text{n}-1)}{2}\text{h}+3\times\frac{(\text{n}-1)\text{n}(2\text{n}-1)}{6}\text{h}^2\bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\bigg[4\text{nh}+6\times\frac{\text{nh}(\text{nh}-\text{h)}}{2}+3\times\frac{(\text{nh}-\text{h})\text{nh}(2\text{nh}-\text{h}}{6}\bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\bigg[4\text{nh}+3\times\text{nh}(\text{nh}-\text{h}+3\times\frac{(\text{nh}-\text{h})\text{nh}(2\text{nh}-\text{h})}{6}\bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\bigg[4\times2+3\times2\times(2-\text{h})+3\times\frac{(2-\text{h)}\times2\times(2\times2-\text{h})}{6}\bigg]$
$= 8 + 6\times(2-0)+\frac{(2-0)\times2\times(4-0)}{2}$
$= 8+12+8$
$= 28$
View full question & answer→Question 725 Marks
Evaluate the following integrals:$\int^\limits3_{0}\big|3\text{x}-1\big|\text{dx}$
Answer$\int^\limits3_{0}\big|3\text{x}-1\big|\text{dx}=\int^\limits{\frac{1}{3}}_0-(3\text{x}-1)\text{dx}+\int^\limits{3}_\frac{1}{3}(3\text{x}-1)\text{dx}$$=-\Big[\frac{3\text{x}^2}{2}-\text{x}\Big]^{\frac{1}{3}}_0+\Big[\frac{3\text{x}^2}{2}-\text{x}\Big]^3_{\frac{1}{3}}$
$=-\bigg[\Big(\frac{3}{9\times2}-\frac{1}{3}\Big)-(0)\Big]+\bigg[\Big(\frac{3\times9}{2}-3\Big)-\Big(\frac{3}{9\times2}-\frac{1}{3}\Big)\bigg]$
$=-\bigg[\Big(\frac{1}{6}-\frac{1}{3}\Big)\bigg]+\bigg[\Big(\frac{27}{2}-3\Big)-\Big(\frac{1}{6}-\frac{1}{3}\Big)\bigg]$
$=-\bigg[\Big(-\frac{1}{6}\Big)\bigg]+\bigg[10\frac{1}{2}+\frac{1}{6}\bigg]$
$=\frac{1}{6}+10\frac{1}{2}+\frac{1}{6}$
$=\frac{1}{3}+\frac{21}{2}$
$=\frac{2+63}{6}$
$=\frac{65}{6}$
View full question & answer→Question 735 Marks
Evaluate the following integrals as limit of sum:$\int\limits^3_1(3\text{x}-2)\text{dx}$
View full question & answer→Question 745 Marks
Evaluate the following integrals:$\int^\limits\frac{\pi}{2}_0\text{x}^2\sin\text{x dx}$
AnswerWe have, Using by parts, we get$\text{x}^2\int\sin\text{x dx}-\int\big(\sin\text{x dx}\big)\frac{\text{dx}^2}{\text{dx}}\text{ dx}$
$=\text{x}^2\cos\text{x}+\int\cos\text{x }2\text{x dx}$
Again applying by parts$=\text{x}^2\cos\text{x}+2\Big[\text{x}\int\cos\text{x dx}-\int\big(\int\cos\text{x dx}\big)\frac{\text{dx}}{\text{dx}}\text{ dx}\Big]$
$=\text{x}^2\cos\text{x}+2\big[\text{x}\sin\text{x}-\int\sin\text{x dx}\big]$
$=\Big[\text{x}^2\cos\text{x}+2\text{x }\sin\text{x}+2\cos\text{x}\Big]^{\frac{\pi}{2}}_0$
$=\pi+0-0-0-2$
$=\pi-2$
View full question & answer→Question 755 Marks
If $\int_{0}^\limits{\text{k}}\frac{1}{2+8\text{x}^2}\text{ dx}=\frac{\pi}{16},$ find the value of k.
AnswerWe have,$\int_{0}^\limits{\text{k}}\frac{1}{2+8\text{x}^2}\text{ dx}=\frac{\pi}{16}$
$\Rightarrow\frac{1}{8}\int_{0}^\limits{\text{k}}\frac{\text{dx}}{\big(\frac{1}{2}\big)^2+\text{x}^2}=\frac{\pi}{16}$
$\Rightarrow\frac{1}{8}\big[2\tan^{-1}2\text{x}\big]^{\text{k}}_0=\frac{\pi}{16}$ $\Big[\because\int\frac{\text{dx}}{\text{a}^2-\text{x}^2}=2\tan^{-1}\frac{\text{x}}{\text{a}}\Big]$
$\Rightarrow\frac{1}{4}\big[\tan^{-1}2\text{k}-\tan^{-1}0\big]=\frac{\pi}{16}$
$\Rightarrow\tan^{-1}2\text{k}-0=\frac{\pi}{4}$
$\Rightarrow\tan^{-1}2\text{k}=\frac{\pi}{4}$
$\Rightarrow\text{k}=\frac{1}{2}$
View full question & answer→Question 765 Marks
Evaluate the following integrals:$\int^\limits{\text{a}}_0\text{x}\sqrt{\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\text{ dx}$
AnswerLet $\text{I}=\int^\limits{\text{a}}_0\text{x}\sqrt{\frac{\text{a}^2-\text{x}^2}{\text{a}^2+\text{x}^2}}\text{ dx}$ Consider, $\text{x}^2=\text{a}^2\cos2\theta$$\Rightarrow2\text{xdx}=-2\text{a}^2\sin2\theta\text{ d}\theta$
$\Rightarrow\text{xdx}=-\text{a}^2\sin2\theta\text{ d}\theta$
When, $\text{x}\rightarrow0;\ \theta\rightarrow\frac{\pi}{4}$ and $\text{x}\rightarrow\text{a};\ \theta\rightarrow0$ Now, integral becomes,$\text{I}=\int^\limits0_\frac{\pi}{4}-\text{a}^2\sin2\theta\sqrt{\frac{\text{a}^2-\text{a}^2\cos2\theta}{\text{a}^2+\text{a}^2\cos2\theta}}\text{ d}\theta$
$=\int^\limits0_\frac{\pi}{4}-\text{a}^2\sin2\theta\tan\theta\text{ d}\theta$
$=\text{a}^2\int^\limits{\frac{\pi}{4}}_02\sin\theta\cos\theta\frac{\sin\theta}{\cos\theta}\text{ d}\theta$
$=\text{a}^2\int^\limits{\frac{\pi}{4}}_02\sin\theta\text{ d}\theta$
$=\text{a}^2\int^\limits{\frac{\pi}{4}}_0\big[1-\cos\theta\big]\text{d}\theta$
$=\text{a}^2\Big[\theta-\frac{\sin2\theta}{2}\Big]^{\frac{\pi}{4}}_0$
$=\text{a}^2\Big[\frac{\pi}{4}-\frac{1}{2}\Big]$
View full question & answer→Question 775 Marks
Evaluate the following integrals as limit of sum:$\int\limits^2_{0}\big(\text{x}^2+4\big)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=2,\text{ f(x)}=\text{x}^2+4,\text{ h}=\frac{2-0}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^2_{0}\big(\text{x}^2+4\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big\{0+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[(0+4)+(\text{h}^2+4)+\ ....\ +\big\{(\text{n}-1)^2\text{h}^2+4\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[4\text{n}+\text{h}^2\big\{1^2+2^2+3^2\ ....\ +(\text{n}-1)^2\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[4\text{n}+\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2}{\text{n}}\bigg[4\text{n}+\frac{2(\text{n}-1)(2\text{n}-1)}{3\text{n}}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\bigg\{4+\frac{2}{3}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)\bigg\}$
$=8+\frac{8}{3}$
$=\frac{32}{3}$
View full question & answer→Question 785 Marks
Evaluate the following integrals:$\int_{0}^\limits{{2\pi}}\sqrt{1-\sin\frac{\text{x}}{2}}\text{ dx}$
Answer$\text{I}=\int_{0}^\limits{{2\pi}}\sqrt{1-\sin\frac{\text{x}}{2}}\text{ dx}$$=\int_{0}^\limits{{2\pi}}\sqrt{\cos^2\frac{\text{x}}{4}+\sin^2\frac{\text{x}}{4}+2\sin\frac{\text{x}}{4}\cos\frac{\text{x}}{4}}\text{ dx}$
$=\int_{0}^\limits{{2\pi}}\sqrt{\Big(\cos\frac{\text{x}}{4}+\sin\frac{\text{x}}{4}\Big)^2}\text{ dx}$
$=\int_{0}^\limits{{2\pi}}\Big[\cos\frac{\text{x}}{4}+\sin\frac{\text{x}}{4}\Big]\text{dx}$
When $0\leq\text{x}\leq2\pi,0\leq\frac{\text{x}}{4}\leq\frac{\pi}{2}$
$\therefore\ \sin\frac{\text{x}}{4}\geq0,\cos\frac{\text{x}}{4}\geq0$
$\Rightarrow\Big[\cos\frac{\text{x}}{4}+\sin\frac{\text{x}}{4}\Big]=\cos\frac{\text{x}}{4}+\sin\frac{\text{x}}{4}$
$\therefore\ \text{I}=\int_{0}^\limits{{2\pi}}\Big(\cos\frac{\text{x}}{4}+\sin\frac{\text{x}}{4}\Big)\text{dx}$
$=\Bigg[\frac{\sin\frac{\text{x}}{4}}{\frac{1}{4}}\Bigg]^{2\pi}_0+\Bigg[\frac{\big(-\cos\frac{\text{x}}{4}\big)}{\frac{1}{4}}\Bigg]^{2\pi}_0$
$=4\Big(\sin\frac{\pi}{2}-\sin0\Big)-4\Big(\cos\frac{\pi}{2}-\cos0\Big)$
$=4(1-0)-4(0-1)$
$=4+4$
$=8$
View full question & answer→Question 795 Marks
Evaluate the following integrals:$\int_{0}^\limits{\pi}\frac{\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$
Answer$\int_{0}^\limits{\pi}\frac{\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$$=\frac{1}{2}\int_{0}^\limits{\pi}\frac{2\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$
$=\frac{1}{2}\int_{0}^\limits{\pi}\frac{(\sin\text{x}+\cos\text{x})-(\cos\text{x}-\sin\text{x})}{\sin\text{x}+\cos\text{x}}\text{ dx}$
$=\frac{1}{2}\int_{0}^\limits{\pi}\text{dx}-\frac{1}{2}\int_{0}^\limits{\pi}\frac{\cos\text{x}-\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$
$=\frac{1}{2}\big[\text{x}\big]^{\pi}_0-\frac{1}{2}\big[\log|\sin\text{x}+\cos\text{x}|\big]^{\pi}_0$
$=\frac{1}{2}\big[\pi-0\big]-\frac{1}{2}\big[\log1-\log1\big]$
$=\frac{\pi}{2}$
View full question & answer→Question 805 Marks
Evaluate the following integrals:$\int\limits^{\frac{\pi}{2}}_0\frac{\text{dx}}{1+\tan\text{x}}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\text{dx}}{1+\tan\text{x}}\ ...(\text{i})$$=\int\limits^{\frac{\pi}{2}}_0\frac{\text{dx}}{1+\tan\big(\frac{\pi}{2}-\text{x}\big)}\text{dx}$ $\Big[\text{Using},\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Big]$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\text{dx}}{1+\cot\text{x}}\ ...(\text{ii})$
Adding (i) and (ii)$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\Big(\frac{1}{1+\tan\text{x}}+\frac{1}{1+\cot\text{x}}\Big)\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\Big[\frac{1+\cot\text{x}+1+\tan\text{x}}{(1+\tan\text{x})(1+\cot\text{x})}\Big]\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{2+\tan\text{x}+\cot\text{x}}{1+\tan\text{x}+\cot\text{x}+\tan\text{x}\cot\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{2+\tan\text{x}+\cot\text{x}}{2+\tan\text{x}+\cot\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\text{dx}$
$=\Big[\text{x}\Big]^{\frac{\pi}{2}}_0$
$2\text{I}=\frac{\pi}{2}$
$\therefore\ \text{I}=\frac{\pi}{4}$
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Evaluate the following integrals:$\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}\ ...(\text{i})$$=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\big(\frac{\pi}{2}-\text{x}\big)}{\sin^\text{n}\big(\frac{\pi}{2}-\text{x}\big)+\cos^\text{n}\big(\frac{\pi}{2}-\text{x}\big)}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\cos^\text{n}\text{x}}{\cos^\text{n}\text{x}+\sin^\text{n}\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\cos^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}+\int\limits^{\frac{\pi}{2}}_0\frac{\cos^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\text{dx}$
$=\Big[\text{x}\Big]^{\frac{\pi}{2}}_0=\frac{\pi}{2}$
Hence, $\text{I}=\frac{\pi}{4}$ i.e., $\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}=\frac{\pi}{4}$
View full question & answer→Question 825 Marks
Evaluate the following integrals:$\int_{0}^\limits{\frac{\pi}{2}}\sqrt{\sin\phi}\cos^5\phi\text{ d}\phi$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\sqrt{\sin\phi}\cos^5\phi\text{ d}\phi$$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\sqrt{\sin\phi}\cos^4\phi\cos\phi\text{ d}\phi$
Also, let $\sin\phi=\text{t}\Rightarrow\cos\phi\text{ d}\phi=\text{dt}$ When, $\phi=0,\text{t}=0$ and when $\phi=\frac{\pi}{2},\text{t}=1$$\therefore\ \text{I}=\int_{0}^\limits{1}\sqrt{\text{t}}\big(1-\text{t}^2\big)^2\text{dt}$
$=\int_{0}^\limits{1}\text{t}^{\frac{1}{2}}\big(1+\text{t}^4-2\text{t}^2\big)\text{dt}$
$=\int_{0}^\limits{1}\Big[\text{t}^{\frac{1}{2}}+\text{t}^{\frac{9}{2}}-2\text{t}^{\frac{5}{2}}\Big]\text{dt}$
$=\Bigg[\frac{\text{t}^{\frac{3}{2}}}{\frac{3}{2}}+\frac{\text{t}^{\frac{11}{2}}}{\frac{11}{2}}-\frac{2\text{t}^{\frac{7}{2}}}{\frac{7}{2}}\Bigg]^1_0$
$=\frac{2}{3}+\frac{2}{11}-\frac{4}{7}$
$=\frac{64}{231}$
View full question & answer→Question 835 Marks
Evaluate the following integrals:$\int\limits^{\text{a}}_{-\text{a}}\frac{1}{1+\text{a}^{\text{x}}}\text{ dx},\text{ a}>0$
AnswerLet $\text{I}=\int\limits^{\text{a}}_{-\text{a}}\frac{1}{1+\text{a}^{\text{x}}}\text{ dx}\ ...(\text{i})$ Then,$\text{I}=\int\limits^{\text{a}}_{-\text{a}}\frac{1}{1+\text{a}^{[\text{a}+(-\text{a})-\text{x}]}}\text{ dx}$
$=\int\limits^{\text{a}}_{-\text{a}}\frac{1}{1+\text{a}^{-\text{x}}}\text{ dx}$
$\text{I}=\int\limits^{\text{a}}_{-\text{a}}\frac{1}{\text{a}^{\text{x}}+1}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get$2\text{I}=\int\limits^{\text{a}}_{-\text{a}}\frac{1+\text{a}^{\text{x}}}{1+\text{a}^{\text{x}}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\text{a}}_{-\text{a}}\text{dx}$
$\Rightarrow2\text{I}=\big[\text{x}\big]^{\text{a}}_{-\text{a}}$
$\Rightarrow2\text{I}=\text{a}-(-\text{a})$
$\Rightarrow2\text{I}=2\text{a}$
$\Rightarrow\text{I}=\text{a}$
View full question & answer→Question 845 Marks
Evaluate the following integrals as limit of sum:$\int\limits^{\text{b}}_{\text{a}}\text{x}\text{ dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=\text{a},\text{ b}=\text{b},\text{ f(x)}=\text{x},\text{ h}=\frac{\text{b}-\text{a}}{\text{n}}$
Therefore, $\text{I}=\int\limits^{\text{b}}_{\text{a}}\text{x}\text{ dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\ ....\ +\text{f}\big\{\text{a}+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{a}+(\text{a}+\text{h})+(\text{a}+2\text{h})+\ ....+\ \big\{\text{a}+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{na}+\text{h}\big\{1+2+3+\ ....\ +(\text{n}-1)\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{na}+\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{b}-\text{a}}{\text{n}}\Big[\text{na}\frac{[\text{b}-\text{a}](\text{n}-1)}{2}\Big]$
$=(\text{b}-\text{a})\text{a}+\frac{(\text{b}-\text{a})^2}{2}$
$=\frac{2\text{ab}-2\text{a}^2+\text{b}^2+\text{a}^2-2\text{ab}}{2}$
$=\frac{\text{b}^2-\text{a}^2}{2}$
View full question & answer→Question 855 Marks
Evaluate the following integrals as limit of sum:$\int\limits^4_{1}\big(\text{x}^2-\text{x}\big)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=1,\text{ b}=4,\text{ f(x)}=\text{x}^2-\text{x},\text{ h}=\frac{4-1}{\text{n}}=\frac{3}{\text{n}}$
Therefore, $\text{I}=\int\limits^4_{1}\big(\text{x}^2-\text{x}\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(1)+\text{f}(1+\text{h})+\ ....\ +\text{f}\big\{1+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[(1-1)+\big\{(1+\text{h}^2)-(1+\text{h})\big\}+\\\ ....\ +\big\{(1+(\text{n}-1)\text{h})^2-(1+(\text{n}-1)\text{h})\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{h}^2\big\{1^2+2^2+3^2\ ....\ +\$\text{n}-1)^2\big\}+1+2\text{h}\big\{1+2+\ ....+\$\text{n}-1)\big\}-\text{n}-\text{h}\big\{1+2+\ .....\ +(\text{n}-1)\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}+\text{h}\frac{(\text{n}-1)}{2}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{3}{\text{n}}\bigg[\frac{9(\text{n}-1)(2\text{n}-1)}{6\text{n}}+\frac{3(\text{n}-1)}{2}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}3\bigg[\frac{3}{2}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)+\frac{3}{2}\Big(1-\frac{1}{\text{n}}\Big)\bigg]$
$=9+\frac{9}{2}$
$=\frac{27}{3}$
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Evaluate the following integrals:$\int_{0}^\limits{2}\text{x}\sqrt{\text{x}+2}\text{ dx}$
Answer$\int_{0}^\limits{2}\text{x}\sqrt{\text{x}+2}\text{ dx}$Let $\text{x}+2=\text{t}^2\Rightarrow\text{dx}=2\text{tdt}$
When $\text{x}=0,\text{t}=\sqrt{2}$ and when $\text{x}=2,\text{t}=2$
$\therefore\ \int_{0}^\limits{2}\text{x}\sqrt{\text{x}+2}\text{ dx}=\int_{\sqrt{2}}^\limits{2}\big(\text{t}^2-2\big)\sqrt{\text{t}^2}2\text{t dt}$
$=2\int_{\sqrt{2}}^\limits{2}\big(\text{t}^2-2\big)\text{t}^2\text{ dt}$
$=2\int_{\sqrt{2}}^\limits{2}\big(\text{t}^4-2\text{t}^2\big)\text{dt}$
$=2\Big[\frac{\text{t}^5}{5}-\frac{2\text{t}^3}{3}\Big]^2_\sqrt{2}$
$=2\Big[\frac{32}{5}-\frac{16}{3}-\frac{4\sqrt{2}}{5}+\frac{4\sqrt{2}}{3}\Big]$
$=2\Big[\frac{96-80-12\sqrt{2}+20\sqrt{2}}{15}\Big]$
$=2\Big[\frac{16+8\sqrt{2}}{15}\Big]$
$=\frac{16\big(2+\sqrt{2}\big)}{15}$
$=\frac{16\sqrt{2}\big(\sqrt{2}+1\big)}{15}$
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Evaluate the following integrals:$\int^\limits{\frac{\pi}{2}}_02\sin\text{x }\cos\text{x}\tan^{-1}(\sin\text{x})\text{dx}$
AnswerLet $\text{I}=\int^\limits{\frac{\pi}{2}}_02\sin\text{x }\cos\text{x}\tan^{-1}(\sin\text{x})\text{dx}$ Then, Let $\sin\text{x}=\text{t}$ Then, $\cos\text{x dx}=\text{dt}$ When $\text{x}=0,\text{t}=0$ and $\text{x}=\frac{\pi}{2},\text{t}=1$$\Rightarrow\text{I}=2\Big[\frac{\text{t}^2\tan^{-1}\text{t}}{2}\Big]^1_0-\int^\limits1_0\frac{\text{t}^2}{1+\text{t}^2}\text{ dt}$
$\Rightarrow\text{I}=2\Big[\frac{\text{t}^2\tan^{-1}\text{t}}{2}\Big]^1_0-\int^\limits1_0\Big(\frac{1+\text{t}^3}{1+\text{t}^2}-\frac{1}{1+\text{t}^2}\Big)\text{dt}$
$\Rightarrow\text{I}=2\Big[\frac{\text{t}^2\tan^{-1}\text{t}}{2}\Big]^1_0-\big[\text{t}-\tan^{-1}\text{t}\big]^1_0$
$\Rightarrow\text{I}=1\tan^{-1}1-0-1+\tan^{-1}1+0$
$\Rightarrow\text{I}=\frac{\pi}{4}-1+\frac{\pi}{4}$
$\Rightarrow\text{I}=\frac{\pi}{2}-1$
View full question & answer→Question 885 Marks
Evaluate the following integrals:$\int_{0}^\limits{\frac{\pi}{4}}\big(\text{a}^2\cos^2\text{x}+\text{b}^2\sin^2\text{x}\big)\text{dx}$
Answer$\int_{0}^\limits{\frac{\pi}{4}}\big(\text{a}^2\cos^2\text{x}+\text{b}^2\sin^2\text{x}\big)\text{dx}$$=\int_{0}^\limits{\frac{\pi}{4}}\Big[\text{a}^2\Big(\frac{1+\cos2\text{x}}{2}\Big)+\text{b}^2\Big(\frac{1-\cos2\text{x}}{2}\Big)\Big]\text{dx}$
$=\int_{0}^\limits{\frac{\pi}{4}}\Big[\Big(\frac{\text{a}^2+\text{b}^2}{2}\Big)+\Big(\frac{\text{a}^2-\text{b}^2}{2}\Big)\cos2\text{x}\Big]\text{dx}$
$=\Big(\frac{\text{a}^2+\text{b}^2}{2}\Big)\int_{0}^\limits{\frac{\pi}{4}}\text{dx}+\Big(\frac{\text{a}^2-\text{b}^2}{2}\Big)\int_{0}^\limits{\frac{\pi}{4}}\cos2\text{x }\text{dx}$
$=\Big(\frac{\text{a}^2+\text{b}^2}{2}\Big)\Big(\frac{\pi}{4}-0\Big)+\Big(\frac{\text{a}^2-\text{b}^2}{2}\Big)\Big(\sin\frac{\pi}{2}-\sin0\Big)$
$=\Big(\frac{\text{a}^2+\text{b}^2}{2}\Big)\frac{\pi}{4}+\Big(\frac{\text{a}^2-\text{b}^2}{2}\Big)(1-0)$
$=\big(\text{a}^2+\text{b}^2\big)\frac{\pi}{8}+\frac{1}{4}\big(\text{a}^2-\text{b}^2\big)$
View full question & answer→Question 895 Marks
Evaluate the following integrals:$\int_{0}^\limits{\pi}\frac{1}{3+2\sin\text{x}+\cos\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\pi}\frac{1}{3+2\sin\text{x}+\cos\text{x}}\text{ dx}$ Then,$\text{I}=\int_{0}^\limits{\pi}\frac{1}{3+2\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)+\frac{1-\tan^{2}\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\pi}\frac{1+\tan^{2}\frac{\text{x}}{2}}{2\tan^{2}\frac{\text{x}}{2}+4\tan\frac{\text{x}}{2}+4}\text{ dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$ Then, $\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{ dx}=\text{dt}$ When $\text{x}=0,\text{t}=0$ and $\text{x}=\pi,\text{t}=\infty$$\therefore\ \text{I}=\int\limits^{\infty}_0\frac{2\text{dt}}{2\text{t}^2+4\text{t}+4}$
$\Rightarrow\text{I}=\int\limits^{\infty}_0\frac{\text{dt}}{(\text{t}+1)^2+1}$
$\Rightarrow\text{I}=\Big[\tan^{-1}\big(\text{t}+1\big)\Big]^{\infty}_0$
$\Rightarrow\text{I}=\frac{\pi}{2}-\frac{\pi}{4}$
$\Rightarrow\text{I}=\frac{\pi}{4}$
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Evaluate the following integrals:$\int\limits^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}\text{x}}{\sin^{\frac{3}{2}}\text{x}+\cos^{\frac{3}{2}}\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}\ ...(\text{i})$$=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\big(\frac{\pi}{2}-\text{x}\big)}{\sin^\text{n}\big(\frac{\pi}{2}-\text{x}\big)+\cos^\text{n}\big(\frac{\pi}{2}-\text{x}\big)}\text{ dx}$ $\Big[\text{Using},\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Big]$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\cos^\text{n}\text{x}}{\cos^\text{n}\text{x}+\sin^\text{n}\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\cos^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}+\int\limits^{\frac{\pi}{2}}_0\frac{\cos^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\text{dx}$
$=\Big[\text{x}\Big]^{\frac{\pi}{2}}_0=\frac{\pi}{2}$
Hence, $\text{I}=\frac{\pi}{4}$ i.e., $\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}=\frac{\pi}{4}$$\therefore\ \int\limits^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}\text{x}}{\sin^{\frac{3}{2}}\text{x}+\cos^{\frac{3}{2}}\text{x}}\text{ dx}=\frac{\pi}{4}$
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If $\text{f}(\text{a}+\text{b}-\text{x})=\text{f(x)},$ then prove that $\int\limits^{\text{b}}_\text{a}\text{x}\text{f(x)}\text{dx}=\frac{\text{a}+\text{b}}{2}\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{x}\text{f(x)}\text{dx}$$=\int\limits^{\text{b}}_\text{a}(\text{a}+\text{b}-\text{x})\text{f}(\text{a}+\text{b}-\text{x})\text{dx}$$\Bigg[\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\int\limits^{\text{b}}_\text{a}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\text{b}}_\text{a}(\text{a}+\text{b}-\text{x})\text{f(x)}\text{dx}$$\Big[\text{f}(\text{a}+\text{b}-\text{x})=\text{f(x)}\Big]$
$\therefore\ \int\limits^{\text{b}}_\text{a}\text{x}\text{f(x)}\text{dx}=\int\limits^{\text{b}}_\text{a}(\text{a}+\text{b})\text{f(x)}\text{dx}-\int\limits^{\text{b}}_\text{a}\text{x}\text{f(x)}\text{dx}$
$\Rightarrow\int\limits^{\text{b}}_\text{a}\text{x}\text{f(x)}\text{dx}+\int\limits^{\text{b}}_\text{a}\text{x}\text{f(x)}\text{dx}=(\text{a}+\text{b})\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}$
$\Rightarrow2\int\limits^{\text{b}}_\text{a}\text{x}\text{f(x)}\text{dx}=(\text{a}+\text{b})\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}$
$\Rightarrow\int\limits^{\text{b}}_\text{a}\text{x}\text{f(x)}\text{dx}=\frac{\text{a}+\text{b}}{2}\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}$
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If f(x) is a continuous function defind on [-a, a], then prove that:$\int\limits^{\text{a}}_{-\text{a}}\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\big\{\text{f(x)}+\text{f}(-\text{x})\big\}\text{dx}$
AnswerLet $\text{I}=\int\limits^{\text{a}}_{-\text{a}}\text{f(x)}\text{dx}$ By Additive property$\text{I}=\int\limits^0_{-\text{a}}\text{f(x)}\text{dx}+\int\limits^{\text{a}}_0\text{f(x)}\text{dx}$
Let $\text{x}=-\text{t},$ then $\text{dx}=-\text{dt}$ When $\text{x}=-\text{a},\text{ t}=\text{a},\text{ x}=0,\text{ t}=0$ Hence, $\int\limits^0_{-\text{a}}\text{f(x)}\text{dx}=-\int\limits^0_{\text{a}}\text{f}(-\text{t})\text{dt}$$=\int\limits_0^{\text{a}}\text{f}(-\text{t})\text{dt}=\int\limits_0^{\text{a}}\text{f}(-\text{x})\text{dx}$ (Changing the varible)
Therefore,$\text{I}=\int\limits_0^{\text{a}}\text{f}(-\text{x})\text{dx}+\int\limits_0^{\text{a}}\text{f}(\text{x})\text{dx}$
$=\int\limits^{\text{a}}_0\big\{\text{f(x)}+\text{f}(-\text{x})\big\}\text{dx}$
Hence, proved.
View full question & answer→Question 935 Marks
Evaluate the following integrals:$\int\limits^{2\pi}_0\frac{\text{e}^{\sin\text{x}}}{\text{e}^{\sin\text{x}}+\text{e}^{-\sin\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{2\pi}_0\frac{\text{e}^{\sin\text{x}}}{\text{e}^{\sin\text{x}}+\text{e}^{-\sin\text{x}}}\text{ dx}\ ...(\text{i})$ Then,$\text{I}=\int\limits^{2\pi}_0\frac{\text{e}^{\sin(2\pi-\text{x})}}{\text{e}^{\sin(2\pi-\text{x})}+\text{e}^{-\sin(2\pi-\text{x})}}\text{ dx}$ $\Bigg(\int\limits^\text{a}_0\text{f(x)}\text{dx}=\int\limits^\text{a}_0\text{f}(\text{a}-\text{x})\text{dx}\Bigg)$
$\text{I}=\int\limits^{2\pi}_0\frac{\text{e}^{-\sin\text{x}}}{\text{e}^{-\sin\text{x}}+\text{e}^{\sin\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get$\text{I}=\int\limits^{2\pi}_0\frac{\text{e}^{\sin\text{x}}+\text{e}^{-\sin\text{x}}}{\text{e}^{\sin\text{x}}+\text{e}^{-\sin\text{x}}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{2\pi}_0\text{dx}$
$\Rightarrow2\text{I}=\big[\text{x}\big]^{2\pi}_0$
$\Rightarrow2\text{I}=2\pi-0$
$\Rightarrow\text{I}=\pi$
View full question & answer→Question 945 Marks
Evaluate the following integrals as limit of sum:$\int\limits^\text{b}_{\text{a}}\text{e}^{\text{x}}\text{ dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=\text{a},\text{ b}=\text{b},\text{ f(x)}=\text{e}^{\text{x}},\text{ h}=\frac{\text{b}-\text{a}}{\text{n}}$
Therefore, $\text{I}=\int\limits^\text{b}_{\text{a}}\text{e}^{\text{x}}\text{ dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\ ....\ +\text{f}\big\{\text{a}+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{e}^\text{a}+\text{e}^{\text{a}+\text{h}}+\ .....\ +\text{e}^{\{\text{a}+(\text{n}-1)\text{h}\}}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[\text{e}^{\text{a}}\bigg\{\frac{(\text{e}^{\text{h}})^{\text{n}}-1}{\text{e}^{\text{h}}-1}\bigg\}\bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\Big[\text{e}^\text{a}\frac{\text{e}^{\text{b}-\text{a}}-1}{\text{e}^{\text{h}}-1}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}=\Bigg[\frac{\text{e}^\text{b}-\text{e}^{\text{a}}}{\frac{\text{e}^\text{h}-1}{\text{h}}}\Bigg]$
$=\frac{\text{e}^\text{b}-\text{e}^\text{a}}{1}$
$=\text{e}^{\text{b}}-\text{e}^\text{a}$
View full question & answer→Question 955 Marks
Evaluate the following definite integrals:$\int_{0}^\limits{\frac{\pi}{2}}\cos^3\text{x}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\cos^3\text{x}\text{ dx}$ Then,$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\cos^3\text{x}\cos\text{x}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\big(1-\sin^2\text{x}\big)\cos\text{x}\text{ dx}$
Let $\text{u}=\sin\text{x},\text{ du}=\cos\text{x dx}$$\Rightarrow\text{I}=\int(1-\text{u}^2)\text{du}$
$\Rightarrow\text{I}=\Big[\text{u}-\frac{\text{u}^3}{3}\Big]$
$\Rightarrow\text{I}=\Big[\sin\text{x}-\frac{\sin^3\text{x}}{3}\Big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=1-\frac{1}{3}-0$
$\Rightarrow\text{I}=\frac{2}{3}$
View full question & answer→Question 965 Marks
Evaluate the following integrals:$\int_{0}^\limits{1}\text{x}\log(1+2\text{x})\text{dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\text{x}\log(1+2\text{x})\text{dx}$Apply integral by part
$\text{I}=\Big[\log(1+2\text{x})\frac{\text{x}^2}{2}\Big]^{1}_0-\int_{0}^\limits{1}\Big(\frac{2}{1+2\text{x}}\Big)\times\frac{\text{x}^2}{2}\text{ dx}$
$=\frac{1}{2}\big(\log3-0\big)-\int_{0}^\limits{1}\frac{\text{x}^2}{1+2\text{x}}\text{ dx}$
$=\frac{1}{2}\log3-\frac{1}{4}\int_{0}^\limits{1}\frac{4\text{x}^2-1+1}{1+2\text{x}}\text{ dx}$
$=\frac{1}{2}\log3-\frac{1}{4}\int_{0}^\limits{1}\frac{(2\text{x}+1)(2\text{x}-1)}{1+2\text{x}}\text{ dx}-\frac{1}{4}\int_{0}^\limits{1}\frac{2}{1+2\text{x}}\text{ dx}$
$=\frac{1}{2}\log3-\frac{1}{4}\int_{0}^\limits{1}(2\text{x}-1)\text{dx}-\frac{1}{4}\int_{0}^\limits{1}\frac{1}{1+2\text{x}}\text{ dx}$
$=\frac{1}{2}\log3-\bigg[\frac{1}{4}\times\frac{(2\text{x}-1)^2}{2\times2}\bigg]^1_0-\bigg[\frac{1}{4}\times\frac{\log(1+2\text{x})}{2}\bigg]^1_0$
$=\frac{1}{2}\log3-\frac{1}{16}(1-1)-\frac{1}{8}\big(\log3-\log1\big)$
$=\frac{1}{2}\log3-0-\frac{1}{8}\log3$ $(\log1=0)$
$=\frac{3}{8}\log3$
View full question & answer→Question 975 Marks
Prove that:$\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}=\frac{\pi}{2}\int\limits^\pi_0\text{f}(\sin\text{x})\text{dx}$
Answer$\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}=\int\limits^\pi_0(\pi-\text{x})\text{f}\big[\sin(\pi-\text{x})\big]\text{dx}$ $\Bigg[\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$$\Rightarrow\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}=\int\limits^\pi_0(\pi-\text{x})\text{f}(\sin\text{x})\text{dx}$
$\Rightarrow\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}=\pi\int\limits^\pi_0\text{f}(\sin\text{x})\text{dx}-\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}$
$\Rightarrow2\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}=\pi\int\limits^\pi_0\text{f}(\sin\text{x})\text{dx}$
$\Rightarrow\int\limits^\pi_0\text{xf}(\sin\text{x})\text{dx}=\frac{\pi}{2}\int\limits^\pi_0\text{f}(\sin\text{x})\text{dx}$
View full question & answer→Question 985 Marks
Evaluate the following definite integrals:$\int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\sin\text{x}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\sin\text{x}\text{ dx}$ Apply integration by part.$\text{I}=\big[\text{x}^2(-\cos\text{x})\big]^{\frac{\pi}{2}}_0-\int_{0}^\limits{\frac{\pi}{2}}2\text{x}(-\cos\text{x})\text{ dx}$
$\Rightarrow\text{I}=(0-0)+2\int_{0}^\limits{\frac{\pi}{2}}\text{x}\cos\text{x}\text{ dx}$ $\Big(\cos\frac{\pi}{2}=0\Big)$
Apply integration by part again,$\text{I}=0+2\Bigg[\big[\text{x}\sin\text{x}\big]_0^{\frac{\pi}{2}}-\int_{0}^\limits{\frac{\pi}{2}}1\times\sin\text{x}\text{ dx}\Bigg]$
$\Rightarrow\text{I}=2\Big(\frac{\pi}{2}\sin\frac{\pi}{2}-0\Big)-2\int_{0}^\limits{\frac{\pi}{2}}\sin\text{x dx}$
$\Rightarrow\text{I}=2\Big(\frac{\pi}{2}-0\Big)-\big[2(-\cos\text{x})\big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=\pi+2\Big(\cos\frac{\pi}{2}-\cos0\Big)$
$\Rightarrow\text{I}=\pi+2(0-1)$
$\Rightarrow\text{I}=\pi-2$
View full question & answer→Question 995 Marks
Evaluate the following integrals as limit of sum:$\int\limits^1_{0}\big(3\text{x}^2+5\text{x}\big)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=1,\text{ f(x)}=3\text{x}^2+5\text{x},\text{ h}=\frac{1-0}{\text{n}}=\frac{1}{\text{n}}$
Therefore, $\text{I}=\int\limits^1_{0}\big(3\text{x}^2+5\text{x}\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big\{0+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[(0+0)+(3\text{h}^2+5\text{h})+\ \\ .....\ +\big\{3(\text{n}-1)^2\text{h}^2+5(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[5\text{h}(1+2+\ ....\ +\text{n})\\+3\text{h}^2\big\{1^2+2^2+3^2\ ....\ +(\text{n}-1)^2\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[5\text{h}\frac{\text{n}(\text{n}-1)}{2}+\text{h}^2\frac{3\text{n}(\text{n}-1)(2\text{n}-1)}{6}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{1}{\text{n}}\bigg[\frac{5(\text{n}-1)}{2}+\frac{(\text{n}-1)(2\text{n}-1)}{2\text{n}}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\bigg[\frac{5}{2}\Big(1-\frac{1}{\text{n}}\Big)+\frac{1}{2}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)\bigg]$
$=\frac{5}{2}+1$
$=\frac{7}{2}$
View full question & answer→Question 1005 Marks
Evaluate the following integrals:$\int\limits^{\infty}_0\frac{\log\text{x}}{1+\text{x}^2}\text{ dx}$
AnswerWe have,$\text{I}=\int\limits^{\infty}_0\frac{\log\text{x}}{1+\text{x}^2}\text{ dx}$
Putting $\text{x}=\tan\theta$$\text{dx}=\sec^2\theta\text{ d}\theta$
When $\text{x}\rightarrow0;\theta\rightarrow0$ And $\text{x}\rightarrow\infty;\theta\rightarrow\frac{\pi}{2}$ Now, integral becomes,$\therefore\ \text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\log(\tan\theta)}{1+\tan^2\theta}\sec^2\theta\text{ d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\log(\tan\theta)\text{d}\theta\ ...(\text{i})$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\log\Big[\tan\Big(\frac{\pi}{2}-\theta\Big)\Big]\text{d}\theta$ $\Bigg[\because\ \int\limits^{\text{a}}_{0}\text{f(x)}\text{dx}=\int\limits^{\text{a}}_{\text{0}}\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\log(\cot\theta)\text{d}\theta\ ....(\text{ii})$
Adding (i) and (ii) we get$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\log(\tan\theta)\text{d}\theta+\int\limits^{\frac{\pi}{2}}_0\log(\cot\theta)\text{d}\theta$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\big[\log(\tan\theta)+\log(\cot\theta)\big]\text{d}\theta$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\big[\log(\tan\theta\times\cot\theta)\big]\text{d}\theta$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\big(\log1\big)\text{d}\theta$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0(0)\text{d}\theta$
$\Rightarrow2\text{I}=0$
$\Rightarrow\text{I}=0$
$\therefore\ \int\limits^{\infty}_0\frac{\log\text{x}}{1+\text{x}^2}\text{ dx}=0$
View full question & answer→Question 1015 Marks
Evaluate the following integrals as limit of sum:$\int\limits^2_{0}\text{e}^{\text{x}}\text{ dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=2,\text{ f(x)}=\text{e}^{\text{x}},\text{ h}=\frac{2-0}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^2_{0}\text{e}^{\text{x}}\text{ dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big\{0+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{e}^0+\text{e}^\text{h}+\text{e}^{2\text{h}}+\ .....\ +\text{e}^{(\text{n}-1)\text{h}}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[\frac{(\text{e}^{\text{h}})^{\text{n}}-1}{\text{e}^{\text{h}}-1}\bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\Bigg[\frac{\text{e}^2-1}{\frac{\text{e}^{\text{h}}-1}{\text{h}}}\Bigg]$
$=\frac{\text{e}^2-1}{1}$
$=\text{e}^2-1$
View full question & answer→Question 1025 Marks
Evaluate the following definite integrals:$\int_{\text{e}}^\limits{\text{e}^2}\Big\{\frac{1}{\log\text{x}}-\frac{1}{(\log\text{x})^2}\Big\}\text{dx}$
AnswerLet $\text{I}=\int_{\text{e}}^\limits{\text{e}^2}\Big\{\frac{1}{\log\text{x}}-\frac{1}{(\log\text{x})^2}\Big\}\text{dx}$ Then,$\text{I}=\int_{\text{e}}^\limits{\text{e}^2}\frac{1}{\log\text{x}}\text{ dx}-\int_{\text{e}}^\limits{\text{e}^2}\frac{1}{(\log\text{x})^2}\text{ dx}$
Integrating by parts$\Rightarrow\text{I}=\Bigg\{\Big[\frac{\text{x}}{\log\text{x}}\Big]^{\text{e}^2}_\text{e}-\int_{\text{e}}^\limits{\text{e}^2}\frac{-1}{\text{x}(\log\text{x})^2}\text{x dx}\Bigg\}-\int_{\text{e}}^\limits{\text{e}^2}\frac{1}{(\log\text{x})^2}\text{ dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{x}}{\log\text{x}}\Big]^{\text{e}^2}_\text{e}+\int_{\text{e}}^\limits{\text{e}^2}\frac{1}{(\log\text{x})^2}\text{ dx}-\int_{\text{e}}^\limits{\text{e}^2}\frac{1}{(\log\text{x})}\text{ dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{x}}{\log\text{x}}\Big]^{\text{e}^2}_\text{e}+0$
$\Rightarrow\text{I}=\frac{\text{e}^2}{\log\text{e}^2}-\frac{\text{e}}{\log\text{e}}$
$\Rightarrow\text{I}=\frac{\text{e}^2}{2\log\text{e}}-\frac{\text{e}}{\log\text{e}}$
$\Rightarrow\text{I}=\frac{\text{e}^2}{2}-\text{e}$
View full question & answer→Question 1035 Marks
Evaluate the following integrals:$\int\limits^4_1\big\{|\text{x}-1|+|\text{x}-2|+|\text{x}-4\big\}\text{dx}$
Answer$\text{I}=\int\limits^4_1\big\{|\text{x}-1|+|\text{x}-2|+|\text{x}-4\big\}\text{dx}$$\Rightarrow\text{I}=\int\limits^4_1|\text{x}-1|\text{dx}+\int\limits^4_1|\text{x}-2|\text{dx}+\int\limits^4_1|\text{x}-4|\text{dx}$
We know that,
$|\text{x}-2|=\begin{cases}-(\text{x}-1),&\text{x}\leq1\\\text{x}-1,&1<\text{x}\leq4\end{cases}$
$|\text{x}-2|=\begin{cases}-(\text{x}-2),&1\leq\text{x}\leq2\\\text{x}-2,&2<\text{x}\leq4\end{cases}$
$|\text{x}-4|=\begin{cases}-(\text{x}-4),&1\leq\text{x}\leq4\\\text{x}-4,&\text{x}>4\end{cases}$
$\therefore\ \text{I}=\int\limits^4_1(\text{x}-1)\text{dx}-\int\limits^2_1(\text{x}-2)\text{dx}+\int\limits^4_2(\text{x}-2)\text{dx}-\int\limits^4_1(\text{x}-4)\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{x}^2}{2}-\text{x}\Big]^4_1-\Big[\frac{\text{x}^2}{2}-\text{x}\Big]^2_1+\Big[\frac{\text{x}^2}{2}-\text{x}\Big]^4_2-\Big[\frac{\text{x}^2}{2}-\text{x}\Big]^4_1$
$\Rightarrow\text{I}=8-4-\frac{1}{2}+1-\Big(2-4-\frac{1}{2}+2\Big)+8-8-2+4-\Big(8-16-\frac{1}{2}+4\Big)$
$\Rightarrow\text{I}=\frac{23}{2}$
View full question & answer→Question 1045 Marks
Evaluate the following integrals:$\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin^4\text{x dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin^4\text{x dx}$$=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\big(\sin^2\text{x}\big)^2\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\Big(\frac{1-\cos2\text{x}}{2}\Big)^2\text{dx}$
$=\frac{1}{4}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\big(1-2\cos2\text{x}+\cos^22\text{x}\big)\text{dx}$
$=\frac{1}{4}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{dx}-\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\cos2\text{x dx}+\frac{1}{8}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}(1+\cos4\text{x})\text{dx}$
$=\frac{1}{4}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{dx}-\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\cos2\text{x dx}+\frac{1}{8}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{dx}+\frac{1}{8}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\cos4\text{x dx}$
$=\frac{1}{4}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{dx}-\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\cos2\text{x dx}+\frac{1}{8}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\cos4\text{x dx}$
$=\frac{3}{8}\Big[\text{x}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}-\frac{1}{4}\Big[\sin2\text{x}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}+\frac{1}{32}\Big[\sin4\text{x}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}$
$=\frac{3}{8}\Big(\frac{\pi}{2}+\frac{\pi}{2}\Big)-\frac{1}{4}(0-0)+\frac{1}{32}(0-0)$
Hence, $\text{I}=\frac{3\pi}{8}$
View full question & answer→Question 1055 Marks
Evaluate the following integrals as limit of sum:$\int\limits^{2}_{0}\big(3\text{x}^2-2\big)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=2,\text{ f(x)}=3\text{x}^2-2,\text{ h}=\frac{2-0}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^{2}_{0}\big(3\text{x}^2-2\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big\{0+(\text{n}-1)\text{h}\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(0-2)+(3\text{h}^2-2)+\ ....\ +\{3(\text{n}-1)^2\text{h}^2-2\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[-2\text{n}+3\text{h}^2\big\{1^2+2^2+3^2+\ ....\ +(\text{n}-1)^2\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[-2\text{n}+3\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2}{\text{n}}\Big[-2\text{n}+\frac{2(\text{n}-1)(2\text{n}-1)}{\text{n}}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\bigg\{-2+2\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)\bigg\}$
$=-4+8$
$=4$
View full question & answer→Question 1065 Marks
Evaluate the following definite integrals:$\int_{-\frac{\pi}{4}}^\limits{\frac{\pi}{4}}\frac{1}{1+\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{-\frac{\pi}{4}}^\limits{\frac{\pi}{4}}\frac{1}{1+\sin\text{x}}\text{ dx}$ Then,$\text{I}=\int_{-\frac{\pi}{4}}^\limits{\frac{\pi}{4}}\frac{1}{1+\sin\text{x}}\times\frac{1-\sin\text{x}}{1-\sin\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\int_{-\frac{\pi}{4}}^\limits{\pi}\frac{1-\sin\text{x}}{1-\sin^2\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\int_{-\frac{\pi}{4}}^\limits{\pi}\frac{1-\sin\text{x}}{\cos^2\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\int_{-\frac{\pi}{4}}^\limits{\pi}\Big(\frac{1}{\cos^2\text{x}}-\frac{\sin\text{x}}{\cos^2\text{x}}\Big)\text{dx}$
$\Rightarrow\text{I}=\int_{-\frac{\pi}{4}}^\limits{\pi}\big(\sec^2\text{x}-\sec\text{x}\tan\text{x}\big)\text{ dx}$
$\Rightarrow\text{I}=\big[\tan\text{x}-\sec\text{x}\big]^\frac{\pi}{4}_{-\frac{\pi}{4}}$
$\Rightarrow\text{I}=\big(1-\sqrt{2}\big)-\big(-1-\sqrt{2}\big)$
$\Rightarrow\text{I}=2$
View full question & answer→Question 1075 Marks
Evaluate the following integrals:$\int\limits^{5}_0\frac{\sqrt[4]{\text{x}+4}}{\sqrt[4]{\text{x}+4}+\sqrt[9]{9-\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{5}_0\frac{\sqrt[4]{\text{x}+4}}{\sqrt[4]{\text{x}+4}-\sqrt[9]{9-\text{x}}}\text{ dx}\ ....(\text{i})$$\text{I}=\int\limits^{5}_0\frac{\sqrt[4]{9-\text{x}}}{\sqrt[4]{9-\text{x}}-\sqrt[4]{\text{x}+4}}\text{ dx}$
$\Big[\text{Using},\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Big]$
$\text{I}=\int\limits^{5}_0\frac{\sqrt[4]{9-\text{x}}}{\sqrt[4]{\text{x}+4}-\sqrt[4]{9-\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii)$2\text{I}=\int\limits^{5}_0\frac{\sqrt[4]{\text{x}+4}}{\sqrt[4]{\text{x}+4}-\sqrt[4]{9-\text{x}}}-\frac{\sqrt[4]{9-\text{x}}}{\sqrt[4]{\text{x}+4}-\sqrt[4]{9-\text{x}}}\text{ dx}$
$=\int\limits^{5}_0\frac{\sqrt[4]{\text{x}+4}-\sqrt[4]{9-\text{x}}}{\sqrt[4]{\text{x}+4}-\sqrt[4]{9-\text{x}}}\text{ dx}$
$=\int\limits^{5}_0\text{dx}$
$=\big[\text{x}\big]^5_0$
$=5$
Hence, $\text{I}=\frac{5}{2}$
View full question & answer→Question 1085 Marks
Evaluate the following integrals:$\int_{1}^\limits{2}\frac{1}{\text{x}\big(1+\log\text{x}\big)^2}\text{ dx}$
AnswerLet $1+\log\text{x}=\text{t}$ Differentiating w.r.t. x, we get$\frac{1}{\text{x}}\text{ dx}=\text{dt}$
Now, $\text{x}=1\Rightarrow\text{t}=1$$=\text{x}=2\Rightarrow\text{t}=1+\log2$
$\therefore\ \int_{1}^\limits{2}\frac{1}{\text{x}\big(1+\log\text{x}\big)^2}\text{ dx}=\int^\limits{1+\log2}_1\frac{\text{dt}}{\text{t}^2}$
$=\Big[\frac{-1}{\text{t}}\Big]^{1+\log2}_1$
$=\bigg[\frac{-1}{1+\log2}+1\bigg]$
$=\bigg[\frac{-1+1+\log2}{1+\log2}\bigg]$
$=\bigg[\frac{\log2}{1+\log2}\bigg]$ $\big[\because\log\text{e}=1\big]$
$=\frac{\log2}{\log\text{e}+\log2}$ $\big[\log\text{a}+\log\text{b}=\log\text{ab}\big]$
$=\frac{\log2}{\log2\text{e}}$
View full question & answer→Question 1095 Marks
Evaluate the following integrals:$\int_{0}^\limits{1}\frac{1}{1+2\text{x}+2\text{x}^2+2\text{x}^3+\text{x}^4}\text{ dx}$
Answer$\int_{0}^\limits{1}\frac{1}{1+2\text{x}+2\text{x}^2+2\text{x}^3+\text{x}^4}\text{ dx}$$=\int_{0}^\limits{1}\frac{1}{\big(\text{x}^2+1\big)^2+2\text{x}\big(\text{x}^2+1\big)}\text{ dx}$
$=\int_{0}^\limits{1}\frac{1}{\big(\text{x}^2+1\big)\big(\text{x}^2+1+2\text{x}\big)}\text{ dx}$
$=\int_{0}^\limits{1}\frac{1}{(\text{x}^2+1)(\text{x}+1)}\text{ dx}$
Let $\frac{1}{(\text{x}+1)^2(\text{x}^2+1)}=\frac{\text{A}}{\text{x}+1}+\frac{\text{B}}{(\text{x}+1)^2}+\frac{\text{Cx}+\text{D}}{\text{x}^2+1}$
$\Rightarrow1=\text{A}(\text{x}+1)(\text{x}^2+1)+\text{B}(\text{x}^2+1)+(\text{Cx}+\text{D})(\text{x}+1)^2$
Putting $x = -1$, we have
$1=2\text{B}$
$\Rightarrow\text{B}=\frac{1}{2}\ ...(\text{i})$
Putting x = 0, we have
$\text{A}+\text{B}+\text{C}=1\ ...(\text{ii})$
Equating co-efficient of $x^3$ on both sides, we have
$\text{A}+\text{C}=0\ ...(\text{iii}) $
Equating co-efficient of $x^2$ on both sides, we have
$\text{A}+\text{B}+2\text{C}+\text{D}=0\ ...(\text{iv})$
$\Rightarrow2\text{C}=-1$ [Using (i)]
$\Rightarrow\text{C}=-\frac{1}{2}$
$\therefore\ \text{A}=\frac{1}{2}$ [Using (iii)]
Putting $\text{A}=\frac{1}{2},\text{ B}=\frac{1}{2}$ and $\text{C}=-\frac{1}{2}$ in (iv), we have
$\text{D}=0$
$\therefore\ \int_{0}^\limits{1}\frac{1}{(\text{x}+1)^2(\text{x}^2+1)}\text{ dx}$
$=\int_{0}^\limits{1}\frac{\frac{1}{2}}{\text{x}+1}\text{ dx}+\int_{0}^\limits{1}\frac{\frac{1}{2}}{(\text{x}+1)^2}\text{ dx}+\int_{0}^\limits{1}\frac{-\frac{1}{2}\text{x}}{\text{x}^2+1}$
$=\Big[\frac{1}{2}\log(\text{x}+1)\Big]^1_0+\Big[\frac{1}{2}\times\Big(-\frac{1}{\text{x}+1}\Big)\Big]^1_0-\frac{1}{4}\int_{0}^\limits{1}\frac{2\text{x}}{\text{x}^2+1}\text{ dx}$
$=\frac{1}{2}\big(\log2-\log1\big)-\frac{1}{2}\Big(\frac{1}{2}-1\Big)-\Big[\frac{1}{4}\log(\text{x}^2+1)\Big]^1_0$
$=\frac{1}{2}\log2+\frac{1}{4}-\frac{1}{4}\big(\log2-\log1\big)$
$(\log1=0)$
$=\frac{1}{2}\log2+\frac{1}{4}\log\text{e}-\frac{1}{4}\log2$
$=\frac{1}{4}\log2+\frac{1}{4}\log_\text{e}$
$=\frac{1}{4}\big(\log2+\log_\text{e}\big)$
$=\frac{1}{4}\log(2\text{e})$
View full question & answer→Question 1105 Marks
Evaluate the following integrals:$\int\limits^{8}_2\frac{\sqrt{10-\text{x}}}{\sqrt{\text{x}}+\sqrt{10-\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{8}_2\frac{\sqrt{10-\text{x}}}{\sqrt{\text{x}}+\sqrt{10-\text{x}}}\text{ dx}\ ...(\text{i})$ Then,$\text{I}=\int\limits^{8}_2\frac{\sqrt{10-(2+8-\text{x}})}{\sqrt{2+8-\text{x}}+\sqrt{10-(2+8-\text{x}})}\text{ dx}$ $\Bigg[\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\int\limits^{\text{b}}_\text{a}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{8}_2\frac{\sqrt{\text{x}}}{\sqrt{10-\text{x}}+\sqrt{\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get$2\text{I}=\int\limits^{8}_2\frac{\sqrt{10-\text{x}}}{\sqrt{\text{x}}+\sqrt{10-\text{x}}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{8}_2\text{dx}$
$\Rightarrow2\text{I}=\big[\text{x}\big]^8_2$
$\Rightarrow2\text{I}=8-2=6$
$\Rightarrow\text{I}=3$
View full question & answer→Question 1115 Marks
Evaluate the following integrals as limit of sum:$\int\limits^{2}_{0}\big(\text{x}^2-\text{x}\big)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=2,\text{ f(x)}=\text{x}^2-\text{x},\text{ h}=\frac{2-0}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^{2}_{0}\big(\text{x}^2-\text{x}\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(0)+\text{f}(0+\text{h})\ ....\ +\text{f}\big\{0+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(0-0)+(\text{h}^2-\text{h})+\ ....+\ \big\{(\text{n}-1)^2\text{h}^2-(\text{n}-1)\text{h}\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{h}^2\big\{1^2+2^2+3^2+\ ....\ +(\text{n}-1)^2\big\}\\-\text{h}\big\{1+2+\ ....+\ (\text{n}-1)\text{h}\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}-\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}\frac{2}{\text{n}}\Big[\frac{2(\text{n}-1)(2\text{n}-1)}{3\text{n}}-\text{n}+1\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}2\Big[\frac{2}{3}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)-1+\frac{1}{\text{n}}\Big]$
$=\frac{8}{3}-2$
$=\frac{2}{3}$
View full question & answer→Question 1125 Marks
Evaluate the following integrals:$\int\limits^0_{-5}\text{f(x)}\text{dx,}$ Where $\text{f(x)}=|\text{x}|+|\text{x}+2|+|\text{x}+5|$
AnswerWe have,$\text{I}=\int\limits^0_{-5}\big(|\text{x}|+|\text{x}+2|+|\text{x}+5|\big)\text{dx}\\=\int\limits^0_{-5}|\text{x}|\text{dx}+\int\limits^0_{-5}|\text{x}+2|\text{dx}+\int\limits^0_{-5}|\text{x}+5|\text{dx}$
$\Rightarrow\text{I}=\int\limits^0_{-5}|\text{x}|\text{dx}+\int\limits^0_{-5}|\text{x}+2|\text{dx}+\int\limits^0_{-5}|\text{x}+5|\text{dx}$
$=\Big[\frac{-\text{x}^2}{2}\Big]^0_{-5}+\Big[\frac{-\text{x}^2}{2}-2\text{x}\Big]^{-2}_{-5}+\Big[\frac{\text{x}^2}{2}+2\text{x}\Big]^0_{-2}+\Big[\frac{\text{x}^2}{2}+5\text{x}\Big]^0_{-5}$
$=\Big[\frac{25}{2}\Big]-\big[\frac{4}{2}-4-\frac{25}{2}+10\Big]+\Big[0+0-\frac{4}{2}+4\Big]+\Big[0+0-\frac{25}{2}+25\big]$
$=\frac{25}{2}-\Big[8-\frac{25}{2}\Big]+\big[2\big]+\Big[25-\frac{25}{2}\Big]$
$=\frac{25}{2}-8+\frac{25}{2}+2+25-\frac{25}{2}$
$=19+\frac{25}{2}$
$=31\frac{1}{2}$
$=\frac{36}{2}$
View full question & answer→Question 1135 Marks
Evaluate the following integrals as limit of sum:$\int\limits^3_0(\text{x}+4)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=3,\text{ f(x)}=\text{x}+4,\text{ h}=\frac{3-0}{\text{n}}=\frac{3}{\text{n}}$
Therefore, $\text{I}=\int\limits^3_0(\text{x}+4)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}(0+(\text{n}-1)\text{h})\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(0+4)+(\text{h}+4)+\ ....\ +((\text{n}-1)\text{h}+4)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[4\text{n}+\text{h}(1+2+\ ....\ +(\text{n}-1))\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[4\text{n}+\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}\frac{3}{\text{n}}\Big[4\text{n}+\frac{3}{\text{n}}\times\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}\Big[12+\frac{9}{2}\Big(1-\frac{1}{\text{n}}\Big)\Big]$
$=12+\frac{9}{2}$
$=\frac{33}{2}$
View full question & answer→Question 1145 Marks
Evaluate the following integrals:$\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{\text{a}^2\sin^2\text{x}+\text{b}^2\cos^2\text{x}}\text{ dx}$
AnswerWe have,$\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{\text{a}^2\sin^2\text{x}+\text{b}^2\cos^2\text{x}}\text{ dx}$
Dividing numerator and denominator by $\cos^2\text{x}$$\int_{0}^\limits{\frac{\pi}{2}}\begin{pmatrix}\frac{\frac{1}{\cos^2\text{x}}}{\text{a}^2\frac{\sin^2\text{x}}{\cos^2\text{x}}+\text{b}^2\frac{\cos^2\text{x}}{\cos^2\text{x}}} \end{pmatrix}\text{dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\Big(\frac{\sec^2\text{x}}{\text{a}^2\tan^{2}\text{x}+\text{b}^2}\Big)\text{dx}$
$=\frac{1}{\text{a}^2}\int_{0}^\limits{\frac{\pi}{2}}\Bigg(\frac{\sec^2\text{x}}{\tan^{2}\text{x}+\big(\frac{\text{b}}{\text{a}}\big)^2}\Bigg)\text{dx}$
Let $\tan\text{x}=\text{t}$ Differentiating w.r.t. x, we get$\sec^2\text{xdx}=\text{dt}$
When $\text{x}=0\Rightarrow\text{t}=0$$\text{x}=\frac{\pi}{2}\Rightarrow\text{t}=\infty$
$\therefore\ \frac{1}{\text{a}^2}\int_{0}^\limits{\frac{\pi}{2}}\Bigg(\frac{\sec^2\text{x}}{\tan^{2}\text{x}+\big(\frac{\text{b}}{\text{a}}\big)^2}\Bigg)\text{dx}$
$=\frac{1}{\text{a}^2}\int_{0}^\limits{\infty}\frac{\text{dt}}{\big(\frac{\text{b}}{\text{a}}\big)^2+\text{t}^2}$
$=\frac{1}{\text{a}^2}\Big[\frac{\text{a}}{\text{b}}\tan^{-1}\frac{\text{at}}{\text{b}}\Big]^{\infty}_0$
$=\frac{1}{\text{a}^2}\frac{\text{a}}{\text{b}}\big[\tan^{-1}\infty-\tan^{-1}0\big]$
$=\frac{1}{\text{ab}}\Big[\tan^{-1}\tan\frac{\pi}{2}\Big]$
$=\frac{\pi}{2\text{ab}}$
View full question & answer→Question 1155 Marks
Evaluate the following integrals:$\int\limits^{\text{b}}_{\text{a}}\frac{\text{x}^{\frac{1}{\text{n}}}}{\text{x}^\frac{1}{\text{n}}+\big(\text{a}+\text{b}-\text{x}\big)^{\frac{1}{\text{n}}}}\text{ dx},\text{ n}\in\text{N},\text{n}\leq2$
AnswerLet $\text{I}=\int\limits^{\text{b}}_{\text{a}}\frac{\text{x}^{\frac{1}{\text{n}}}}{\text{x}^\frac{1}{\text{n}}+\big(\text{a}+\text{b}-\text{x}\big)^{\frac{1}{\text{n}}}}\text{ dx}\ ...(\text{i})$ Then,$\text{I}=\int\limits^{\text{b}}_{\text{a}}\frac{\big(\text{a}+\text{b}-\text{x}\big)^{\frac{1}{\text{n}}}}{\big(\text{a}+\text{b}-\text{x}\big)^\frac{1}{\text{n}}+\big[\text{a}+\text{b}-\big(\text{a}+\text{b}-\text{x}\big)\big]^{\frac{1}{\text{n}}}}\text{ dx}$ $\Bigg[\int\limits^{\text{b}}_{\text{a}}\text{f(x)}\text{dx}=\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\text{b}}_{\text{a}}\frac{\big(\text{a}+\text{b}-\text{x}\big)^{\frac{1}{\text{n}}}}{\big(\text{a}+\text{b}-\text{x}\big)^\frac{1}{\text{n}}+\text{x}^{\frac{1}{\text{n}}}}\text{ dx}$
Adding (i) and (ii) we get$2\text{I}=\int\limits^{\text{b}}_{\text{a}}\frac{\text{x}^{^{\frac{1}{\text{n}}}}+\big(\text{a}+\text{b}-\text{x}\big)^{\frac{1}{\text{n}}}}{\text{x}^\frac{1}{\text{n}}+\big(\text{a}+\text{b}-\text{x}\big)^{\frac{1}{\text{n}}}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\text{b}}_{\text{a}}\text{dx}$
$\Rightarrow2\text{I}=\big[\text{x}\big]^{\text{b}}_{\text{a}}=(\text{b}-\text{a})$
$\Rightarrow\text{I}=\frac{\text{b}-\text{a}}{2}$
View full question & answer→Question 1165 Marks
Evaluate the following integrals as limit of sum:$\int\limits^{4}_{1}\big(\text{x}^2-\text{x}\big)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=1,\text{ b}=4,\text{ f(x)}=\text{x}^2-\text{x},\text{ h}=\frac{4-1}{\text{n}}=\frac{3}{\text{n}}$
Therefore, $\text{I}=\int\limits^{4}_{1}\big(\text{x}^2-\text{x}\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(1)+\text{f}(1+\text{h})\ ....\ +\text{f}\big\{1+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(1-1)+(1+\text{h})^2-(1+\text{h})+\\\ ....\ +\big\{(\text{n}-1)\text{h}+1\big\}^2-\big\{(\text{n}-1)\text{h}+1\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{h}^2\big\{1^2+2^2+3^2+\ ....\ +(\text{n}-1)^2+4\text{h}\big\}\\-\text{h}\big\{1+2+\ ....+\ (\text{n}-1)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}-\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}\frac{3}{\text{n}}\Big[\frac{3(\text{n}-1)(2\text{n}-1)}{2\text{n}}+\frac{3(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}3\Big[\frac{3}{2}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)-\frac{3}{2}\Big(1-\frac{1}{\text{n}}\Big)\Big]$
$=9+\frac{9}{3}$
$=\frac{38}{3}$
View full question & answer→Question 1175 Marks
Evaluate the following integrals:$\int\limits^{\frac{\pi}{2}}_0\frac{\text{a}\sin\text{x}+\text{b}\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\text{a}\sin\text{x}+\text{b}\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}\ ...(\text{i})$ Then,$\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sin\big(\frac{\pi}{2}-\text{x}\big)+\text{b}\cos\big(\frac{\pi}{2}-\text{x}\big)}{\sin\big(\frac{\pi}{2}-\text{x}\big)+\cos\big(\frac{\pi}{2}-\text{x}\big)}\text{ dx}$ $\Bigg[\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\text{a}\cos\text{x}+\text{b}\sin\text{x}}{\cos\text{x}+\sin\text{x}}\text{ dx}\ ....(\text{ii})$
Adding (i) and (ii) we get$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\Big(\frac{\text{a}\sin\text{x}+\text{b}\cos\text{x}}{\cos\text{x}+\sin\text{x}}+\frac{\text{a}\cos\text{x}+\text{b}\sin\text{x}}{\sin\text{x}+\cos\text{x}}\Big)\text{dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\Big(\frac{\text{a}\sin\text{x}+\text{b}\cos\text{x}+\text{a}\cos\text{x}+\text{b}\sin\text{x}}{\sin\text{x}+\cos\text{x}}\Big)\text{dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{(\text{a}+\text{b})\sin\text{x}+(\text{a}+\text{b})\cos\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{(\text{a}+\text{b})(\sin\text{x}+\cos\text{x})}{\sin\text{x}+\cos\text{x}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0(\text{a}+\text{b})\text{dx}$
$\Rightarrow2\text{I}=(\text{a}+\text{b})\times\big[\text{x}\big]^{\frac{\pi}{2}}_0$
$\Rightarrow2\text{I}=(\text{a}+\text{b})\times\Big(\frac{\pi}{2}-0\Big)$
$\Rightarrow2\text{I}=\frac{\pi}{2}(\text{a}+\text{b})$
$\Rightarrow\text{I}=\frac{\pi}{4}(\text{a}+\text{b})$
View full question & answer→Question 1185 Marks
Evaluate the following integrals as limit of sum:$\int\limits^3_{1}(2\text{x}+3)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=1,\text{ b}=3,\text{ f(x)}=2\text{x}+3,\text{ h}=\frac{3-1}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^3_{1}(2\text{x}+3)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(1)+\text{f}(1+\text{h})+\ ....\ +\text{f}\big\{1+(\text{n}-1)\text{h}\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(2+3)+(2+2\text{h}+3)+\\\ ....\ +\{2+2(\text{n}-1)\text{h}+3\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[5\text{n}+2\text{h}\big\{1+2+3+\ ....\ +(\text{n}-1)\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[5\text{n}+2\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2}{\text{n}}\Big[5\text{n}+2\text{n}-2\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\Big({7}-\frac{2}{\text{n}}\Big)$
$=14$
View full question & answer→Question 1195 Marks
If $\text{f}(\text{a}+\text{b}-\text{x})=\text{f(x)},$ then prove that, $\int\limits^{\text{b}}_{\text{a}}\text{xf}(\text{x})\text{dx}=\Big(\frac{\text{a}+\text{b}}{2}\Big)\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{x})\text{dx}$
Answer$\int\limits^{\text{b}}_{\text{a}}\text{xf}(\text{x})\text{dx}=\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{a}+\text{b}-\text{x})(\text{a}+\text{b}-\text{x}){}\text{dx}$ $\Bigg[\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\int\limits^{\text{b}}_\text{a}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}\Bigg]$$\Rightarrow\int\limits^{\text{b}}_{\text{a}}\text{xf}(\text{x})\text{dx}=\int\limits^{\text{b}}_{\text{a}}(\text{a}+\text{b}-\text{x})\text{f}(\text{x})\text{dx}$ $\big[\text{f}(\text{a}+\text{b}-\text{x})=\text{f}(\text{x})\big]$
$\Rightarrow\int\limits^{\text{b}}_{\text{a}}\text{xf}(\text{x})\text{dx}=\int\limits^{\text{b}}_{\text{a}}(\text{a}+\text{b})\text{f}(\text{x})\text{dx}-\int\limits^{\text{b}}_{\text{a}}\text{x}\text{f}(\text{x})\text{dx}$
$\Rightarrow2\int\limits^{\text{b}}_{\text{a}}\text{xf}(\text{x})\text{dx}=(\text{a}+\text{b})\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{x})\text{dx}$
$\Rightarrow\int\limits^{\text{b}}_{\text{a}}\text{xf}(\text{x})\text{dx}=\Big(\frac{\text{a}+\text{b}}{2}\Big)\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{x})\text{dx}$
View full question & answer→Question 1205 Marks
Evaluate the following integrals:$\int^\limits{\frac{\pi}{4}}_{0}\frac{\sin^2\text{x}\cos^2\text{x}}{(\sin^3\text{x}\cos^3\text{x})}\text{ dx}$
AnswerLet $\text{I}=\int^\limits{\frac{\pi}{4}}_{0}\frac{\sin^2\text{x}\cos^2\text{x}}{(\sin^3\text{x}\cos^3\text{x})}\text{ dx}$$=\int^\limits{\frac{\pi}{4}}_{0}\frac{\sin^2\text{x}\cos^2\text{x}}{\cos^6\text{x}(\tan^3\text{x}+1)^2}\text{ dx}=\int^\limits{\frac{\pi}{4}}_{0}\frac{\tan^2\text{x}\sec^2\text{x}}{(\tan^3\text{x}+1)}\text{ dx}$
Put $\tan^3\text{x}+1=\text{z}$$\therefore\ 3\tan^{2}\text{x}\sec^2\text{x dx}=\text{dt}$
$\Rightarrow\tan^{2}\text{x}\sec^2\text{x dx}=\frac{\text{dz}}{3}$
When $\text{x}\rightarrow0,\text{z}\rightarrow1$ When $\text{x}\rightarrow\frac{\pi}{4},\text{z}\rightarrow2$$\therefore\ \text{I}=\frac{1}{3}\int^\limits{2}_1\frac{\text{dz}}{\text{z}^2}$
$=\frac{1}{3}\times-\Big[\frac{1}{\text{z}}\Big]^2_1$
$=-\frac{1}{3}\Big(\frac{1}{2}-1\Big)$
$=-\frac{1}{3}\times\Big(-\frac{1}{2}\Big)$
$=\frac{1}{6}$
View full question & answer→Question 1215 Marks
Evaluate the following integrals:$\int^\limits2_{0}\big|\text{x}^2-3\text{x}+2\big|\text{dx}$
Answer$\int^\limits2_{0}\big|\text{x}^2-3\text{x}+2\big|\text{dx}$We know that,
$\big|\text{x}^2-3\text{x}+2\big|\text{dx}=\begin{cases}-(\text{x}^2-3\text{x}+2),&(\text{x}-1)(\text{x}-2)\leq0\text{ or },&1\leq\text{x}\leq2\$\text{x}^2-3\text{x}+2),&\text{x}^2-3\text{x}+2\leq0\text{ or },&\text{x}\in(-\infty,1)(2,\infty)\end{cases}$
$\therefore\ \text{I}=\ \int^\limits2_{0}\big(\text{x}^2-3\text{x}+2\big)\text{dx}$
$\Rightarrow\text{I}=\int^\limits1_{0}\big(\text{x}^2-3\text{x}+2\big)\text{dx}-\int^\limits2_1\big(\text{x}^2-3\text{x}+2\big)\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{x}^3}{3}-\frac{3\text{x}^2}{2}+2\text{x}\Big]^1_0-\Big[\frac{\text{x}^3}{3}-\frac{3\text{x}^2}{2}+2\text{x}\Big]^2_1$
$\Rightarrow\text{I}=\frac{1}{3}-\frac{3}{2}+2-\Big[\frac{8}{3}-6+4-\frac{1}{3}+\frac{3}{2}-2\Big]$
$\Rightarrow\text{I}=\frac{1}{3}-\frac{3}{2}+2-\frac{8}{3}+6-2+\frac{1}{3}-\frac{3}{2}$
$\Rightarrow\text{I}=1$
View full question & answer→Question 1225 Marks
Evaluate the following integrals:$\int^\limits{\frac{\pi}{2}}_0\sin2\text{x }\tan^{-1}(\sin\text{x})\text{dx}$
AnswerLet $\text{I}=\int^\limits{\frac{\pi}{2}}_0\sin2\text{x }\tan^{-1}(\sin\text{x})\text{dx}$ Differentiating w.r.t. x, we get$\cos\text{x dx}=\text{dt}$
Now, $\text{x}=0,\text{t}=0$$\text{x}=\frac{\pi}{2},\text{t}=1$
$\therefore\ \int^\limits{\frac{\pi}{2}}_0\sin2\text{x }\tan^{-1}(\sin\text{x})\text{dx}=2\int^\limits{1}_0\text{t }\tan^{-1}\text{t dt}$ $\big[\because\sin2\text{x}=2\sin\text{x }\cos\text{x}\big]$
Using by parts$=2\Big\{\tan^{-1}\text{t}\int\text{t dt}-\int\big(\int\text{t dt}\big)\frac{\text{d}\tan^{-1}\text{t}}{\text{dt}}\Big\}$
$=2\Big\{\frac{\text{t}^2}{2}\tan^{-1}\text{t}-\frac{1}{2}\int\frac{\text{t}^2}{1+\text{t}^2}\text{ dt}\Big\}$
$=2\bigg\{\frac{\text{t}^2}{2}\tan^{-1}\text{t}-\frac{1}{2}\Big(\int\text{dt}-\int\frac{\text{dt}}{1+\text{t}^2}\text{ dt}\Big)\bigg\}$
$=2\Big[\frac{\text{t}^2}{2}\tan^{-1}\text{t}-\frac{1}{2}\big(\text{t}-\tan^{-1}\text{t}\big)\Big]^1_0$
$=2\bigg\{\frac{1}{2}\frac{\pi}{4}-\frac{1}{2}\Big(1-\frac{\pi}{4}\Big)\bigg\}$
$=2\Big\{\frac{\pi}{8}-\frac{1}{2}+\frac{\pi}{8}\Big\}$
$=2\Big(\frac{\pi}{4}-\frac{1}{2}\Big)$
$=\frac{\pi}{2}-1$
View full question & answer→Question 1235 Marks
Evaluate the following integrals:$\int\limits^{\frac{3}{2}}_0\big|\text{x}\cos\pi\text{x}\big|\text{dx}$
Answer$\int\limits^\frac{3}{2}_{0}|\text{x} \cos\pi \text{ x}| \text{dx} = \int\limits^{1/2}_{0}\text{x}\cos\pi \text{x dx}{-}\int\limits^{3/2}_{1/2}\text{x}\cos\pi \text{x dx}$$= \left\{\frac{\text{x}\sin\pi \text{x}}{\pi} + \frac{\cos\pi\text{x}}{\pi^{2}}\right\}^{1/2}_{0}= \left\{\frac{\text{x}\sin\pi \text{x}}{\pi} + \frac{\cos\pi\text{x}}{\pi^{2}}\right\}^{3/2}_{1/2}$
$=\frac{1}{2\pi} - \frac{1}{\pi^{2}} - \bigg(-\frac{3}{2\pi}-\frac{1}{2\pi}\bigg) = \frac{5}{2\pi}-\frac{1}{\pi^{2}}$
View full question & answer→Question 1245 Marks
Evaluate the following integrals:$\int\limits^{\frac{\pi}{2}}_0\frac{\sin^{\text{n}}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx},\text{ n}\in\text{N}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^{\text{n}}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}\ ...(\text{i})$$=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^{\text{n}}\big(\frac{\pi}{2}-\text{x}\big)}{\sin^{\text{n}}\big(\frac{\pi}{2}-\text{x}\big)+\cos^{\text{n}}\big(\frac{\pi}{2}-\text{x}\big)}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\cos^\text{n}\text{x}}{\cos^\text{n}\text{x}+\sin^\text{n}\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\cos^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}\ ....(\text{ii})$
Adding (i) and (ii)$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\bigg[\frac{\sin^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}+\frac{\cos^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\bigg]\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\text{dx}$
$=\big[\text{x}\big]^{\frac{\pi}{2}}_0$
$=\frac{\pi}{2}$
Hence, $\text{I}=\frac{\pi}{4}$
View full question & answer→Question 1255 Marks
Evaluate the following integrals:$\int^\limits{\frac{\pi}{2}}_0\cos^5\text{x dx}$
Answer$\int^\limits{\frac{\pi}{2}}_0\cos^5\text{x dx}=\int^\limits{\frac{\pi}{2}}_0\big(1-\sin^2\text{x}\big)^2\cos\text{x dx}$Let $\sin\text{x}=\text{t}$
Differentiating w.r.t. x, we get
$\cos\text{x dx}=\text{dt}$
When $\text{x}=0\Rightarrow\text{t}=0$
$\text{x}=\frac{\pi}{2}\Rightarrow\text{t}=1$
$=\int^\limits{\frac{\pi}{2}}_0\big(1-\sin^2\text{x}\big)^2\cos\text{x dx}$
$=\int^\limits{1}_0\big(1-\text{t}^2\big)^2\text{ dt}$
$=\int^\limits{1}_0\big(1-2\text{t}^2+\text{t}^4\big)\text{dt}$
$=\Big[\text{t}-\frac{2}{3}\text{t}^3+\frac{\text{t}^5}{5}\Big]^1_0$
$=1-\frac{2}{3}+\frac{1}{5}$
$=\frac{8}{15}$
View full question & answer→Question 1265 Marks
Evaluate the following definite integrals:$\int_{\frac{\pi}{2}}^\limits{\pi}\text{e}^{\text{x}}\Big(\frac{1-\sin\text{x}}{1-\cos\text{x}}\Big)\text{dx}$
AnswerLet $\text{I}=\int_{\frac{\pi}{2}}^\limits{\pi}\text{e}^{\text{x}}\Big(\frac{1-\sin\text{x}}{1-\cos\text{x}}\Big)\text{dx}$ Then,$\text{I}=\int_{\frac{\pi}{2}}^\limits{\pi}\text{e}^{\text{x}}\bigg(\frac{1-2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}{2\sin^2\frac{\text{x}}{2}}\bigg)\text{dx}$
$\Rightarrow\text{I}=\int_{\frac{\pi}{2}}^\limits{\pi}\text{e}^{\text{x}}\Big(\frac{1}{2}\text{cosec}^2\frac{\text{x}}{2}-\cot\frac{\text{x}}{2}\Big)\text{dx}$
$\Rightarrow\text{I}=\int_{\frac{\pi}{2}}^\limits{\pi}\frac{1}{2}\text{e}^{\text{x}}\text{cosec}^2\frac{\text{x}}{2}\text{ dx}-\int_{\frac{\pi}{2}}^\limits{\pi}\text{e}^{\text{x}}\cot\frac{\text{x}}{2}\text{ dx}$
Integrating second term by parts,$\text{I}=\bigg\{-\Big[\text{e}^{\text{x}}\cot\frac{\text{x}}{2}\Big]^{\pi}_\frac{\pi}{2}-\int_{\frac{\pi}{2}}^\limits{\pi}\frac{1}{2}\text{ e}^{\text{x}}\text{cosec}^2\frac{\text{x}}{2}\text{ dx}\bigg\}+\int_{\frac{\pi}{2}}^\limits{\pi}\frac{1}{2}\text{ e}^{\text{x}}\text{cosec}^2\frac{\text{x}}{2}\text{ dx}$
$\Rightarrow\text{I}=-\Big[0-\text{e}^{\frac{\pi}{2}}\Big]$
$\Rightarrow\text{I}=\text{e}^{\frac{\pi}{2}}$
View full question & answer→Question 1275 Marks
Evaluate the following integrals:$\int\limits^{\frac{3}{2}}_0\big|\text{x}\sin\pi\text{x}\big|\text{dx}$
AnswerFor $0<\text{x}<1,\text{ x}>0$ and $\sin\pi\text{x}>0\Rightarrow\text{x}\sin\pi\text{x}>0$ For $1<\text{x}<\frac{3}{2},\text{ x}>0$ and $\sin\pi\text{x}>0\Rightarrow\text{x}\sin\pi\text{x}>0$$\therefore\ \int\limits^{\frac{3}{2}}_0\big|\text{x}\sin\pi\text{x}\big|\text{dx}=\int\limits^1_0\text{x}\sin\pi\text{x dx}-\int\limits^{\frac{3}{2}}_1\text{x}\sin\pi\text{x dx}$
Let $\text{I}=\int\text{x}\sin\pi\text{ dx}$$=\text{x}\int\sin\pi\text{x}-\int\Big(\frac{\text{d}}{\text{dx}}\text{x}\int\sin\pi\text{x dx}\Big)\text{dx}$
$=\text{x}\Big(\frac{-\cos\pi\text{x}}{\pi}\Big)-\int\Big(\frac{-\cos\pi\text{x}}{\pi}\Big)\text{dx}$
$=\frac{-\text{x}\cos\pi\text{x}}{\pi}+\frac{\sin\pi\text{x}}{\pi^2}$
Applying the limits, we get$\int\limits^{\frac{3}{2}}_0|\text{x}\sin\pi\text{x}|\text{dx}=\Big[\frac{-\text{x}\cos\pi\text{x}}{\pi}+\frac{\sin\pi\text{x}}{\pi^2}\Big]^1_0-\Big[\frac{-\text{x}\cos\pi\text{x}}{\pi}+\frac{\sin\pi\text{x}}{\pi^2}\Big]^{\frac{3}{2}}_1$
$=\Big[\Big(\frac{\cos\text{x}}{\pi}+\frac{\sin\pi}{\pi^2}\Big)-(0+0)\Big]-\Bigg[\bigg(\frac{-\frac{3}{2}\cos\frac{3\pi}{2}}{\pi}+\frac{\sin\frac{3\pi}{2}}{\pi^2}\bigg)-\Big(\frac{-\cos\pi}{\pi}+\frac{\sin\pi}{\pi^2}\Big)\Bigg]$
$=\bigg[\Big(\frac{1}{\pi}+0\Big)\bigg]-\bigg[\Big(0-\frac{1}{\pi^2}\Big)-\Big(\frac{1}{\pi}+0\Big)\bigg]$
$=\frac{1}{\pi}+\frac{1}{\pi^2}+\frac{1}{\pi}$
$=\frac{2\pi+1}{\pi^2}$
View full question & answer→Question 1285 Marks
Evaluate the following integrals:$\int\limits^4_{0}\big(|\text{x}|+|\text{x}+2|+|\text{x}+4|\big)\text{dx}$
Answer$\text{I}=\int\limits^4_{0}\big\{|\text{x}|+|\text{x}+2|+|\text{x}+4|\big\}\text{dx}$$\Rightarrow\text{I}=\int\limits^4_0|\text{x}|\text{dx}+\int\limits^4_0|\text{x}-2|\text{dx}+\int\limits^4_0|\text{x}-4|\text{dx}$
We know that,
$|\text{x}|=\begin{cases}-\text{x},&-5\leq\text{x}\leq0\\\text{x},&\text{x}>0\end{cases}$
$|\text{x}-2|=\begin{cases}-(\text{x}-2),&0\leq\text{x}\leq2\\\text{x}-2,&2<\text{x}\leq4\end{cases}$
$|\text{x}-4|=\begin{cases}-(\text{x}-4),&0\leq\text{x}\leq4\\\text{x}-4,&\text{x}>4\end{cases}$
$\therefore\ \text{I}=\int\limits^4_{0}\text{x dx}-\int\limits^2_{0}(\text{x}-2)\text{dx}+\int\limits^4_{2}(\text{x}-2)\text{dx}-\int\limits^4_{0}(\text{x}-4)\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{x}^2}{2}\Big]^4_0-\Big[\frac{\text{x}^2}{2}-2\text{x}\Big]^2_0+\Big[\frac{\text{x}^2}{2}-2\text{x}\Big]^4_2-\Big[\frac{\text{x}^2}{2}-2\text{x}\Big]^4_0$
$\Rightarrow\text{I}=8-(2-4)+8-8-2+4-(8-16)$
$\Rightarrow\text{I}=20$
View full question & answer→Question 1295 Marks
Evaluate the following integrals:$\int\limits^1_0\log\Big(\frac{1}{\text{x}}-1\Big)\text{dx}$
AnswerLet $\text{I}=\int\limits^1_0\log\Big(\frac{1}{\text{x}}-1\Big)\text{dx}\ ....(\text{i})$$=\int\limits^1_0\log\Big(\frac{1}{1-\text{x}}-1\Big)\text{dx}$ $\Big[\text{Using},\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Big]$
$\text{I}=\int\limits^1_0\log\Big(\frac{\text{x}}{\text{x}-1}\Big)\text{dx}\ ....(\text{ii})$
Adding (i) and (ii)$2\text{I}=\int\limits^1_0\log\Big(\frac{1-\text{x}}{\text{x}}\Big)\log\Big(\frac{1-\text{x}}{\text{x}}\Big)\text{dx}$
$=\int\limits^1_0\log\Big(\frac{1-\text{x}}{\text{x}}\times\frac{\text{x}}{1-\text{x}}\Big)\text{dx}$
$=\int\limits^1_0\log1\text{ dx}$
$=0$
Hence, $\text{I}=0$
View full question & answer→Question 1305 Marks
Evaluate the following integrals:$\int\limits^{\frac{\pi}{2}}_0\frac{\sqrt{\cot\text{x}}}{\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sqrt{\cot\text{x}}}{\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}\text{ dx}\ ...(\text{i})$$=\int\limits^{\frac{\pi}{2}}_0\frac{\sqrt{\cot\big(\frac{\pi}{2}-\text{x}\big)}}{\sqrt{\cot\big(\frac{\pi}{2}-\text{x}\big)}+\sqrt{\tan\big(\frac{\pi}{2}-\text{x}\big)}}\text{ dx}$ $\Big[\text{Using},\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Big]$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\sqrt{\tan\text{x}}}{\sqrt{\tan\text{x}}+\sqrt{\cot\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii)$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\bigg(\frac{\sqrt{\cot\text{x}}}{\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}+\frac{\sqrt{\tan\text{x}}}{\sqrt{\tan\text{x}}+\sqrt{\cot\text{x}}}\bigg)\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\text{dx}$
$=\Big[\text{x}\Big]^{\frac{\pi}{2}}_0=\frac{\pi}{2}$
Hence, $\text{I}=\frac{\pi}{4}$
View full question & answer→Question 1315 Marks
Evaluate the following integrals:$\int^\limits{\frac{\pi}{6}}_{0}\cos^{-3}2\theta\sin2\theta\text{ d}\theta$
AnswerWe have,$\int^\limits{\frac{\pi}{6}}_{0}\cos^{-3}2\theta\sin2\theta\text{ d}\theta$
$=\int^\limits{\frac{\pi}{6}}_{0}\frac{\sin2\theta}{\cos^32\theta}\text{ d}\theta$
$=\int^\limits{\frac{\pi}{6}}_{0}\tan2\theta\cdot\sec^22\theta\text{ d}\theta$
Let $\tan2\theta=\text{t}$ Differentiating w.r.t. x, we get$2\sec^22\theta\text{d}\theta=\text{dt}$
Now, $\theta=0\Rightarrow\text{t}=0$$\theta=\frac{\pi}{6}\Rightarrow\text{t}=\sqrt{3}$
$\therefore\ \int^\limits{\frac{\pi}{6}}_{0}\tan2\theta\cdot\sec^22\theta\text{ d}\theta=\frac{1}{2}\int^\limits{\sqrt{3}}_0\text{t dt}=\frac{1}{2}\Big[\frac{\text{t}^2}{2}\Big]^{\sqrt{\text{3}}}_0$
$=\frac{3}{4}$
View full question & answer→Question 1325 Marks
Evaluate the following integrals as limit of sum:$\int\limits^1_{-1}(\text{x}+3)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=-1,\text{ b}=1,\text{ f(x)}=\text{x}+3,\text{ h}=\frac{1+1}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^1_{-1}(\text{x}+3)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(-1)+\text{f}(-1+\text{h})+\\\ ....\ +\text{f}\big\{-1+(\text{n}-1)\text{h}\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(-1+3)+(-1+\text{h}+3)+\\ ....\ +\{-1+(\text{n}-1)\text{h}+3\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[2\text{n}+\text{h}\big\{1+2+3+\ ....\ +(\text{n}-1)\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[2\text{n}+\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2}{\text{n}}\Big[\text{n}+\text{n}-1\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\Big(3-\frac{1}{\text{n}}\Big)$
$=6$
View full question & answer→Question 1335 Marks
Evaluate the following integrals as limit of sum:$\int\limits^{3}_{1}\big(2\text{x}^2+5\text{x}\big)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=1,\text{ b}=3,\text{ f(x)}=2\text{x}^2+5\text{x},\text{ h}=\frac{3-1}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^{3}_{1}\big(2\text{x}^2+5\text{x}\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(1)+\text{f}(1+\text{h})\ ....\ +\text{f}\big\{1+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(2+5)+\big\{2(1+\text{h})^2+5(1+\text{h})\big\}+\ \\....+\ \big\{2(1+(\text{n}-1)\text{h}^2+5(1+(\text{n}-1)\text{h})\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[2\Big\{1^2+(1+\text{h}^2)+\ ....+\ \big\{1+(\text{n}-1)\text{h}\big\}^2\Big\}+\\5\big\{1+(1+\text{h})+(1+2\text{h}+\ ....+\ (1+(\text{n}+1)\text{h}))\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[2\text{n}+2\text{h}^2(1^2+2^2+3^2+\ ....\ +(\text{n}-1)^2+\\4\text{h}\big\{1+2+\ ....+ (\text{n}-1)\big\}+5\text{n}+5\text{h}\big\{1+2+\ ...+\ (\text{n}-1)\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[7\text{n}+2\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}+9\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}\frac{2}{\text{n}}\Big[7\text{n}+\frac{4(\text{n}-1)(2\text{n}-1)}{3\text{n}}+9\text{n}-9\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}2\Big[16+\frac{4}{3}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)-\frac{9}{\text{n}}\Big]$
$=32+\frac{16}{3}$
$=\frac{112}{3}$
View full question & answer→Question 1345 Marks
Evaluate the following integrals:$\int\limits^{\pi}_0\text{x}\cos^2\text{x dx}$
AnswerLet $\text{I}=\int\limits^{\pi}_0\text{x}\cos^2\text{x dx}\ ...(\text{i})$$=\int\limits^{\pi}_0(\pi-\text{x})\cos^2(\pi-\text{x})\text{dx}$
$=\int\limits^{\pi}_0(\pi-\text{x})\cos^2\text{x}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get$2\text{I}=\int\limits^{\pi}_0(\text{x}+\pi-\text{x})\cos^2\text{x}\text{ dx}$
$=\int\limits^{\pi}_0\pi\cos^2\text{x}\text{ dx}$
$=\pi\int\limits^{\pi}_0\frac{1+\cos2\text{x}}{2}\text{ dx}$
$=\frac{\pi}{2}\int\limits^{\pi}_0\big(1+\cos2\text{x}\big)\text{dx}$
$=\frac{\pi}{2}\Big[\text{x}+\frac{\sin2\text{x}}{2}\Big]^{\pi}_0$
$=\frac{\pi}{2}(\pi-0)$
Hence, $\text{I}=\frac{\pi^2}{4}$
View full question & answer→Question 1355 Marks
Evaluate the following integrals:$\int^\limits2_{-2}|2\text{x}+3|\text{dx}$
Answer$\int^\limits2_{-2}|2\text{x}+3|\text{dx}$We know that,
$|2\text{x}+3|=\begin{cases}-(2\text{x}+3),&-2\leq\text{x}\leq-\frac{3}{2}\$2\text{x}+3),&-\frac{3}{2}<\text{x}\leq2\end{cases}$
$\therefore\ \text{I}=\int^\limits{\frac{-3}{2}}_{-2}-\big(2\text{x}+3\big)\text{dx}+\int^\limits2_{-\frac{3}{2}}\big(2\text{x}+3\big)\text{dx}$
$\Rightarrow\text{I}=-\Big[\text{x}^3+3\text{x}\Big]^{\frac{-3}{2}}_{-2}+\Big[\text{x}^2+3\text{x}\Big]^2_{-\frac{3}{2}}$
$\Rightarrow\text{I}=-\frac{9}{4}+\frac{9}{2}+4-6+4+6-\frac{9}{4}+\frac{9}{2}$
$\Rightarrow\text{I}=\frac{25}{2}$
View full question & answer→Question 1365 Marks
Evaluate the following integrals:$\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\sin^2\text{x dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\sin^2\text{x dx}$ Here $\text{f(x)}=\sin^2\text{x}$$\text{f}(-\text{x})=\sin^2(-\text{x})=\sin^2\text{x}=\text{f(x)}$
Hence $\sin^2\text{x}$ is an even function Therefore,$\text{I}=2\int\limits^{\frac{\pi}{4}}_{0}\sin^2\text{x dx}$
$=2\int\limits^{\frac{\pi}{4}}_{0}\Big(\frac{1-\cos2\text{x}}{2}\Big)\text{dx}$
$=\int\limits^{\frac{\pi}{4}}_{0}(1-\cos2\text{x})\text{dx}$
$=\Big[\text{x}-\frac{\sin^2\text{x}}{2}\Big]^{\frac{\pi}{4}}_0$
$=\frac{\pi}{4}-\frac{1}{2}$
View full question & answer→Question 1375 Marks
Evaluate the following definite integrals:$\int_{0}^\limits{\frac{\pi}{6}}\cos\text{x }\cos2\text{x}\text{ dx}$
AnswerWe have,$\int_{0}^\limits{\frac{\pi}{6}}\cos\text{x }\cos2\text{x}\text{ dx}$ $\big[\because2\cos\text{C}\cos\text{D}=\cos(\text{C}+\text{D})-\cos(\text{C}-\text{D})\big]$
$=\frac{1}{2}\int_{0}^\limits{\frac{\pi}{6}}2\cos\text{x }\cos2\text{x dx}$
$=\frac{1}{2}\int_{0}^\limits{\frac{\pi}{6}}(\cos3\text{x}+\cos\text{x})\text{dx}$
$=\frac{1}{2}\int\Big[\frac{\sin3\text{x}}{3}+\sin\text{x}\Big]_0^{\frac{\pi}{6}}$
$=\frac{1}{2}\Bigg[\bigg(\frac{\sin3\frac{\pi}{6}}{3}+\sin\frac{\pi}{6}\bigg)-(\sin0-\sin0)\Bigg]$
$=\frac{1}{2}\bigg[\frac{\sin\frac{\pi}{2}}{3}+\sin\frac{\pi}{6}\bigg]$
$=\frac{1}{2}\Big(\frac{1}{3}+\frac{1}{2}\Big)$
$=\frac{1}{2}\Big(\frac{5}{6}\Big)$
$=\frac{5}{12}$
$\therefore\ \int_{0}^\limits{\frac{\pi}{6}}\cos\text{x }\cos2\text{x}\text{ dx}=\frac{5}{12}$
View full question & answer→Question 1385 Marks
Evaluate the following integrals:$\int^\limits\frac{\pi}{2}_{0}\frac{\cos^2\text{x}}{1+3\sin^3\text{x}}\text{ dx}$
Answer$\text{I}=\int^\limits\frac{\pi}{2}_{0}\frac{\cos^2\text{x}}{1+3\sin^3\text{x}}\text{ dx}$$\text{I}=\int^\limits\frac{\pi}{2}_{0}\frac{\sec^2\text{x}}{\sec^2\text{x}\big(\sec^2\text{x}+3\tan^2\text{x}\big)}\text{ dx}$
Put $\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\text{x}=0\Rightarrow\text{t}=0$ and $\text{x}=\frac{\pi}{2}\Rightarrow\text{t}=\infty$
$\Rightarrow\text{I}=\int^{\infty}\limits_0\frac{1}{\big(1+\text{t}^2\big)\big(1+4\text{t}^2\big)}\text{ dt}$
$\Rightarrow\text{I}=-\frac{1}{3}\int^{\infty}\limits_0\bigg[\frac{1}{(1+\text{t}^2)-{(1+4\text{t}^2)}}\bigg]\text{dt}$
$\Rightarrow\text{I}=-\frac{1}{3}\Big[\tan^{-1}\text{t}-2\tan^{-1}2\text{t}\Big]^{\infty}_0$
$\Rightarrow\text{I}=\frac{\pi}{6}$
View full question & answer→Question 1395 Marks
Evaluate the following integrals:$\int\limits^{\frac{\pi}{3}}_\frac{\pi}{6}\frac{1}{1+\cot^{\frac{3}{2}}\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{3}}_\frac{\pi}{6}\frac{1}{1+\cot^{\frac{3}{2}}\text{x}}\text{ dx}\ ...(\text{i})$ Then,$\text{I}=\int\limits^{\frac{\pi}{3}}_\frac{\pi}{6}\frac{1}{1+\cot^{\frac{3}{2}}\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}\text{ dx}$ $\Bigg[\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\int\limits^{\text{b}}_\text{a}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\frac{\pi}{3}}_\frac{\pi}{6}\frac{1}{1+\cot^{\frac{3}{2}}\big(\frac{\pi}{2}-\text{x}\big)}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_\frac{\pi}{6}\frac{1}{1+\tan^{\frac{3}{2}}\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_\frac{\pi}{6}\frac{\cot^{\frac{3}{2}}\text{x}}{\cot^{\frac{3}{2}}\text{x}+1}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii)$2\text{I}=\int\limits^{\frac{\pi}{3}}_\frac{\pi}{6}\frac{1+\cot^{\frac{3}{2}}\text{x}}{1+\cot^{\frac{3}{2}}\text{x}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{3}}_\frac{\pi}{6}\text{dx}$
$\Rightarrow2\text{I}=\big[\text{x}\big]^{\frac{\pi}{3}}_\frac{\pi}{6}$
$\Rightarrow2\text{I}=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}$
$\Rightarrow\text{I}=\frac{\pi}{12}$
View full question & answer→Question 1405 Marks
Evaluate the following integrals:$\int^\limits{\pi}_0\sin^3\text{x}(1+2\cos\text{x})(1+\cos\text{x})^2\text{ dx}$
AnswerLet $\text{I}=\int^\limits{\pi}_0\sin^3\text{x}(1+2\cos\text{x})(1+\cos\text{x})^2\text{ dx}$ Then,$\text{I}=\int^\limits{\pi}_0\sin\text{x }\sin^2\text{x}(1+2\cos\text{x})(1+\cos\text{x})^2\text{ dx}$
$\Rightarrow\text{I}=\int^\limits{\pi}_0\sin\text{x}(1-2\cos^2\text{x})(1+\cos\text{x})(1+\cos\text{x})^2\text{ dx}$
$\Rightarrow\text{I}=\int^\limits{\pi}_0\sin\text{x}(1-\cos\text{x})(1+2\cos\text{x})(1+\cos\text{x})^3\text{ dx}$
Let $\cos\text{x}=\text{t}$ Then, $-\sin\text{x dx}=\text{dt}$ When $\text{x}=0,\text{t}=1$ and $\text{x}=\pi,\text{t}=-1$$\therefore\ \text{I}=-\int^\limits{-1}_1(1-\text{t})(1+2\text{t})(1+\text{t})^3\text{ dt}$
$\Rightarrow\text{I}=\int^\limits{-1}_1\big(1+\text{t}-2\text{t}^2\big)\big(1+\text{t}^3+3\text{t}+3\text{t}^2\big)\text{dt}$
$\Rightarrow\text{I}=\int^\limits{-1}_1\big(1+\text{t}^3+3\text{t}+3\text{t}^3+\text{t}+\text{t}^4\\+3\text{t}^2+3\text{t}^3-2\text{t}^2-2\text{t}^5-6\text{t}^3-6\text{t}^4\big)\text{dt}$
$\Rightarrow\text{I}=\int^\limits{-1}_1\big(1+4\text{t}+4\text{t}^2-2\text{t}^3-5\text{t}^4-2\text{t}^5\big)\text{dt}$
$\Rightarrow\text{I}=\Big[\text{t}+2\text{t}^2+\frac{4\text{t}^3}{3}-\frac{\text{t}^4}{2}-\text{t}^5-\frac{\text{t}^6}{3}\Big]^1_{-1}$
$\Rightarrow\text{I}=1+2+\frac{4}{3}-\frac{1}{2}-1-\frac{1}{3}\\+1-2+\frac{4}{3}+\frac{1}{2}-1+\frac{1}{3}$
$\Rightarrow\text{I}=\frac{8}{3}$
View full question & answer→Question 1415 Marks
Evaluate the following definite integrals:$\int_{1}^\limits{2}\Big(\frac{\text{x}-1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{1}^\limits{2}\Big(\frac{\text{x}-1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{ dx}$ Then,$\text{I}=\int_{1}^\limits{2}\Big(\frac{\text{e}^{\text{x}}}{\text{x}}-\frac{\text{e}^{\text{x}}}{\text{x}^2}\Big)\text{dx}$
$\Rightarrow\text{I}=\int_{1}^\limits{2}\frac{\text{e}^\text{x}}{\text{x}}\text{ dx}-\int_{1}^\limits{2}\frac{\text{e}^\text{x}}{\text{x}^2}\text{ dx}$
Integrating first term by parts,$\text{I}=\bigg\{\Big[\frac{\text{e}^\text{x}}{\text{x}}\Big]^2_1-\int_{1}^\limits{2}\frac{-1}{\text{x}^2}\text{e}^{\text{x}}\text{ dx}\bigg\}-\int_{1}^\limits{2}\frac{\text{e}^\text{x}}{\text{x}^2}\text{ dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{e}^\text{x}}{\text{x}}\Big]^2_1+\int_{1}^\limits{2}\frac{-1}{\text{x}^2}\text{e}^{\text{x}}\text{ dx}-\int_{1}^\limits{2}\frac{\text{e}^\text{x}}{\text{x}^2}\text{ dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{e}^\text{x}}{\text{x}}\Big]^2_1$
$\Rightarrow\text{I}=\frac{\text{e}^2}{2}-\text{e}$
View full question & answer→Question 1425 Marks
Evaluate the following integrals:$\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\tan\text{x}}}{\sqrt{\tan\text{x}}+\sqrt{\cot\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\tan\text{x}}}{\sqrt{\tan\text{x}}+\sqrt{\cot\text{x}}}\text{ dx}\ ...(\text{i})$ Then,$\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}}{\sqrt{\sin\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}+\sqrt{\cos\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}}\text{ dx}$ $\Bigg[\int\limits^{\text{b}}_{\text{a}}\text{f(x)}\text{dx}=\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\big(\frac{\pi}{2}-\text{x}\big)}}{\sqrt{\sin\big(\frac{\pi}{2}-\text{x}\big)}+\sqrt{\cos\big(\frac{\pi}{2}-\text{x}\big)}}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\cos\text{x}}}{\sqrt{\cos\text{x}}+\sqrt{\sin\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get$2\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\text{dx}$
$\Rightarrow2\text{I}=\Big[\text{x}\Big]^{\frac{\pi}{3}}_{\frac{\pi}{6}}$
$\Rightarrow2\text{I}=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}$
$\Rightarrow\text{I}=\frac{\pi}{12}$
View full question & answer→Question 1435 Marks
Evaluate the following integrals:$\int_{0}^\limits{\frac{\pi}{4}}\frac{\tan^{3}\text{x}}{1+\cos2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\frac{\tan^{3}\text{x}}{1+\cos2\text{x}}\text{ dx}$ Then,$\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\frac{\tan^{3}\text{x}}{1+\cos2\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\frac{1}{2}\int_{0}^\limits{\frac{\pi}{4}}\tan^{3}\text{x}\sec^2\text{x}\text{ dx}$
Let $\tan\text{x}=\text{t}$ Then, $\sec^2\text{x dx}=\text{dt}$ When $\text{x}=0,\text{t}=0$ and $\text{x}=\frac{\pi}{4},\text{t}=1$$\therefore\ \text{I}=\frac{1}{2}\int_{0}^\limits{1}\text{t}^3\text{ d}t$
$\Rightarrow\text{I}=\frac{1}{2}\Big[\frac{\text{t}^4}{4}\Big]^1_0$
$\Rightarrow\text{I}=\frac{1}{2}\Big(\frac{1}{4}-0\Big)$
$\Rightarrow\text{I}=\frac{1}{8}$
View full question & answer→Question 1445 Marks
Evaluate the following integrals:$\int\limits^{\pi}_0\text{x}\sin^3\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\pi}_0\text{x}\sin^3\text{x}\text{ dx}\ ...(\text{i})$$=\int\limits^{\pi}_0(\pi-\text{x})\sin^3(\pi-\text{x})\text{dx}$
$=\int\limits^{\pi}_0(\pi-\text{x})\sin^3\text{x}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\pi}_0(\text{x}+\pi-\text{x})\sin^3\text{x}\text{ dx}$
$=\int\limits^{\pi}_0\pi\sin^3\text{x}\text{ dx}$
$=\int\limits^{\pi}_0\pi\frac{3\sin\text{x}-\sin3\text{x}}{4}\text{ dx}$
$=\frac{\pi}{4}\int\limits^{\pi}_0\big(3\sin\text{x}-\sin3\text{x}\big)\text{dx}$
$=\frac{\pi}{4}\Big[-3\cos\text{x}+\frac{\cos3\text{x}}{3}\Big]^{\pi}_0$
$=\frac{\pi}{4}\Big[-3\cos\pi+3\cos0+\frac{\cos3\pi}{3}-\frac{\cos0}{3}\Big]$
$=\frac{\pi}{4}\Big[3+3+\frac{-1}{3}-\frac{1}{3}\Big]$
$=\frac{\pi}{4}\Big[3-\frac{1}{3}\Big]$
$=\frac{\pi}{2}\times\frac{8}{3}$
$=\frac{4\pi}{3}$
$\therefore\ \text{I}=\frac{2\pi}{3}$
View full question & answer→Question 1455 Marks
Evaluate the following integrals:$\int_{0}^\limits{\text{a}}\frac{\text{x}}{\sqrt{\text{a}^2+\text{x}^2}}\text{ dx}$
AnswerLet $\text{a}^2+\text{x}^2=\text{t}^2$ Differentiating w.r.t. x, we get$2\text{xdx}=2\text{tdt}$
$\text{xdx}=\text{tdt}$
Now, $\text{x}=0\Rightarrow\text{t}=0$$\text{x}=\text{a}\Rightarrow\text{t}=\sqrt{2}\text{a}$
$\therefore\ \int_{0}^\limits{\text{a}}\frac{\text{x}}{\sqrt{\text{a}^2+\text{x}^2}}\text{ dx}=\int_{\text{a}}^\limits{\sqrt{2\text{a}}}\frac{\text{t dt}}{\text{t}}$
$=\int_{\text{a}}^\limits{\sqrt{2\text{a}}}\text{dt}$
$=\big[\text{t}\big]^{\sqrt{2}\text{a}}_\text{a}$
$=\big[\sqrt{2}\text{a}-\text{a}\big]$
$=\text{a}\big(\sqrt{2}-1\big)$
View full question & answer→Question 1465 Marks
Evaluate the following integrals:$\int^\limits{\frac{1}{2}}_{0}\frac{1}{(1+\text{x}^2)\sqrt{1-\text{x}^2}}\text{ dx}$
Answer$\text{I}=\int^\limits{\frac{1}{2}}_{0}\frac{1}{(1+\text{x}^2)\sqrt{1-\text{x}^2}}\text{ dx}$Let $\text{x}=\sin\text{u}$
$\text{dx}=\cos\text{u du}$
$\text{I}=\int^\limits{\frac{\pi}{6}}_{0}\frac{1}{(1+\sin^2\text{u})}\text{ du}$
$\text{I}=\int^\limits{\frac{\pi}{6}}_{0}\frac{\sec^2\text{u}}{(1+2\tan^2\text{u})}\text{ du}$
Let $\tan\text{u}=\text{v}$
$\text{dv}=\sec^2\text{u du}$
$\text{I}=\int^\limits{\frac{1}{\sqrt{3}}}_{0}\frac{1}{(1+2\text{v}^2)}\text{ dv}$
$\text{I}=\frac{1}{\sqrt{2}}\Big[\tan^{-1}\big(\sqrt{2}\text{v}\big)\Big]^{\frac{1}{\sqrt{3}}}_0$
$\text{I}=\frac{1}{\sqrt{2}}\Big[\tan^{-1}\Big(\sqrt{\frac{2}{3}}\Big)\Big]$
View full question & answer→Question 1475 Marks
Evaluate the following definite integrals:$\int_{0}^\limits{1}\Big(\text{xe}^{\text{x}}+\cos\frac{\pi\text{x}}{4}\Big)\text{dx}$
AnswerWe have,$\int_{0}^\limits{1}\Big(\text{xe}^{\text{x}}+\cos\frac{\pi\text{x}}{4}\Big)\text{dx}$
$=\int_{0}^\limits{1}\text{xe}^{\text{x}}\text{ dx}+\int_{0}^\limits{1}\cos\frac{\pi\text{x}}{4}\text{ dx}$
Applying by parts in $1^{st}$ integral we get,$=\text{x}\int_{0}^\limits{1}\text{e}^{\text{x}}\text{ dx}-\int_{0}^\limits{1}\big(\int\text{e}^{\text{x}}\text{ dx}\big)+\int_{0}^\limits{1}\cos\frac{\pi\text{x}}{4}\text{ dx}$
$=\big[\text{xe}^{\text{x}}\big]^1_0-\int_{0}^\limits{1}\text{e}^{\text{x}}\text{ dx}+\bigg[\frac{\sin\frac{\pi\text{x}}{4}}{\frac{\pi}{4}}\bigg]^1_0$
$=\big[\text{xe}^{\text{x}}-\text{e}^{\text{x}}\big]^1_0+\frac{4}{\pi}\Big[\frac{1}{\sqrt{2}}-0\Big]$
$=\big[\text{e}^{\text{x}}(\text{x}-1)\big]^1_0+\frac{4}{\pi}\Big[\frac{1}{\sqrt{2}}\Big]$
$=0+1+\frac{4}{\pi\sqrt{2}}$
$=1+\frac{2\sqrt{2}}{\pi}$
View full question & answer→Question 1485 Marks
Evaluate the following definite integrals:$\int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\cos^2\text{x}\text{ dx}$
AnswerWe have,$\int\text{x}^2\cos^2\text{x dx}=\int\text{x}^2\Big(\frac{1+\cos2\text{x}}{2}\Big)\text{dx}\\=\frac{1}{2}\int(\text{x}^2+\text{x}^2\cos2\text{x})\text{dx}=\frac{1}{2}\big[\int\text{x}^2\text{dx}+\int\text{x}^2\cos2\text{x dx}\big]\ ...(\text{A})$
Now, $\int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\text{ dx}=\Big[\frac{\text{x}^3}{3}\Big]^{\frac{\pi}{2}}_0=\frac{\pi^3}{24}\ ....(\text{B})$
$\int\text{x}^2\cos2\text{x dx}=\text{x}^2\int\cos2\text{x dx}-\int2\text{x}\big(\int\cos2\text{x dx}\big)$
$=\frac{\text{x}^2\sin2\text{x}}{2}-\int\frac{\sin2\text{x}}{2}2\text{x dx}$
$=\frac{\text{x}^2\sin2\text{x}}{2}-\Big[\text{x}\int\sin2\text{x}-\int\big(\int\sin2\text{x dx}\big)\text{dx}\Big]$
$=\frac{\text{x}^2\sin2\text{x}}{2}+\frac{\text{x}\cos2\text{x}}{2}-\int\frac{\cos2\text{x}}{2}\text{ dx}$
$\therefore\ \int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\cos2\text{x dx}=\Big[\frac{\text{x}^2\sin2\text{x}}{2}+\frac{\text{x}\cos2\text{x}}{2}-\frac{\sin2\text{x}}{4}\Big]^{\frac{\pi}{2}}_0=\frac{-\pi}{4}\ ...(\text{C})$
Now, put (B) & (C) in (A), we get$\int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\cos^2\text{x}\text{ dx}=\int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\text{ dx}+\int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\cos2\text{x dx}\\=\frac{1}{2}\Big[\frac{\pi^2}{24}-\frac{\pi}{4}\Big]=\frac{\pi^3}{48}-\frac{\pi}{8}$
View full question & answer→Question 1495 Marks
Evaluate the following definite integrals:$\int_{0}^\limits{\pi}\text{e}^{2\text{x}}\sin\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\pi}\text{e}^{2\text{x}}\sin\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}$ Integrating by parts, we get$\text{I}=\frac{1}{2}\Big[\text{e}^{2\text{x}}\sin\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}\Big]^\pi_0-\frac{1}{2}\int_{0}^\limits{\pi}\text{e}^{2\text{x}}\cos\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}$
Now, integrating the second term by parts, we get$\Rightarrow\text{I}=\frac{1}{2}\Big[\text{e}^{2\text{x}}\sin\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}\Big]^\pi_0-\frac{1}{2}\bigg\{\Big[\frac{1}{2}\text{e}^{2\text{x}}\cos\Big(\frac{\pi}{4}+{\text{x}}\Big)\Big]^{\pi}_0\\+\frac{1}{2}\int_{0}^\limits{\pi}\text{e}^{2\text{x}}\sin\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}\bigg\}$
$\Rightarrow\text{I}=\frac{1}{2}\Big[\text{e}^{2\text{x}}\sin\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}\Big]^\pi_0-\frac{1}{4}\Big[\text{e}^{2\text{x}}\cos\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}\Big]^{\pi}_0-\frac{1}{4}\text{I}$
$\Rightarrow\frac{5}{4}\text{I}=\frac{1}{2}\Big[\text{e}^{2\pi}\sin\Big(\pi+\frac{\pi}{4}\Big)-\sin\Big(\frac{\pi}{4}\Big)\Big]\\-\frac{1}{4}\Big[\text{e}^{2\pi}\cos\Big(\pi+\frac{\pi}{4}\Big)-\cos\Big(\frac{\pi}{4}\Big)\Big]$
$\Rightarrow\frac{5}{4}\text{I}=\frac{1}{2}\Big[-\text{e}^{2\pi}\times\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\Big]-\frac{1}{4}\Big[-\text{e}^{2\pi}\times\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\Big]$
$\Rightarrow\frac{5}{4}\text{I}=-\frac{1}{2\sqrt{2}}\text{e}^{2\pi}-\frac{1}{2\sqrt{2}}+\frac{1}{4\sqrt{2}}\text{e}^{2\pi}+\frac{1}{4\sqrt{2}}$
$\Rightarrow\text{I}=-\frac{1}{5\sqrt{2}}\big(\text{e}^{2\pi}+1\big)$
View full question & answer→Question 1505 Marks
If f is an integrable function such that f(2a - x) = f(x), then prove that:$\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=2\int\limits^{\text{a}}_0\text{f(x)}\text{dx}$
AnswerLet $\text{I}=\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}$ By Additive property$\text{I}=\int\limits^{\text{a}}_0\text{f(x)}\text{dx}+\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}$
Consider the integral $\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}$ Let $\text{x}=2\text{a}-\text{t},$ then $\text{dx}=-\text{dt}$ When $\text{x}=\text{a},\text{ t}=\text{a},\text{ x}=2\text{x},\text{t}=0$ Hence $\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=-\int\limits^{0}_\text{a}\text{f(2a}-\text{t})\text{dt}$$=\int\limits_{0}^\text{a}\text{f(2a}-\text{t})\text{dt}$
$=\int\limits_{0}^\text{a}\text{f(2a}-\text{x})\text{dx}$ (Changeing the varible)
Therefore,$\text{I}=\int\limits_{0}^\text{a}\text{f(x})\text{dx}+\int\limits_{0}^\text{a}\text{f(2a}-\text{x})\text{dx}$
$=\int\limits_{0}^\text{a}\text{f(x})\text{dx}+\int\limits_{0}^\text{a}\text{f(x})\text{dx}$ $\Bigg[\text{Given}\int\limits_{0}^\text{a}\text{f(x})\text{dx}+\int\limits_{0}^\text{a}\text{f(2a}-\text{x})\text{dx}\Bigg]$
$=2\int\limits_{0}^\text{a}\text{f(x})\text{dx}$
View full question & answer→Question 1515 Marks
Evaluate the following integrals:$\int_{0}^\limits{\text{a}}\sqrt{\text{a}^2-\text{x}^2}\text{ dx}$
AnswerLet $\text{x}=\text{a}\sin\theta$ Differentiating w.r.t. x, we get$\text{dx}=\text{a}\cos\theta\text{ d}\theta$
Now, $\text{x}=0\Rightarrow\theta=0$$\text{x}=\text{a}\Rightarrow\theta=\frac{\pi}{2}$
$\therefore\ \int_{0}^\limits{\text{a}}\sqrt{\text{a}^2-\text{x}^2}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\sqrt{\text{a}^2(1-\sin^2\theta)}\text{a}\cos\theta\text{ d}\theta$
$=\text{a}^2\int_{0}^\limits{\frac{\pi}{2}}\cos^2\theta\text{ d}\theta$ $\Big[\because(1-\sin^2\theta)=\cos^2\theta\text{ and }\frac{1+\cos2\theta}{2}=\cos2\theta\Big]$
$=\frac{\text{a}^2}{2}\int_{0}^\limits{\frac{\pi}{2}}\big(1+\cos2\theta\big)\text{d}\theta$
$=\frac{\text{a}^2}{2}\Big[\frac{\pi}{2}+0-0-0\big]$
$=\frac{\pi\text{a}^2}{4}$
View full question & answer→Question 1525 Marks
Evaluate the following integrals:$\int\limits^{\pi}_0\frac{\text{x}\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\pi}_0\frac{\text{x}\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}\ ....(\text{i})$$=\int\limits^{\pi}_0\frac{(\pi-\text{x})\tan\text{x}}{\sec(\pi-\text{x})\text{ cosec}(\pi-\text{x})}\text{ dx}$ $\Bigg[\text{Using}\ \int\limits^{\text{a}}_{0}\text{f(x)}\text{dx}=\int\limits^{\text{a}}_{\text{0}}\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\pi}_0\frac{-(\pi-\text{x})\tan\text{x}}{-\sec\text{x}\text{ cosec}\text{x}}\text{ dx}$
$=\int\limits^{\pi}_0\frac{(\pi-\text{x})\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii)$2\text{I}=\int\limits^{\pi}_0\frac{\text{x}\tan\text{x}}{\sec\text{x}\text{ cosecx}}+\frac{(\pi-\text{x})\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}$
$=\int\limits^{\pi}_0(\text{x}+\pi-\text{x})\frac{\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}$
$=\int\limits^{\pi}_0\frac{\pi\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}$
$=\int\limits^{\pi}_0\pi\sin^2\text{x dx}$
$=\pi\int\limits^{\pi}_0\big(1-\cos^2\text{x}\big)\text{dx}$
$=\pi\big[\text{x}\big]^{\pi}_0-\frac{\pi}{2}\int\limits^{\pi}_0\big(1-\cos^2\text{x}\big)\text{dx}$
$=\frac{\pi}{2}\big[\text{x}\big]^{\pi}_0-\frac{\pi}{2}\Big[\frac{\sin2\text{x}}{2}\Big]^{\pi}_0$
$=\frac{\pi^2}{4}$
Hence, $\text{I}=\frac{\pi^2}{4}$
View full question & answer→Question 1535 Marks
Evaluate the following definite integrals:$\int_{0}^\limits{\frac{\pi}{2}}\cos^4\text{x}\text{ dx}$
AnswerWe have,$\int_{0}^\limits{\frac{\pi}{2}}\cos^4\text{x}\text{ dx}$
$=\frac{1}{4}\int_{0}^\limits{\frac{\pi}{2}}(1+\cos2\text{x})^2\text{dx}$ $\big[\because2\cos^2\text{x}=1+\cos2\text{x}\big]$
$=\frac{1}{4}\int_{0}^\limits{\frac{\pi}{2}}\big(1+\cos^22\text{x}+2\cos2\text{x}\big)\text{dx}$
$=\frac{1}{4}\int_{0}^\limits{\frac{\pi}{2}}\Big(1+\frac{1+\cos4\text{x}}{2}+2\cos2\text{x}\Big)\text{dx}$
$=\frac{1}{4}\Big[\text{x}+\frac{1}{2}\text{x}+\frac{\sin4\text{x}}{8}+\sin2\text{x}\Big]^{\frac{\pi}{2}}_0$ $\Big[\because\int\cos4\text{x dx}=\frac{\sin4\text{x}}{4}\Big]$
$=\frac{1}{4}\Big[\frac{\pi}{2}+\frac{\pi}{4}+0+0-0-0-0-0\Big]$
$=\frac{1}{4}\times\frac{3\pi}{4}$
$=\frac{3\pi}{16}$
View full question & answer→Question 1545 Marks
Evaluate the following integrals:$\int_{0}^\limits{1}\text{x}\tan^{-1}\text{x}\text{ dx} $
AnswerWe have,$\int_{0}^\limits{1}\text{x}\tan^{-1}\text{x}\text{ dx}=\tan^{-1}\text{x}\int_{0}^\limits{1}\text{x dx}-\int_{0}^\limits{1}\big(\int\text{dx}\big)\frac{\text{d}}{\text{dx}}\big(\tan^{-1}\text{x}\big)\text{dx}$
$=\Big[\frac{\text{x}^2}{2}\tan^{-1}\text{x}\Big]^1_0-\frac{1}{2}\int_{0}^\limits{1}\frac{\text{x}^2}{1+\text{x}^2}\text{ dx}$
$=\Big[\frac{\text{x}^2}{2}\tan^{-1}\text{x}\Big]^1_0-\frac{1}{2}\int_{0}^\limits{1}\frac{1+\text{x}^2-1}{1+\text{x}^2}\text{ dx}$
$=\frac{1}{2}\Big(\frac{\pi}{4}\Big)-\frac{1}{2}\Bigg[\int_{0}^\limits{1}\text{dx}-\int_{0}^\limits{1}\frac{\text{dx}}{1+\text{x}^2}\Bigg]$
$=\frac{\pi}{8}-\frac{1}{2}\big[\text{x}-\tan^{-1}\text{x}\big]^{1}_0$
$=\frac{\pi}{8}-\frac{1}{2}\Big[1-\frac{\pi}{4}\Big]$
$=\frac{\pi}{8}-\frac{1}{2}+\frac{\pi}{8}$
$=\frac{\pi}{4}-\frac{1}{2}$
View full question & answer→Question 1555 Marks
Evaluate the following integrals:$\int_{0}^\limits{\frac{\pi}{2}}\frac{\sin\text{x}\cos\text{x}}{1+\sin^4\text{x}}\text{ dx}$
AnswerLet $\sin^2\text{x}=\text{t}$ Differentiating w.r.t. x, we get$2\sin\text{x}\cos\text{x dx}=\text{dt}$
Now, $\text{x}=0\Rightarrow\text{t}=0$$\text{x}=\frac{\pi}{2}\Rightarrow\text{t}=1$
$\int_{0}^\limits{\frac{\pi}{2}}\frac{\sin\text{x}\cos\text{x}}{1+\sin^4\text{x}}\text{ dx}$
$=\frac{1}{2}\int\limits^1_0\frac{\text{dt}}{1+\text{t}^2}$
$=\frac{1}{2}\big[\tan^{-1}\text{t}\big]^1_0$
$=\frac{1}{2}\Big[\tan^{-1}(1)-\tan^{-1}(0)\Big]$
$=\frac{1}{2}\Big[\tan^{-1}\Big(\tan\frac{\pi}{4}\Big)-\tan^{-1}(\tan0)\Big]$
$=\frac{1}{2}\times\frac{\pi}{4}$
$=\frac{\pi}{8}$
View full question & answer→Question 1565 Marks
Evaluate the following integrals:$\int^\limits1_{-1}|2\text{x}+1|\text{dx}$
AnswerWe know that,$\int^\limits1_{-1}|2\text{x}+1|\text{dx}$
$=\int^\limits\frac{1}{2}_{-1}-(2\text{x}+1)\text{dx}+\int\limits_{-\frac{1}{2}}^{1}(2\text{x}+1)\text{dx}$
$=-\Big[\frac{2\text{x}^2}{2}+\text{x}\Big]^{-\frac{1}{2}}_{-1}+\Big[\frac{2\text{x}^2}{2}+\text{x}\Big]^{1}_{-\frac{1}{2}}$
$=-\bigg[\Big(\frac{2}{8}-\frac{1}{2}\Big)-\Big(\frac{2}{2}-1\Big)\bigg]+\bigg[\Big(\frac{2}{2}+1\Big)-\Big(\frac{2}{8}-\frac{1}{2}\Big)\bigg]$
$=-\bigg[\Big(\frac{1}{4}-\frac{1}{2}\Big)-(1-1)\bigg]+\bigg[(1+1)+\Big(\frac{1}{4}-\frac{1}{2}\Big)\bigg]$
$=-\Big[-\frac{1}{4}\Big]++\Big[2+\frac{1}{4}\Big]$
$=\frac{1}{4}+2+\frac{1}{4}$
$=2\frac{1}{2}$
View full question & answer→Question 1575 Marks
Evaluate the following integrals:$\int\limits^{\text{a}}_0\frac{1}{\text{x}+\sqrt{\text{a}^2-\text{x}^2}}\text{ dx}$
AnswerWe have,$\text{I}=\int\limits^{\text{a}}_0\frac{1}{\text{x}+\sqrt{\text{a}^2-\text{x}^2}}\text{ dx}$
Putting $\text{x}=\text{a}\sin\theta$$\text{dx}=\text{a}\cos\theta\text{ d}\theta$
When $\text{x}\rightarrow0;\theta\rightarrow0$ And $\text{x}\rightarrow\text{a};\theta\rightarrow\frac{\pi}{2}$$\therefore\ \text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\text{a}\cos\theta}{\text{a}\sin\theta+\sqrt{\text{a}^2-(\text{a}\sin\theta)^2}}\text{ d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\text{a}\cos\theta}{\text{a}\sin\theta+\text{a}\cos\theta}\text{ d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\cos\theta}{\sin\theta+\cos\theta}\text{ dx}\ ....(\text{i})$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\cos\big(\frac{\pi}{2}-\theta\big)}{\sin\big(\frac{\pi}{2}-\theta\big)+\cos\big(\frac{\pi}{2}-\theta\big)}\text{ d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sin\theta}{\cos\theta+\sin\theta}\text{ d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sin\theta}{\sin\theta+\cos\theta}\text{ d}\theta\ ....(\text{ii})$
By adding (i) and (ii) we get$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\cos\theta+\sin\theta}{\sin\theta+\cos\theta}\text{ d}\theta$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\text{d}\theta$
$\Rightarrow2\text{I}=\Big[\theta\Big]^{\frac{\pi}{2}}_0$
$\Rightarrow2\text{I}=\frac{\pi}{2}$
$\Rightarrow\text{I}=\frac{\pi}{4}$
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Evaluate the following definite integrals:$\int_{0}^\limits{2}\frac{1}{4+\text{x}-\text{x}^2}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{2}\frac{1}{4+\text{x}-\text{x}^2}\text{ dx}$ Then$\text{I}=\int_{0}^\limits{2}\frac{1}{\text{x}^2-\text{x}-4}\text{ dx}$
$\Rightarrow\text{I}=-\int_{0}^\limits{2}\frac{1}{\Big(\text{x}^2-\text{x}+\frac{1}{4}\Big)-\frac{1}{4}-4}\text{ dx}$
$\Rightarrow\text{I}=-\int_{0}^\limits{2}\frac{1}{\Big(\text{x}-\frac{1}{2}\Big)^2-\frac{17}{4}}\text{ dx}$
$\Rightarrow\text{I}=-\int_{0}^\limits{2}\frac{1}{\Big(\text{x}-\frac{1}{2}\Big)^2-\Big(\frac{\sqrt{17}}{2}\Big)^2}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{2}\frac{1}{-\big(\frac{2\text{x}-1}{2}\big)^2+\big(\frac{\sqrt{17}}{2}\big)^2}\text{ dx}$
$\Rightarrow\text{I}=\frac{1}{\sqrt{17}}\Big[\log\Big(\frac{\sqrt{17}+2\text{x}-1}{\sqrt{17}-2\text{x}+1}\Big)\Big]^2_0$
$\Rightarrow\text{I}=\frac{1}{\sqrt{17}}\Big\{\log\frac{\sqrt{17}+3}{\sqrt{17}-3}-\log\frac{\sqrt{17}-1}{\sqrt{17}+1}\Big\}$
$\Rightarrow\text{I}=\frac{1}{\sqrt{17}}\Big\{\log\frac{26+6\sqrt{17}}{8}-\log\frac{18-2\sqrt{17}}{16}\Big\}$
$\Rightarrow\text{I}=\frac{1}{\sqrt{17}}\Big\{\log\frac{52+12\sqrt{17}}{18-2\sqrt{17}}\Big\}$
$\Rightarrow\text{I}=\frac{1}{\sqrt{17}}\Big\{\log\frac{52+12\sqrt{17}}{18-2\sqrt{17}}\times\frac{18+2\sqrt{17}}{18+2\sqrt{17}}\Big\}$
$\Rightarrow\text{I}=\frac{1}{\sqrt{17}}\log\frac{1344+320\sqrt{17}}{256}$
$\Rightarrow\text{I}=\frac{1}{\sqrt{17}}\log\frac{21+5\sqrt{17}}{4}$
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Evaluate the following integrals:$\int^\limits{\frac{\pi}{2}}_{0}\frac{\tan\text{x}}{1+\text{m}^2\tan^2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int^\limits{\frac{\pi}{2}}_{0}\frac{\tan\text{x}}{1+\text{m}^2\tan^2\text{x}}\text{ dx}$$=\int^\limits{\frac{\pi}{2}}_{0}\frac{\frac{\sin\text{x}}{\cos\text{x}}}{1+\text{m}^2\frac{\sin^2\text{x}}{\cos^2\text{x}}}\text{ dx}=\int^\limits{\frac{\pi}{2}}_{0}\frac{{\sin\text{x}}{\cos\text{x}}}{\cos^2\text{x}+\text{m}^2{\sin^2\text{x}}}\text{ dx}$
Put $\cos^2\text{x}+\text{m}^2\sin^2\text{x}=\text{z}$$\therefore\ 2\cos\text{x}(-\sin\text{x})\text{dx}+\text{m}^2\times2\sin\text{x}\cos\text{x dx}=\text{ dz}$
$\Rightarrow2(\text{m}^2-1)\sin\text{x}\cos\text{x dx}=\text{dz}$
$\Rightarrow\sin\text{x}\cos\text{x dx}=\frac{\text{dz}}{2(\text{m}^2-1)}$
When $\text{x}\rightarrow0,\text{ z}\rightarrow1$ $\big(\text{z}=\cos^2\text{x}+\text{m}^2\sin^2\text{x}=1+\text{m}^2\times0=1\big)$ When $\text{x}\rightarrow\frac{\pi}{2},\text{ z}\rightarrow\text{m} ^2$ $\big(\text{z}=\cos^2\text{x}+\text{m}^2\sin^2\text{x}=0+\text{m}^2\times0=\text{m}^2\big)$$\therefore\ \text{I}=-\frac{1}{2(\text{m}^2-1)}\int\limits^{\text{m}^2}_1\frac{\text{dz}}{\text{z}}$
$=\frac{1}{2(\text{m}^2-1)}\big[\log\text{z}\big]^{\text{m}^2}_1$
$=\frac{1}{2(\text{m}^2-1)}\big(\log\text{m}^2-\log1\big)$
$=\frac{1}{2(\text{m}^2-1)}\big(2\log|\text{m}|-0\big)$
$=\frac{\log|\text{m}|}{\text{m}^2-1}$
View full question & answer→Question 1605 Marks
Evaluate the following integrals as limit of sum:$\int\limits^{3}_{2}\text{x}^2\text{ dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=2,\text{ b}=3,\text{ f(x)}=\text{x}^2,\text{ h}=\frac{3-2}{\text{n}}=\frac{1}{\text{n}}$
Therefore, $\text{I}=\int\limits^{3}_{2}\text{x}^2\text{ dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(2)+\text{f}(2+\text{h})\ ....\ +\text{f}\big\{2+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[2^2+(2+\text{h})^2+\ ....+\ \big\{2(\text{n}-1)\text{h}\big\}^2\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[4\text{n}+\text{h}\big\{1^2+2^2+3^2+\ ....\ +(\text{n}-1)^2+4\text{h}\big\}\\+4\text{h}\big\{1+2+\ ....+\ (\text{n}-1)\text{h}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[4\text{n}+\text{h}^2\frac{\text{n}(\text{n}-1)(\text{n}-2)}{6}+4\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}\frac{1}{\text{n}}\Big[4\text{n}+\frac{(\text{n}-1)(2\text{n}-1)}{6\text{n}}+2\text{n}-2\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}5\Big[6+\frac{1}{6}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)-\frac{2}{\text{n}}\Big]$
$=6+\frac{1}{3}$
$=\frac{19}{3}$
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Evaluate the following integrals:$\int\limits^{1}_0\frac{\log(1+\text{x})}{1+\text{x}^2}\text{ dx}$
AnswerWe have,$\text{I}=\int\limits^{1}_0\frac{\log(1+\text{x})}{1+\text{x}^2}\text{ dx}$
Putting $\text{x}=\tan\theta$$\text{dx}=\sec^2\theta\text{ d}\theta$
When $\text{x}\rightarrow0;\theta\rightarrow0$ And $\text{x}\rightarrow1;\theta\rightarrow\frac{\pi}{4}$ Now, integral becomes,$\text{I}=\int\limits^{\frac{\pi}{4}}_0\frac{\log(1+\tan\theta)}{\sec^2\theta}\sec^2\theta\text{ d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\big[\log(\tan\theta)\big]\text{d}\theta\ ...(\text{i})$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\bigg[\log\Big\{1+\tan\Big(\frac{\pi}{4}-\theta\Big)\Big\}\bigg]\text{d}\theta$ $\Bigg[\because\ \int\limits^{\text{a}}_{0}\text{f(x)}\text{dx}=\int\limits^{\text{a}}_{\text{0}}\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\Bigg[\log\bigg\{1+\frac{\tan\frac{\pi}{4}-\tan\theta}{1+\tan\frac{\pi}{4}\tan\theta}\bigg\}\Bigg]\text{d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\bigg[\log\Big\{1+\frac{1-\tan\theta}{1+\tan\theta}\Big\}\bigg]\text{d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\bigg[\log\Big\{\frac{2}{1+\tan\theta}\Big\}\bigg]\text{d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\big[\log2-\log(1+\tan\theta)\big]\text{d}\theta\ ...(\text{ii})$
Adding (i) and (ii) we get$2\text{I}=\int\limits^{\frac{\pi}{4}}_0\big(\log2\big)\text{d}\theta$
$\Rightarrow2\text{I}=\big(\log2\big)\Big[\theta\Big]^{\frac{\pi}{4}}_0$
$\Rightarrow2\text{I}=\frac{\pi}{4}\log2$
$\Rightarrow\text{I}=\frac{\pi}{8}\log2$
$\therefore\ \int\limits^{1}_0\frac{\log(1+\text{x})}{1+\text{x}^2}\text{ dx}=\frac{\pi}{8}\log2$
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Evaluate the following integrals:$\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\frac{1}{1+\text{e}^{\tan\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\frac{1}{1+\text{e}^{\tan\text{x}}}\text{ dx}\ ...(\text{i})$ Then,$\text{I}=\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\frac{1}{1+\text{e}^{\tan\big[\frac{\pi}{3}+\big(-\frac{\pi}{3}\big)-\text{x}\big]}}\text{ dx}$ $\Bigg[\int\limits^{\text{a}}_{0}\text{f(x)}\text{dx}=\int\limits^{\text{a}}_{\text{0}}\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\frac{1}{1+\text{e}^{\tan(-\text{x})}}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\frac{1}{1+\text{e}^{-\tan\text{x}}}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\frac{\text{e}^{\tan\text{x}}}{\text{e}^{\tan\text{x}}+1}\text{ dx}\ ...{\text{(ii)}}$
Adding (i) and (ii) we get$2\text{I}=\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\frac{\text{e}^{1+\tan\text{x}}}{\text{e}^{1+\tan\text{x}}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\text{dx}$
$\Rightarrow2\text{I}=\big[\text{x}\big]^{\frac{\pi}{3}}_{-\frac{\pi}{3}}$
$\Rightarrow2\text{I}=\frac{\pi}{3}-\Big(-\frac{\pi}{3}\Big)$
$\Rightarrow2\text{I}=\frac{2\pi}{3}$
$\Rightarrow\text{I}=\frac{\pi}{3}$
View full question & answer→Question 1635 Marks
Evaluate the following integrals:$\int^\limits{\frac{\pi}{2}}_{0}\frac{\sin\text{x}\cos\text{x}}{\cos^2\text{x}+3\cos\text{x}+2}\text{ dx}$
AnswerLet $\text{I}=\int^\limits{\frac{\pi}{2}}_{0}\frac{\sin\text{x}\cos\text{x}}{\cos^2\text{x}+3\cos\text{x}+2}\text{ dx}$ Then, Let $\cos\text{x}=\text{t},$ Then, $-\sin\text{x dx}=\text{dt}$ When, $\text{x}=0,\text{ t}=1$ and $\text{x}=\frac{\pi}{2},\text{ t}=0$$\therefore\ \text{I}=-\int^\limits0_1\frac{\text{t dt}}{\text{t}^2+3\text{t}+2}$
$\Rightarrow\text{I}=\int\limits^0_1\frac{-\text{t dt}}{\text{t}^3+3\text{t}+2}$
$\Rightarrow\text{I}=\int\limits^0_1\Big(\frac{1}{(\text{t}+1)}-\frac{2}{(\text{t}+2)}\Big)\text{dt}$
$\Rightarrow\text{I}=\Big[\log(\text{t}+1)-2\log(\text{t}+2)\Big]^0_1$
$\Rightarrow\text{I}=\bigg[\log\frac{(\text{t}+1)}{(\text{t}+2)^2}\bigg]^1_0$
$\Rightarrow\text{I}=\bigg[\log\Big(\frac{1}{4}\Big)-\log\Big(\frac{2}{9}\Big)\bigg]^1_0$
$\Rightarrow\text{I}=\log\frac{9}{8}$
View full question & answer→Question 1645 Marks
Evaluate the following integrals:$\int_{0}^\limits{1}\frac{24\text{x}^3}{(1+\text{x}^2)^4}\text{ dx}$
AnswerLet $1+\text{x}^2=\text{t}$ Differentiating w.r.t. x, we get$2\text{xdx}=\text{dt}$
Now, $\text{x}=0\Rightarrow\text{t}=1$$\text{x}=1\Rightarrow\text{t}=2$
$\int_{0}^\limits{1}\frac{24\text{x}^3}{(1+\text{x}^2)^4}\text{ dx}=\int_{1}^\limits{2}\frac{12(\text{t}-1)}{\text{t}^4}\text{ dt}$
$=12\int_{1}^\limits{2}\Big(\frac{1}{\text{t}^3}-\frac{1}{\text{t}^4}\Big)\text{dt}$
$=12\Big[-\frac{1}{2\text{t}^2}-\frac{1}{3\text{t}^3}\Big]^2_1$
$=12\Big[-\frac{1}{8}+\frac{1}{24}+\frac{1}{2}-\frac{1}{3}\Big]$
$=12\Big[\frac{-3+1+12-8}{24}\Big]$
$=\frac{12\times2}{24}=1$
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If f(2a - x) = -f(x), prove that $\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=0$
AnswerLet $\text{I}=\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=0$ Using additive property$\text{I}=\int\limits^{\text{a}}_0\text{f(x)}\text{dx}+\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}$
Consider the integral $\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}$ Let $\text{x}=2\text{a}-\text{t},$ Then $\text{dx}=-\text{dt}$ When $\text{x}=\text{a},\text{t}=\text{a}$ and $\text{x}=2\text{a},\text{t}=0$ Threrfore,$=\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=-\int\limits^0_\text{a}\text{f}(2\text{a}-\text{t})\text{dt}$
$=\int\limits^\text{a}_0\text{f}(2\text{a}-\text{t})\text{dt}$
$=\int\limits^\text{a}_0\text{f}(2\text{a}-\text{x})\text{dx}$ (Chang ing the variable)
We have$\text{f}(2\text{a}-\text{x})=-\text{f(x)}$
Therefore,$\text{I}=\int\limits^\text{a}_0\text{f(x)}\text{dx}-\int\limits^\text{a}_0\text{f(x)}\text{dx}=0$
View full question & answer→Question 1665 Marks
Evaluate the following definite integrals:$\int_{0}^\limits{\infty}\frac{1}{\text{a}^2+\text{b}^2\text{x}^2} \text{ dx}$
AnswerWe have,$\int_{0}^\limits{\infty}\frac{1}{\text{a}^2+\text{b}^2\text{x}^2} \text{ dx}=\frac{1}{\text{b}^2}\int_{0}^\limits{\infty}\frac{1}{\big(\frac{\text{a}}{\text{b}}\big)^2+\text{x}^2}\text{ dx}$
We have that,$\int\frac{1}{\text{a}^2+\text{x}^2}=\frac{1}{\text{a}}\tan^{-1}\frac{\text{x}}{\text{a}}$
$\therefore\ \frac{1}{\text{b}^2}\int_{0}^\limits{\infty}\frac{1}{\big(\frac{\text{a}}{\text{b}}\big)^2+\text{x}^2}\text{ dx}=\frac{1}{\text{b}^2}\Big[\frac{\text{b}}{\text{a}}\tan^{-1}\Big(\frac{\text{bx}}{\text{a}}\Big)\Big]^{\infty}_0$
$=\frac{1}{\text{ab}}\Big[\tan^{-1}\Big(\frac{\text{bx}}{\text{a}}\Big)\Big]^{\infty}_0$
$=\frac{1}{\text{ab}}\big[\tan^{-1}\infty-\tan^{-1}0\big]$
$=\frac{1}{\text{ab}}\Big[\frac{\pi}{2}-0\Big]$
$=\frac{\pi}{2\text{ab}}$
$\int_{0}^\limits{\infty}\frac{1}{\text{a}^2+\text{b}^2\text{x}^2} \text{ dx}=\frac{\pi}{2\text{ab}}$
View full question & answer→Question 1675 Marks
Evaluate the following integrals:$\int\limits^{1}_0\big|\text{x}\sin\pi\text{x}\big|\text{dx}$
AnswerFor $0<\text{x}<1,\text{ x}>0$ and $\sin\pi\text{x}>0\Rightarrow\text{x}\sin\pi\text{x}>0$$\therefore\ \int\limits^{1}_0|\text{x}\sin\pi\text{x}|\text{ dx}=\int\limits^{1}_0\text{x}\sin\pi\text{x}\text{ dx}$
Let $\text{I}=\int\text{x}\sin\pi\text{x}\text{ dx}$$=\text{x}\int\sin\pi\text{x}\text{ dx}-\int\Big(\frac{\text{d}}{\text{dx}}\text{x}\int\sin\pi\text{x dx}\Big)\text{dx}$
$=\text{x}\Big(\frac{-\cos\pi\text{x}}{\pi}\Big)-\int\Big(\frac{-\text{cos}\pi\text{x}}{\pi}\Big)\text{dx}$
$=\frac{-\text{x}\cos\pi\text{x}}{\pi}+\frac{\sin\pi\text{x}}{\pi^2}$
Applying the limits, we get$\int\limits^{1}_0|\text{x}\sin\pi\text{x}|\text{ dx}=\Big[\frac{-\text{x}\cos\pi\text{x}}{\pi}+\frac{\sin\pi\text{x}}{\pi^2}\Big]^1_0$
$=\Big(\frac{-\cos\pi}{\pi}+\frac{\sin\pi}{\pi^2}\big)-(0+0)$
$=\frac{1}{\pi}+0-0$
$=\frac{1}{\pi}$
View full question & answer→Question 1685 Marks
Evaluate the following integrals as limit of sum:$\int\limits^2_{0}\big(\text{x}^2+1\big)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=2,\text{ f(x)}=\text{x}^2+1,\text{ h}=\frac{2-0}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^2_{0}\big(\text{x}^2+1\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big(0+(\text{n}-1)\text{h}\big)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(0+1)+(\text{h}^2+1)+\ ....\ +\big\{(\text{n}-1)^2\text{h}^2+1\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+\text{h}^2\big\{1^2+2^2+3^2+\ ....\ +(\text{n}-1)^2\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2}{\text{n}}\Big[\text{n}+\frac{2(\text{n}-1)(2\text{n}-1)}{3\text{n}}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\bigg\{1+\frac{2}{3}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)\bigg\}$
$=2+\frac{8}{3}$
$=\frac{14}{3}$
View full question & answer→Question 1695 Marks
Evaluate the following integrals:$\int\limits^{\pi}_0\frac{\text{x}\sin\text{x}}{1+\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\pi}_0\frac{\text{x}\sin\text{x}}{1+\sin\text{x}}\text{ dx}\ ...(\text{i})$$=\int\limits^{\pi}_0\frac{(\pi-\text{x})\sin(\pi-\text{x})}{1+\sin(\pi-\text{x})}\text{ dx}$
$=\int\limits^{\pi}_0\frac{(\pi-\text{x})\sin\text{x}}{1+\sin\text{x}}\text{ dx}\ ....(\text{ii})$
Adding (i) and (ii) we get$2\text{I}=\int\limits^{\pi}_0(\text{x}+\pi-\text{x})\frac{\sin\text{x}}{1+\sin\text{x}}\text{ dx}$
$=\int\limits^{\pi}_0\frac{\pi\sin\text{x}}{1+\sin\text{x}}\text{ dx}$
$=\pi\int\limits^{\pi}_0\frac{1+\sin\text{x}-1}{1+\sin\text{x}}\text{ dx}$
$=\pi\int\limits^{\pi}_0\text{dx}-\pi\int\limits^{\pi}_0\frac{1}{1+\sin\text{x}}\text{ dx}$
$=\pi\int\limits^{\pi}_0\text{dx}-\pi\int\limits^{\pi}_0\frac{(1-\sin\text{x})}{(1+\sin\text{x})(1-\sin\text{x})}\text{ dx}$
$=\pi\int\limits^{\pi}_0\text{dx}-\pi\int\limits^{\pi}_0\frac{(1-\sin\text{x})}{1-\sin}$
$=\pi\int\limits^{\pi}_0\text{dx}-\pi\int\limits^{\pi}_0\big(\sec^2\text{x}-\sec\text{x}\tan\text{x}\big)\text{dx}$
$=\pi\big[\text{x}\big]^{\pi}_0-\pi\big[\tan\text{x}-\sec\text{x}\big]^{\pi}_0$
$=\pi^2-\pi(0+1-0+1)$
$=\pi^2-2\pi$
Hence, $\text{I}=\pi\Big(\frac{\pi}{2}-1\Big)$
View full question & answer→Question 1705 Marks
Evaluate the following integrals:$\int_{0}^\limits{1}\tan^{-1}\text{x dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\tan^{-1}\text{x dx}$ Then,$\text{I}=\int_{0}^\limits{1}\tan^{-1}\text{x dx}$
Integrating by parts,$\text{I}=\big[\text{x}\tan^{-1}\text{x}\big]^1_0-\int_{0}^\limits{1}\frac{\text{x}}{1+\text{x}^2}\text{ dx}$
$\Rightarrow\text{I}=\big[\text{x}\tan^{-1}\text{x}\big]^1_0-\frac{1}{2}\big[\log\big(\text{x}^2+1\big)\big]^1_0$
$\Rightarrow\text{I}=\frac{\pi}{4}-0-\frac{1}{2}\log2+0$
$\Rightarrow\text{I}=\frac{\pi}{4}-\frac{1}{2}\log2$
View full question & answer→Question 1715 Marks
Evaluate the following integrals:$\int_{0}^\limits{\frac{\pi}{4}}\Big(\sqrt{\tan\text{x}}+\sqrt{\cot}\text{x}\Big)\text{dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\Big(\sqrt{\tan\text{x}}+\sqrt{\cot}\text{x}\Big)\text{dx}$ Then,$\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\bigg(\sqrt{\frac{\sin\text{x}}{\cos\text{x}}}+\sqrt{\frac{\cos\text{x}}{\sin\text{x}}}\bigg)\text{dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{\sqrt{\sin\text{x}\cos\text{x}}}\text{ dx}$
$\Rightarrow\text{I}=\sqrt{2}\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{\sqrt{2\sin\text{x}\cos\text{x}}}\text{ dx}$
$\Rightarrow\text{I}=\sqrt{2}\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{\sqrt{1-(\sin\text{x}+\cos\text{x})^2}}\text{ dx}$
Let $\sin\text{x}-\cos\text{x}=\text{t}$ Then, $\cos\text{x}+\sin\text{x dx}=\text{dt}$ When $\text{x}=0,\text{t}=1$ and $\text{x}=\frac{\pi}{4},\text{t}=0$$\therefore\ \text{I}=\sqrt{2}\int^\limits0_{-1}\frac{\text{dt}}{\sqrt{1-\text{t}^2}}$
$\Rightarrow\text{I}=\sqrt{2}\Big[\sin^{-1}\text{t}\Big]^0_{-1}$
$\Rightarrow\text{I}=\frac{\pi}{\sqrt{2}}$
View full question & answer→Question 1725 Marks
Evaluate the following integrals:$\int^\limits{\frac{\pi}{2}}_\frac{\pi}{3}\frac{\sqrt{1+\cos\text{x}}}{(1-\cos\text{x})^{\frac{3}{2}}}\text{ dx}$
Answer$\int^\limits{\frac{\pi}{2}}_\frac{\pi}{3}\frac{\sqrt{1+\cos\text{x}}}{(1-\cos\text{x})^{\frac{3}{2}}}\text{ dx}$$=\int^\limits{\frac{\pi}{2}}_\frac{\pi}{3}\frac{\sqrt{1+\cos\text{x}}}{(1-\cos\text{x})^{\frac{3}{2}}}\text{ dx}\times\frac{\sqrt{1-\cos\text{x}}}{\sqrt{1-\cos\text{x}}}\text{ dx}$
$=\int^\limits{\frac{\pi}{2}}_\frac{\pi}{3}\frac{\sqrt{1-\cos^2\text{x}}}{(1-\cos\text{x})^2}\text{ dx}$
$=\int^\limits{\frac{\pi}{2}}_\frac{\pi}{3}\frac{\sin\text{x}}{(1-\cos\text{x})^2}\text{ dx}$
Let $1-\cos\text{x}=\text{t},$ Then $\sin\text{x dx}=\text{dt}$
When $\text{x}=\frac{\pi}{3},\text{ t}=\frac{1}{2}$ and $\text{x}=\frac{\pi}{2},\text{ t}=1$
Therefore the integral becomes
$=\int_{1}^{\frac{1}{2}}\frac{\text{dt}}{\text{t}^2}$
$=\Big[-\frac{1}{\text{t}}\Big]^1_\frac{1}{2}$
$=-1+2$
$=1$
View full question & answer→Question 1735 Marks
Evaluate the following integrals:$\int^\limits1_0\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{ dx}$
AnswerLet $\text{x}=\cos2\theta$ Differentiating w.r.t. x, we get$\text{dx}=-2\sin2\theta\text{ d}\theta$
Now, $\text{x}=0\Rightarrow\theta=\frac{\pi}{4}$$\text{x}=1\Rightarrow\theta=0$
$\therefore\ \int^\limits1_0\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{ dx}=\int^0\limits_\frac{\pi}{4}\sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}\big(-2\sin2\theta\big)\text{d}\theta$
$=\int_0\limits^\frac{\pi}{4}\sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}\big(2\sin2\theta\big)\text{d}\theta$ $\Big[\because\sin2\theta=2\sin\theta\cos\theta;\text{ and }\sin^2\theta=\frac{1-\cos2\theta}{2}\Big]$
$=2=\int_0\limits^\frac{\pi}{4}\frac{\sin\theta}{\cos\theta}\cdot\sin2\theta\text{ d}\theta$
$=4\int_0\limits^\frac{\pi}{4}\sin^2\theta\text{ d}\theta$
$=2\int_0\limits^\frac{\pi}{4}\big(1-\cos2\theta\big)\text{d}\theta$
$=2\Big[\theta-\frac{\sin^2\theta}{2}\Big]^{\frac{\pi}{4}}_0$
$=2\Big[\frac{\pi}{4}-\frac{1}{2}\Big]$
$=\frac{\pi}{2}-1$
View full question & answer→Question 1745 Marks
Evaluate the following integrals:$\int\limits^1_{-1}\log\Big(\frac{2-\text{x}}{2+\text{x}}\Big)\text{dx}$
AnswerLet $\text{I}=\int\limits^1_{-1}\log\Big(\frac{2-\text{x}}{2+\text{x}}\Big)\text{dx}$ Here $\text{f(x)}=\log\Big(\frac{2-\text{x}}{2+\text{x}}\Big)$$\text{f}(-\text{x})=\log\Big(\frac{2+\text{x}}{2-\text{x}}\Big)$
$=-\log\Big(\frac{2-\text{x}}{2+\text{x}}\Big)$
$=-\text{f(x)}$
Hence f(x) is an odd function, Therefore,$\text{I}=0$
View full question & answer→Question 1755 Marks
Evaluate the following integrals:$\int\limits^\pi_0\frac{\text{x}}{1+\sin\alpha\sin\text{x}}\text{ dx}$
AnswerWe have,$\text{I}=\int\limits^\pi_0\frac{\text{x}}{1+\sin\alpha\sin\text{x}}\text{ dx}\ ...(\text{i})$
$=\int\limits^\pi_0\frac{\pi-\text{x}}{1+\cos\alpha\sin(\pi-\text{x})}\text{ dx}$
$=\int\limits^\pi_0\frac{\pi-\text{x}}{1+\cos\alpha\sin\text{x}}\text{ dx}\ ....(\text{ii})$
Adding (i) and (ii) we get$2\text{I}=\int\limits^\pi_0\frac{\text{x}+\pi-\text{x}}{1+\cos\alpha\sin\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\frac{\pi}{2}\int\limits^\pi_0\frac{1}{1+\cos\alpha\sin\text{x}}\text{ dx}$
$=\frac{\pi}{2}\int\limits^\pi_0\frac{1}{1+\cos\alpha\sin\text{x}}$
$=\frac{\pi}{2}\int\limits^\pi_0\frac{1}{1+\cos\alpha\frac{2\tan\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}}}\text{ dx}$
$=\frac{\pi}{2}\int\limits^\pi_0\frac{1+\tan^{2}\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}+2\cos\alpha\tan^{2}\frac{\text{x}}{2}}\text{ dx}$
$=\frac{\pi}{2}\int\limits^\pi_0\frac{\sec^{2}\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}+2\cos\alpha\tan^{2}\frac{\text{x}}{2}}\text{ dx}$
Putting $\tan\frac{\text{x}}{2}=\text{t}$$\Rightarrow\frac{1}{2}\sec^2\text{x dx}=\text{dt}$
When $\text{x}\rightarrow0;\text{ t}\rightarrow0$ and $\text{x}\rightarrow\pi;\text{ t}\rightarrow\infty$$\therefore\ \text{I}=\frac{\pi}{2}\int\limits^{\infty}_0\frac{2}{1+\text{t}^2+2\cos\alpha+1}\text{ dt}$
$=\frac{\pi}{2}\int\limits^{\infty}_0\frac{2}{(\text{t}+\cos\alpha)-\cos^2\alpha+1}\text{ dt}$
$={\pi}\int\limits^{\infty}_0\frac{1}{(\text{t}+\cos\alpha)+\sin^2\alpha}\text{ dt}$
$=\pi\Big[\frac{1}{\sin\alpha}\tan^{-1}\Big(\frac{1+\cos\alpha}{\sin\alpha}\Big)\Big]^1_0$
$=\frac{\pi}{\sin\alpha}\Big[\tan^{-1}(\infty)-\tan^{-1}(\cot\alpha)\Big]$
$=\frac{\pi}{\sin\alpha}\Big[\frac{\pi}{2}-\tan^{-1}\Big(\tan\Big(\frac{\pi}{2}-\alpha\Big)\Big)\Big]$
$=\frac{\pi\alpha}{\sin\alpha}$
View full question & answer→Question 1765 Marks
Evaluate the following integrals as limit of sum:$\int\limits^2_0(\text{x}+3)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=2,\text{ f(x)}=\text{x}+3,\text{ h}=\frac{2-0}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^2_0(\text{x}+3)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}(0+(\text{n}-1)\text{h})\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(0+3)+(0+\text{h}+3)+\ ....\ +(0+(\text{n}-1)\text{h}+3)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[3\text{n}+\text{h}\{1+2+3+\ ....\ +(\text{n}-1)\}\Big]$
$=\lim\limits_{\text{n}\rightarrow0}\text{h}\Big[3\text{n}+\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2}{\text{n}}\Big[3\text{n}+\text{n}-1\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\Big(4-\frac{1}{\text{n}}\Big)$
$=8$
View full question & answer→Question 1775 Marks
Evaluate the following integrals:$\int\limits^{\pi}_0\text{x}\sin\text{x}\cos^4\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\pi}_0\text{x}\sin\text{x}\cos^4\text{x}\text{ dx}\ ...(\text{i})$$=\int\limits^{\pi}_0(\pi-\text{x})\sin(\pi-\text{x})\cos^4(\pi-\text{x})\text{dx}$
$=\int\limits^{\pi}_0(\pi-\text{x})\sin\text{x}\cos^4\text{x dx}\ ...(\text{ii})$
Adding (i) and (ii) we get$2\text{I}=\int\limits^{\pi}_0(\text{x}+\pi-\text{x})\sin\text{x}\cos^4\text{x dx}$
$=\pi\int\limits^{\pi}_0\sin\text{x}\cos^4\text{x dx}$
Let $\cos\text{x}=\text{t},$ Then $-\sin\text{x dx}=\text{dt}$ When $\text{x}=0,\text{t}=1,\text{x}=\pi,\text{t}=-1$ Therefore, $2\text{I}=-\pi\int\limits^{-1}_1\text{t}^4\text{ dt}$$=\pi\int\limits^{1}_{-1}\text{t}^4\text{ dt}$
$=\pi\Big[\frac{\text{t}^5}{5}\Big]^{1}_{-1}$
$=\frac{\pi}{5}+\frac{\pi}{5}$
$=\frac{2\pi}{5}$
Hence, $\text{I}=\frac{\pi}{5}$
View full question & answer→Question 1785 Marks
Evaluate the following integrals:$\int_{0}^\limits{\frac{\pi}{2}}\frac{\text{dx}}{\text{a}\cos\text{x}+\text{b}\sin\text{x}}\text{ a},\text{b}>0$
Answer$\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{\text{a}\cos\text{x}+\text{b}\sin\text{x}}\text{ dx}$$=\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{\text{a}\Bigg(\frac{1-\tan^{2}\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}}\Bigg)+\text{b}\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}}\Bigg)}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\frac{\big(1+\tan^{2}\frac{\text{x}}{2}\big)}{\text{a}-\text{a}\tan^{2}\frac{\text{x}}{2}+2\text{b}\tan\frac{\text{x}}{2}}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\frac{\sec^2\frac{\text{x}}{2}}{\text{a}-\text{a}\tan^{2}\frac{\text{x}}{2}+2\text{b}\tan\frac{\text{x}}{2}}\text{ dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$ Then, $\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{ dx}=\text{dt}$
When $\text{x}=0,\text{t}=0,\text{x}=\frac{\pi}{2},\text{t}=1$
Therefore the integral becomes
$\text{I}=\int^\limits1_0\frac{2\text{dt}}{\text{a}-\text{a}\text{t}^2+2\text{bt}}$
$=\int^\limits1_0\frac{2\text{dt}}{-\text{a}\big[\text{t}^2-\frac{2\text{bt}}{\text{a}-1}\big]}$
$=\frac{2}{\text{a}}\int^\limits1_0\frac{\text{dt}}{-\Big[\big(\text{t}-\frac{\text{b}}{\text{a}}\big)^2-1-\frac{\text{b}^2}{\text{a}^2}\Big]}$
$=\frac{2}{\text{a}}\int^\limits1_0\frac{\text{dt}}{\Big(\frac{\text{b}^2}{\text{a}^2}+1\Big)-\big(\text{t}-\frac{\text{b}}{\text{a}}\big)^2}$
$=\frac{2}{\text{a}}\begin{bmatrix}\frac{1}{2\sqrt{\frac{\text{a}^2+\text{b}^2}{\text{a}^2}}}\begin{pmatrix}\log\begin{vmatrix}\frac{2\sqrt{\frac{\text{a}^2+\text{b}^2}{\text{a}^2}}+\big(\text{t}-\frac{\text{b}}{\text{a}}\big)}{\sqrt{\frac{\text{a}^2+\text{b}^2}{\text{a}^2}-\big(\text{t}-\frac{\text{b}}{\text{a}}\big)}}\end{vmatrix} \end{pmatrix}^1_0\end{bmatrix}$
$=\frac{1}{\sqrt{\text{a}^2+\text{b}^2}}\log\bigg(\frac{\text{a}+\text{b}+\sqrt{\text{a}^2+\text{b}^2}}{\text{a}+\text{b}-\sqrt{\text{a}^2+\text{b}^2}}\bigg)$
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