Question 15 Marks
Evaluate the following integrals:$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}(\tan x+\cot x)^2 d x$
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:
$\int\limits_{1}^{4}\frac{\text{x}^2+\text{x}}{\sqrt{2\text{x}+1}}\text{ dx}$
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:
$\int\limits_{0}^{\frac{\pi}{2}}\frac{\text{x}+\sin\text{x}}{1+\cos\text{x}}\text{ dx}$
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:
$\int\limits_{0}^{\frac{\pi}{4}}\sec\text{x}\text{ dx}$
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:
$\int\limits_{0}^{\frac{\pi}{2}}\frac{1}{5\cos\text{x}+3\sin\text{x}}\text{ dx}$
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:
$\int\limits_{0}^{{\pi}}\frac{1}{5+3\cos\text{x}}\text{ dx}$
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:
$\int\limits^1_0\frac{1-\text{x}^2}{\text{x}^4+\text{x}^2+1}\text{ dx}$
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:
$\int\limits_{0}^{\frac{\pi}{2}}\frac{1}{5+4\sin\text{x}}\text{ dx}$
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:
$\int\limits_{0}^{1}\frac{1-\text{x}}{1+\text{x}}\text{ dx}$
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:
$\int\limits^2_1\frac{1}{\text{x}(1+\log\text{x})^2}\text{ dx}$
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:
$\int\limits_{0}^{1}\frac{1}{\sqrt{(\text{x}-1)(2-\text{x})}}\text{ dx}$
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:
$\int\limits_{0}^{\frac{\pi}{2}}\sin^3\text{x}\text{ dx}$
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:
$\int\limits_{0}^{1}\sqrt{\text{x}(1-\text{x})}\text{ dx}$
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:
$\int\limits_{0}^{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{3+\sin2\text{x}}\text{ dx}$
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:
$\int\limits_{1}^{2}\frac{\text{x}+3}{\text{x}(\text{x}+2)}\text{ dx}$
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]
$=\big[\log(\text{x}+2)\big]^2_1+\Big[\frac{3}{2}\log\text{x}-\frac{3}{2}\log(\text{x}+2)\Big]^2_1$
$=\Big[\frac{3}{2}\log\text{x}-\frac{1}{2}\log(\text{x}+2)\Big]^2_1$
$=\frac{1}{2}\big[3\log2-\log4+\log3\big]$
$=\frac{1}{2}\big[3\log2-2\log2+\log3\big]$ $\big[\because\log4=2\log2\big]$
$=\frac{1}{2}\big[\log2+\log3\big]$
$=\frac{1}{2}\big[\log6\big]$
$=\frac{1}{2}\log6$
$\therefore\ \int_{1}^\limits{2}\frac{\text{x}+3}{\text{x}(\text{x}+2)}\text{dx}=\frac{1}{2}\log6$
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