Question 1015 Marks
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
$2\text{I}=\int\limits^{\frac{\pi}{4}}_0\big(\log2\big)\text{d}\theta$
$\Rightarrow2\text{I}=\big(\log2\big)\Big[\theta\Big]^{\frac{\pi}{4}}_0$
$\Rightarrow2\text{I}=\frac{\pi}{4}\log2$
$\Rightarrow\text{I}=\frac{\pi}{8}\log2$
$\therefore\ \int\limits^{1}_0\frac{\log(1+\text{x})}{1+\text{x}^2}\text{ dx}=\frac{\pi}{8}\log2$
View full question & answer→Question 1025 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 1035 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
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\cos\theta+\sin\theta}{\sin\theta+\cos\theta}\text{ d}\theta$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\text{d}\theta$
$\Rightarrow2\text{I}=\Big[\theta\Big]^{\frac{\pi}{2}}_0$
$\Rightarrow2\text{I}=\frac{\pi}{2}$
$\Rightarrow\text{I}=\frac{\pi}{4}$
View full question & answer→Question 1045 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 1055 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 1065 Marks
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 1075 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 1085 Marks
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 1095 Marks
Evaluate the following integrals:
$\int\limits_{0}^{1}\text{x}\log(1+2\text{x})\text{dx}$
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 1105 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 1115 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 1125 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 1135 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 1145 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 1155 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 1165 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 1175 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 1185 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 1195 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 1205 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 1215 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 1225 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 1235 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 1245 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 1255 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 1265 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 1275 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 1285 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 1295 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 1305 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 1315 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 1325 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 1335 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 1345 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 1355 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 1365 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 1375 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 1385 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 1395 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 1405 Marks
Evaluate the following definite integrals:
$\int\limits_{\frac{\pi}{3}}^{\frac{\pi}{4}}(\tan\text{x}+\cot\text{x})^2\text{ dx}$
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 1415 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]
$=\big[\log(\text{x}+2)\big]^2_1+\Big[\frac{3}{2}\log\text{x}-\frac{3}{2}\log(\text{x}+2)\Big]^2_1$
$=\Big[\frac{3}{2}\log\text{x}-\frac{1}{2}\log(\text{x}+2)\Big]^2_1$
$=\frac{1}{2}\big[3\log2-\log4+\log3\big]$
$=\frac{1}{2}\big[3\log2-2\log2+\log3\big]$ $\big[\because\log4=2\log2\big]$
$=\frac{1}{2}\big[\log2+\log3\big]$
$=\frac{1}{2}\big[\log6\big]$
$=\frac{1}{2}\log6$
$\therefore\ \int_{1}^\limits{2}\frac{\text{x}+3}{\text{x}(\text{x}+2)}\text{dx}=\frac{1}{2}\log6$
View full question & answer→Question 1425 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 1435 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 1445 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 1455 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 1465 Marks
Evaluate the following integrals:
$\int\limits_{0}^{\frac{\pi}{3}}\frac{\cos\text{x}}{3+4\sin\text{x}}\text{ dx}$
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 1475 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 1485 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 1495 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 1505 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, $\text{I}=\int\limits^3_{2}\big(2\text{x}^2+1\big)\text{ dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(2)+\text{f}(2+\text{h})+\ ....\ +\text{f}\big\{2+(\text{n}-1)\text{h}\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[2(2.2)^2+1+\big\{2(2+\text{h})^2+1\big\}+\\\ ....\ +\big\{2((2+\text{n}-1)\text{h})^2+1\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+2\Big\{2^2+(2+\text{h})^2+\ .....\big((2+\text{n}-1\big)\text{h}\big)^2\Big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+8\text{n}+2\text{h}^2\Big\{1^2+2^2+3^3+\ .....\ +(\text{n}-1)^2\Big\}\\+8\text{h}\big\{1+2+\ ....\ +(\text{n}-1)\big\}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\text{h}\bigg[9\text{n}+\text{h}^2\frac{2\text{n}(\text{n}-1)(2\text{n}-1)}{6}+8\text{h}\frac{\text{n}(\text{n}-1)}{2}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{1}{\text{n}}\bigg[9\text{n}+\frac{(\text{n}-1)(2\text{n}-1)}{3\text{n}}+4\text{n}-4\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\bigg\{13+\frac{1}{3}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)-\frac{4}{\text{n}}\bigg\}$
$=13+\frac{2}{3}$
$=\frac{41}{3}$
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