Class 12: Definite Integrals – ISC Board Problems

$\displaystyle \textbf{Question 1: } \text{The value of } \int_{1}^{\sqrt{3}} \frac{dx}{1+x^{2}} \text{ is } \quad \text{ISC 2024}$
$\displaystyle \text{(a) } \frac{\pi}{2} \quad \text{(b) } \frac{2\pi}{3} \quad \text{(c) } \frac{\pi}{6} \quad \text{(d) } \frac{\pi}{12}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(d) Let } I=\int_{1}^{\sqrt{3}} \frac{dx}{1+x^{2}}$
$\displaystyle =\left[\tan^{-1}x\right]_{1}^{\sqrt{3}}$
$\displaystyle =\tan^{-1}(\sqrt{3})-\tan^{-1}(1)$
$\displaystyle =\frac{\pi}{3}-\frac{\pi}{4}$
$\displaystyle =\frac{\pi}{12}$
$\\$

$\displaystyle \textbf{Question 2: } \text{Evaluate } \int_{0}^{6}|x+3|\,dx. \quad \text{ISC 2024}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, } \int_{0}^{6} |x+3|\,dx$
$\displaystyle \text{Here, } f(x)=\begin{cases} x+3, & x\geq -3 \\ -(x+3), & x<-3 \end{cases}$
$\displaystyle \text{By the property of definite integral, we get}$
$\displaystyle \int_{0}^{6} f(x)\,dx=\int_{0}^{-3} f(x)\,dx+\int_{-3}^{6} f(x)\,dx$
$\displaystyle \Rightarrow \int_{0}^{6} |x+3|\,dx=\int_{0}^{-3} -(x+3)\,dx+\int_{-3}^{6} (x+3)\,dx$
$\displaystyle =\left[-\frac{x^{2}}{2}-3x\right]_{0}^{-3}+\left[\frac{x^{2}}{2}+3x\right]_{-3}^{6}$
$\displaystyle =\left[-\frac{9}{2}+9\right]+\left[18+18-\left(\frac{9}{2}-9\right)\right]$
$\displaystyle =\frac{9}{2}+45-\frac{9}{2}=45$
$\\$

$\displaystyle \textbf{Question 3: } \text{Evaluate } \int_{0}^{2\pi} \frac{1}{1+e^{\sin x}}\,dx. \quad \text{ISC 2024}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{2\pi} \frac{1}{1+e^{\sin x}}\,dx \quad (i)$
$\displaystyle \text{We know that } \int_{0}^{a} f(x)\,dx=\int_{0}^{a} f(a-x)\,dx$
$\displaystyle \therefore I=\int_{0}^{2\pi} \frac{1}{1+e^{\sin(2\pi-x)}}\,dx$
$\displaystyle =\int_{0}^{2\pi} \frac{1}{1+e^{-\sin x}}\,dx$
$\displaystyle =\int_{0}^{2\pi} \frac{e^{\sin x}}{1+e^{\sin x}}\,dx \quad (ii)$
$\displaystyle \text{On adding (i) and (ii), we get } 2I=\int_{0}^{2\pi} 1\,dx$
$\displaystyle =2\pi$
$\displaystyle \therefore I=\pi$
$\\$

$\displaystyle \textbf{Question 4: } \text{Evaluate } \int_{0}^{\frac{\pi}{2}} \frac{a\sin x+b\cos x}{\sin x+\cos x}\,dx. \quad \text{ISC 2023}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{2}} \frac{a\sin x+b\cos x}{\sin x+\cos x}\,dx \quad (i)$
$\displaystyle \text{Using } \int_{0}^{a} f(x)\,dx=\int_{0}^{a} f(a-x)\,dx$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}} \frac{a\sin\left(\frac{\pi}{2}-x\right)+b\cos\left(\frac{\pi}{2}-x\right)}{\sin\left(\frac{\pi}{2}-x\right)+\cos\left(\frac{\pi}{2}-x\right)}\,dx$
$\displaystyle =\int_{0}^{\frac{\pi}{2}} \frac{a\cos x+b\sin x}{\cos x+\sin x}\,dx \quad (ii)$
$\displaystyle \text{On adding Eqs. (i) and (ii), we get}$
$\displaystyle 2I=\int_{0}^{\frac{\pi}{2}} \frac{(a+b)(\sin x+\cos x)}{\sin x+\cos x}\,dx$
$\displaystyle =(a+b)\int_{0}^{\frac{\pi}{2}} dx$
$\displaystyle =(a+b)\cdot \frac{\pi}{2}$
$\displaystyle \therefore I=\frac{\pi}{4}(a+b)$
$\\$

$\displaystyle \textbf{Question 5: } \text{Evaluate } \int_{-1}^{1}x^{17}\cos^{4}x\,dx. \quad \text{ISC 2023}$
$\displaystyle \text{(a) } \infty \quad \text{(b) } 1 \quad \text{(c) } -1 \quad \text{(d) } 0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(d) Let } I=\int_{-1}^{1} x^{17}\cos^{4}x\,dx$
$\displaystyle f(x)=x^{17}\cos^{4}x$
$\displaystyle \therefore f(-x)=(-x)^{17}\cos^{4}(-x)=-x^{17}\cos^{4}x=-f(x)$
$\displaystyle \text{Hence, } f(x)\text{ is an odd function}$
$\displaystyle \therefore I=0$
$\\$

$\displaystyle \textbf{Question 6: } \text{Evaluate } \int_{0}^{1}|2x+1|\,dx. \quad \text{ISC 2023}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{1} |2x+1|\,dx$
$\displaystyle =\int_{0}^{1} (2x+1)\,dx \quad [\because 2x+1>0,\ x\in(0,1)]$
$\displaystyle =\left[x^{2}+x\right]_{0}^{1}$
$\displaystyle =(1+1)-0=2$
$\\$

$\displaystyle \textbf{Question 7: } \text{Evaluate } \int_{4}^{5}|x-5|\,dx. \quad \text{ISC 2020}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{4}^{5} |x-5|\,dx$
$\displaystyle =-\int_{4}^{5} (x-5)\,dx \quad [\because x-5<0,\ x\in(4,5)]$
$\displaystyle =-\left[\frac{x^{2}}{2}-5x\right]_{4}^{5}$
$\displaystyle =-\left[\left(\frac{25}{2}-25\right)-\left(\frac{16}{2}-20\right)\right]$
$\displaystyle =-\left[-\frac{25}{2}-(-12)\right]$
$\displaystyle =-\left[-\frac{25}{2}+12\right]=\frac{1}{2}$
$\\$

$\displaystyle \textbf{Question 8: } \text{Evaluate } \int_{-6}^{1}|x+3|\,dx. \quad \text{ISC 2019}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{-6}^{1}|x+3|\,dx$
$\displaystyle =\int_{-6}^{-3}-(x+3)\,dx+\int_{-3}^{1}(x+3)\,dx$
$\displaystyle =-\left[\frac{x^{2}}{2}+3x\right]_{-6}^{-3}+\left[\frac{x^{2}}{2}+3x\right]_{-3}^{1}$
$\displaystyle =-\left[\left(\frac{9}{2}-9\right)-\left(\frac{36}{2}-18\right)\right]+\left[\left(\frac{1}{2}+3\right)-\left(\frac{9}{2}-9\right)\right]$
$\displaystyle =-\left[-\frac{9}{2}-0\right]+\left[\frac{7}{2}+\frac{9}{2}\right]$
$\displaystyle =\frac{9}{2}+8=\frac{25}{2}$
$\\$

$\displaystyle \textbf{Question 9: } \text{Evaluate } \int_{0}^{\pi} \frac{x\tan x}{\sec x+\tan x}\,dx. \quad \text{ISC 2019}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\pi}\frac{x\tan x}{\sec x+\tan x}\,dx \quad (i)$
$\displaystyle \text{Using } \int_{0}^{a} f(x)\,dx=\int_{0}^{a} f(a-x)\,dx$
$\displaystyle I=\int_{0}^{\pi}\frac{(\pi-x)\tan(\pi-x)}{\sec(\pi-x)+\tan(\pi-x)}\,dx$
$\displaystyle =\int_{0}^{\pi}\frac{(\pi-x)\tan x}{\sec x+\tan x}\,dx \quad (ii)$
$\displaystyle \text{On adding (i) and (ii), we get}$
$\displaystyle 2I=\pi\int_{0}^{\pi}\frac{\tan x}{\sec x+\tan x}\,dx$
$\displaystyle =\pi\int_{0}^{\pi}\frac{\tan x(\sec x-\tan x)}{\sec^{2}x-\tan^{2}x}\,dx$
$\displaystyle =\pi\int_{0}^{\pi}(\tan x\sec x-\tan^{2}x)\,dx$
$\displaystyle =\pi\left[\int_{0}^{\pi}\sec x\tan x\,dx-\int_{0}^{\pi}(\sec^{2}x-1)\,dx\right]$
$\displaystyle \text{Since } \tan x \text{ is not defined at } x=\frac{\pi}{2}, \text{ split the integral}$
$\displaystyle 2I=\pi\left[\int_{0}^{\frac{\pi}{2}}\sec x\tan x\,dx+\int_{\frac{\pi}{2}}^{\pi}\sec x\tan x\,dx-\int_{0}^{\pi}(\sec^{2}x-1)\,dx\right]$
$\displaystyle =\pi\left[[\sec x]_{0}^{\frac{\pi}{2}}+[\sec x]_{\frac{\pi}{2}}^{\pi}-[\tan x-x]_{0}^{\pi}\right]$
$\displaystyle =\pi\left[(\infty-1)+(-1-\infty)-(0-\pi)\right]$
$\displaystyle =\pi(\pi-2)$
$\displaystyle \therefore I=\frac{\pi}{2}(\pi-2)$
$\\$

$\displaystyle \textbf{Question 10: } \text{Evaluate } \int_{0}^{\frac{\pi}{2}} \frac{\cos^{2}x}{1+\sin x\cos x}\,dx. \quad \text{ISC 2018}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{2}}\frac{\cos^{2}x}{1+\sin x\cos x}\,dx \quad (i)$
$\displaystyle \text{Using } \int_{0}^{a} f(x)\,dx=\int_{0}^{a} f(a-x)\,dx$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\sin^{2}x}{1+\cos x\sin x}\,dx \quad (ii)$
$\displaystyle \text{On adding (i) and (ii), we get}$
$\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\frac{\sin^{2}x+\cos^{2}x}{1+\sin x\cos x}\,dx$
$\displaystyle =\int_{0}^{\frac{\pi}{2}}\frac{dx}{1+\sin x\cos x}$
$\displaystyle =\int_{0}^{\frac{\pi}{2}}\frac{\sec^{2}x\,dx}{\sec^{2}x+\tan x}$
$\displaystyle \text{On putting } \tan x=t \Rightarrow \sec^{2}x\,dx=dt$
$\displaystyle \therefore 2I=\int_{0}^{\infty}\frac{dt}{t^{2}+t+1}$
$\displaystyle =\int_{0}^{\infty}\frac{dt}{\left(t+\frac{1}{2}\right)^{2}+\frac{3}{4}}$
$\displaystyle =\frac{2}{\sqrt{3}}\left[\tan^{-1}\left(\frac{2t+1}{\sqrt{3}}\right)\right]_{0}^{\infty}$
$\displaystyle =\frac{2}{\sqrt{3}}\left(\frac{\pi}{2}-\tan^{-1}\frac{1}{\sqrt{3}}\right)$
$\displaystyle =\frac{2}{\sqrt{3}}\left(\frac{\pi}{2}-\frac{\pi}{6}\right)$
$\displaystyle =\frac{2\pi}{3\sqrt{3}}$
$\displaystyle \therefore I=\frac{\pi}{3\sqrt{3}}$
$\\$

$\displaystyle \textbf{Question 11: } \text{Evaluate } \int_{0}^{\frac{\pi}{4}} \log(1+\tan\theta)\,d\theta. \quad \text{ISC 2017, 07}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{4}} \log(1+\tan\theta)\,d\theta \quad (i)$
$\displaystyle \text{Using } \int_{0}^{a} f(x)\,dx=\int_{0}^{a} f(a-x)\,dx$
$\displaystyle I=\int_{0}^{\frac{\pi}{4}} \log\left[1+\tan\left(\frac{\pi}{4}-\theta\right)\right]\,d\theta$
$\displaystyle =\int_{0}^{\frac{\pi}{4}} \log\left(\frac{2}{1+\tan\theta}\right)\,d\theta$
$\displaystyle =\int_{0}^{\frac{\pi}{4}} [\log 2-\log(1+\tan\theta)]\,d\theta \quad (ii)$
$\displaystyle \text{On adding (i) and (ii), we get}$
$\displaystyle 2I=\int_{0}^{\frac{\pi}{4}} \log 2\,d\theta$
$\displaystyle =(\log 2)\cdot \frac{\pi}{4}$
$\displaystyle \therefore I=\frac{\pi}{8}\log 2$
$\\$

$\displaystyle \textbf{Question 12: } \text{Using properties of definite integrals, evaluate } \int_{0}^{\frac{\pi}{2}}\frac{\sin x-\cos x}{1+\sin x\cos x}\,dx. \quad \text{ISC 2016}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x\cos x}\,dx$
$\displaystyle \text{Using } \int_{0}^{a} f(x)\,dx=\int_{0}^{a} f(a-x)\,dx$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}} \frac{\sin\left(\frac{\pi}{2}-x\right)-\cos\left(\frac{\pi}{2}-x\right)}{1+\sin\left(\frac{\pi}{2}-x\right)\cos\left(\frac{\pi}{2}-x\right)}\,dx$
$\displaystyle =\int_{0}^{\frac{\pi}{2}} \frac{\cos x-\sin x}{1+\cos x\sin x}\,dx$
$\displaystyle =-\int_{0}^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x\cos x}\,dx$
$\displaystyle =-I$
$\displaystyle \therefore 2I=0 \Rightarrow I=0$
$\\$

$\displaystyle \textbf{Question 13: } \text{Evaluate } \int_{0}^{3}f(x)\,dx, \text{ where } f(x)= \begin{cases} \cos 2x, & 0 \leq x \leq \frac{\pi}{2}\\ 3, & \frac{\pi}{2}<x\leq 3 \end{cases} \quad \text{ISC 2015}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, } f(x)=\begin{cases} \cos 2x, & 0\leq x\leq \frac{\pi}{2} \\ 3, & \frac{\pi}{2}<x\leq 3 \end{cases}$
$\displaystyle \text{Let } I=\int_{0}^{3} f(x)\,dx$
$\displaystyle =\int_{0}^{\frac{\pi}{2}} \cos 2x\,dx+\int_{\frac{\pi}{2}}^{3} 3\,dx$
$\displaystyle =\left[\frac{\sin 2x}{2}\right]_{0}^{\frac{\pi}{2}}+3\left[x\right]_{\frac{\pi}{2}}^{3}$
$\displaystyle =\frac{1}{2}[\sin \pi-\sin 0]+3\left(3-\frac{\pi}{2}\right)$
$\displaystyle =0+\frac{3}{2}(6-\pi)$
$\displaystyle \therefore I=\frac{3}{2}(6-\pi)$
$\\$

$\displaystyle \textbf{Question 14: } \text{Using properties of definite integrals, evaluate } \int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx. \quad \text{ISC 2014}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx \quad (i)$
$\displaystyle \text{Using } \int_{0}^{a} f(x)\,dx=\int_{0}^{a} f(a-x)\,dx$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin\left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin\left(\frac{\pi}{2}-x\right)}+\sqrt{\cos\left(\frac{\pi}{2}-x\right)}}\,dx$
$\displaystyle =\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx \quad (ii)$
$\displaystyle \text{On adding Eqs. (i) and (ii), we get}$
$\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx$
$\displaystyle =\int_{0}^{\frac{\pi}{2}}1\,dx=\left[x\right]_{0}^{\frac{\pi}{2}}$
$\displaystyle =\frac{\pi}{2}-0=\frac{\pi}{2}$
$\displaystyle \therefore I=\frac{\pi}{4}$
$\\$

$\displaystyle \textbf{Question 15: } \text{Evaluate } \int_{0}^{1} \log\left(\frac{1}{x}-1\right)\,dx. \quad \text{ISC 2013}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{1}\log\left(\frac{1}{x}-1\right)\,dx$
$\displaystyle =\int_{0}^{1}\log\left(\frac{1-x}{x}\right)\,dx \quad (i)$
$\displaystyle \text{Using } \int_{0}^{a} f(x)\,dx=\int_{0}^{a} f(a-x)\,dx$
$\displaystyle I=\int_{0}^{1}\log\left(\frac{1-(1-x)}{1-x}\right)\,dx$
$\displaystyle =\int_{0}^{1}\log\left(\frac{x}{1-x}\right)\,dx \quad (ii)$
$\displaystyle \text{On adding Eqs. (i) and (ii), we get}$
$\displaystyle 2I=\int_{0}^{1}\log\left(\frac{1-x}{x}\right)dx+\int_{0}^{1}\log\left(\frac{x}{1-x}\right)dx$
$\displaystyle =\int_{0}^{1}\log\left[\left(\frac{1-x}{x}\right)\left(\frac{x}{1-x}\right)\right]dx$
$\displaystyle =\int_{0}^{1}\log 1\,dx=0$
$\displaystyle \therefore I=0$
$\\$

$\displaystyle \textbf{Question 16: } \text{Evaluate } \int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}}\,dx. \quad \text{ISC 2012}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{1}^{2}\frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}}\,dx \quad (i)$
$\displaystyle \text{Using } \int_{a}^{b} f(x)\,dx=\int_{a}^{b} f(a+b-x)\,dx$
$\displaystyle I=\int_{1}^{2}\frac{\sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}}\,dx \quad (ii)$
$\displaystyle \text{On adding Eqs. (i) and (ii), we get}$
$\displaystyle 2I=\int_{1}^{2}\frac{\sqrt{x}+\sqrt{3-x}}{\sqrt{3-x}+\sqrt{x}}\,dx$
$\displaystyle =\int_{1}^{2}1\,dx=[x]_{1}^{2}=1$
$\displaystyle \therefore I=\frac{1}{2}$
$\displaystyle I=\int_{1}^{2}\frac{\sqrt{3-x}}{\sqrt{3-x}+\sqrt{x}}\,dx \quad (ii)$
$\displaystyle \text{On adding Eqs. (i) and (ii), we get}$
$\displaystyle 2I=\int_{1}^{2}\frac{\sqrt{x}+\sqrt{3-x}}{\sqrt{3-x}+\sqrt{x}}\,dx$
$\displaystyle =\int_{1}^{2}1\,dx=[x]_{1}^{2}=1$
$\displaystyle \therefore I=\frac{1}{2}$
$\\$

$\displaystyle \textbf{Question 17: } \text{Evaluate } \int_{0}^{1} \frac{xe^{x}}{(1+x)^{2}}\,dx. \quad \text{ISC 2011}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{1}\frac{xe^{x}}{(1+x)^{2}}\,dx$
$\displaystyle =\int_{0}^{1}\frac{[(1+x)-1]e^{x}}{(1+x)^{2}}\,dx$
$\displaystyle =\int_{0}^{1}e^{x}\left(\frac{1}{1+x}-\frac{1}{(1+x)^{2}}\right)dx$
$\displaystyle =\int_{0}^{1}e^{x}\left[f(x)+f'(x)\right]dx \quad \left[\text{where } f(x)=\frac{1}{1+x}\right]$
$\displaystyle =\left[e^{x}f(x)\right]_{0}^{1}$
$\displaystyle =\left[\frac{e^{x}}{1+x}\right]_{0}^{1}$
$\displaystyle =\frac{e}{2}-1$
$\\$

$\displaystyle \textbf{Question 18: } \text{Evaluate } \int_{0}^{9}f(x)\,dx, \text{ where } f(x) \text{ is defined by }$
$\displaystyle f(x)= \begin{cases} \sin x, & \text{if } 0 \leq x \leq \frac{\pi}{2}\\ 1, & \text{if } \frac{\pi}{2}<x\leq 5\\ e^{x-5}, & \text{if } 5<x\leq 9 \end{cases} \quad \text{ISC 2008}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, integrand is }$
$\displaystyle f(x)=\begin{cases} \sin x, & 0\leq x\leq \frac{\pi}{2} \\ 1, & \frac{\pi}{2}<x\leq 5 \\ e^{x-5}, & 5<x\leq 9 \end{cases}$
$\displaystyle \text{Now, } \int_{0}^{9}f(x)\,dx=\int_{0}^{\frac{\pi}{2}}\sin x\,dx+\int_{\frac{\pi}{2}}^{5}1\,dx+\int_{5}^{9}e^{x-5}\,dx$
$\displaystyle =[-\cos x]_{0}^{\frac{\pi}{2}}+[x]_{\frac{\pi}{2}}^{5}+[e^{x-5}]_{5}^{9}$
$\displaystyle =[-\cos\frac{\pi}{2}+\cos 0]+\left(5-\frac{\pi}{2}\right)+(e^{4}-1)$
$\displaystyle =(0+1)+5-\frac{\pi}{2}+e^{4}-1$
$\displaystyle =5-\frac{\pi}{2}+e^{4}$
$\\$