\displaystyle \textbf{Question 1: } \text{Evaluate:}
\displaystyle \text{(i)}\ \int_{0}^{\pi/4}\sqrt{1+\sin 2x}\,dx \qquad \text{(ii)}\ \int_{0}^{\pi/4}\sqrt{1-\sin 2x}\,dx \hspace{2.0cm} \text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{}\ \text{(i) Let }I=\int_{0}^{\pi/4}\sqrt{1+\sin 2x}\,dx.
\displaystyle I=\int_{0}^{\pi/4}\sqrt{\sin^{2}x+\cos^{2}x+2\sin x\cos x}\,dx
\displaystyle I=\int_{0}^{\pi/4}\sqrt{(\sin x+\cos x)^{2}}\,dx
\displaystyle I=\int_{0}^{\pi/4}|\cos x+\sin x|\,dx
\displaystyle =\int_{0}^{\pi/4}(\cos x+\sin x)\,dx
\displaystyle =[\sin x-\cos x]_{0}^{\pi/4}
\displaystyle =(\sin\frac{\pi}{4}-\cos\frac{\pi}{4})-(\sin 0-\cos 0)
\displaystyle =\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)-(0-1)=1

\displaystyle \text{(ii) Let }I=\int_{0}^{\pi/4}\sqrt{1-\sin 2x}\,dx.
\displaystyle I=\int_{0}^{\pi/4}\sqrt{\sin^{2}x+\cos^{2}x-2\sin x\cos x}\,dx
\displaystyle I=\int_{0}^{\pi/4}\sqrt{(\cos x-\sin x)^{2}}\,dx
\displaystyle I=\int_{0}^{\pi/4}|\cos x-\sin x|\,dx
\displaystyle \because 0<x<\frac{\pi}{4}\Rightarrow \cos x>\sin x\Rightarrow \cos x-\sin x>0
\displaystyle \Rightarrow |\cos x-\sin x|=\cos x-\sin x
\displaystyle I=\int_{0}^{\pi/4}(\cos x-\sin x)\,dx
\displaystyle =[\sin x+\cos x]_{0}^{\pi/4}
\displaystyle =\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)-(0+1)=\sqrt{2}-1

\displaystyle \textbf{Question 2: } \text{Evaluate:}
\displaystyle \text{(i)}\ \int_{1}^{2}\frac{5x^{2}}{x^{2}+4x+3}\,dx \qquad \text{(ii)}\ \int_{1}^{3}\frac{1}{x^{2}(x+1)}\,dx \hspace{2.0cm} \text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let }I=\int_{1}^{2}\frac{5x^{2}}{x^{2}+4x+3}\,dx.
\displaystyle I=5\int_{1}^{2}\frac{x^{2}}{x^{2}+4x+3}\,dx =5\int_{1}^{2}\left(1-\frac{4x+3}{x^{2}+4x+3}\right)\,dx
\displaystyle =5\int_{1}^{2}1\,dx-5\int_{1}^{2}\frac{4x+3}{x^{2}+4x+3}\,dx
\displaystyle =5\int_{1}^{2}1\,dx-5\int_{1}^{2}\frac{2(2x+4)-5}{x^{2}+4x+3}\,dx
\displaystyle =5\int_{1}^{2}1\,dx-10\int_{1}^{2}\frac{2x+4}{x^{2}+4x+3}\,dx +25\int_{1}^{2}\frac{1}{x^{2}+4x+3}\,dx
\displaystyle =5[x]_{1}^{2}-10[\log(x^{2}+4x+3)]_{1}^{2} +\frac{25}{2}\left[\log\frac{x+2-1}{x+2+1}\right]_{1}^{2}
\displaystyle =5(2-1)-10(\log 15-\log 8) +\frac{25}{2}\left(\log\frac{3}{5}-\log\frac{2}{4}\right)
\displaystyle =5-10\log\frac{15}{8}+\frac{25}{2}\log\frac{6}{5}

\displaystyle \text{(ii) Let }\frac{1}{x^{2}(x+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^{2}}
\displaystyle \text{Then }1=A x^{2}+(Bx+C)(x+1)
\displaystyle \text{Putting }x=0,x=-1\text{ respectively, we get: }C=1\text{ and }A=1
\displaystyle \text{Equating coefficients of }x^{2}\text{ on both sides, we get: }0=A+B \Rightarrow B=-1
\displaystyle \text{Substituting values of }A,B,C\text{ we obtain}
\displaystyle \frac{1}{x^{2}(x+1)}=\frac{1}{x+1}-\frac{1}{x}+\frac{1}{x^{2}}
\displaystyle \int_{1}^{3}\frac{1}{x^{2}(x+1)}\,dx =\int_{1}^{3}\left(\frac{1}{x+1}-\frac{1}{x}+\frac{1}{x^{2}}\right)dx
\displaystyle =[\log|x+1|-\log|x|-\frac{1}{x}]_{1}^{3}
\displaystyle =(\log 4-\log 3-\frac{1}{3})-(\log 2-\log 1-1)
\displaystyle =\log\frac{4}{3}-\frac{1}{3}+1=\log\frac{2}{3}+\frac{2}{3}

\displaystyle \textbf{Question 3: } \text{Evaluate:}\qquad \int_{\pi/4}^{\pi/2}\cos 2x\log\sin x\,dx \qquad [\text{CBSE 2003}]
\displaystyle \text{Answer:}
\displaystyle  \text{Let }I=\int_{\pi/4}^{\pi/2}\cos 2x\log\sin x\,dx.
\displaystyle I=\left[\frac{1}{2}(\log\sin x)\sin 2x\right]_{\pi/4}^{\pi/2} -\int_{\pi/4}^{\pi/2}\frac{1}{2}\cot x\sin 2x\,dx
\displaystyle \Rightarrow I=\left[0-\frac{1}{2}\log\frac{1}{\sqrt{2}}\right] -\int_{\pi/4}^{\pi/2}\cos^{2}x\,dx
\displaystyle \Rightarrow I=\frac{1}{4}\log 2-\frac{1}{2} \int_{\pi/4}^{\pi/2}(1+\cos 2x)\,dx
\displaystyle \Rightarrow I=\frac{1}{4}\log 2-\frac{1}{2} \left[x+\frac{1}{2}\sin 2x\right]_{\pi/4}^{\pi/2}
\displaystyle =\frac{1}{4}\log 2-\frac{1}{2} \left(\frac{\pi}{2}+0\right)-\left(\frac{\pi}{4}+\frac{1}{2}\right)
\displaystyle =\frac{1}{4}\log 2-\frac{\pi}{8}-\frac{1}{4}

\displaystyle \textbf{Question 4: } \text{Evaluate:}
\displaystyle \text{(i)}\ \int_{0}^{\pi/2}\frac{\cos\theta}{(1+\sin\theta)(2+\sin\theta)}\,d\theta \hspace{4.0cm}[\text{CBSE 2004}]
\displaystyle \text{(ii)}\ \int_{0}^{1/2}\frac{1}{(1+x^{2})\sqrt{1-x^{2}}}\,dx \hspace{5.0cm}[\text{CBSE 2015}]
\displaystyle \text{(iii)}\ \int_{0}^{\pi/4}\frac{1}{\cos^{3}x\sqrt{2\sin 2x}}\,dx \hspace{5.0cm}[\text{CBSE 2015}]
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let }\sin\theta=t.\text{ Then }d(\sin\theta)=dt \Rightarrow \cos\theta\,d\theta=dt
\displaystyle \text{Also, }\theta=0\Rightarrow t=0 \text{ and }\theta=\frac{\pi}{2}\Rightarrow t=1
\displaystyle I=\int_{0}^{\pi/2}\frac{\cos\theta}{(1+\sin\theta)(2+\sin\theta)}\,d\theta =\int_{0}^{1}\frac{1}{(1+t)(2+t)}\,dt
\displaystyle =\int_{0}^{1}\left(\frac{1}{1+t}-\frac{1}{2+t}\right)dt \qquad [\text{By using partial fractions}]
\displaystyle I=[\log(1+t)-\log(2+t)]_{0}^{1}
\displaystyle =( \log 2-\log 1)-(\log 3-\log 2) =\log 2-\log 3+\log 2=2\log 2-\log 3=\log\frac{4}{3}

\displaystyle \text{(ii) Let }I=\int_{0}^{1/2}\frac{1}{(1+x^{2})\sqrt{1-x^{2}}}\,dx. \text{ Let }x=\sin\theta
\displaystyle \text{Then }dx=\cos\theta\,d\theta \text{ and }x=0\Rightarrow\theta=0,\ x=\frac{1}{2}\Rightarrow\theta=\frac{\pi}{6}
\displaystyle I=\int_{0}^{\pi/6}\frac{\cos\theta}{(1+\sin^{2}\theta)\sqrt{1-\sin^{2}\theta}}\,d\theta =\int_{0}^{\pi/6}\frac{1}{1+\sin^{2}\theta}\,d\theta
\displaystyle I=\int_{0}^{\pi/6}\frac{\sec^{2}\theta}{\sec^{2}\theta+\tan^{2}\theta}\,d\theta \qquad [\text{Dividing N and D by }\cos^{2}\theta]
\displaystyle =\int_{0}^{\pi/6}\frac{\sec^{2}\theta}{1+2\tan^{2}\theta}\,d\theta
\displaystyle \text{Let }\tan\theta=t.\text{ Then }\sec^{2}\theta\,d\theta=dt
\displaystyle \theta=0\Rightarrow t=0,\ \theta=\frac{\pi}{6}\Rightarrow t=\frac{1}{\sqrt{3}}
\displaystyle I=\int_{0}^{1/\sqrt{3}}\frac{1}{1+2t^{2}}\,dt =\frac{1}{2}\int_{0}^{1/\sqrt{3}}\frac{1}{(1/\sqrt{2})^{2}+t^{2}}\,dt
\displaystyle =\frac{1}{\sqrt{2}}\tan^{-1}(\sqrt{2}t)\Big|_{0}^{1/\sqrt{3}} =\frac{1}{\sqrt{2}}\tan^{-1}\sqrt{\frac{2}{3}}

\displaystyle \text{(iii) Let }I=\int_{0}^{\pi/4}\frac{1}{\cos^{3}x\sqrt{2\sin 2x}}\,dx
\displaystyle I=\int_{0}^{\pi/4}\frac{1}{\cos^{3}x\sqrt{4\sin x\cos x}}\,dx =\frac{1}{2}\int_{0}^{\pi/4}\sin^{-1/2}x\cos^{-7/2}x\,dx
\displaystyle \text{Divide numerator and denominator by }\cos^{4}x
\displaystyle I=\frac{1}{2}\int_{0}^{\pi/4}\frac{\sec^{4}x}{\sqrt{\tan x}}\,dx =\frac{1}{2}\int_{0}^{\pi/4}\frac{1+\tan^{2}x}{\sqrt{\tan x}}\sec^{2}x\,dx
\displaystyle \text{Let }\tan x=t.\text{ Then }dt=\sec^{2}x\,dx
\displaystyle x=0\Rightarrow t=0,\ x=\frac{\pi}{4}\Rightarrow t=1
\displaystyle I=\frac{1}{2}\int_{0}^{1}\frac{1+t^{2}}{\sqrt{t}}\,dt =\frac{1}{2}\int_{0}^{1}(t^{-1/2}+t^{3/2})\,dt
\displaystyle =\frac{1}{2}\left[2t^{1/2}+\frac{2}{5}t^{5/2}\right]_{0}^{1} =\frac{1}{2}\times\frac{12}{5}=\frac{6}{5}

\displaystyle \textbf{Question 5: } \text{Evaluate:}
\displaystyle \text{(i)}\ \int_{0}^{1/\sqrt{2}}\frac{\sin^{-1}x}{(1-x^{2})^{3/2}}\,dx \qquad [\text{CBSE 2007}]
\displaystyle \text{(ii)}\ \int_{0}^{1}\sin^{-1}\left(\frac{2x}{1+x^{2}}\right)dx \qquad [\text{CBSE 2002}]
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let }\sin^{-1}x=\theta,\ x=\sin\theta. \text{ Then }dx=\cos\theta\,d\theta
\displaystyle x=0\Rightarrow\theta=0,\ x=\frac{1}{\sqrt{2}}\Rightarrow\theta=\frac{\pi}{4}
\displaystyle I=\int_{0}^{\pi/4}\frac{\theta}{\cos^{3}\theta}\cos\theta\,d\theta =\int_{0}^{\pi/4}\theta\sec^{2}\theta\,d\theta
\displaystyle I=[\theta\tan\theta]_{0}^{\pi/4}-\int_{0}^{\pi/4}\tan\theta\,d\theta
\displaystyle =\frac{\pi}{4}-\frac{1}{2}\log 2

\displaystyle \text{(ii) Let }I=\int_{0}^{1}\sin^{-1}\left(\frac{2x}{1+x^{2}}\right)dx
\displaystyle \because \sin^{-1}\frac{2x}{1+x^{2}}=2\tan^{-1}x
\displaystyle I=2\int_{0}^{1}\tan^{-1}x\,dx
\displaystyle =2\left[x\tan^{-1}x-\frac{1}{2}\log(1+x^{2})\right]_{0}^{1}
\displaystyle =2\left(\frac{\pi}{4}-\frac{1}{2}\log 2\right) =\frac{\pi}{2}-\log 2

\displaystyle \textbf{Question 6: } \text{Evaluate:}\ \int_{0}^{\pi/4}\tan^{3}x\,dx \qquad [\text{CBSE 2004}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{0}^{\pi/4}\tan^{3}x\,dx
\displaystyle I=\int_{0}^{\pi/4}(\sec^{2}x-1)\tan x\,dx
\displaystyle \text{Let }\tan x=t,\ dt=\sec^{2}x\,dx
\displaystyle x=0\Rightarrow t=0,\ x=\frac{\pi}{4}\Rightarrow t=1
\displaystyle I=\int_{0}^{1}t^{2}\,dt-\int_{0}^{1}\tan x\,dx
\displaystyle =\frac{1}{2}-\log\sqrt{2}+1 =\frac{1}{2}-\log 2

\displaystyle \textbf{Question 7: } \text{Evaluate:}
\displaystyle \text{}\ \int_{0}^{\pi}\frac{1}{5+4\cos x}\,dx \qquad [\text{CBSE 2005}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{0}^{\pi}\frac{1}{5+4\cos x}\,dx
\displaystyle =\int_{0}^{\pi}\frac{1+\tan^{2}\frac{x}{2}} {5(1+\tan^{2}\frac{x}{2})+4(1-\tan^{2}\frac{x}{2})}\,dx
\displaystyle =\int_{0}^{\pi}\frac{\sec^{2}\frac{x}{2}} {9+\tan^{2}\frac{x}{2}}\,dx
\displaystyle \text{Let }\tan\frac{x}{2}=t,\ dx=\frac{2dt}{\sec^{2}\frac{x}{2}}
\displaystyle I=2\int_{0}^{\infty}\frac{dt}{9+t^{2}} =\frac{2}{3}\tan^{-1}\infty-\tan^{-1}0=\frac{\pi}{3}

\displaystyle \textbf{Question 8: } \text{Evaluate:}
\displaystyle \text{(i)}\ \int_{0}^{\pi/2}\frac{\cos x}{1+\cos x+\sin x}\,dx \qquad [\text{CBSE 2014}]
\displaystyle \text{(ii)}\ \int_{0}^{\pi/4}\frac{\sin x+\cos x}{9+16\sin 2x}\,dx \qquad [\text{CBSE 2010}]
\displaystyle \text{(iii)}\ \int_{0}^{\pi/2}\sqrt{\tan x+\cot x}\,dx \qquad [\text{CBSE 2002, 2003}]
\displaystyle \text{Answer:}

\displaystyle \text{(i) We have }I=\int_{0}^{\pi/2}\frac{\cos x}{1+\cos x+\sin x}\,dx
\displaystyle =\int_{0}^{\pi/2}\frac{\cos x}{(1+\cos x)+\sin x}\,dx
\displaystyle =\int_{0}^{\pi/2}\frac{\cos^{2}\frac{x}{2}-\sin^{2}\frac{x}{2}} {2\cos^{2}\frac{x}{2}+2\sin^{2}\frac{x}{2}}\,dx
\displaystyle =\int_{0}^{\pi/2}\frac{1-\tan^{2}\frac{x}{2}}{2+2\tan^{2}\frac{x}{2}}\,dx
\displaystyle \text{[Dividing numerator and denominator by }\cos^{2}\frac{x}{2}]
\displaystyle =\frac{1}{2}\int_{0}^{\pi/2}\frac{(1-\tan\frac{x}{2})(1+\tan\frac{x}{2})} {1+\tan^{2}\frac{x}{2}}\,dx
\displaystyle =\frac{1}{2}\int_{0}^{\pi/2}(1-\tan\frac{x}{2})\,dx
\displaystyle =\frac{1}{2}\left[x+2\log\cos\frac{x}{2}\right]_{0}^{\pi/2}
\displaystyle =\frac{1}{2}\left(\frac{\pi}{2}+2\log\frac{1}{\sqrt{2}}\right) =\frac{1}{2}\left(\frac{\pi}{2}-\log 2\right)

\displaystyle \text{(ii) Let }I=\int_{0}^{\pi/4}\frac{\sin x+\cos x}{9+16\sin 2x}\,dx
\displaystyle \sin 2x=1-(\sin x-\cos x)^{2}
\displaystyle I=\int_{0}^{\pi/4}\frac{\sin x+\cos x} {25-16(\sin x-\cos x)^{2}}\,dx
\displaystyle \text{Let }\sin x-\cos x=t,\ dt=(\cos x+\sin x)dx
\displaystyle x=0\Rightarrow t=-1,\ x=\frac{\pi}{4}\Rightarrow t=0
\displaystyle I=\int_{-1}^{0}\frac{dt}{25-16t^{2}} =\frac{1}{40}\left[\log\frac{5+4t}{5-4t}\right]_{-1}^{0}
\displaystyle =\frac{1}{40}(\log 1-\log\frac{9}{1}) =-\frac{1}{40}\log 9

\displaystyle \text{(iii) Let }I=\int_{0}^{\pi/2}\sqrt{\tan x+\cot x}\,dx
\displaystyle =\sqrt{2}\int_{0}^{\pi/2}\frac{\sin x+\cos x} {\sqrt{\sin x\cos x}}\,dx
\displaystyle \text{Let }\sin x-\cos x=t,\ dt=(\cos x+\sin x)dx
\displaystyle x=0\Rightarrow t=-1,\ x=\frac{\pi}{2}\Rightarrow t=1
\displaystyle I=\sqrt{2}\int_{-1}^{1}\frac{dt}{\sqrt{1-t^{2}}} =\sqrt{2}\left[\sin^{-1}t\right]_{-1}^{1}=\sqrt{2}\pi

\displaystyle \textbf{Question 9: } \text{Evaluate:}
\displaystyle \text{(i)}\ \int_{0}^{\pi/2}\frac{\sin 2x}{\sin^{4}x+\cos^{4}x}\,dx \qquad [\text{CBSE 2003C}]
\displaystyle \text{(ii)}\ \int_{0}^{\pi/4}\frac{\sin 2x}{\cos^{4}x+\sin^{4}x}\,dx \qquad [\text{CBSE 2013}]
\displaystyle \text{Answer:}

\displaystyle \text{(i) Let }I=\int_{0}^{\pi/2}\frac{\sin 2x}{\sin^{4}x+\cos^{4}x}\,dx =\int_{0}^{\pi/2}\frac{2\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx
\displaystyle \text{Dividing numerator and denominator by }\cos^{4}x\text{ we obtain}
\displaystyle I=\int_{0}^{\pi/2}\frac{2\tan x\sec^{2}x}{\tan^{4}x+1}\,dx
\displaystyle \text{Let }t=\tan^{2}x.\text{ Then }dt=d(\tan^{2}x)=2\tan x\sec^{2}x\,dx
\displaystyle \text{Also, }x=0\Rightarrow t=\tan^{2}0=0,\ x=\frac{\pi}{2} \Rightarrow t=\tan^{2}\frac{\pi}{2}=\infty
\displaystyle I=\int_{0}^{\infty}\frac{dt}{t^{2}+1} =\left[\tan^{-1}t\right]_{0}^{\infty} =\tan^{-1}\infty-\tan^{-1}0=\frac{\pi}{2}

\displaystyle \text{(ii) Let }I=\int_{0}^{\pi/4}\frac{\sin 2x}{\cos^{4}x+\sin^{4}x}\,dx =\int_{0}^{\pi/4}\frac{2\sin x\cos x}{\cos^{4}x+\sin^{4}x}\,dx
\displaystyle \text{Dividing numerator and denominator by }\cos^{4}x\text{ we get}
\displaystyle I=\int_{0}^{\pi/4}\frac{2\tan x\sec^{2}x}{1+\tan^{4}x}\,dx
\displaystyle \text{Let }t=\tan^{2}x.\text{ Then }dt=2\tan x\sec^{2}x\,dx
\displaystyle \text{Also, }x=0\Rightarrow t=\tan^{2}0=0,\ x=\frac{\pi}{4} \Rightarrow t=\tan^{2}\frac{\pi}{4}=1
\displaystyle I=\int_{0}^{1}\frac{dt}{1+t^{2}} =\left[\tan^{-1}t\right]_{0}^{1} =\tan^{-1}1-\tan^{-1}0=\frac{\pi}{4}

\displaystyle \textbf{Question 10: } \text{Evaluate:}\ \int_{0}^{\pi/2} \frac{\cos^{2}x}{\cos^{2}x+4\sin^{2}x}\,dx\qquad [\text{CBSE 2012}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{0}^{\pi/2} \frac{\cos^{2}x}{\cos^{2}x+4\sin^{2}x}\,dx
\displaystyle I=\int_{0}^{\pi/2} \frac{\cos^{2}x}{\cos^{2}x+4(1-\cos^{2}x)}\,dx
\displaystyle =\int_{0}^{\pi/2}\frac{\cos^{2}x}{4-3\cos^{2}x}\,dx
\displaystyle =-\frac{1}{3}\int_{0}^{\pi/2}\frac{3\cos^{2}x}{4-3\cos^{2}x}\,dx
\displaystyle =-\frac{1}{3}\int_{0}^{\pi/2} \frac{(4-3\cos^{2}x)-4}{4-3\cos^{2}x}\,dx
\displaystyle =-\frac{1}{3}\int_{0}^{\pi/2} \left(1-\frac{4}{4-3\cos^{2}x}\right)dx
\displaystyle =-\frac{1}{3}\int_{0}^{\pi/2}1\,dx +\frac{4}{3}\int_{0}^{\pi/2}\frac{1}{4-3\cos^{2}x}\,dx
\displaystyle =-\frac{1}{3}\int_{0}^{\pi/2}1\,dx +\frac{4}{3}\int_{0}^{\pi/2} \frac{\sec^{2}x}{4(1+\tan^{2}x)-3}\,dx
\displaystyle \text{[Dividing N and D by }\cos^{2}x]
\displaystyle =-\frac{1}{3}\int_{0}^{\pi/2}1\,dx +\frac{4}{3}\int_{0}^{\pi/2}\frac{\sec^{2}x}{1+4\tan^{2}x}\,dx
\displaystyle =-\frac{1}{3}\left[x\right]_{0}^{\pi/2} +\frac{4}{3}\int_{0}^{\infty}\frac{dt}{1+4t^{2}},\ \text{where }t=\tan x
\displaystyle =-\frac{\pi}{6}+\frac{4}{3}\times\frac{1}{2} \left[\tan^{-1}(2t)\right]_{0}^{\infty}
\displaystyle =-\frac{\pi}{6}+\frac{2}{3} \left(\frac{\pi}{2}-0\right)=\frac{\pi}{6}

\displaystyle \textbf{Question 11: } \text{Evaluate:}
\displaystyle \text{}\ \int_{-1}^{2}|x^{3}-x|\,dx \qquad [\text{CBSE 2012, 2013, 2016}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{-1}^{2}|x^{3}-x|\,dx\text{ and }f(x)=x^{3}-x
\displaystyle \text{Clearly, }f(x)=x^{3}-x=x(x-1)(x+1)
\displaystyle \text{The signs of }f(x)\text{ for different values of }x
\displaystyle \text{We observe that }f(x)>0\text{ for }x\in(-1,0)\cup(1,2) \text{ and }f(x)<0\text{ for }x\in(0,1)
\displaystyle |f(x)|=\begin{cases} x^{3}-x, & x\in(-1,0)\cup(1,2)\\ -(x^{3}-x), & x\in(0,1) \end{cases}
\displaystyle \Rightarrow |x^{3}-x|= \begin{cases} x^{3}-x, & x\in(-1,0)\cup(1,2)\\ -(x^{3}-x), & x\in(0,1) \end{cases}
\displaystyle I=\int_{-1}^{0}|x^{3}-x|\,dx+\int_{0}^{1}|x^{3}-x|\,dx +\int_{1}^{2}|x^{3}-x|\,dx\qquad [\text{Using additive property}]
\displaystyle I=\int_{-1}^{0}(x^{3}-x)\,dx-\int_{0}^{1}(x^{3}-x)\,dx +\int_{1}^{2}(x^{3}-x)\,dx
\displaystyle I=\left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]_{-1}^{0} -\left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]_{0}^{1} +\left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]_{1}^{2}
\displaystyle I=-\left(\frac{1}{4}-\frac{1}{2}\right) -\left(\frac{1}{4}-\frac{1}{2}\right) +\left(\frac{16}{4}-\frac{4}{2}\right)-\left(\frac{1}{4}-\frac{1}{2}\right)
\displaystyle I=\frac{3}{4}+(4-2)=\frac{11}{4}

\displaystyle \textbf{Question 12: } \text{Evaluate:}
\displaystyle \text{}\ \int_{\pi/6}^{\pi/3}\frac{1}{1+\sqrt{\cot x}}\,dx \qquad [\text{CBSE 2014}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{\pi/6}^{\pi/3}\frac{1}{1+\sqrt{\cot x}}\,dx =\int_{\pi/6}^{\pi/3}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx
\displaystyle \text{Then,}
\displaystyle I=\int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin(\frac{\pi}{2}-x)}}{\sqrt{\sin(\frac{\pi}{2}-x)}+ \sqrt{\cos(\frac{\pi}{2}-x)}}\,dx
\displaystyle \Rightarrow I=\int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx
\displaystyle \text{Adding (i) and (ii), we get}
\displaystyle 2I=\int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx =\int_{\pi/6}^{\pi/3}1\,dx
\displaystyle =\left[x\right]_{\pi/6}^{\pi/3} =\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}
\displaystyle I=\frac{\pi}{12}

\displaystyle \textbf{Question 13 } \text{Prove that:}\ \int_{0}^{\pi/2}\frac{\sin x}{\sin x+\cos x}\,dx=\frac{\pi}{4} \qquad [\text{CBSE 2002C, 2007}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{0}^{\pi/2} \frac{\sin x}{\sin x+\cos x}\,dx
\displaystyle \text{Then,}
\displaystyle I=\int_{0}^{\pi/2} \frac{\sin(\frac{\pi}{2}-x)}{\sin(\frac{\pi}{2}-x)+\cos(\frac{\pi}{2}-x)}\,dx \qquad [\text{Using }\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx]
\displaystyle \Rightarrow I=\int_{0}^{\pi/2} \frac{\cos x}{\cos x+\sin x}\,dx
\displaystyle \text{Adding (i) and (ii), we get}
\displaystyle 2I=\int_{0}^{\pi/2}\frac{\sin x}{\sin x+\cos x}\,dx +\int_{0}^{\pi/2}\frac{\cos x}{\sin x+\cos x}\,dx
\displaystyle \Rightarrow 2I=\int_{0}^{\pi/2} \frac{\sin x+\cos x}{\sin x+\cos x}\,dx =\int_{0}^{\pi/2}1\,dx
\displaystyle =\left[x\right]_{0}^{\pi/2} =\frac{\pi}{2}-0=\frac{\pi}{2}
\displaystyle \Rightarrow I=\frac{\pi}{4}

\displaystyle \textbf{Question 14: } \text{Evaluate:}
\displaystyle \text{(i)}\ \int_{0}^{\pi/2}\log\tan x\,dx\qquad [\text{CBSE 2007}]
\displaystyle \text{(ii)}\ \int_{0}^{\pi/4}\log(1+\tan x)\,dx \qquad [\text{CBSE 2002C,03,04,11,13}]
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let }I=\int_{0}^{\pi/2}\log\tan x\,dx
\displaystyle \text{Then,}
\displaystyle I=\int_{0}^{\pi/2}\log\tan\left(\frac{\pi}{2}-x\right)dx \qquad [\text{Using }\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx]
\displaystyle \Rightarrow I=\int_{0}^{\pi/2}\log\cot x\,dx
\displaystyle \text{Adding (i) and (ii), we get}
\displaystyle 2I=\int_{0}^{\pi/2}(\log\tan x+\log\cot x)\,dx
\displaystyle \Rightarrow 2I=\int_{0}^{\pi/2}\log(\tan x\cot x)\,dx =\int_{0}^{\pi/2}\log 1\,dx=\int_{0}^{\pi/2}0\,dx=0
\displaystyle \Rightarrow I=0

\displaystyle \text{(ii) Let }I=\int_{0}^{\pi/4}\log(1+\tan x)\,dx
\displaystyle \text{Then,}
\displaystyle I=\int_{0}^{\pi/4}\log\!\left(1+\tan\left(\frac{\pi}{4}-x\right)\right)dx \qquad [\text{Using }\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx]
\displaystyle \Rightarrow I=\int_{0}^{\pi/4} \log\!\left(1+\frac{\tan\frac{\pi}{4}-\tan x}{1+\tan\frac{\pi}{4}\tan x}\right)dx
\displaystyle =\int_{0}^{\pi/4}\log\!\left(1+\frac{1-\tan x}{1+\tan x}\right)dx
\displaystyle =\int_{0}^{\pi/4}\log\!\left(\frac{2}{1+\tan x}\right)dx
\displaystyle =\int_{0}^{\pi/4}\log 2\,dx-\int_{0}^{\pi/4}\log(1+\tan x)\,dx
\displaystyle \Rightarrow I=(\log 2)[x]_{0}^{\pi/4}-I
\displaystyle \Rightarrow 2I=\frac{\pi}{4}\log 2\Rightarrow I=\frac{\pi}{8}\log 2

\displaystyle \textbf{Question 15: } \text{Evaluate:}
\displaystyle \text{}\ \int_{0}^{\pi/2}(2\log\sin x-\log\sin 2x)\,dx \qquad [\text{CBSE 2009}]
\displaystyle \text{Answer:}
\displaystyle \text{We have }I=\int_{0}^{\pi/2}(2\log\sin x-\log\sin 2x)\,dx
\displaystyle =\int_{0}^{\pi/2}\{2\log\sin x-\log(2\sin x\cos x)\}dx
\displaystyle =\int_{0}^{\pi/2}\{2\log\sin x-\log 2-\log\sin x-\log\cos x\}dx
\displaystyle =\int_{0}^{\pi/2}\log\sin x\,dx-\int_{0}^{\pi/2}\log 2\,dx -\int_{0}^{\pi/2}\log\cos x\,dx
\displaystyle =\int_{0}^{\pi/2}\log\sin x\,dx-(\log 2)\int_{0}^{\pi/2}1\,dx -\int_{0}^{\pi/2}\log\cos\left(\frac{\pi}{2}-x\right)dx
\displaystyle =\int_{0}^{\pi/2}\log\sin x\,dx-(\log 2)[x]_{0}^{\pi/2} -\int_{0}^{\pi/2}\log\sin x\,dx
\displaystyle =-(\log 2)\frac{\pi}{2}=-\frac{\pi}{2}\log 2

\displaystyle \textbf{Question 16: } \text{Evaluate:}\ \int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}}\,dx\qquad [\text{CBSE 2009}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}}\,dx
\displaystyle \text{Then,}
\displaystyle I=\int_{0}^{\pi} \frac{e^{\cos(\pi-x)}}{e^{\cos(\pi-x)}+e^{-\cos(\pi-x)}}\,dx \qquad [\text{Using }\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx]
\displaystyle \Rightarrow I=\int_{0}^{\pi} \frac{e^{-\cos x}}{e^{-\cos x}+e^{\cos x}}\,dx
\displaystyle \text{Adding (i) and (ii), we get}
\displaystyle 2I=\int_{0}^{\pi}1\,dx=\pi\Rightarrow I=\frac{\pi}{2}

\displaystyle \textbf{Question 17: } \text{Prove that:}\ \int_{0}^{2a}f(x)\,dx =\int_{0}^{2a}f(2a-x)\,dx\qquad [\text{CBSE 2002C}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{0}^{2a}f(x)\,dx
\displaystyle \text{Let }2a-x=t.\text{ Then }d(2a-x)=dt\Rightarrow -dx=dt \Rightarrow dx=-dt
\displaystyle \text{Also, }x=0\Rightarrow t=2a-0=2a\text{ and }x=a\Rightarrow t=2a-a=a
\displaystyle I=\int_{2a}^{a}f(2a-t)(-dt)=-\int_{2a}^{a}f(2a-t)\,dt
\displaystyle \Rightarrow I=\int_{0}^{2a}f(2a-t)\,dt
\displaystyle \Rightarrow I=\int_{0}^{2a}f(2a-x)\,dx
\displaystyle \text{Hence, }\int_{0}^{2a}f(x)\,dx=\int_{0}^{2a}f(2a-x)\,dx

\displaystyle \textbf{Question 18: } \text{Evaluate:}
\displaystyle \text{(i)}\ \int_{0}^{\pi/2}\frac{\sin^{2}x}{\sin x+\cos x}\,dx \qquad [\text{CBSE 2002, 2003, 2016}]
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let }I=\int_{0}^{\pi/2} \frac{\sin^{2}x}{\sin x+\cos x}\,dx
\displaystyle \Rightarrow I=\int_{0}^{\pi/2} \frac{\sin^{2}(\frac{\pi}{2}-x)}{\sin(\frac{\pi}{2}-x)+\cos(\frac{\pi}{2}-x)}\,dx \\ \qquad [\text{Using }\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx]
\displaystyle \Rightarrow I=\int_{0}^{\pi/2} \frac{\cos^{2}x}{\cos x+\sin x}\,dx
\displaystyle \text{Adding (i) and (ii), we get}
\displaystyle 2I=\int_{0}^{\pi/2} \frac{\sin^{2}x}{\sin x+\cos x}\,dx +\int_{0}^{\pi/2}\frac{\cos^{2}x}{\sin x+\cos x}\,dx
\displaystyle =\int_{0}^{\pi/2}\frac{1}{\sin x+\cos x}\,dx
\displaystyle 2I=\int_{0}^{\pi/2} \frac{1} {\frac{2\tan\frac{x}{2}}{1+\tan^{2}\frac{x}{2}} +\frac{1-\tan^{2}\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}}\,dx
\displaystyle =\int_{0}^{\pi/2} \frac{1+\tan^{2}\frac{x}{2}} {2\tan\frac{x}{2}+1-\tan^{2}\frac{x}{2}}\,dx
\displaystyle =\int_{0}^{\pi/2} \frac{\sec^{2}\frac{x}{2}} {2\tan\frac{x}{2}+1-\tan^{2}\frac{x}{2}}\,dx
\displaystyle \text{Let }\tan\frac{x}{2}=t. \text{ Then }d\!\left(\tan\frac{x}{2}\right)=dt \Rightarrow \sec^{2}\frac{x}{2}\frac{1}{2}dx=dt \Rightarrow \sec^{2}\frac{x}{2}dx=2dt
\displaystyle \text{Also, }x=0\Rightarrow t=\tan 0=0, \ x=\frac{\pi}{2}\Rightarrow t=\tan\frac{\pi}{4}=1
\displaystyle 2I=\int_{0}^{1}\frac{2dt}{2t+1-t^{2}} =2\int_{0}^{1}\frac{dt}{(\sqrt{2})^{2}-(t-1)^{2}}
\displaystyle =2\times\frac{1}{2\sqrt{2}} \left[\log\frac{\sqrt{2}+t-1}{\sqrt{2}-t+1}\right]_{0}^{1}
\displaystyle =\frac{1}{\sqrt{2}} \left\{\log\frac{\sqrt{2}}{\sqrt{2}} -\log\frac{\sqrt{2}-1}{\sqrt{2}+1}\right\}
\displaystyle =-\frac{1}{\sqrt{2}} \log\frac{(\sqrt{2}-1)^{2}}{(\sqrt{2}+1)(\sqrt{2}-1)} =-\frac{2}{\sqrt{2}}\log(\sqrt{2}-1)
\displaystyle I=-\frac{1}{\sqrt{2}}\log(\sqrt{2}-1)

\displaystyle \textbf{Question 19: } \text{Evaluate:}\ \int_{0}^{1}\cot^{-1}(1-x+x^{2})\,dx \qquad [\text{CBSE 2008}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{0}^{1}\cot^{-1}(1-x+x^{2})\,dx
\displaystyle I=\int_{0}^{1}\tan^{-1}\!\left(\frac{1}{1-x+x^{2}}\right)dx \qquad [\because\ \cot^{-1}x=\tan^{-1}\frac{1}{x},\ x>0]
\displaystyle I=\int_{0}^{1}\tan^{-1}\!\left(\frac{1}{1-x(1-x)}\right)dx
\displaystyle I=\int_{0}^{1}\tan^{-1}\!\left(\frac{x+(1-x)}{1-x(1-x)}\right)dx
\displaystyle I=\int_{0}^{1}\{\tan^{-1}x+\tan^{-1}(1-x)\}\,dx
\displaystyle I=\int_{0}^{1}\tan^{-1}x\,dx+\int_{0}^{1}\tan^{-1}(1-x)\,dx
\displaystyle I=\int_{0}^{1}\tan^{-1}x\,dx+\int_{0}^{1}\tan^{-1}\{1-(1-x)\}\,dx \\ \qquad [\because\ \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx]
\displaystyle I=\int_{0}^{1}\tan^{-1}x\,dx+\int_{0}^{1}\tan^{-1}x\,dx
\displaystyle I=2\int_{0}^{1}\tan^{-1}x\,dx
\displaystyle I=2\int_{0}^{1}\tan^{-1}x\cdot 1\,dx
\displaystyle I=2\left[x\tan^{-1}x\right]_{0}^{1} -2\int_{0}^{1}\frac{x}{1+x^{2}}\,dx
\displaystyle I=2\left[x\tan^{-1}x\right]_{0}^{1} -\int_{0}^{1}\frac{2x}{1+x^{2}}\,dx
\displaystyle I=2\left[x\tan^{-1}x\right]_{0}^{1} -\left[\log(1+x^{2})\right]_{0}^{1}
\displaystyle I=2\left(\frac{\pi}{4}-0\right)-(\log 2-\log 1) =\frac{\pi}{2}-\log 2

\displaystyle \textbf{Question 20: } \text{Evaluate:}\ \int_{0}^{2\pi}\frac{1}{1+e^{\sin x}}\,dx \qquad [\text{CBSE 2013}]

\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{0}^{2\pi}\frac{1}{1+e^{\sin x}}\,dx
\displaystyle I=\int_{0}^{\pi}\left\{\frac{1}{1+e^{\sin x}} +\frac{1}{1+e^{\sin(2\pi-x)}}\right\}dx \\ \qquad [\text{Using }\int_{0}^{2a}f(x)\,dx=\int_{0}^{a}\{f(x)+f(2a-x)\}dx]
\displaystyle \Rightarrow I=\int_{0}^{\pi}\left\{\frac{1}{1+e^{\sin x}} +\frac{1}{1+e^{-\sin x}}\right\}dx
\displaystyle \Rightarrow I=\int_{0}^{\pi}\left\{\frac{1}{1+e^{\sin x}} +\frac{e^{\sin x}}{1+e^{\sin x}}\right\}dx=\int_{0}^{\pi}1\,dx
\displaystyle =\left[x\right]_{0}^{\pi}=\pi

\displaystyle \textbf{Question 21: } \text{Evaluate:}
\displaystyle \text{(i)}\ \int_{0}^{\pi}\frac{x}{1+\sin x}\,dx \qquad [\text{CBSE 2001, 2004, 2010, 2012}]
\displaystyle \text{(ii)}\ \int_{0}^{\pi}\frac{x\tan x}{\sec x+\tan x}\,dx \qquad [\text{CBSE 2008, 2010, 2014}]
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let }I=\int_{0}^{\pi}\frac{x}{1+\sin x}\,dx
\displaystyle \Rightarrow I=\int_{0}^{\pi}\frac{\pi-x}{1+\sin(\pi-x)}\,dx \\ \qquad [\because\ \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx]
\displaystyle \Rightarrow I=\int_{0}^{\pi}\frac{\pi-x}{1+\sin x}\,dx
\displaystyle \text{Adding (i) and (ii), we get}
\displaystyle 2I=\int_{0}^{\pi}\frac{x+\pi-x}{1+\sin x}\,dx
\displaystyle \Rightarrow 2I=\pi\int_{0}^{\pi}\frac{1}{1+\sin x}\,dx
\displaystyle \Rightarrow 2I=\pi\int_{0}^{\pi/2} \left\{\frac{1}{1+\sin x}+\frac{1}{1+\sin(\pi-x)}\right\}dx \\ \qquad [\because\ \int_{0}^{2a}f(x)\,dx=\int_{0}^{a}\{f(x)+f(2a-x)\}dx]
\displaystyle \Rightarrow 2I=2\pi\int_{0}^{\pi/2}\frac{1}{1+\sin x}\,dx
\displaystyle \Rightarrow 2I=2\pi\int_{0}^{\pi/2} \frac{1-\sin x}{1-\sin^{2}x}\,dx
\displaystyle \Rightarrow 2I=2\pi\int_{0}^{\pi/2} (\sec^{2}x-\tan x\sec x)\,dx
\displaystyle \Rightarrow 2I=2\pi[\tan x-\sec x]_{0}^{\pi/2} =2\pi\left[\frac{\sin x-1}{\cos x}\right]_{0}^{\pi/2}
\displaystyle \Rightarrow 2I=2\pi\left[\frac{\sin^{2}x-1} {\cos x(\sin x+1)}\right]_{0}^{\pi/2} =2\pi\left[\frac{-\cos x}{1+\sin x}\right]_{0}^{\pi/2} =2\pi(0+1)=2\pi
\displaystyle \Rightarrow I=\pi

\displaystyle \text{(ii) Let }I=\int_{0}^{\pi} \frac{x\tan x}{\sec x+\tan x}\,dx=\int_{0}^{\pi}\frac{x\sin x}{1+\sin x}\,dx
\displaystyle \Rightarrow I=\int_{0}^{\pi} \frac{(\pi-x)\sin(\pi-x)}{1+\sin(\pi-x)}\,dx
\displaystyle \Rightarrow I=\int_{0}^{\pi} \frac{(\pi-x)\sin x}{1+\sin x}\,dx
\displaystyle \text{Adding (i) and (ii), we get}
\displaystyle 2I=\int_{0}^{\pi}\frac{\pi\sin x}{1+\sin x}\,dx
\displaystyle \Rightarrow 2I=\pi\int_{0}^{\pi/2} \left\{\frac{\sin x}{1+\sin x} +\frac{\sin(\pi-x)}{1+\sin(\pi-x)}\right\}dx \\ \qquad [\because\ \int_{0}^{2a}f(x)\,dx=\int_{0}^{a}\{f(x)+f(2a-x)\}dx]
\displaystyle \Rightarrow 2I=\pi\int_{0}^{\pi/2} \left(\frac{\sin x}{1+\sin x}+\frac{\sin x}{1+\sin x}\right)dx
\displaystyle \Rightarrow 2I=2\pi\int_{0}^{\pi/2} \frac{\sin x}{1+\sin x}\,dx
\displaystyle \Rightarrow 2I=2\pi\int_{0}^{\pi/2} \frac{\sin x(1-\sin x)}{1-\sin^{2}x}\,dx
\displaystyle \Rightarrow 2I=2\pi\int_{0}^{\pi/2} \frac{\sin x-\sin^{2}x}{\cos^{2}x}\,dx
\displaystyle \Rightarrow 2I=2\pi\int_{0}^{\pi/2} (\tan x\sec x-\tan^{2}x)\,dx
\displaystyle \Rightarrow I=\pi\int_{0}^{\pi/2} (\tan x\sec x-\tan^{2}x)\,dx
\displaystyle \Rightarrow I=\pi\int_{0}^{\pi/2} (\tan x\sec x-(\sec^{2}x-1))\,dx
\displaystyle \Rightarrow I=\pi\int_{0}^{\pi/2} (\sec x\tan x-\sec^{2}x+1)\,dx
\displaystyle \Rightarrow I=\pi[\sec x-\tan x+x]_{0}^{\pi/2}
\displaystyle \Rightarrow I=\pi\left[\frac{1-\sin x}{\cos x}+x\right]_{0}^{\pi/2} \\ =\pi\left[\frac{\cos x}{1+\sin x}+x\right]_{0}^{\pi/2} \\ =\pi\left(0+\frac{\pi}{2}-(1+0)\right) \\ =\frac{\pi}{2}(\pi-2)

\displaystyle \textbf{Question 22: } \text{Evaluate:}\ \int_{0}^{\pi/2} \frac{x\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx \qquad [\text{CBSE 2010, 2011, 2014}]

\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{0}^{\pi/2} \frac{x\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx
\displaystyle I=\int_{0}^{\pi/2} \frac{(\frac{\pi}{2}-x)\sin(\frac{\pi}{2}-x)\cos(\frac{\pi}{2}-x)} {\sin^{4}(\frac{\pi}{2}-x)+\cos^{4}(\frac{\pi}{2}-x)}\,dx
\displaystyle \Rightarrow I=\int_{0}^{\pi/2} \frac{(\frac{\pi}{2}-x)\sin x\cos x}{\cos^{4}x+\sin^{4}x}\,dx
\displaystyle \Rightarrow I=\frac{\pi}{2}\int_{0}^{\pi/2} \frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx -\int_{0}^{\pi/2} \frac{x\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx
\displaystyle \Rightarrow I=\frac{\pi}{2}\int_{0}^{\pi/2} \frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx-I
\displaystyle \Rightarrow 2I=\frac{\pi}{2}\int_{0}^{\pi/2} \frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx
\displaystyle \Rightarrow 2I=\frac{\pi}{2}\int_{0}^{\pi/2} \frac{\tan x\sec^{2}x}{1+\tan^{4}x}\,dx \qquad [\text{Dividing numerator and denominator by }\cos^{4}x]
\displaystyle \Rightarrow 2I=\frac{\pi}{4}\int_{0}^{\pi/2} \frac{2\tan x\sec^{2}x}{1+(\tan^{2}x)^{2}}\,dx
\displaystyle \text{Let }t=\tan^{2}x.\text{ Then }dt=2\tan x\sec^{2}x\,dx
\displaystyle \text{Also, }x=0\Rightarrow t=\tan^{2}0=0,\ x=\frac{\pi}{2} \Rightarrow t=\tan^{2}\frac{\pi}{2}=\infty
\displaystyle 2I=\frac{\pi}{4}\int_{0}^{\infty}\frac{dt}{1+t^{2}}
\displaystyle \Rightarrow 2I=\frac{\pi}{4}[\tan^{-1}t]_{0}^{\infty} =\frac{\pi}{4}\left(\frac{\pi}{2}-0\right)
\displaystyle \Rightarrow I=\frac{\pi^{2}}{16}

\displaystyle \textbf{Question 23: } \text{Evaluate:}\ \int_{-\pi/2}^{\pi/2} \frac{\cos x}{1+e^{x}}\,dx\qquad [\text{CBSE 2015}]

\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{-\pi/2}^{\pi/2} \frac{\cos x}{1+e^{x}}\,dx
\displaystyle I=\int_{-\pi/2}^{\pi/2} \left\{\frac{\cos x}{1+e^{x}}+\frac{\cos(-x)}{1+e^{-x}}\right\}dx \\ \qquad [\because\ \int_{-a}^{a}f(x)\,dx=\int_{0}^{a}\{f(x)+f(-x)\}dx]
\displaystyle \Rightarrow I=\int_{0}^{\pi/2} \left\{\frac{\cos x}{1+e^{x}}+\frac{\cos x}{1+e^{-x}}\right\}dx
\displaystyle \Rightarrow I=\int_{0}^{\pi/2} \left\{\frac{1}{1+e^{x}}+\frac{e^{x}}{1+e^{x}}\right\}\cos x\,dx
\displaystyle \Rightarrow I=\int_{0}^{\pi/2}\cos x\,dx =[\sin x]_{0}^{\pi/2}=1

\displaystyle \textbf{Question 24: } \text{Evaluate:}
\displaystyle \text{}\ \int_{-a}^{a}\sqrt{\frac{a-x}{a+x}}\,dx \qquad [\text{CBSE 2002, 2008}]

\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{-a}^{a}\sqrt{\frac{a-x}{a+x}}\,dx
\displaystyle I=\int_{-a}^{a}\sqrt{\frac{a-x}{a+x}\times\frac{a-x}{a-x}}\,dx =\int_{-a}^{a}\frac{a-x}{\sqrt{a^{2}-x^{2}}}\,dx
\displaystyle \Rightarrow I=\int_{-a}^{a}\frac{a}{\sqrt{a^{2}-x^{2}}}\,dx -\int_{-a}^{a}\frac{x}{\sqrt{a^{2}-x^{2}}}\,dx=aI_{1}-I_{2}
\displaystyle \text{where }I_{1}=\int_{-a}^{a}\frac{1}{\sqrt{a^{2}-x^{2}}}\,dx \text{ and }I_{2}=\int_{-a}^{a}\frac{x}{\sqrt{a^{2}-x^{2}}}\,dx
\displaystyle \text{Let }f(x)=\frac{1}{\sqrt{a^{2}-x^{2}}} \text{ and }g(x)=\frac{x}{\sqrt{a^{2}-x^{2}}}
\displaystyle f(-x)=\frac{1}{\sqrt{a^{2}-(-x)^{2}}} =\frac{1}{\sqrt{a^{2}-x^{2}}}=f(x)
\displaystyle g(-x)=\frac{-x}{\sqrt{a^{2}-(-x)^{2}}} =-\frac{x}{\sqrt{a^{2}-x^{2}}}=-g(x)
\displaystyle \Rightarrow f(x)\text{ is even and }g(x)\text{ is odd}
\displaystyle \Rightarrow I_{1}=2\int_{0}^{a}\frac{1}{\sqrt{a^{2}-x^{2}}}\,dx \text{ and }I_{2}=0
\displaystyle I_{1}=2\left[\sin^{-1}\frac{x}{a}\right]_{0}^{a} \text{ and }I_{2}=0
\displaystyle I_{1}=2\left(\sin^{-1}1-\sin^{-1}0\right) =2\left(\frac{\pi}{2}-0\right)=\pi
\displaystyle \text{Substituting values of }I_{1}\text{ and }I_{2}\text{ in (i), we get}
\displaystyle I=a\pi-0=a\pi

\displaystyle \textbf{Question 25: } \text{Evaluate:}\ \int_{-\pi}^{\pi}(\cos ax-\sin bx)^{2}\,dx \qquad [\text{CBSE 2015}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{-\pi}^{\pi}(\cos ax-\sin bx)^{2}\,dx
\displaystyle I=\int_{-\pi}^{\pi}(\cos^{2}ax+\sin^{2}bx-2\cos ax\sin bx)\,dx
\displaystyle \Rightarrow I=\int_{-\pi}^{\pi}\cos^{2}ax\,dx +\int_{-\pi}^{\pi}\sin^{2}bx\,dx-2\int_{-\pi}^{\pi}\cos ax\sin bx\,dx
\displaystyle \Rightarrow I=2\int_{0}^{\pi}\cos^{2}ax\,dx +2\int_{0}^{\pi}\sin^{2}bx\,dx-2\times 0 \qquad [\because\ \cos^{2}ax,\sin^{2}bx\text{ even; }\cos ax\sin bx\text{ odd}]
\displaystyle \Rightarrow I=\int_{0}^{\pi}(1+\cos 2ax)\,dx +\int_{0}^{\pi}(1-\cos 2bx)\,dx
\displaystyle \Rightarrow I=\left[x+\frac{1}{2a}\sin 2ax\right]_{0}^{\pi} +\left[x-\frac{1}{2b}\sin 2bx\right]_{0}^{\pi}
\displaystyle \Rightarrow I=\left(\pi+\frac{1}{2a}\sin 2a\pi-0\right) +\left(\pi-\frac{1}{2b}\sin 2b\pi-0\right)
\displaystyle \Rightarrow I=2\pi+\frac{1}{2a}\sin 2a\pi-\frac{1}{2b}\sin 2b\pi
\displaystyle I= \begin{cases} 2\pi,& a,b\text{ integers}\\[4pt] 2\pi+\frac{1}{2a}\sin 2a\pi-\frac{1}{2b}\sin 2b\pi,& a,b\text{ not integers} \end{cases}

\displaystyle \textbf{Question 26: } \text{Prove that:}\ \int_{0}^{\pi/2}\log\sin x\,dx=\int_{0}^{\pi/2}\log\cos x\,dx =-\frac{\pi}{2}\log 2\qquad [\text{CBSE 2008}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{0}^{\pi/2}\log\sin x\,dx
\displaystyle \text{Then, }I=\int_{0}^{\pi/2}\log\sin\left(\frac{\pi}{2}-x\right)\,dx
\displaystyle \Rightarrow I=\int_{0}^{\pi/2}\log\cos x\,dx
\displaystyle \text{Adding (i) and (ii), we get}
\displaystyle 2I=\int_{0}^{\pi/2}\log\sin x\,dx +\int_{0}^{\pi/2}\log\cos x\,dx
\displaystyle \Rightarrow 2I=\int_{0}^{\pi/2} (\log\sin x+\log\cos x)\,dx
\displaystyle \Rightarrow 2I=\int_{0}^{\pi/2}\log(\sin x\cos x)\,dx
\displaystyle \Rightarrow 2I=\int_{0}^{\pi/2} \log\left(\frac{2\sin x\cos x}{2}\right)\,dx
\displaystyle \Rightarrow 2I=\int_{0}^{\pi/2}\log\left(\frac{\sin 2x}{2}\right)\,dx
\displaystyle \Rightarrow 2I=\int_{0}^{\pi/2}\log\sin 2x\,dx -\int_{0}^{\pi/2}\log 2\,dx
\displaystyle \Rightarrow 2I=\int_{0}^{\pi/2}\log\sin 2x\,dx -\frac{\pi}{2}\log 2
\displaystyle \text{Let }I_{1}=\int_{0}^{\pi/2}\log\sin 2x\,dx
\displaystyle \text{Putting }2x=t,\text{ we get } I_{1}=\int_{0}^{\pi}\log\sin t\,\frac{dt}{2}
\displaystyle \Rightarrow I_{1}=\frac{1}{2}\int_{0}^{\pi}\log\sin t\,dt
\displaystyle \Rightarrow I_{1}=\frac{1}{2}\times 2\int_{0}^{\pi/2}\log\sin t\,dt \qquad [\text{Using property X}]
\displaystyle \Rightarrow I_{1}=\int_{0}^{\pi/2}\log\sin x\,dx=I \qquad [\text{Using Property I}]
\displaystyle \text{Putting }I_{1}=I\text{ in (iii), we get }2I=I-\frac{\pi}{2}\log 2 \Rightarrow I=-\frac{\pi}{2}\log 2
\displaystyle \text{Hence, }\int_{0}^{\pi/2}\log\sin x\,dx =\int_{0}^{\pi/2}\log\cos x\,dx=-\frac{\pi}{2}\log 2

\displaystyle \textbf{Question 27: } \text{Evaluate the following integrals as limit of sums:}
\displaystyle \text{(i)}\ \int_{0}^{2}(x^{2}+3)\,dx\qquad [\text{CBSE 2001C}]
\displaystyle \text{(ii)}\ \int_{1}^{3}(2x^{2}+5)\,dx\qquad [\text{CBSE 2010}]
\displaystyle \text{Answer:}
\displaystyle \text{(i) We have,}
\displaystyle \int_{a}^{b}f(x)\,dx=\lim_{h\to 0}h \left[f(a)+f(a+h)+f(a+2h)+\cdots+f\{a+(n-1)h\}\right], \text{ where }h=\frac{b-a}{n}
\displaystyle \text{Here, }a=0,\ b=2,\ f(x)=x^{2}+3\text{ and }h=\frac{2-0}{n}=\frac{2}{n}
\displaystyle I=\int_{0}^{2}(x^{2}+3)\,dx
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[f(0)+f(0+h)+f(0+2h)+\cdots+f\{0+(n-1)h\}\right]
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[f(0)+f(h)+f(2h)+\cdots+f\{(n-1)h\}\right]
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[(0+3)+(h^{2}+3)+(2^{2}h^{2}+3)+\cdots+\{(n-1)^{2}h^{2}+3\}\right]
\displaystyle \Rightarrow I=\lim_{h\to 0} \left[3n+h^{2}(1^{2}+2^{2}+\cdots+(n-1)^{2})\right]
\displaystyle \Rightarrow I=\lim_{h\to 0} \left\{3n+h^{2}\frac{n(n-1)(2n-1)}{6}\right\}
\displaystyle \Rightarrow I=\lim_{n\to\infty}\frac{2}{n} \left\{3n+\frac{4}{n^{2}}\times\frac{n(n-1)(2n-1)}{6}\right\}
\displaystyle \Rightarrow I=\lim_{n\to\infty} \left\{6+\frac{8}{6}\frac{(n-1)(2n-1)}{n^{2}}\right\}
\displaystyle \Rightarrow I=\lim_{n\to\infty} \left\{6+\frac{8}{6}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)\right\}
\displaystyle \Rightarrow I=6+\frac{8}{6}(1-0)(2-0) =6+\frac{8}{3}=\frac{26}{3}

\displaystyle \text{(ii) We have,}
\displaystyle \int_{a}^{b}f(x)\,dx=\lim_{h\to 0}h \left[f(a)+f(a+h)+f(a+2h)+\cdots+f\{a+(n-1)h\}\right], \text{ where }h=\frac{b-a}{n}
\displaystyle \text{Here, }a=1,\ b=3,\ f(x)=2x^{2}+5\text{ and }h=\frac{3-1}{n}=\frac{2}{n}
\displaystyle I=\int_{1}^{3}(2x^{2}+5)\,dx
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[f(1)+f(1+h)+f(1+2h)+\cdots+f\{1+(n-1)h\}\right]
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[2(1)^{2}+5+2(1+h)^{2}+5+\cdots+2\{1+(n-1)h\}^{2}+5\right]
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left\{2\left[1^{2}+(1+h)^{2}+(1+2h)^{2}+\cdots+\{1+(n-1)h\}^{2}\right]+5n\right\}
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left\{2\left[n+2h(1+2+\cdots+(n-1)) +h^{2}(1^{2}+2^{2}+\cdots+(n-1)^{2})\right]+5n\right\}
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left\{2\left[n+2h\frac{n(n-1)}{2} +h^{2}\frac{n(n-1)(2n-1)}{6}\right]+5n\right\}
\displaystyle \Rightarrow I=\lim_{n\to\infty}\frac{2}{n} \left\{7n+2\times\frac{2}{n}n(n-1) +2\times\frac{4}{n^{2}}\frac{n(n-1)(2n-1)}{6}\right\}
\displaystyle \Rightarrow I=\lim_{n\to\infty} \left[14+8\left(\frac{n-1}{n}\right) +\frac{8}{3}\frac{(n-1)(2n-1)}{n^{2}}\right]
\displaystyle \Rightarrow I=\lim_{n\to\infty} \left[14+8\left(1-\frac{1}{n}\right) +\frac{8}{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)\right]
\displaystyle \Rightarrow I=14+8(1-0)+\frac{8}{3}(1-0)(2-0) =14+8+\frac{16}{3}=\frac{82}{3}

\displaystyle \textbf{Question 28: } \text{Evaluate the following integrals as limit of sums:}
\displaystyle \text{(i)}\ \int_{1}^{3}(x^{2}+x)\,dx\qquad [\text{CBSE 2000C}]
\displaystyle \text{(ii)}\ \int_{1}^{3}(x^{2}+5x)\,dx\qquad [\text{CBSE 2010}]
\displaystyle \text{Answer:}
\displaystyle \text{(i) We have,}
\displaystyle \int_{a}^{b}f(x)\,dx=\lim_{h\to 0}h \left[f(a)+f(a+h)+f(a+2h)+\cdots+f\{a+(n-1)h\}\right], \text{ where }h=\frac{b-a}{n}
\displaystyle \text{Here, }a=1,\ b=3,\ f(x)=x^{2}+x\text{ and }h=\frac{3-1}{n} =\frac{2}{n}
\displaystyle I=\int_{1}^{3}(x^{2}+x)\,dx
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[f(1)+f(1+h)+f(1+2h)+\cdots+f\{1+(n-1)h\}\right]
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[\{1^{2}+1\}+\{(1+h)^{2}+(1+h)\}+\{(1+2h)^{2}+(1+2h)\} +\cdots+\{(1+(n-1)h)^{2}+(1+(n-1)h)\}\right]
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[\{1^{2}+(1+h)^{2}+(1+2h)^{2}+\cdots+\{1+(n-1)h\}^{2}\} +\{1+(1+h)+(1+2h)+\cdots+(1+(n-1)h)\}\right]
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left\{n+2h(1+2+\cdots+(n-1)) +h^{2}(1^{2}+2^{2}+\cdots+(n-1)^{2}) +n+h(1+2+\cdots+(n-1))\right\}
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[n+2h\frac{n(n-1)}{2} +h^{2}\frac{n(n-1)(2n-1)}{6} +n+h\frac{n(n-1)}{2}\right]
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[2n+3h\frac{n(n-1)}{2} +h^{2}\frac{n(n-1)(2n-1)}{6}\right]
\displaystyle \Rightarrow I=\lim_{n\to\infty}\frac{2}{n} \left[2n+\frac{6}{n}\frac{n(n-1)}{2} +\frac{4}{n^{2}}\frac{n(n-1)(2n-1)}{6}\right] \qquad [\because\ h=\frac{2}{n}]
\displaystyle \Rightarrow I=\lim_{n\to\infty} \left[4+6\left(\frac{n-1}{n}\right) +\frac{4}{3}\frac{(n-1)(2n-1)}{n^{2}}\right]
\displaystyle \Rightarrow I=\lim_{n\to\infty} \left[4+6\left(1-\frac{1}{n}\right) +\frac{4}{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)\right]
\displaystyle \Rightarrow I=4+6(1-0)+\frac{4}{3}(1-0)(2-0) =4+6+\frac{8}{3}=\frac{38}{3}

\displaystyle \text{(ii) We have,}
\displaystyle \int_{a}^{b}f(x)\,dx=\lim_{h\to 0}h \left[f(a)+f(a+h)+f(a+2h)+\cdots+f\{a+(n-1)h\}\right], \text{ where }h=\frac{b-a}{n}
\displaystyle \text{Here, }a=1,\ b=3,\ f(x)=x^{2}+5x\text{ and }h=\frac{3-1}{n} =\frac{2}{n}
\displaystyle I=\int_{1}^{3}(x^{2}+5x)\,dx
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[f(1)+f(1+h)+f(1+2h)+\cdots+f\{1+(n-1)h\}\right]
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[\{1^{2}+5(1)\}+\{(1+h)^{2}+5(1+h)\} +\{(1+2h)^{2}+5(1+2h)\} +\cdots+\{(1+(n-1)h)^{2}+5(1+(n-1)h)\}\right]
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[\{1^{2}+(1+h)^{2}+(1+2h)^{2}+\cdots+\{1+(n-1)h\}^{2}\} +5\{1+(1+h)+(1+2h)+\cdots+(1+(n-1)h)\}\right]
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left\{n+2h(1+2+\cdots+(n-1)) +h^{2}(1^{2}+2^{2}+\cdots+(n-1)^{2}) +5n+5h(1+2+\cdots+(n-1))\right\}
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[6n+7h\frac{n(n-1)}{2} +h^{2}\frac{n(n-1)(2n-1)}{6}\right]
\displaystyle \Rightarrow I=\lim_{n\to\infty}\frac{2}{n} \left[6n+\frac{14}{n}\frac{n(n-1)}{2} +\frac{4}{n^{2}}\frac{n(n-1)(2n-1)}{6}\right]
\displaystyle \Rightarrow I=\lim_{n\to\infty} \left[12+14\left(\frac{n-1}{n}\right) +\frac{8}{3}\frac{(n-1)(2n-1)}{n^{2}}\right]
\displaystyle \Rightarrow I=\lim_{n\to\infty} \left[12+14\left(1-\frac{1}{n}\right) +\frac{8}{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)\right]
\displaystyle \Rightarrow I=12+14(1-0)+\frac{8}{3}(1-0)(2-0) =12+14+\frac{8}{3}=\frac{86}{3}

\displaystyle \textbf{Question 29: } \text{Evaluate the following integrals as a limit of sums.}
\displaystyle \text{}\ \int_{1}^{3}(e^{2-3x}+x^{2}+1)\,dx \qquad [\text{CBSE 2015}]
\displaystyle \text{Answer:}
\displaystyle \text{Here, }a=1,\ b=3\text{ and }f(x)=e^{2-3x}+x^{2}+1. \text{ Therefore, }h=\frac{3-1}{n}=\frac{2}{n}\text{ and }nh=2
\displaystyle \text{Substituting these values in } \\ \int_{a}^{b}f(x)\,dx=\lim_{h\to 0}h \left[f(a)+f(a+h)+f(a+2h)+\cdots+f\{a+(n-1)h\}\right],\text{ we obtain}
\displaystyle I=\int_{1}^{3}(e^{2-3x}+x^{2}+1)\,dx
\displaystyle =\lim_{h\to 0}h\left[f(1)+f(1+h)+f(1+2h)+\cdots+f\{1+(n-1)h\}\right]
\displaystyle \Rightarrow I=\lim_{h\to 0}h \Big[e^{2-3\cdot1}+1^{2}+1+e^{2-3(1+h)} \\ +(1+h)^{2}+1 +e^{2-3(1+2h)}+(1+2h)^{2}+1+\cdots
\displaystyle \qquad\qquad +e^{2-3\{1+(n-1)h\}}+\{1+(n-1)h\}^{2}+1\Big]
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[\{e^{-1}+e^{-1-3h}+e^{-1-3(2h)}+\cdots+e^{-1-3(n-1)h}\}\right.
\displaystyle \qquad\qquad \left.+\{1^{2}+(1+h)^{2}+(1+2h)^{2}+\cdots+(1+(n-1)h)^{2}\}+n\right]
\displaystyle \Rightarrow I=\lim_{h\to 0}h\Bigg[ e^{-1}\frac{1-e^{-3nh}}{1-e^{-3h}}  \\ +\left\{n+2h(1+2+\cdots+(n-1)) +h^{2}(1^{2}+2^{2}+\cdots+(n-1)^{2})\right\}+n\Bigg]
\displaystyle \Rightarrow I=\lim_{h\to 0}h\Bigg[ e^{-1}\frac{1-e^{-3nh}}{1-e^{-3h}} +2n+2h\frac{n(n-1)}{2} +h^{2}\frac{n(n-1)(2n-1)}{6}\Bigg]
\displaystyle \Rightarrow I=\lim_{h\to 0}\Bigg[ h e^{-1}\frac{1-e^{-3nh}}{1-e^{-3h}} +2nh+h^{2}n(n-1) +\frac{1}{6}h^{3}n(n-1)(2n-1)\Bigg]
\displaystyle \Rightarrow I=\lim_{h\to 0}\Bigg[ e^{-1}\frac{e^{-6}-1}{-3} +4+2(2-h)+\frac{2}{6}(2-h)(4-h)\Bigg] \qquad [\because\ nh=2]
\displaystyle \Rightarrow I=\frac{e^{-1}}{3}(1-e^{-6}) +4+4+\frac{4}{3} =-\frac{1}{3}(e^{-7}-e^{-1})+\frac{32}{3}

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