\displaystyle \textbf{Question 1: }\quad \text{Evaluate:}
\displaystyle \int \frac{\cos 2x-\cos 2\alpha}{\cos x-\cos \alpha}\,dx \hspace{8.0cm} \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int \frac{\cos 2x-\cos 2\alpha}{\cos x-\cos \alpha}\,dx.\ \text{Then,}
\displaystyle I=\int \frac{(2\cos^{2}x-1)-(2\cos^{2}\alpha-1)}{\cos x-\cos \alpha}\,dx
\displaystyle \Rightarrow\ I=\int \frac{2(\cos^{2}x-\cos^{2}\alpha)}{\cos x-\cos \alpha}\,dx
\displaystyle \Rightarrow\ I=2\int (\cos x+\cos \alpha)\,dx=2\int \cos x\,dx+2\int \cos \alpha\,dx
\displaystyle \Rightarrow\ I=2\int \cos x\,dx+2\cos \alpha\int 1\,dx=2\sin x+2x\cos \alpha+C

\displaystyle \textbf{Question 2: }\quad \text{Evaluate:}
\displaystyle \int \frac{\sin^{6}x+\cos^{6}x}{\sin^{2}x\cos^{2}x}\,dx \hspace{8.0cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int \frac{\sin^{6}x+\cos^{6}x}{\sin^{2}x\cos^{2}x}\,dx.\ \text{Then,}
\displaystyle I=\int \frac{(\sin^{2}x+\cos^{2}x)^{3}-3\sin^{2}x\cos^{2}x(\sin^{2}x+\cos^{2}x)}{\sin^{2}x\cos^{2}x}\,dx
\displaystyle \text{Using: }a^{3}+b^{3}=(a+b)^{3}-3ab(a+b)
\displaystyle \Rightarrow\ I=\int \frac{1-3\sin^{2}x\cos^{2}x}{\sin^{2}x\cos^{2}x}\,dx=\int \left\{\frac{1}{\sin^{2}x\cos^{2}x}-3\right\}\,dx
\displaystyle \Rightarrow\ I=\int \left\{\frac{\sin^{2}x+\cos^{2}x}{\sin^{2}x\cos^{2}x}-3\right\}\,dx=\int (\sec^{2}x+\mathrm{cosec}^{2}x-3)\,dx
\displaystyle =\tan x-\cot x-3x+C

\displaystyle \textbf{Question 3: }\quad \text{Evaluate:}
\displaystyle \int \tan^{-1}(\sec x+\tan x)\,dx,\ -\frac{\pi}{2}<x<\frac{\pi}{2} \hspace{5.0cm} \text{[CBSE 2003]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int \tan^{-1}(\sec x+\tan x)\,dx.\ \text{Then,}
\displaystyle I=\int \tan^{-1}\left(\frac{1+\sin x}{\cos x}\right)dx
\displaystyle \Rightarrow\ I=\int \tan^{-1}\left\{\frac{1-\cos\left(\frac{\pi}{2}+x\right)}{\sin\left(\frac{\pi}{2}+x\right)}\right\}dx
\displaystyle =\int \tan^{-1}\left\{\frac{2\sin^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)}{2\sin\left(\frac{\pi}{4}+\frac{x}{2}\right)\cos\left(\frac{\pi}{4}+\frac{x}{2}\right)}\right\}dx
\displaystyle \Rightarrow\ I=\int \tan^{-1}\left\{\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)\right\}dx
\displaystyle =\int \left(\frac{\pi}{4}+\frac{x}{2}\right)dx=\frac{\pi}{4}\int 1\,dx+\frac{1}{2}\int x\,dx
\displaystyle =\frac{\pi}{4}x+\frac{x^{2}}{4}+C

\displaystyle \textbf{Question 4: }\quad \text{Evaluate:}\ \int \tan^{-1}\sqrt{\frac{1-\sin x}{1+\sin x}}\,dx,\ -\frac{\pi}{2}<x<\frac{\pi}{2} \hspace{1.0cm} \text{[CBSE 2003, 2006]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int \tan^{-1}\sqrt{\frac{1-\sin x}{1+\sin x}}\,dx.\ \text{Then,}
\displaystyle I=\int \tan^{-1}\sqrt{\frac{1-\cos\left(\frac{\pi}{2}-x\right)}{1+\cos\left(\frac{\pi}{2}-x\right)}}\,dx
\displaystyle =\int \tan^{-1}\sqrt{\frac{2\sin^{2}\left(\frac{\pi}{4}-\frac{x}{2}\right)}{2\cos^{2}\left(\frac{\pi}{4}-\frac{x}{2}\right)}}\,dx
\displaystyle \Rightarrow\ I=\int \tan^{-1}\left\{\tan\left(\frac{\pi}{4}-\frac{x}{2}\right)\right\}dx
\displaystyle =\int \left(\frac{\pi}{4}-\frac{x}{2}\right)dx=\frac{\pi}{4}\int 1\,dx-\frac{1}{2}\int x\,dx
\displaystyle =\frac{\pi}{4}x-\frac{x^{2}}{4}+C

\displaystyle \textbf{Question 5: }\quad \text{Evaluate:}
\displaystyle \int \sec^{2}(7-4x)\,dx \hspace{8.0cm} \text{[CBSE 2009]}
\displaystyle \text{Answer:}
\displaystyle  \quad \int \sec^{2}(7-4x)\,dx=-\frac{1}{4}\tan(7-4x)+C

\displaystyle \textbf{Question 6: }\quad \text{Evaluate:}
\displaystyle \int \frac{x^{2}+1}{(x+1)^{2}}\,dx \hspace{8.0cm} \text{[CBSE 2006]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int \frac{x^{2}+1}{(x+1)^{2}}\,dx.\ \text{Then,}
\displaystyle I=\int \frac{x^{2}+1+2x-2x}{(x+1)^{2}}\,dx
\displaystyle \Rightarrow\ I=\int \frac{(x+1)^{2}-2x}{(x+1)^{2}}\,dx
\displaystyle \Rightarrow\ I=\int \left\{1-\frac{2x}{(x+1)^{2}}\right\}dx
\displaystyle \Rightarrow\ I=\int 1\cdot dx-2\int \frac{x}{(x+1)^{2}}\,dx
\displaystyle \Rightarrow\ I=\int 1\cdot dx-2\int \frac{(x+1)-1}{(x+1)^{2}}\,dx
\displaystyle \Rightarrow\ I=\int 1\cdot dx-2\int \left\{\frac{1}{x+1}-\frac{1}{(x+1)^{2}}\right\}dx
\displaystyle \Rightarrow\ I=\int 1\cdot dx-2\int \frac{1}{x+1}\,dx+2\int \frac{1}{(x+1)^{2}}\,dx
\displaystyle =x-2\log|x+1|-\frac{2}{x+1}+C

\displaystyle \textbf{Question 7: }\quad \text{Evaluate:}
\displaystyle (i)\ \int \sin^{4}x\,dx \hspace{8.0cm} \text{[CBSE 2000, 2004]}
\displaystyle (ii)\ \int \cos^{4}x\,dx \hspace{8.0cm} \text{[CBSE 2000, 2003]}
\displaystyle \text{Answer:}
\displaystyle  \quad (i)\ \text{Let }I=\int \sin^{4}x\,dx.\ \text{Then,}
\displaystyle I=\int \left(\frac{1-\cos 2x}{2}\right)^{2}dx\qquad \left[\because\ \sin^{2}\theta=\frac{1-\cos 2\theta}{2}\right]
\displaystyle \Rightarrow\ I=\frac{1}{4}\int (1-2\cos 2x+\cos^{2}2x)\,dx
\displaystyle \Rightarrow\ I=\frac{1}{4}\int \left(1-2\cos 2x+\frac{1+\cos 4x}{2}\right)dx
\displaystyle \Rightarrow\ I=\frac{1}{8}\int (2-4\cos 2x+1+\cos 4x)\,dx
\displaystyle \Rightarrow\ I=\frac{1}{8}\int (3-4\cos 2x+\cos 4x)\,dx=\frac{1}{8}\left(3x-2\sin 2x+\frac{\sin 4x}{4}\right)+C
\displaystyle (ii)\ \text{Let }I=\int \cos^{4}x\,dx.\ \text{Then,}
\displaystyle I=\int \left(\frac{1+\cos 2x}{2}\right)^{2}dx\qquad \left[\because\ \cos^{2}\theta=\frac{1+\cos 2\theta}{2}\right]
\displaystyle \Rightarrow\ I=\frac{1}{4}\int (1+2\cos 2x+\cos^{2}2x)\,dx
\displaystyle \Rightarrow\ I=\frac{1}{4}\int \left(1+2\cos 2x+\frac{1+\cos 4x}{2}\right)dx
\displaystyle \Rightarrow\ I=\frac{1}{8}\int (3+4\cos 2x+\cos 4x)\,dx=\frac{1}{8}\left(3x+2\sin 2x+\frac{\sin 4x}{4}\right)+C

\displaystyle \textbf{Question 8: }\quad \text{Evaluate:}
\displaystyle \int \sin x\sin 2x\sin 3x\,dx \hspace{8.0cm} \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int \sin x\sin 2x\sin 3x\,dx.\ \text{Then,}
\displaystyle I=\frac{1}{2}\int (2\sin 2x\sin x)\sin 3x\,dx
\displaystyle \Rightarrow\ I=\frac{1}{2}\int (\cos x-\cos 3x)\sin 3x\,dx
\displaystyle \Rightarrow\ I=\frac{1}{4}\int (2\sin 3x\cos x-2\sin 3x\cos 3x)\,dx
\displaystyle \Rightarrow\ I=\frac{1}{4}\int (\sin 4x+\sin 2x-\sin 6x)\,dx
\displaystyle =\frac{1}{4}\left(-\frac{\cos 4x}{4}-\frac{\cos 2x}{2}+\frac{\cos 6x}{6}\right)+C

\displaystyle \textbf{Question 9: }\quad \text{Evaluate:}
\displaystyle \int \frac{\sin x}{\sin(x-a)}\,dx \hspace{8.0cm} \text{[CBSE 2004]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int \frac{\sin x}{\sin(x-a)}\,dx.\ \text{Then,}
\displaystyle I=\int \frac{\sin\{(x-a)+a\}}{\sin(x-a)}\,dx
\displaystyle \Rightarrow\ I=\int \frac{\sin(x-a)\cos a+\cos(x-a)\sin a}{\sin(x-a)}\,dx
\displaystyle \Rightarrow\ I=\int \{\cos a+\cot(x-a)\sin a\}\,dx
\displaystyle \Rightarrow\ I=\cos a\int 1\cdot dx+\sin a\int \cot(x-a)\,dx=x\cos a+\sin a\log|\sin(x-a)|+C

\displaystyle \textbf{Question 10: }\quad \text{Evaluate:}
\displaystyle (i)\ \int \frac{\sin 2x}{a^{2}\sin^{2}x+b^{2}\cos^{2}x}\,dx \hspace{6.0cm} \text{[CBSE 2005]}
\displaystyle (ii)\ \int \frac{\sin x-x\cos x}{x(x+\sin x)}\,dx \hspace{8.0cm} \text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle (i)\ \text{Let }I=\int \frac{\sin 2x}{a^{2}\sin^{2}x+b^{2}\cos^{2}x}\,dx
\displaystyle \text{Let }a^{2}\sin^{2}x+b^{2}\cos^{2}x=t.\ \text{Then,}
\displaystyle d(a^{2}\sin^{2}x+b^{2}\cos^{2}x)=dt
\displaystyle \Rightarrow\ (a^{2}-b^{2})\sin 2x\,dx=dt
\displaystyle \Rightarrow\ dx=\frac{dt}{(a^{2}-b^{2})\sin 2x}
\displaystyle \text{Putting }a^{2}\sin^{2}x+b^{2}\cos^{2}x=t\ \text{and }dx=\frac{dt}{(a^{2}-b^{2})\sin 2x},\ \text{we get}
\displaystyle I=\int \frac{\sin 2x}{t}\times \frac{dt}{2\sin x\cos x\,(a^{2}-b^{2})}
\displaystyle \Rightarrow\ I=\frac{1}{(a^{2}-b^{2})}\int \frac{1}{t}\,dt
\displaystyle =\frac{1}{(a^{2}-b^{2})}\log|t|+C
\displaystyle =\frac{1}{(a^{2}-b^{2})}\log|a^{2}\sin^{2}x+b^{2}\cos^{2}x|+C
\displaystyle (ii)\ \text{Let }I=\int \frac{\sin x-x\cos x}{x(x+\sin x)}\,dx.\ \text{Then,}
\displaystyle I=\int \frac{(x+\sin x)-x-x\cos x}{x(x+\sin x)}\,dx
\displaystyle \Rightarrow\ I=\int \frac{(x+\sin x)-x(1+\cos x)}{x(x+\sin x)}\,dx
\displaystyle \Rightarrow\ I=\int \left\{\frac{1}{x}-\frac{1+\cos x}{x+\sin x}\right\}dx
\displaystyle \Rightarrow\ I=\log|x|-\log(x+\sin x)+C
\displaystyle \Rightarrow\ I=\log\left|\frac{x}{x+\sin x}\right|+C

\displaystyle \textbf{Question 11: }\quad \text{Evaluate:}
\displaystyle \int \frac{(1+\log x)^{2}}{x}\,dx \hspace{8.0cm} \text{[CBSE 2009]}
\displaystyle \text{Answer:}
\displaystyle  \quad \text{Let }I=\int \frac{(1+\log x)^{2}}{x}\,dx
\displaystyle \text{Let }1+\log x=t.\ \text{Then, }d(1+\log x)=dt\Rightarrow \frac{1}{x}dx=dt\Rightarrow dx=x\,dt
\displaystyle \text{Putting }1+\log x=t\ \text{and }dx=x\,dt,\ \text{we get}
\displaystyle I=\int \frac{t^{2}}{x}\times x\,dt=\int t^{2}\,dt=\frac{t^{3}}{3}+C=\frac{(1+\log x)^{3}}{3}+C

\displaystyle \textbf{Question 12: }\quad \text{Evaluate:}\ \int \frac{x^{2}-3x+1}{\sqrt{1-x^{2}}}\,dx \hspace{3.0cm} \text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }I=\int \frac{x^{2}-3x+1}{\sqrt{1-x^{2}}}\,dx\ \text{and }x=\sin\theta.\ \text{Then, }dx=d(\sin\theta)=\cos\theta\,d\theta
\displaystyle \therefore\ I=\int \frac{\sin^{2}\theta-3\sin\theta+1}{\sqrt{1-\sin^{2}\theta}}\cos\theta\,d\theta=\int (\sin^{2}\theta-3\sin\theta+1)\,d\theta
\displaystyle \Rightarrow\ I=\int \left(\frac{1-\cos 2\theta}{2}-3\sin\theta+1\right)d\theta=\frac{1}{2}\int (3-6\sin\theta-\cos 2\theta)\,d\theta
\displaystyle \Rightarrow\ I=\frac{1}{2}\left(3\theta+6\cos\theta-\frac{1}{2}\sin 2\theta\right)+C
\displaystyle \Rightarrow\ I=\frac{1}{2}\left(3\theta+6\sqrt{1-\sin^{2}\theta}-\sin\theta\sqrt{1-\sin^{2}\theta}\right)+C
\displaystyle \Rightarrow\ I=\frac{1}{2}\left(3\sin^{-1}x+6\sqrt{1-x^{2}}-x\sqrt{1-x^{2}}\right)+C

\displaystyle \textbf{Question 13: }\quad \text{Evaluate:}
\displaystyle \int \frac{1}{x^{2}+4x+8}\,dx \hspace{8.0cm} \text{[CBSE 2002, 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int \frac{1}{x^{2}+4x+8}\,dx.\ \text{Then,}
\displaystyle I=\int \frac{1}{x^{2}+4x+4+4}\,dx=\int \frac{1}{(x+2)^{2}+2^{2}}\,dx =\frac{1}{2}\tan^{-1}\left(\frac{x+2}{2}\right)+C

\displaystyle \textbf{Question 14: }\quad \text{Evaluate:}
\displaystyle \int \frac{1}{x(x^{n}+1)}\,dx \hspace{8.0cm} \text{[CBSE 2000C]}
\displaystyle \text{Answer:}
\displaystyle  \text{We have,}
\displaystyle I=\int \frac{1}{x(x^{n}+1)}\,dx=\int \frac{x^{n-1}}{x^{n}(x^{n}+1)}\,dx
\displaystyle \text{Let }x^{n}+1=t.\ \text{Then, }d(x^{n}+1)=dt\Rightarrow n x^{n-1}dx=dt
\displaystyle \Rightarrow\ dx=\frac{dt}{n x^{n-1}}
\displaystyle \therefore\ I=\int \frac{1}{n x^{n}t}\,dt=\frac{1}{n}\int \frac{1}{(t-1)t}\,dt
\displaystyle \left[\because\ x^{n}+1=t\ \therefore\ x^{n}=t-1\right]
\displaystyle \Rightarrow\ I=\frac{1}{n}\int \frac{1}{t^{2}-t}\,dt=\frac{1}{n}\int \frac{dt}{t^{2}-t+\frac{1}{4}-\frac{1}{4}}
\displaystyle =\frac{1}{n}\int \frac{dt}{\left(t-\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}
\displaystyle \Rightarrow\ I=\frac{1}{n}\times\frac{1}{2\left(\frac{1}{2}\right)} \log\left|\frac{t-\frac{1}{2}-\frac{1}{2}}{t-\frac{1}{2}+\frac{1}{2}}\right|+C
\displaystyle =\frac{1}{n}\log\left|\frac{t-1}{t}\right|+C=\frac{1}{n}\log\left|\frac{x^{n}}{x^{n}+1}\right|+C

\displaystyle \textbf{Question 15: }\quad \text{Evaluate:}
\displaystyle \int \frac{1}{\sqrt{(x-a)(x-b)}}\,dx \hspace{8.0cm} \text{[CBSE 2001]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }I=\int \frac{1}{\sqrt{(x-a)(x-b)}}\,dx.\ \text{Then,}
\displaystyle I=\int \frac{1}{\sqrt{x^{2}-x(a+b)+ab}}\,dx
\displaystyle \Rightarrow\ I=\int \frac{1}{\sqrt{x^{2}-x(a+b)+\left(\frac{a+b}{2}\right)^{2}-\left(\frac{a+b}{2}\right)^{2}+ab}}\,dx
\displaystyle \Rightarrow\ I=\int \frac{1}{\sqrt{\left\{x-\left(\frac{a+b}{2}\right)\right\}^{2}-\left(\frac{a-b}{2}\right)^{2}}}\,dx
\displaystyle \Rightarrow\ I=\log\left|x-\left(\frac{a+b}{2}\right)+\sqrt{\left\{x-\left(\frac{a+b}{2}\right)\right\}^{2}-\left(\frac{a-b}{2}\right)^{2}}\right|+C
\displaystyle \Rightarrow\ I=\log\left|\frac{2x-a-b}{2}+\sqrt{(x-a)(x-b)}\right|+C
\displaystyle \Rightarrow\ I=\log\left|\frac{(x-a)+(x-b)+2\sqrt{(x-a)(x-b)}}{2}\right|+C
\displaystyle \Rightarrow\ I=\log\left|\left(\sqrt{x-a}+\sqrt{x-b}\right)^{2}\right|-\log 2+C
\displaystyle \Rightarrow\ I=2\log\left|\sqrt{x-a}+\sqrt{x-b}\right|+C_{1},\ \text{where }C_{1}=C-\log 2

\displaystyle \textbf{Question 16:}\quad \text{Evaluate:}
\displaystyle (i)\ \int \frac{2x}{\sqrt{1-x^{2}-x^{4}}}\,dx \hspace{8.0cm} \text{[CBSE 2005]}
\displaystyle (ii)\ \int \frac{e^{x}}{\sqrt{5-4e^{x}-e^{2x}}}\,dx \hspace{8.0cm} \text{[CBSE 2009]}
\displaystyle (iii)\ \int \sqrt{\frac{x}{a^{3}-x^{3}}}\,dx \hspace{8.0cm} \text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle (i)\ \text{Let }I=\int \frac{2x}{\sqrt{1-x^{2}-x^{4}}}\,dx
\displaystyle =\int \frac{2x}{\sqrt{1-x^{2}-(x^{2})^{2}}}\,dx
\displaystyle \text{Let }x^{2}=t.\ \text{Then, }d(x^{2})=dt\Rightarrow 2x\,dx=dt
\displaystyle \Rightarrow\ dx=\frac{dt}{2x}
\displaystyle \therefore\ I=\int \frac{1}{\sqrt{1-t-t^{2}}}\,dt
\displaystyle =\int \frac{1}{\sqrt{-(t^{2}+t-1)}}\,dt
\displaystyle =\int \frac{1}{\sqrt{-\left(t^{2}+t+\frac{1}{4}-\frac{1}{4}-1\right)}}\,dt
\displaystyle =\int \frac{1}{\sqrt{\frac{5}{4}-\left(t+\frac{1}{2}\right)^{2}}}\,dt
\displaystyle =\int \frac{1}{\sqrt{\left(\frac{\sqrt{5}}{2}\right)^{2}-\left(t+\frac{1}{2}\right)^{2}}}\,dt
\displaystyle =\sin^{-1}\left(\frac{t+\frac{1}{2}}{\frac{\sqrt{5}}{2}}\right)+C
\displaystyle =\sin^{-1}\left(\frac{2t+1}{\sqrt{5}}\right)+C =\sin^{-1}\left(\frac{2x^{2}+1}{\sqrt{5}}\right)+C
\displaystyle (ii)\ \text{Let }I=\int \frac{e^{x}}{\sqrt{5-4e^{x}-e^{2x}}}\,dx
\displaystyle =\int \frac{e^{x}}{\sqrt{5-4e^{x}-(e^{x})^{2}}}\,dx
\displaystyle \text{Let }e^{x}=t.\ \text{Then, }d(e^{x})=dt\Rightarrow e^{x}dx=dt
\displaystyle \Rightarrow\ dx=\frac{dt}{e^{x}}
\displaystyle \therefore\ I=\int \frac{1}{\sqrt{5-4t-t^{2}}}\,dt
\displaystyle =\int \frac{1}{\sqrt{-(t^{2}+4t-5)}}\,dt
\displaystyle =\int \frac{1}{\sqrt{-\left(t^{2}+4t+4-9\right)}}\,dt
\displaystyle =\int \frac{1}{\sqrt{9-(t+2)^{2}}}\,dt =\sin^{-1}\left(\frac{t+2}{3}\right)+C
\displaystyle =\sin^{-1}\left(\frac{e^{x}+2}{3}\right)+C
\displaystyle (iii)\ \text{Let }I=\int \sqrt{\frac{x}{a^{3}-x^{3}}}\,dx
\displaystyle =\int \frac{\sqrt{x}}{\sqrt{(a^{\frac{3}{2}})^{2}-(x^{\frac{3}{2}})^{2}}}\,dx
\displaystyle \text{Let }x^{\frac{3}{2}}=t.\ \text{Then, }d(x^{\frac{3}{2}})=dt \Rightarrow \frac{3}{2}x^{\frac{1}{2}}dx=dt\Rightarrow dx=\frac{2}{3\sqrt{x}}\,dt
\displaystyle \therefore\ I=\frac{2}{3}\int \frac{1}{\sqrt{(a^{\frac{3}{2}})^{2}-t^{2}}}\,dt
\displaystyle =\frac{2}{3}\sin^{-1}\left(\frac{t}{a^{\frac{3}{2}}}\right)+C =\frac{2}{3}\sin^{-1}\left(\frac{x^{\frac{3}{2}}}{a^{\frac{3}{2}}}\right)+C

\displaystyle \textbf{Question 17: }\quad \text{Evaluate:}
\displaystyle \int \frac{x+2}{\sqrt{x^{2}+5x+6}}\,dx \hspace{8.0cm} \text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }x+2=\lambda \frac{d}{dx}(x^{2}+5x+6)+\mu.\ \text{Then, }x+2=\lambda(2x+5)+\mu
\displaystyle \text{Comparing the coefficients of like powers of }x,\ \text{we get}
\displaystyle 1=2\lambda\ \text{and }5\lambda+\mu=2\ \Rightarrow\ \lambda=\frac{1}{2}\ \text{and }\mu=-\frac{1}{2}
\displaystyle \therefore\ I=\int \frac{x+2}{\sqrt{x^{2}+5x+6}}\,dx
\displaystyle \Rightarrow\ I=\int \frac{\frac{1}{2}(2x+5)-\frac{1}{2}}{\sqrt{x^{2}+5x+6}}\,dx
\displaystyle \Rightarrow\ I=\frac{1}{2}\int \frac{2x+5}{\sqrt{x^{2}+5x+6}}\,dx-\frac{1}{2}\int \frac{1}{\sqrt{x^{2}+5x+6}}\,dx
\displaystyle \Rightarrow\ I=\frac{1}{2}\int \frac{dt}{\sqrt{t}}-\frac{1}{2}\int \frac{1}{\sqrt{\left(x+\frac{5}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}}\,dx,\ \text{where }t=x^{2}+5x+6
\displaystyle \Rightarrow\ I=\sqrt{t}-\frac{1}{2}\log\left|\left(x+\frac{5}{2}\right)+\sqrt{x^{2}+5x+6}\right|+C
\displaystyle \Rightarrow\ I=\sqrt{x^{2}+5x+6}-\frac{1}{2}\log\left|\left(x+\frac{5}{2}\right)+\sqrt{x^{2}+5x+6}\right|+C

\displaystyle \textbf{Question 18: }\quad \text{Evaluate:}
\displaystyle \int (\sin^{-1}x)^{2}\,dx \hspace{8.0cm} \text{[CBSE 2004]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\sin^{-1}x=t.\ \text{Then, }x=\sin t\Rightarrow dx=\cos t\,dt
\displaystyle \therefore\ I=\int (\sin^{-1}x)^{2}\,dx
\displaystyle \Rightarrow\ I=\int t^{2}\cos t\,dt=t^{2}(\sin t)-\int 2t\sin t\,dt=t^{2}\sin t-2\int t\sin t\,dt
\displaystyle \Rightarrow\ I=t^{2}\sin t-2\left\{t(-\cos t)-\int 1\times(-\cos t)\,dt\right\}
\displaystyle \Rightarrow\ I=t^{2}\sin t-2\left\{-t\cos t+\int \cos t\,dt\right\}
\displaystyle \Rightarrow\ I=t^{2}\sin t-2(-t\cos t+\sin t)+C
\displaystyle \Rightarrow\ I=t^{2}\sin t-2\left\{-t\sqrt{1-\sin^{2}t}+\sin t\right\}+C
\displaystyle \Rightarrow\ I=x(\sin^{-1}x)^{2}-2\left\{-\sqrt{1-x^{2}}\sin^{-1}x+x\right\}+C

\displaystyle \textbf{Question 19: }\quad \text{Evaluate:}
\displaystyle \int x\sin^{-1}x\,dx \hspace{8.0cm} \text{[CBSE 2000, 2009]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int x\sin^{-1}x\,dx.\ \text{Then,}
\displaystyle I=(\sin^{-1}x)\frac{x^{2}}{2}-\int \frac{1}{\sqrt{1-x^{2}}}\times\frac{x^{2}}{2}\,dx
\displaystyle \Rightarrow\ I=\frac{x^{2}}{2}\sin^{-1}x+\frac{1}{2}\int \frac{-x^{2}}{\sqrt{1-x^{2}}}\,dx
\displaystyle =\frac{x^{2}}{2}\sin^{-1}x+\frac{1}{2}\int \frac{1-x^{2}-1}{\sqrt{1-x^{2}}}\,dx
\displaystyle \Rightarrow\ I=\frac{x^{2}}{2}\sin^{-1}x+\frac{1}{2}\left\{\int \frac{1-x^{2}}{\sqrt{1-x^{2}}}\,dx-\int \frac{1}{\sqrt{1-x^{2}}}\,dx\right\}
\displaystyle \Rightarrow\ I=\frac{x^{2}}{2}\sin^{-1}x+\frac{1}{2}\left\{\int \sqrt{1-x^{2}}\,dx-\int \frac{1}{\sqrt{1-x^{2}}}\,dx\right\}
\displaystyle \Rightarrow\ I=\frac{x^{2}}{2}\sin^{-1}x+\frac{1}{2}\left\{\frac{1}{2}x\sqrt{1-x^{2}}+\frac{1}{2}\sin^{-1}x\right\}-\sin^{-1}x+C
\displaystyle \Rightarrow\ I=\frac{1}{2}x^{2}\sin^{-1}x+\frac{1}{4}x\sqrt{1-x^{2}}-\frac{1}{4}\sin^{-1}x+C

\displaystyle \textbf{Question 20: }\quad \text{Evaluate:}\ \int \frac{\sqrt{x^{2}+1}\{\log(x^{2}+1)-2\log x\}}{x^{4}}\,dx\qquad \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let}
\displaystyle I=\int \frac{\sqrt{x^{2}+1}\{\log(x^{2}+1)-2\log x\}}{x^{4}}\,dx
\displaystyle \Rightarrow\ I=\int \frac{\sqrt{x^{2}+1}}{x^{4}}\{\log(x^{2}+1)-\log x^{2}\}\,dx
\displaystyle \Rightarrow\ I=\int \sqrt{1+\frac{1}{x^{2}}}\ \log\left(1+\frac{1}{x^{2}}\right)\frac{1}{x^{3}}\,dx
\displaystyle \text{Let }1+\frac{1}{x^{2}}=t.\ \text{Then, }d\left(1+\frac{1}{x^{2}}\right)=dt\Rightarrow -\frac{2}{x^{3}}dx=dt
\displaystyle \Rightarrow\ \frac{1}{x^{3}}dx=-\frac{1}{2}dt
\displaystyle \therefore\ I=-\frac{1}{2}\int \sqrt{t}\ \log t\,dt
\displaystyle \Rightarrow\ I=-\frac{1}{2}\left\{\frac{2}{3}(\log t)t^{\frac{3}{2}}-\frac{2}{3}\int \frac{1}{t}\times t^{\frac{3}{2}}\,dt\right\} =-\frac{1}{2}\left\{\frac{2}{3}(\log t)t^{\frac{3}{2}}-\frac{4}{9}t^{\frac{3}{2}}\right\}+C
\displaystyle \Rightarrow\ I=-\frac{1}{3}t^{\frac{3}{2}}\left\{\log t-\frac{2}{3}\right\}+C =-\frac{1}{3}\left(1+\frac{1}{x^{2}}\right)^{\frac{3}{2}}\left\{\log\left(1+\frac{1}{x^{2}}\right)-\frac{2}{3}\right\}+C

\displaystyle \textbf{Question 21: }\quad \text{Evaluate:}
\displaystyle \int \left\{\log(\log x)+\frac{1}{(\log x)^{2}}\right\}dx \hspace{5.0cm} \text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int \left\{\log(\log x)+\frac{1}{(\log x)^{2}}\right\}dx.\ \text{Let }\log x=t.\ \text{Then, }x=e^{t}\Rightarrow dx=d(e^{t})=e^{t}\,dt
\displaystyle \therefore\ I=\int \left\{\log t+\frac{1}{t^{2}}\right\}e^{t}\,dt
\displaystyle \Rightarrow\ I=\int \left\{\log t+\frac{1}{t}-\frac{1}{t}+\frac{1}{t^{2}}\right\}e^{t}\,dt
\displaystyle \Rightarrow\ I=\int \left(\log t+\frac{1}{t}\right)e^{t}\,dt+\int \left(-\frac{1}{t}+\frac{1}{t^{2}}\right)e^{t}\,dt
\displaystyle \Rightarrow\ I=\int e^{t}\log t\,dt+\int e^{t}\frac{1}{t}\,dt+\int e^{t}\left(-\frac{1}{t}\right)dt+\int e^{t}\frac{1}{t^{2}}\,dt
\displaystyle \Rightarrow\ I=(\log t)e^{t}-\int \frac{1}{t}\cdot e^{t}\,dt+\int e^{t}\frac{1}{t}\,dt+\left(-\frac{1}{t}\right)e^{t}-\int \frac{1}{t^{2}}\cdot e^{t}\,dt+\int e^{t}\frac{1}{t^{2}}\,dt+C
\displaystyle \Rightarrow\ I=e^{t}\cdot\log t-\frac{1}{t}e^{t}+C=x\log(\log x)-\frac{x}{\log x}+C

\displaystyle \textbf{Question 22: }\quad \text{Evaluate:}\ \int e^{2x}\left(\frac{1+\sin 2x}{1+\cos 2x}\right)dx \hspace{3.0cm} \text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }I=\int e^{2x}\left(\frac{1+\sin 2x}{1+\cos 2x}\right)dx
\displaystyle \Rightarrow\ I=\int e^{2x}\left\{\frac{1+2\sin x\cos x}{2\cos^{2}x}\right\}dx
\displaystyle \Rightarrow\ I=\int e^{2x}\left\{\frac{1}{2}\sec^{2}x+\tan x\right\}dx=\int e^{2x}\left\{2\left(\frac{1}{2}\tan x\right)'+\frac{1}{2}\sec^{2}x\right\}dx
\displaystyle \Rightarrow\ I=\int e^{2x}\cdot\tan x\,dx+\frac{1}{2}\int e^{2x}\cdot\sec^{2}x\,dx
\displaystyle \Rightarrow\ I=(\tan x)\frac{e^{2x}}{2}-\int \sec^{2}x\cdot\frac{e^{2x}}{2}\,dx+\frac{1}{2}\int e^{2x}\sec^{2}x\,dx+C
\displaystyle \Rightarrow\ I=\frac{1}{2}e^{2x}\tan x-\frac{1}{2}\int e^{2x}\sec^{2}x\,dx+\frac{1}{2}\int e^{2x}\sec^{2}x\,dx+C=\frac{1}{2}e^{2x}\tan x+C

\displaystyle \textbf{Question 23: }\quad \text{Prove that:}
\displaystyle \int e^{ax}\sin bx\,dx=\frac{e^{ax}}{a^{2}+b^{2}}(a\sin bx-b\cos bx)+C \hspace{3.0cm} \text{[CBSE 2002]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }I=\int e^{ax}\sin bx\,dx.\ \text{Then,}
\displaystyle I=\int e^{ax}\sin bx\,dx
\displaystyle \Rightarrow\ I=-e^{ax}\frac{\cos bx}{b}-\int ae^{ax}\left(-\frac{\cos bx}{b}\right)dx
\displaystyle \Rightarrow\ I=-\frac{1}{b}e^{ax}\cos bx+\frac{a}{b}\int e^{ax}\cos bx\,dx
\displaystyle \Rightarrow\ I=-\frac{1}{b}e^{ax}\cos bx+\frac{a}{b}\left\{\frac{e^{ax}\sin bx}{b}-\int ae^{ax}\frac{\sin bx}{b}\,dx\right\}
\displaystyle \Rightarrow\ I=-\frac{1}{b}e^{ax}\cos bx+\frac{a}{b^{2}}e^{ax}\sin bx-\frac{a^{2}}{b^{2}}\int e^{ax}\sin bx\,dx
\displaystyle \Rightarrow\ I=-\frac{1}{b}e^{ax}\cos bx+\frac{a}{b^{2}}e^{ax}\sin bx-\frac{a^{2}}{b^{2}}I
\displaystyle \Rightarrow\ I+I\cdot\frac{a^{2}}{b^{2}}=\frac{e^{ax}}{b^{2}}(a\sin bx-b\cos bx)
\displaystyle \Rightarrow\ I\left(\frac{a^{2}+b^{2}}{b^{2}}\right)=\frac{e^{ax}}{b^{2}}(a\sin bx-b\cos bx)
\displaystyle \Rightarrow\ I=\frac{e^{ax}}{a^{2}+b^{2}}(a\sin bx-b\cos bx)+C

\displaystyle \textbf{Question 24: }\quad \text{Evaluate:}
\displaystyle  \int (3x-2)\sqrt{x^{2}+x+1}\,dx \hspace{5.0cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }3x-2=\lambda\frac{d}{dx}(x^{2}+x+1)+\mu\ \text{i.e. }3x-2=\lambda(2x+1)+\mu
\displaystyle \text{Comparing the coefficients of like powers of }x,\ \text{we get}
\displaystyle 2\lambda=3\ \text{and }\lambda+\mu=-2\Rightarrow \lambda=\frac{3}{2}\ \text{and }\mu=-\frac{7}{2}
\displaystyle \therefore\ I=\int (3x-2)\sqrt{x^{2}+x+1}\,dx
\displaystyle \Rightarrow\ I=\int \left\{\frac{3}{2}(2x+1)-\frac{7}{2}\right\}\sqrt{x^{2}+x+1}\,dx
\displaystyle \Rightarrow\ I=\frac{3}{2}\int (2x+1)\sqrt{x^{2}+x+1}\,dx-\frac{7}{2}\int \sqrt{x^{2}+x+1}\,dx
\displaystyle \Rightarrow\ I=\frac{3}{2}\int \sqrt{t}\,dt-\frac{7}{2}\int \sqrt{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\,dx,\ \text{where }t=x^{2}+x+1
\displaystyle \Rightarrow\ I=t^{\frac{3}{2}}-\frac{7}{4}\left\{\left(x+\frac{1}{2}\right)\sqrt{x^{2}+x+1}+\left(\frac{\sqrt{3}}{2}\right)^{2}\log\left|x+\frac{1}{2}+\sqrt{x^{2}+x+1}\right|\right\}+C
\displaystyle \Rightarrow\ I=(x^{2}+x+1)^{\frac{3}{2}}-\frac{7}{2}\left\{\left(x+\frac{1}{2}\right)\sqrt{x^{2}+x+1}+\frac{3}{8}\log\left|x+\frac{1}{2}+\sqrt{x^{2}+x+1}\right|\right\}+C

\displaystyle \textbf{Question 25: }\quad \text{Evaluate:}
\displaystyle \int \frac{2x-1}{(x-1)(x+2)(x-3)}\,dx \hspace{5.0cm} \text{[CBSE 2005]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\frac{2x-1}{(x-1)(x+2)(x-3)}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{x-3}\qquad \cdots(i)
\displaystyle \Rightarrow\ 2x-1=A(x+2)(x-3)+B(x-1)(x-3)+C(x-1)(x+2)\qquad \cdots(ii)
\displaystyle \text{Putting }x+2=0\ \text{or, }x=-2\ \text{in (ii), we get}
\displaystyle -5=B(-3)(-5)\Rightarrow B=-\frac{1}{3}
\displaystyle \text{Putting }x-3=0\ \text{or, }x=3\ \text{in (ii), we get}
\displaystyle 5=C(2)(5)\Rightarrow C=\frac{1}{2}
\displaystyle \text{Putting }x-1=0\ \text{or, }x=1\ \text{in (ii), we get}
\displaystyle 1=A(3)(-2)\Rightarrow A=-\frac{1}{6}
\displaystyle \text{Substituting the values of }A,B\ \text{and }C\ \text{in (i), we obtain}
\displaystyle \frac{2x-1}{(x-1)(x+2)(x-3)}=-\frac{1}{6}\cdot\frac{1}{x-1}-\frac{1}{3}\cdot\frac{1}{x+2}+\frac{1}{2}\cdot\frac{1}{x-3}
\displaystyle \therefore\ I=\int \frac{2x-1}{(x-1)(x+2)(x-3)}\,dx
\displaystyle \Rightarrow\ I=-\frac{1}{6}\int \frac{1}{x-1}\,dx-\frac{1}{3}\int \frac{1}{x+2}\,dx+\frac{1}{2}\int \frac{1}{x-3}\,dx
\displaystyle \Rightarrow\ I=-\frac{1}{6}\log|x-1|-\frac{1}{3}\log|x+2|+\frac{1}{2}\log|x-3|+C

\displaystyle \textbf{Question 26: }\quad \text{Evaluate:}
\displaystyle  \int \frac{1}{\sin x-\sin 2x}\,dx \hspace{8.0cm} \text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle  \text{We have,}
\displaystyle I=\int \frac{1}{\sin x-\sin 2x}\,dx
\displaystyle \Rightarrow\ I=\int \frac{1}{(\sin x-2\sin x\cos x)}\,dx=\int \frac{1}{\sin x(1-2\cos x)}\,dx=\int \frac{\sin x}{\sin^{2}x(1-2\cos x)}\,dx
\displaystyle \Rightarrow\ I=\int \frac{\sin x}{(1-\cos^{2}x)(1-2\cos x)}\,dx
\displaystyle \text{Putting }\cos x=t,\ \text{and }-\sin x\,dx=dt\ \text{or, }\sin x\,dx=-dt,\ \text{we get}
\displaystyle I=\int \frac{-dt}{(1-t^{2})(1-2t)}=\int \frac{-1}{(1-t)(1+t)(1-2t)}\,dt
\displaystyle \text{Let }\frac{-1}{(1-t)(1+t)(1-2t)}=\frac{A}{1-t}+\frac{B}{1+t}+\frac{C}{1-2t}.\ \text{Then,}
\displaystyle -1=A(1+t)(1-2t)+B(1-t)(1-2t)+C(1-t)(1+t)\qquad \cdots(i)
\displaystyle \text{Putting }t+1=0\ \text{or, }t=-1\ \text{in (i), we get}
\displaystyle -1=6B\Rightarrow B=-\frac{1}{6}
\displaystyle \text{Putting }1-t=0\ \text{or, }t=1\ \text{in (i), we get}
\displaystyle -1=-2A\Rightarrow A=\frac{1}{2}
\displaystyle \text{Putting }1-2t=0\ \text{or, }t=\frac{1}{2}\ \text{in (i), we get}
\displaystyle -1=C\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)\Rightarrow C=-\frac{4}{3}
\displaystyle \therefore\ \frac{-1}{(1-t)(1+t)(1-2t)}=\frac{1}{2}\cdot\frac{1}{1-t}-\frac{1}{6}\cdot\frac{1}{1+t}-\frac{4}{3}\cdot\frac{1}{1-2t}
\displaystyle \Rightarrow\ I=\int \frac{-dt}{(1-t)(1+t)(1-2t)}
\displaystyle \Rightarrow\ I=\frac{1}{2}\int \frac{1}{1-t}\,dt-\frac{1}{6}\int \frac{1}{1+t}\,dt-\frac{4}{3}\int \frac{1}{1-2t}\,dt
\displaystyle \Rightarrow\ I=-\frac{1}{2}\log|1-t|-\frac{1}{6}\log|1+t|-\frac{4}{3}\times\left(-\frac{1}{2}\right)\log|1-2t|+C
\displaystyle \Rightarrow\ I=-\frac{1}{2}\log|1-\cos x|-\frac{1}{6}\log|1+\cos x|+\frac{2}{3}\log|1-2\cos x|+C

\displaystyle \textbf{Question 27: }\quad \text{Evaluate:}
\displaystyle (i)\ \int \frac{3x+1}{(x-2)^{2}(x+2)}\,dx \hspace{8.0cm} \text{[CBSE 2007]}
\displaystyle (ii)\ \int \frac{x^{2}+1}{(x-1)^{2}(x+3)}\,dx \hspace{8.0cm} \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle  (i)\ \text{Let }\frac{3x+1}{(x-2)^{2}(x+2)}=\frac{A}{x-2}+\frac{B}{(x-2)^{2}}+\frac{C}{x+2}\qquad \cdots(i)
\displaystyle \Rightarrow\ 3x+1=A(x-2)(x+2)+B(x+2)+C(x-2)^{2}\qquad \cdots(ii)
\displaystyle \text{Putting }x-2=0\ \text{i.e. }x=2\ \text{in (ii), we get}
\displaystyle 7=4B\Rightarrow B=\frac{7}{4}
\displaystyle \text{Putting }x+2=0\ \text{i.e. }x=-2\ \text{in (ii), we get}
\displaystyle -5=16C\Rightarrow C=-\frac{5}{16}
\displaystyle \text{Comparing coefficients of }x^{2}\ \text{on both sides of the identity (ii), we get}
\displaystyle A+C=0\Rightarrow A=-C=\frac{5}{16}
\displaystyle \text{Substituting the values of }A,B\ \text{and }C\ \text{in (i), we get}
\displaystyle \frac{3x+1}{(x-2)^{2}(x+2)}=\frac{5}{16}\cdot\frac{1}{x-2}+\frac{7}{4}\cdot\frac{1}{(x-2)^{2}}-\frac{5}{16}\cdot\frac{1}{x+2}
\displaystyle \therefore\ I=\int \frac{3x+1}{(x-2)^{2}(x+2)}\,dx
\displaystyle \Rightarrow\ I=\frac{5}{16}\int \frac{1}{x-2}\,dx+\frac{7}{4}\int \frac{1}{(x-2)^{2}}\,dx-\frac{5}{16}\int \frac{1}{x+2}\,dx
\displaystyle \Rightarrow\ I=\frac{5}{16}\log|x-2|-\frac{7}{4(x-2)}-\frac{5}{16}\log|x+2|+C
\displaystyle (ii)\ \text{We have,}
\displaystyle I=\int \frac{x^{2}+1}{(x-1)^{2}(x+3)}\,dx
\displaystyle \text{Let }\frac{x^{2}+1}{(x-1)^{2}(x+3)}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{x+3}\qquad \cdots(i)
\displaystyle \Rightarrow\ x^{2}+1=A(x-1)(x+3)+B(x+3)+C(x-1)^{2}\qquad \cdots(ii)
\displaystyle \text{Putting }x-1=0\ \text{i.e. }x=1\ \text{in (ii), we get}
\displaystyle 2=4B\Rightarrow B=\frac{1}{2}
\displaystyle \text{Putting }x+3=0\ \text{i.e. }x=-3\ \text{in (ii), we get}
\displaystyle 10=16C\Rightarrow C=\frac{5}{8}
\displaystyle \text{Equating the coefficients of }x^{2}\ \text{on both sides of the identity (ii), we get}
\displaystyle 1=A+C\Rightarrow A=1-C=1-\frac{5}{8}=\frac{3}{8}
\displaystyle \text{Substituting the values of }A,B\ \text{and }C\ \text{in (i), we get}
\displaystyle \frac{x^{2}+1}{(x-1)^{2}(x+3)}=\frac{3}{8}\cdot\frac{1}{x-1}+\frac{1}{2}\cdot\frac{1}{(x-1)^{2}}+\frac{5}{8}\cdot\frac{1}{x+3}
\displaystyle \Rightarrow\ I=\int \frac{x^{2}+1}{(x-1)^{2}(x+3)}\,dx=\frac{3}{8}\int \frac{1}{x-1}\,dx+\frac{1}{2}\int \frac{1}{(x-1)^{2}}\,dx+\frac{5}{8}\int \frac{1}{x+3}\,dx
\displaystyle \Rightarrow\ I=\frac{3}{8}\log|x-1|-\frac{1}{2(x-1)}+\frac{5}{8}\log|x+3|+C

\displaystyle \textbf{Question 28: }\quad \text{Evaluate:}
\displaystyle \int \frac{8}{(x+2)(x^{2}+4)}\,dx \hspace{8.0cm} \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\frac{8}{(x+2)(x^{2}+4)}=\frac{A}{x+2}+\frac{Bx+C}{x^{2}+4}\qquad \cdots(i)
\displaystyle \text{Then, }8=A(x^{2}+4)+(Bx+C)(x+2)\qquad \cdots(ii)
\displaystyle \text{Putting }x+2=0\ \text{i.e. }x=-2\ \text{in (ii), we get}
\displaystyle 8=8A\Rightarrow A=1
\displaystyle \text{Putting }x=0\ \text{and }1\ \text{respectively in (ii), we get}
\displaystyle 8=4A+2C\ \text{and }8=5A+3B+3C
\displaystyle \text{Solving these equations, we obtain}
\displaystyle A=1,\ C=2\ \text{and }B=-1
\displaystyle \text{Substituting the values of }A,B\ \text{and }C\ \text{in (i), we obtain}
\displaystyle \frac{8}{(x+2)(x^{2}+4)}=\frac{1}{x+2}+\frac{-x+2}{x^{2}+4}
\displaystyle \therefore\ I=\int \frac{8}{(x+2)(x^{2}+4)}\,dx
\displaystyle \Rightarrow\ I=\int \frac{1}{x+2}\,dx+\int \frac{-x+2}{x^{2}+4}\,dx
\displaystyle \Rightarrow\ I=\int \frac{1}{x+2}\,dx-\int \frac{x}{x^{2}+4}\,dx+2\int \frac{1}{x^{2}+4}\,dx
\displaystyle \Rightarrow\ I=\log|x+2|-\frac{1}{2}\int \frac{1}{t}\,dt+2\times\frac{1}{2}\tan^{-1}\frac{x}{2}+C,\ \text{where }t=x^{2}+4
\displaystyle \Rightarrow\ I=\log|x+2|-\frac{1}{2}\log t+\tan^{-1}\frac{x}{2}+C =\log|x+2|-\frac{1}{2}\log(x^{2}+4)+\tan^{-1}\frac{x}{2}+C

\displaystyle \textbf{Question 29: }\quad \text{Evaluate:}\ \int \left\{\sqrt{\tan \theta}+\sqrt{\cot \theta}\right\}d\theta\qquad \text{[CBSE 2010, 2013, 2014]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }I=\int \left\{\sqrt{\tan \theta}+\sqrt{\cot \theta}\right\}d\theta.\ \text{Then,}
\displaystyle I=\int \left\{\sqrt{\tan \theta}+\frac{1}{\sqrt{\tan \theta}}\right\}d\theta=\int \frac{\tan \theta+1}{\sqrt{\tan \theta}}\,d\theta
\displaystyle \text{Let }\tan \theta=x^{2}.\ \text{Then, }d(\tan \theta)=d(x^{2})\Rightarrow \sec^{2}\theta\,d\theta=2x\,dx
\displaystyle \Rightarrow\ d\theta=\frac{2x\,dx}{\sec^{2}\theta}=\frac{2x\,dx}{1+\tan^{2}\theta}=\frac{2x\,dx}{1+x^{4}}
\displaystyle \therefore\ I=\int \frac{x^{2}+1}{x}\cdot\frac{2x\,dx}{1+x^{4}}=2\int \frac{x^{2}+1}{1+x^{4}}\,dx=2\int \frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}}\,dx
\displaystyle \Rightarrow\ I=2\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+2}\,dx
\displaystyle \text{Let }u=x-\frac{1}{x}.\ \text{Then, }du=\left(1+\frac{1}{x^{2}}\right)dx
\displaystyle \Rightarrow\ I=2\int \frac{du}{u^{2}+(\sqrt{2})^{2}}=\frac{2}{\sqrt{2}}\tan^{-1}\left(\frac{u}{\sqrt{2}}\right)+C
\displaystyle \Rightarrow\ I=\sqrt{2}\tan^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)+C =\sqrt{2}\tan^{-1}\left(\frac{x^{2}-1}{\sqrt{2}x}\right)+C
\displaystyle \Rightarrow\ I=\sqrt{2}\tan^{-1}\left(\frac{\tan \theta-1}{\sqrt{2\tan \theta}}\right)+C

\displaystyle \textbf{Question 30: }\quad \text{Evaluate:}\ \int \frac{1}{\sin^{4}x+\cos^{4}x}\,dx \hspace{5.0cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }I=\int \frac{1}{\sin^{4}x+\cos^{4}x}\,dx.\ \text{Then,}
\displaystyle I=\int \frac{\frac{1}{\cos^{4}x}}{\frac{\sin^{4}x+\cos^{4}x}{\cos^{4}x}}\,dx=\int \frac{\sec^{4}x}{\tan^{4}x+1}\,dx
\displaystyle =\int \frac{\sec^{2}x\cdot\sec^{2}x}{\tan^{4}x+1}\,dx=\int \frac{1+\tan^{2}x}{1+\tan^{4}x}\sec^{2}x\,dx
\displaystyle \text{Putting }\tan x=t\ \text{and }\sec^{2}x\,dx=dt,\ \text{we get}
\displaystyle I=\int \frac{1+t^{2}}{1+t^{4}}\,dt=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t^{2}-1}{\sqrt{2}t}\right)+C
\displaystyle \Rightarrow\ I=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\tan^{2}x-1}{\sqrt{2}\tan x}\right)+C

\displaystyle \textbf{Question 31. }\text{If }\frac{d}{dx}f(x)=\log x\text{, then }f(x)\text{ equals} \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{(a) }-\frac{1}{x}+C \hspace{2cm} \text{(b) }x(\log x-1)+C
\displaystyle \text{(c) }x(\log x+x)+C \hspace{1.5cm} \text{(d) }\frac{1}{x}+C
\displaystyle \text{Answer: (b) Given, }\frac{d}{dx}[f(x)]=\log x 
\displaystyle \text{On integrating both sides, we get}
\displaystyle \Rightarrow f(x)=\int\log x\,dx
\displaystyle =\log x\int dx-\int\left[\frac{d}{dx}(\log x)\int dx\right]dx
\displaystyle \Rightarrow f(x)=\log x\cdot x-\int\frac{1}{x}\cdot x\,dx
\displaystyle \Rightarrow f(x)=\log x\cdot x-\int dx
\displaystyle \Rightarrow f(x)=x\log x-x+C
\displaystyle \Rightarrow f(x)=x(\log x-1)+C
\\

\displaystyle \textbf{Question 32. }\int\frac{\sec x}{\sec x-\tan x}\,dx\text{ equals} \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{(a) }\sec x-\tan x+C \hspace{2cm} \text{(b) }\sec x+\tan x+C
\displaystyle \text{(c) }\tan x-\sec x+C \hspace{2cm} \text{(d) }-(\sec x+\tan x)+C
\displaystyle \text{Answer: (b) }\int\frac{\sec x}{\sec x-\tan x}\,dx=\int\frac{\sec x(\sec x+\tan x)}{(\sec^2x-\tan^2x)}\,dx
\displaystyle =\frac{\int(\sec^2x+\sec x\tan x)dx}{1}
\displaystyle =\int\sec^2x\,dx+\int\sec x\tan x\,dx=\tan x+\sec x+C
\\

\displaystyle \textbf{Question 33. }\int 2^{x+2}\,dx\text{ is equal to} \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{(a) }2^{x+2}+C \hspace{2cm} \text{(b) }2^{x+2}\log 2+C
\displaystyle \text{(c) }\frac{2^{x+2}}{\log 2}+C \hspace{2cm} \text{(d) }2\cdot\frac{2^x}{\log 2}+C
\displaystyle \text{Answer: (c) We have }\int 2^{x+2}\,dx
\displaystyle \text{Let }x+2=t\Rightarrow dx=dt\qquad\text{[using the substitution]}
\displaystyle \therefore\int 2^t\,dt=\frac{2^t}{\log 2}+C=\frac{2^{(x+2)}}{\log 2}+C
\\

\displaystyle \textbf{Question 34. }\int e^{5\log x}\,dx\text{ is equal to} \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{(a) }\frac{x^5}{5}+C \hspace{2cm} \text{(b) }\frac{x^6}{6}+C
\displaystyle \text{(c) }5x^4+C \hspace{2cm} \text{(d) }6x^5+C
\displaystyle \text{Answer: (b) }\int e^{5\log x}\,dx=\int e^{\log x^5}\,dx=\int x^5\,dx\qquad[\because e^{\log p}=p]
\displaystyle =\frac{x^6}{6}+C
\\

\displaystyle \textbf{Question 35. }\text{If }f'(x)=x+\frac{1}{x}\text{, then }f(x)\text{ is} \hspace{1.2cm} \text{[CBSE Sample Paper 2023]}
\displaystyle \text{(a) }x^2+\log|x|+C \hspace{2cm} \text{(b) }\frac{x^2}{2}+\log|x|+C
\displaystyle \text{(c) }\frac{x}{2}+\log|x|+C \hspace{2cm} \text{(d) }\frac{x}{2}-\log|x|+C
\displaystyle \text{Answer: (b) Given, }f'(x)=x+\frac{1}{x}\qquad\ldots\text{(i)}
\displaystyle \text{On integrating both sides of Eq. (i), we get}
\displaystyle \int f'(x)\,dx=\int\left(x+\frac{1}{x}\right)dx
\displaystyle \Rightarrow f(x)=\int x\,dx+\int\frac{1}{x}\,dx
\displaystyle \Rightarrow f(x)=\frac{x^2}{2}+\log|x|+C
\\

\displaystyle \textbf{Question 36. }\text{Find }\int\frac{e^x}{\sqrt{e^{2x}-4e^x-5}}\,dx. \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{e^x}{\sqrt{e^{2x}-4e^x-5}}\,dx
\displaystyle \text{Put }e^x=t\Rightarrow e^x\,dx=dt
\displaystyle \therefore\ I=\int\frac{dt}{\sqrt{t^2-4t-5}}
\displaystyle =\int\frac{dt}{\sqrt{(t-2)^2-(3)^2}}
\displaystyle =\log|(t-2)+\sqrt{(t-2)^2-(3)^2}|+C
\displaystyle =\log|(e^x-2)+\sqrt{(e^x-2)^2-(3)^2}|+C
\displaystyle =\log|(e^x-2)+\sqrt{e^{2x}-4e^x-5}|+C
\\

\displaystyle \textbf{Question 37. }\text{Find }\int x^2\log(x^2+1)\,dx. \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int x^2\log(x^2+1)\,dx
\displaystyle =\log(x^2+1)\int x^2\,dx-\int\left(\frac{d}{dx}\log(x^2+1)\int x^2\,dx\right)dx
\displaystyle \qquad\text{[integration by parts]}
\displaystyle =\log(x^2+1)\left(\frac{x^3}{3}\right)-\int\frac{2x}{x^2+1}\cdot\frac{x^3}{3}\,dx
\displaystyle =\frac{x^3}{3}\log(x^2+1)-\frac{2}{3}\int\frac{x^4}{x^2+1}\,dx
\displaystyle =\frac{x^3}{3}\log(x^2+1)-\frac{2}{3}\int\left(x^2-1+\frac{1}{x^2+1}\right)dx
\displaystyle =\frac{x^3}{3}\log(x^2+1)-\frac{2}{3}\left[\int x^2\,dx-\int dx+\int\frac{1}{x^2+1}\,dx\right]
\displaystyle =\frac{x^3}{3}\log(x^2+1)-\frac{2x^3}{9}+\frac{2}{3}x-\tan^{-1}x+C
\\

\displaystyle \textbf{Question 38. }\text{Find }\int\frac{1}{\sqrt{x}(\sqrt{x}+1)(\sqrt{x}+2)}\,dx. \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{1}{\sqrt{x}(\sqrt{x}+1)(\sqrt{x}+2)}\,dx
\displaystyle \text{Put }\sqrt{x}=t\Rightarrow\frac{1}{2\sqrt{x}}\,dx=dt\Rightarrow\frac{1}{\sqrt{x}}\,dx=2\,dt
\displaystyle \therefore\ I=2\int\frac{dt}{(t+1)(t+2)}
\displaystyle \text{Again, let }\frac{1}{(t+1)(t+2)}=\frac{A}{t+1}+\frac{B}{t+2}
\displaystyle \Rightarrow 1=A(t+2)+B(t+1)\qquad\ldots\text{(i)}
\displaystyle \text{On putting }t+1=0\text{, we get }1=A(-1+2)+B(0)\Rightarrow A=1
\displaystyle \text{On putting }t+2=0\Rightarrow t=-2\text{ in Eq. (i), we get }1=A(0)+B(-2+1)\Rightarrow B=-1
\displaystyle \therefore\ \frac{1}{(t+1)(t+2)}=\frac{1}{t+1}-\frac{1}{t+2}
\displaystyle \therefore\ I=2\int\left(\frac{1}{t+1}-\frac{1}{t+2}\right)dt
\displaystyle =2\left[\int\frac{1}{t+1}\,dt-\int\frac{1}{t+2}\,dt\right]
\displaystyle =2[\log(t+1)-\log(t+2)]+C
\displaystyle =2\left[\log\left(\frac{t+1}{t+2}\right)\right]+C
\displaystyle =2\log\left(\frac{\sqrt{x}+1}{\sqrt{x}+2}\right)+C\qquad[\text{put }t=\sqrt{x}]
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\displaystyle \textbf{Question 39. }\text{Find }\int\frac{1}{\cos(x-a)\cos(x-b)}\,dx. \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }\int\frac{1}{\cos(x-a)\cos(x-b)}\,dx
\displaystyle =\frac{1}{\sin(b-a)}\int\frac{\sin\{(x-a)-(x-b)\}}{\cos(x-a)\cos(x-b)}\,dx
\displaystyle =\frac{1}{\sin(b-a)}\int\frac{\sin(x-a)\cos(x-b)-\cos(x-a)\sin(x-b)}{\cos(x-a)\cos(x-b)}\,dx
\displaystyle =\frac{1}{\sin(b-a)}\int[\tan(x-a)-\tan(x-b)]\,dx
\displaystyle =\frac{1}{\sin(b-a)}\left[\int\tan(x-a)\,dx-\int\tan(x-b)\,dx\right]
\displaystyle =\frac{1}{\sin(b-a)}[-\log|\cos(x-a)|+\log|\cos(x-b)|]+C
\displaystyle =\frac{1}{\sin(b-a)}\left[\log\left|\frac{\cos(x-b)}{\cos(x-a)}\right|\right]+C
\\

\displaystyle \textbf{Question 40. }\text{Find }\int\frac{2}{(1-x)(1+x^2)}\,dx. \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }\int\frac{2}{(1-x)(1+x^2)}\,dx
\displaystyle \text{Let }\frac{2}{(1-x)(1+x^2)}=\frac{A}{1-x}+\frac{Bx+C}{1+x^2}
\displaystyle \Rightarrow\frac{2}{(1-x)(1+x^2)}=\frac{A(1+x^2)+(Bx+C)(1-x)}{(1-x)(1+x^2)}
\displaystyle \Rightarrow 2=A+Ax^2+Bx-Bx^2+C-Cx
\displaystyle \Rightarrow 2=(A+C)+x(B-C)+x^2(A-B)
\displaystyle \text{On equating coefficients of constant term, coefficient of }x\text{ and coefficients of }x^2\text{,}
\displaystyle \therefore\ A+C=2\qquad\ldots\text{(i)}
\displaystyle \text{and }B-C=0\qquad\ldots\text{(ii)}
\displaystyle \text{and }A-B=0\qquad\ldots\text{(iii)}
\displaystyle \text{On solving Eqs. (i), (ii) and (iii), we get }A=1,\ B=1\text{ and }C=1
\displaystyle \therefore\ \frac{2}{(1-x)(1+x^2)}=\frac{1}{1-x}+\frac{x+1}{1+x^2}
\displaystyle \therefore\ \int\frac{2}{(1-x)(1+x^2)}\,dx=\int\frac{1}{1-x}\,dx+\int\frac{x+1}{1+x^2}\,dx
\displaystyle =-\log(1-x)+\frac{1}{2}\int\frac{2x}{1+x^2}\,dx+\int\frac{dx}{1+x^2}
\displaystyle =-\log(1-x)+\frac{1}{2}\log(1+x^2)+\tan^{-1}x+C
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\displaystyle \textbf{Question 41. }\text{Find }\int\frac{e^x}{\sqrt{5-4e^x-e^{2x}}}\,dx. \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }\int\frac{e^x}{\sqrt{5-4e^x-e^{2x}}}\,dx
\displaystyle \text{Put }e^x=t\Rightarrow e^x\,dx=dt\text{, we get}
\displaystyle \int\frac{dt}{\sqrt{5-4t-t^2}}=\int\frac{dt}{\sqrt{-(t^2+4t-5+4-4)}}
\displaystyle =\int\frac{dt}{\sqrt{-(t^2+4t+4)+9}}=\int\frac{dt}{\sqrt{9-(t+2)^2}}
\displaystyle =\sin^{-1}\left(\frac{t+2}{3}\right)+C=\sin^{-1}\left(\frac{e^x+2}{3}\right)+C
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\displaystyle \textbf{Question 42. }\text{Find }\int\frac{x^2+x+1}{(x+1)^2(x+2)}\,dx. \hspace{1.2cm} \text{[CBSE 2023; CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }\int\frac{x^2+x+1}{(x+1)^2(x+2)}\,dx
\displaystyle \text{Given integrand }\frac{x^2+x+1}{(x+1)^2(x+2)}\text{ is a proper rational function.}
\displaystyle \text{Now, by using partial fraction,}
\displaystyle \text{let }\frac{x^2+x+1}{(x+1)^2(x+2)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+2}\qquad\ldots\text{(i)}
\displaystyle \Rightarrow x^2+x+1=A(x+1)(x+2)+B(x+2)+C(x+1)^2
\displaystyle \Rightarrow x^2+x+1=A(x^2+3x+2)+B(x+2)+C(x^2+2x+1)
\displaystyle \Rightarrow x^2+x+1=(A+C)x^2+(3A+B+2C)x+(2A+2B+C)
\displaystyle \text{On comparing the coefficients of like powers of }x\text{ from both sides, we get}
\displaystyle A+C=1,\ 3A+B+2C=1\text{ and }2A+2B+C=1
\displaystyle \text{On solving these equations, we get }A=-2,\ B=1\text{ and }C=3
\displaystyle \text{From Eq. (i), we get}
\displaystyle \frac{x^2+x+1}{(x+1)^2(x+2)}=\frac{-2}{x+1}+\frac{1}{(x+1)^2}+\frac{3}{x+2}
\displaystyle \therefore\ \int\frac{x^2+x+1}{(x+1)^2(x+2)}\,dx=-2\int\frac{1}{x+1}\,dx+\int\frac{dx}{(x+1)^2}+3\int\frac{dx}{x+2}
\displaystyle =-2\log|x+1|-\frac{1}{x+1}+3\log|x+2|+C
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\displaystyle \textbf{Question 43. }\text{Find }\int\frac{dx}{\sqrt{3-2x-x^2}}. \hspace{1.2cm} \text{[CBSE Sample Paper 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{dx}{\sqrt{3-2x-x^2}}
\displaystyle \therefore\ I=\int\frac{dx}{\sqrt{3-(2x+x^2)}}
\displaystyle =\int\frac{dx}{\sqrt{3-(1+2x+x^2-1)}}
\displaystyle =\int\frac{dx}{\sqrt{3-[(x+1)^2-1]}}
\displaystyle =\int\frac{dx}{\sqrt{3-(x+1)^2+1}}
\displaystyle \Rightarrow I=\int\frac{dx}{\sqrt{4-(x+1)^2}}
\displaystyle \text{Let }(x+1)=t\Rightarrow dx=dt
\displaystyle \therefore\ I=\int\frac{dt}{\sqrt{4-t^2}}=\int\frac{dt}{\sqrt{(2)^2-t^2}}
\displaystyle \Rightarrow I=\sin^{-1}\left(\frac{t}{2}\right)+C
\displaystyle \left[\because\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\left(\frac{x}{a}\right)+C\right]
\displaystyle \therefore\ I=\sin^{-1}\left(\frac{x+1}{2}\right)+C\qquad[\because t=x+1]
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\displaystyle \textbf{Question 44. }\text{Find }\int\frac{(x^3+x+1)}{(x^2-1)}\,dx. \hspace{1.2cm} \text{[CBSE Sample Paper 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{(x^3+x+1)}{(x^2-1)}\,dx
\displaystyle =\int\frac{x^3-x+x+x+1}{(x^2-1)}\,dx
\displaystyle =\int\frac{x(x^2-1)+(2x+1)}{(x^2-1)}\,dx
\displaystyle \Rightarrow I=\int\left(x+\frac{2x+1}{x^2-1}\right)dx\qquad\ldots\text{(i)}
\displaystyle \text{Now, resolving }\frac{2x+1}{x^2-1}\text{ into partial fraction as}
\displaystyle \frac{2x+1}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}\qquad\ldots\text{(ii)}
\displaystyle \Rightarrow 2x+1=A(x+1)+B(x-1)
\displaystyle \Rightarrow 2x+1=x(A+B)+A-B
\displaystyle \Rightarrow A+B=2\text{ and }A-B=1
\displaystyle \text{On solving, we get }A=\frac{3}{2}\text{ and }B=\frac{1}{2}
\displaystyle \text{On substituting the values of }A\text{ and }B\text{ in Eq. (ii), we get}
\displaystyle \frac{2x+1}{(x-1)(x+1)}=\frac{3}{2(x-1)}+\frac{1}{2(x+1)}
\displaystyle \text{From Eq. (i), we get}
\displaystyle I=\int x\,dx+\frac{3}{2}\int\frac{1}{x-1}\,dx+\frac{1}{2}\int\frac{1}{x+1}\,dx
\displaystyle I=\frac{x^2}{2}+\frac{3}{2}\log|x-1|+\frac{1}{2}\log|x+1|+C
\displaystyle I=\frac{x^2}{2}+\frac{1}{2}\log|x-1|^3+\frac{1}{2}\log|x+1|+C
\displaystyle I=\frac{x^2}{2}+\frac{1}{2}[\log|(x-1)^3(x+1)|]+C
\\

\displaystyle \textbf{Question 45. }\text{Find }\int\frac{dx}{x^2-6x+13}. \hspace{1.2cm} \text{[CBSE 2022 (Term II)]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{dx}{x^2-6x+13}=\int\frac{dx}{(x-3)^2+13-9}
\displaystyle =\int\frac{dx}{(x-3)^2+4}=\int\frac{dx}{(x-3)^2+2^2}
\displaystyle =\frac{1}{2}\tan^{-1}\frac{x-3}{2}+C
\\

\displaystyle \textbf{Question 46. }\text{Find }\int\frac{\log x}{(1+\log x)^2}\,dx. \hspace{1.2cm} \text{[CBSE Sample Paper 2022 (Term II)]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{\log x}{(1+\log x)^2}\,dx
\displaystyle =\int\frac{\log x+1-1}{(1+\log x)^2}\,dx
\displaystyle =\int\left[\frac{1+\log x}{(1+\log x)^2}-\frac{1}{(1+\log x)^2}\right]dx
\displaystyle =\int\left[\frac{1}{(1+\log x)}-\frac{1}{(1+\log x)^2}\right]dx
\displaystyle \text{On putting }\log x=t\Rightarrow x=e^t\Rightarrow dx=e^t\,dt
\displaystyle \therefore\ I=\int\left[\frac{1}{1+t}-\frac{1}{(1+t)^2}\right]e^t\,dt
\displaystyle \text{Now, we have a formula }\int e^x[f(x)+f'(x)]\,dx=e^x f(x)+C
\displaystyle \text{So, here }f(t)=\frac{1}{1+t}\Rightarrow f'(t)=\frac{-1}{(1+t)^2}
\displaystyle \text{So, it is from Eq. (i).}
\displaystyle \text{The given integral }=e^t\cdot\frac{1}{1+t}+C=\frac{x}{1+\log x}+C
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\displaystyle \textbf{Question 47. }\text{Find }\int\frac{\sin 2x}{\sqrt{9-\cos^4x}}\,dx. \hspace{1.2cm} \text{[CBSE Sample Paper 2022 (Term II)]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{\sin 2x}{\sqrt{9-\cos^4x}}\,dx=\int\frac{\sin 2x}{\sqrt{(3)^2-(\cos^2x)^2}}\,dx
\displaystyle \text{Put }\cos^2x=t\Rightarrow-2\cos x\sin x\,dx=dt
\displaystyle \Rightarrow\sin 2x\,dx=-dt
\displaystyle \therefore\ I=-\int\frac{dt}{\sqrt{3^2-t^2}}=-\sin^{-1}\left(\frac{t}{3}\right)+C
\displaystyle I=-\sin^{-1}\left(\frac{\cos^2x}{3}\right)+C
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\displaystyle \textbf{Question 48. }\text{Find }\int\frac{x+1}{(x^2+1)x}\,dx. \hspace{1.2cm} \text{[CBSE Sample Paper 2022 (Term II)]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{(x+1)dx}{(x^2+1)x}
\displaystyle \text{Now, by partial fraction, we get}
\displaystyle \frac{x+1}{(x^2+1)x}=\frac{Ax+B}{x^2+1}+\frac{C}{x}\qquad\ldots\text{(i)}
\displaystyle =\frac{(Ax+B)x+C(x^2+1)}{x(x^2+1)}
\displaystyle \Rightarrow x+1=(Ax+B)x+C(x^2+1)
\displaystyle \Rightarrow x+1=x^2(A+C)+xB+C
\displaystyle \text{Now, equating the coefficients, we get}
\displaystyle A+C=0,\ B=1\text{ and }C=1
\displaystyle \text{Hence, }A=-1,\ B=1\text{ and }C=1
\displaystyle \text{On substituting the values of }A,\ B\text{ and }C\text{ in Eq. (i), we get}
\displaystyle I=\int\frac{-x+1}{x^2+1}\,dx+\int\frac{1}{x}\,dx\qquad\text{[integrating both sides]}
\displaystyle =-\frac{1}{2}\int\frac{2x-2}{x^2+1}\,dx+\int\frac{1}{x}\,dx
\displaystyle =-\frac{1}{2}\int\frac{2x}{x^2+1}\,dx+\int\frac{1}{x^2+1}\,dx+\int\frac{1}{x}\,dx
\displaystyle =-\frac{1}{2}\log(x^2+1)+\tan^{-1}x+\log|x|+C
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\displaystyle \textbf{Question 49. }\text{Find }\int\frac{x^2}{(x^2+1)(3x^2+4)}\,dx. \hspace{1.2cm} \text{[CBSE 2022 (Term II)]}
\displaystyle \text{Answer:}
\displaystyle \text{We will solve this integral by partial fraction.}
\displaystyle \text{Let }\frac{x^2}{(x^2+1)(3x^2+4)}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{3x^2+4}
\displaystyle \Rightarrow x^2=(Ax+B)(3x^2+4)+(Cx+D)(x^2+1)
\displaystyle =(3A+C)x^3+(3B+D)x^2+(4A+C)x+4B+D
\displaystyle \text{Equating similar terms, we get}
\displaystyle 3A+C=0,\ 3B+D=1,\ 4A+C=0\text{ and }4B+D=0
\displaystyle \text{On solving, we get }A=0,\ B=-1,\ C=0\text{ and }D=4
\displaystyle \text{Thus, }I=\int\frac{-dx}{(x^2+1)}+\int\frac{4\,dx}{(3x^2+4)}
\displaystyle =-\int\frac{dx}{(x^2+1)}+\frac{4}{3}\int\frac{dx}{\left(x^2+\dfrac{4}{3}\right)}
\displaystyle =-\int\frac{dx}{x^2+1}+\frac{4}{3}\int\frac{dx}{x^2+\left(\dfrac{2}{\sqrt{3}}\right)^2}
\displaystyle =-\tan^{-1}x+\frac{4}{3}\cdot\frac{\sqrt{3}}{2}\tan^{-1}\left(\frac{\sqrt{3}x}{2}\right)+C
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\displaystyle \textbf{Question 50. }\text{Find }\int e^x(1-\cot x+\text{cosec}^2x)\,dx. \hspace{1.2cm} \text{[CBSE Sample Paper 2021]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int e^x(1-\cot x+\text{cosec}^2x)\,dx
\displaystyle I=\int e^x(1-\cot x)\,dx+\int e^x\text{cosec}^2x\,dx
\displaystyle =(1-\cot x)e^x-\int e^x(-\text{cosec}^2x)\,dx+\int e^x\text{cosec}^2x\,dx
\displaystyle =(1-\cot x)e^x+C
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\displaystyle \textbf{Question 51. }\text{Find }\int\frac{1}{\cos^2x(1-\tan x)^2}\,dx. \hspace{1.2cm} \text{[CBSE Sample Paper 2021]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{1}{\cos^2x(1-\tan x)^2}\,dx=\int\frac{\sec^2x}{(1-\tan x)^2}\,dx
\displaystyle \text{Now, put }(1-\tan x)=t
\displaystyle \Rightarrow-\sec^2x\,dx=dt\Rightarrow\sec^2x\,dx=-dt
\displaystyle \therefore\ I=-\int\frac{dt}{t^2}=\frac{1}{t}+C=\frac{1}{1-\tan x}+C
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\displaystyle \textbf{Question 52. }\text{Find }\int\frac{x^2+1}{(x^2+2)(x^2+3)}\,dx. \hspace{1.2cm} \text{[CBSE Sample Paper 2021]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{x^2+1}{(x^2+2)(x^2+3)}\,dx
\displaystyle \text{Let }x^2=y
\displaystyle \text{Then, }\frac{x^2+1}{(x^2+2)(x^2+3)}=\frac{y+1}{(y+2)(y+3)}
\displaystyle \text{Again, let }\frac{y+1}{(y+2)(y+3)}=\frac{A}{y+2}+\frac{B}{y+3}\qquad\ldots\text{(i)}
\displaystyle \Rightarrow y+1=A(y+3)+B(y+2)\qquad\ldots\text{(ii)}
\displaystyle \text{Putting }y=-2\text{ and }y=-3\text{ successively in Eq. (ii), we get}
\displaystyle A=-1\text{ and }B=2
\displaystyle \text{On substituting the values of }A\text{ and }B\text{ in Eq. (i), we get}
\displaystyle \frac{y+1}{(y+2)(y+3)}=\frac{-1}{y+2}+\frac{2}{y+3}
\displaystyle \therefore\ \int\frac{x^2+1}{(x^2+2)(x^2+3)}\,dx=\int\frac{-1}{x^2+2}\,dx+\int\frac{2}{x^2+3}\,dx
\displaystyle =-\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{x}{\sqrt{2}}\right)+\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right)+C
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\displaystyle \textbf{Question 53. }\text{Find }\int\frac{x}{x^2+3x+2}\,dx. \hspace{1.2cm} \text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{x}{x^2+3x+2}\,dx
\displaystyle \text{Again, let }x=A\frac{d}{dx}(x^2+3x+2)+B
\displaystyle \Rightarrow x=(2x+3)A+B\Rightarrow x=2Ax+(3A+B)
\displaystyle \therefore\ 2A=1\text{ and }3A+B=0
\displaystyle \Rightarrow A=\frac{1}{2}\text{ and }B=-\frac{3}{2}
\displaystyle \therefore\ I=\frac{1}{2}\int\frac{2x+3}{x^2+3x+2}\,dx-\frac{3}{2}\int\frac{dx}{x^2+3x+2}
\displaystyle \Rightarrow I=\frac{1}{2}I_1-\frac{3}{2}I_2\text{,}
\displaystyle \text{where }I_1=\int\frac{2x+3}{x^2+3x+2}\,dx\text{ and }I_2=\int\frac{dx}{x^2+3x+2}
\displaystyle \text{Now, }I_1=\int\frac{2x+3}{x^2+3x+2}\,dx
\displaystyle \text{Put }x^2+3x+2=t\Rightarrow(2x+3)\,dx=dt
\displaystyle \therefore\ I_1=\int\frac{dt}{t}=\log|t|+C_1=\log|x^2+3x+2|+C_1
\displaystyle \text{and }I_2=\int\frac{dx}{x^2+3x+2}=\int\frac{dx}{\left(x+\dfrac{3}{2}\right)^2+2-\dfrac{9}{4}}
\displaystyle =\int\frac{dx}{\left(x+\dfrac{3}{2}\right)^2-\left(\dfrac{1}{2}\right)^2}
\displaystyle =\frac{1}{2\times\frac{1}{2}}\log\left|\frac{x+\frac{3}{2}-\frac{1}{2}}{x+\frac{3}{2}+\frac{1}{2}}\right|+C_2=\log\left|\frac{x+1}{x+2}\right|+C_2
\displaystyle \therefore\ I=\frac{1}{2}\log|x^2+3x+2|+\frac{1}{2}C_1-\frac{3}{2}\log\left|\frac{x+1}{x+2}\right|-\frac{3}{2}C_2
\displaystyle =\frac{1}{2}\log|x^2+3x+2|-\frac{3}{2}\log\left|\frac{x+1}{x+2}\right|+C
\displaystyle \text{where }C=\frac{1}{2}C_1-\frac{3}{2}C_2
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\displaystyle \textbf{Question 54. }\text{Find }\int\sin^5\left(\frac{x}{2}\right)\cdot\cos\frac{x}{2}\,dx. \hspace{1.2cm} \text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\sin^5\left(\frac{x}{2}\right)\cos\frac{x}{2}\,dx
\displaystyle \text{Put }\sin\frac{x}{2}=t\Rightarrow\frac{1}{2}\cos\frac{x}{2}\,dx=dt
\displaystyle \therefore\ I=2\int t^5\,dt
\displaystyle I=2\cdot\frac{t^6}{6}+C
\displaystyle I=\frac{1}{3}\left(\sin^6\frac{x}{2}\right)+C
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\displaystyle \textbf{Question 55. }\text{Find }\int\frac{2^{x+1}-5^{x-1}}{10^x}\,dx. \hspace{1.2cm} \text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{2^{x+1}-5^{x-1}}{10^x}\,dx
\displaystyle \Rightarrow I=\int\left\{2\left(\frac{2}{10}\right)^x-\frac{1}{5}\left(\frac{5}{10}\right)^x\right\}dx
\displaystyle \Rightarrow I=\int\left\{2(5)^{-x}-\frac{1}{5}(2)^{-x}\right\}dx
\displaystyle \Rightarrow I=-2\cdot 5^{-x}\log 5+\frac{1}{5}\cdot 2^{-x}\log 2+C
\displaystyle \Rightarrow I=\frac{1}{5}\log 2(2^{-x})-2\log 5(5^{-x})+C
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\displaystyle \textbf{Question 56. }\text{Find }\int\frac{dx}{\sqrt{x}+x}. \hspace{1.2cm} \text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{dx}{\sqrt{x}+x}\Rightarrow I=\int\frac{dx}{\sqrt{x}(1+\sqrt{x})}
\displaystyle \text{Put }\sqrt{x}+1=t\text{, then }\frac{1}{2\sqrt{x}}\,dx=dt
\displaystyle \therefore\ I=2\int\frac{dt}{t}=2\log|t|+C
\displaystyle =2\log|\sqrt{x}+1|+C\qquad[\because t=\sqrt{x}+1]
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\displaystyle \textbf{Question 57. }\text{Find }\int\frac{1}{x(1+x^2)}\,dx. \hspace{1.2cm} \text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{dx}{x(1+x^2)}
\displaystyle I=\int\frac{dx}{x^3\left(\dfrac{1}{x^2}+1\right)}
\displaystyle \text{Put }\frac{1}{x^2}+1=t\Rightarrow\frac{-2}{x^3}\,dx=dt
\displaystyle \therefore\ I=-\frac{1}{2}\int\frac{dt}{t}=-\frac{1}{2}\log|t|+C
\displaystyle \Rightarrow I=-\frac{1}{2}\log\left|\frac{1}{x^2}+1\right|+C
\displaystyle \Rightarrow I=-\frac{1}{2}\log\left|\frac{x^2+1}{x^2}\right|+C
\displaystyle \Rightarrow I=\frac{1}{2}\log\left|\frac{x^2}{x^2+1}\right|+C
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\displaystyle \textbf{Question 58. }\text{Find }\int\frac{\sec^2x}{\sqrt{\tan^2x+4}}\,dx. \hspace{1.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\int\frac{\sec^2x}{\sqrt{\tan^2x+4}}\,dx
\displaystyle \text{Let }\tan x=t\Rightarrow\sec^2x=\frac{dt}{dx}\Rightarrow dx=\frac{dt}{\sec^2x}
\displaystyle \therefore\ \int\frac{\sec^2x}{\sqrt{\tan^2x+4}}\,dx=\int\frac{\sec^2x}{\sqrt{t^2+4\sec^2x}}\cdot\frac{dt}{\sec^2 x}
\displaystyle =\int\frac{dt}{\sqrt{t^2+2^2}}=\log|t+\sqrt{t^2+4}|+C
\displaystyle \left[\because\int\frac{dx}{\sqrt{x^2+a^2}}=\log|x+\sqrt{x^2+a^2}|+C\right]
\displaystyle =\log|\tan x+\sqrt{\tan^2x+4}|+C\qquad[\because t=\tan x]
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\displaystyle \textbf{Question 59. }\text{Find }\int\sqrt{1-\sin 2x}\,dx,\ \frac{\pi}{4}<x<\frac{\pi}{2}. \hspace{1.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\int\sqrt{1-\sin 2x}\,dx
\displaystyle =\int\sqrt{\sin^2x+\cos^2x-2\sin x\cos x}\,dx
\displaystyle [\because\sin^2\theta+\cos^2\theta=1\text{ and }2\sin\theta\cos\theta=\sin 2\theta]
\displaystyle =\int\sqrt{(\sin x-\cos x)^2}\,dx=\int(\sin x-\cos x)\,dx
\displaystyle \left[\because\text{ in the interval }\frac{\pi}{4}<x<\frac{\pi}{2},\ \sin x>\cos x\right]
\displaystyle =-\cos x-\sin x+C=−(\cos x+\sin x)+C
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\displaystyle \textbf{Question 60. }\text{Find }\int\sin^{-1}(2x)\,dx. \hspace{1.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\sin^{-1}(2x)\,dx
\displaystyle \text{Let }2x=y\Rightarrow x=\frac{y}{2}\Rightarrow dx=\frac{dy}{2}
\displaystyle \therefore\ I=\frac{1}{2}\int\sin^{-1}(y)\,dy
\displaystyle =\frac{1}{2}\left[\sin^{-1}(y)\cdot y-\int\frac{1}{\sqrt{1-y^2}}\cdot y\,dy\right]\qquad\text{[integrating by parts]}
\displaystyle =\frac{1}{2}\left[y\sin^{-1}y+\frac{1}{2}\int\frac{-2y}{\sqrt{1-y^2}}\,dy\right]
\displaystyle =\frac{1}{2}\left[y\sin^{-1}y+\sqrt{1-y^2}+C\right]\qquad\left[\because\int\frac{dy}{\sqrt{x}}=2\sqrt{x}+C\right]
\displaystyle =\frac{1}{2}[2x\sin^{-1}2x+\sqrt{1-4x^2}+C]\qquad[\because y=2x]
\\

\displaystyle \textbf{Question 61. }\text{Find the values of }\int\frac{\tan^2x\cdot\sec^2x}{1-\tan^6x}\,dx. \hspace{1.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{\tan^2x\cdot\sec^2x}{1-\tan^6x}\,dx
\displaystyle \text{Let }\tan^3x=t
\displaystyle \Rightarrow 3\tan^2x\sec^2x\,dx=dt
\displaystyle \Rightarrow\tan^2x\sec^2x\,dx=\frac{1}{3}\cdot dt
\displaystyle \therefore\ I=\frac{1}{3}\int\frac{dt}{1-t^2}=\frac{1}{3}\left[\frac{1}{2}\log\left|\frac{1+t}{1-t}\right|\right]+C
\displaystyle \left[\because\int\frac{dx}{a^2-x^2}=\frac{1}{2a}\log\left|\frac{a+x}{a-x}\right|+C\right]
\displaystyle =\frac{1}{6}\log\left|\frac{1+t}{1-t}\right|+C
\displaystyle \text{Now, put the value of }t\text{, we get}
\displaystyle I=\frac{1}{6}\log\left|\frac{1+\tan^3x}{1-\tan^3x}\right|+C
\\

\displaystyle \textbf{Question 62. }\text{Find the value of }\int\sin x\cdot\log\cos x\,dx. \hspace{1.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\sin x\cdot\log(\cos x)\,dx
\displaystyle \text{Put }\cos x=t\Rightarrow-\sin x\,dx=dt
\displaystyle \therefore\ I=-\int\log t\,dt=-\int(\log t)\cdot 1\,dt
\displaystyle =-\left[\log t\int 1\,dt-\int\left\{\frac{d}{dt}(\log t)\int 1\,dt\right\}dt\right]
\displaystyle =-\left[(\log t)\cdot t-\int\frac{1}{t}\cdot t\,dt\right]
\displaystyle =-[t\cdot\log t-\int 1\,dt]
\displaystyle =-[t\cdot\log t-t]+C
\displaystyle =-t\cdot\log t+t+C
\displaystyle =-\cos x\log(\cos x)+\cos x+C
\\

\displaystyle \textbf{Question 63. }\text{Find }\int\sqrt{3-2x-x^2}\,dx. \hspace{1.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\sqrt{3-2x-x^2}\,dx
\displaystyle =\int\sqrt{-(x^2+2x-3)}\,dx
\displaystyle =\int\sqrt{-(x^2+2x+1-4)}\,dx
\displaystyle =\int\sqrt{-\{(x+1)^2-2^2\}}\,dx
\displaystyle =\int\sqrt{2^2-(x+1)^2}\,dx
\displaystyle \text{Now, put }x+1=t\Rightarrow dx=dt
\displaystyle \therefore\ I=\int\sqrt{2^2-t^2}\,dt
\displaystyle =\frac{1}{2}\left[t\sqrt{2^2-t^2}+2^2\sin^{-1}\left(\frac{t}{2}\right)\right]+C
\displaystyle \left[\because\int\sqrt{a^2-x^2}\,dx=\frac{1}{2}\left(x\sqrt{a^2-x^2}+a^2\sin^{-1}\frac{x}{a}\right)+C\right]
\displaystyle =\frac{1}{2}\left[(x+1)\sqrt{3-2x-x^2}+4\sin^{-1}\left(\frac{x+1}{2}\right)\right]+C
\displaystyle [\because t=x+1]
\\

\displaystyle \textbf{Question 64. }\text{Find }\int\frac{\sin^3x+\cos^3x}{\sin^2x\cos^2x}\,dx. \hspace{1.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\left(\frac{\sin^3x+\cos^3x}{\sin^2x\cos^2x}\right)dx
\displaystyle =\int\left(\frac{\sin^3x}{\sin^2x\cos^2x}+\frac{\cos^3x}{\sin^2x\cos^2x}\right)dx
\displaystyle =\int\left(\frac{\sin x}{\cos^2x}+\frac{\cos x}{\sin^2x}\right)dx
\displaystyle =\int[(\tan x\cdot\sec x)+(\cot x\cdot\text{cosec }x)]\,dx
\displaystyle =\int\sec x\cdot\tan x\,dx+\int\cot x\cdot\text{cosec }x\,dx
\displaystyle =\sec x+(-\text{cosec }x)+C=\sec x-\text{cosec }x+C
\\

\displaystyle \textbf{Question 65. }\text{Find }\int\frac{x-3}{(x-1)^3}\,e^x\,dx. \hspace{1.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{x-3}{(x-1)^3}\,e^x\,dx
\displaystyle =\int\frac{e^x(x-1-2)}{(x-1)^3}\,dx
\displaystyle =\int e^x\left\{\frac{(x-1)}{(x-1)^3}-\frac{2}{(x-1)^3}\right\}dx
\displaystyle =\int e^x\left\{\frac{1}{(x-1)^2}-\frac{2}{(x-1)^3}\right\}dx
\displaystyle =\int e^x\cdot\{f(x)+f'(x)\}\,dx\text{,}
\displaystyle \text{where }f(x)=\frac{1}{(x-1)^2}\text{ and }f'(x)=\frac{-2}{(x-1)^3}
\displaystyle =e^x\cdot f(x)+C
\displaystyle \left[\because\int e^x\{f(x)+f'(x)\}\,dx=e^x f(x)+C\right]
\displaystyle =e^x\cdot\frac{1}{(x-1)^2}+C=\frac{e^x}{(x-1)^2}+C
\\

\displaystyle \textbf{Question 66. }\text{Find }\int \frac{x-5}{(x-3)^{3}}\,e^{x}\,dx. \qquad \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \int \frac{x-5}{(x-3)^{3}}\,e^{x}\,dx
\displaystyle =\int \frac{(x-3)-2}{(x-3)^{3}}\,e^{x}\,dx
\displaystyle =\int \left(\frac{1}{(x-3)^{2}}-\frac{2}{(x-3)^{3}}\right)e^{x}\,dx
\displaystyle =e^{x}\cdot\frac{1}{(x-3)^{2}}+C
\displaystyle \therefore \int \frac{x-5}{(x-3)^{3}}\,e^{x}\,dx=\frac{e^{x}}{(x-3)^{2}}+C
\\

\displaystyle \textbf{Question 67. }\text{Find }\int\frac{2\cos x}{(1-\sin x)(2-\cos^2x)}\,dx. \hspace{1.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{2\cos x}{(1-\sin x)(2-\cos^2x)}\,dx
\displaystyle =\int\frac{2\cos x}{(1-\sin x)(1+\sin^2x)}\,dx
\displaystyle \text{Put }\sin x=t\Rightarrow\cos x\,dx=dt
\displaystyle \therefore\ I=\int\frac{2}{(1-t)(1+t^2)}\,dt
\displaystyle \text{Now, let }\frac{2}{(1-t)(1+t^2)}=\frac{A}{1-t}+\frac{Bt+C}{1+t^2}
\displaystyle \Rightarrow 2=A(1+t^2)+(Bt+C)(1-t)\qquad\ldots\text{(i)}
\displaystyle \text{Putting }t=1\text{ in Eq. (i), we get }2=2A\Rightarrow A=1
\displaystyle \text{Putting }t=0\text{ in Eq. (i), we get }2=A+C\Rightarrow 2=1+C\Rightarrow C=1
\displaystyle \text{and putting }t=-1\text{ in Eq. (i), we get}
\displaystyle 2=2A+(-B+C)(2)\Rightarrow 2=2-2B+2\Rightarrow 2B=2\Rightarrow B=1
\displaystyle \therefore\ I=\int\frac{1}{1-t}\,dt+\int\frac{t+1}{1+t^2}\,dt
\displaystyle =\int\frac{1}{1-t}\,dt+\frac{1}{2}\int\frac{2t}{1+t^2}\,dt+\int\frac{1}{1+t^2}\,dt
\displaystyle =-\log|1-t|+\frac{1}{2}\log|1+t^2|+\tan^{-1}t+C
\displaystyle =-\log|1-\sin x|+\frac{1}{2}\log|1+\sin^2x|+\tan^{-1}(\sin x)+C\qquad[\because t=\sin x]
\displaystyle =\tan^{-1}(\sin x)+\log\sqrt{1+\sin^2x}+\log\frac{1}{1-\sin x}+C
\displaystyle \left[\because\log m-\log n=\log\left(\frac{m}{n}\right)\text{ and }n\log m=\log m^n\right]
\\

\displaystyle \textbf{Question 68. }\text{Find }\int\frac{3x+5}{x^2+3x-18}\,dx. \hspace{1.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{3x+5}{x^2+3x-18}\,dx\qquad\ldots\text{(i)}
\displaystyle \text{Also, let }3x+5=A\frac{d}{dx}(x^2+3x-18)+B
\displaystyle \Rightarrow 3x+5=A(2x+3)+B\qquad\ldots\text{(ii)}
\displaystyle \text{On comparing the coefficient of }x\text{, we get }2A=3\Rightarrow A=\frac{3}{2}
\displaystyle \text{and on comparing the constant terms, we get }B=5-3A\Rightarrow B=5-3\left(\frac{3}{2}\right)=\frac{1}{2}
\displaystyle \text{From Eq. (ii), we get}
\displaystyle 3x+5=\frac{3}{2}(2x+3)+\frac{1}{2}\qquad\ldots\text{(iii)}
\displaystyle \text{From Eqs. (i) and (iii), we get}
\displaystyle I=\int\frac{\dfrac{3}{2}(2x+3)+\dfrac{1}{2}}{x^2+3x-18}\,dx
\displaystyle =\frac{3}{2}\int\frac{2x+3}{x^2+3x-18}\,dx+\frac{1}{2}\int\frac{1}{x^2+3x-18}\,dx
\displaystyle =\frac{3}{2}\log|x^2+3x-18|+\frac{1}{2}\int\frac{1}{\left(x+\dfrac{3}{2}\right)^2-\dfrac{81}{4}}\,dx
\displaystyle =\frac{3}{2}\log|x^2+3x-18|+\frac{1}{2}\int\frac{dx}{\left(x+\dfrac{3}{2}\right)^2-\left(\dfrac{9}{2}\right)^2}
\displaystyle =\frac{3}{2}\log|x^2+3x-18|+\frac{1}{2}\cdot\frac{1}{2\cdot\dfrac{9}{2}}\log\left|\frac{x+\dfrac{3}{2}-\dfrac{9}{2}}{x+\dfrac{3}{2}+\dfrac{9}{2}}\right|+C
\displaystyle \left[\because\int\frac{dx}{x^2-a^2}=\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|+C\right]
\displaystyle =\frac{3}{2}\log|x^2+3x-18|+\frac{1}{18}\log\left|\frac{x-3}{x+6}\right|+C
\\

\displaystyle \textbf{Question 69. }\text{Find the value of }\int\frac{\cos x}{(1+\sin x)(2+\sin x)}\,dx. \hspace{1.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{\cos x}{(1+\sin x)(2+\sin x)}\,dx
\displaystyle \text{Put }\sin x=t\Rightarrow\cos x\,dx=dt
\displaystyle \therefore\ I=\int\frac{dt}{(1+t)(2+t)}=\int\left[\frac{A}{1+t}+\frac{B}{2+t}\right]dt\qquad\ldots\text{(i)}
\displaystyle \therefore\ \frac{1}{(1+t)(2+t)}=\frac{A(2+t)+B(1+t)}{(1+t)(2+t)}
\displaystyle 1=2A+tA+B+tB
\displaystyle 1=1(2A+B)+t(A+B)
\displaystyle \text{On comparing the coefficients of }t\text{ and constant terms from both sides, we get}
\displaystyle 2A+B=1\text{ and }A+B=0
\displaystyle \Rightarrow A=1\text{ and }B=-1
\displaystyle \therefore\ I=\int\left(\frac{1}{1+t}-\frac{1}{2+t}\right)dt
\displaystyle =\int\frac{1}{(1+t)}\,dt-\int\frac{1}{(2+t)}\,dt
\displaystyle =\log|1+t|-\log|2+t|+C
\displaystyle =\log\left|\frac{1+t}{2+t}\right|+C
\displaystyle =\log\left|\frac{1+\sin x}{2+\sin x}\right|+C\qquad[\because t=\sin x]
\\

\displaystyle \textbf{Question 70. }\text{Find }\int\frac{x^2+x+1}{(x+2)(x^2+1)}\,dx. \hspace{1.2cm} \text{[CBSE 2019, 2016, 2015, 2009C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{x^2+x+1}{(x+2)(x^2+1)}\,dx
\displaystyle \text{By partial fraction,}
\displaystyle \frac{x^2+x+1}{(x^2+1)(x+2)}=\frac{A}{(x+2)}+\frac{Bx+C}{x^2+1}
\displaystyle \Rightarrow x^2+x+1=A(x^2+1)+(Bx+C)(x+2)
\displaystyle \text{Putting }x=-2\text{,}
\displaystyle 4-2+1=A(5)+0\Rightarrow 5A=3\Rightarrow A=\frac{3}{5}
\displaystyle \text{Putting }x=0\text{,}
\displaystyle 0+0+1=A(0+1)+(0+C)(0+2)
\displaystyle \Rightarrow 1=A+2C\Rightarrow 1=\frac{3}{5}+2C\Rightarrow 2C=\frac{2}{5}\Rightarrow C=\frac{1}{5}
\displaystyle \text{and putting }x=1\text{,}
\displaystyle 1+1+1=2A+(B+C)(3)
\displaystyle \Rightarrow 3=2A+3(B+C)
\displaystyle \Rightarrow 3=2\left(\frac{3}{5}\right)+3\left(B+\frac{1}{5}\right)
\displaystyle \Rightarrow 3-\frac{6}{5}=3\left(B+\frac{1}{5}\right)
\displaystyle \Rightarrow \frac{9}{5}=3\left(B+\frac{1}{5}\right)\Rightarrow\frac{3}{5}-\frac{1}{5}=B\Rightarrow B=\frac{2}{5}
\displaystyle \text{Thus, }\frac{x^2+x+1}{(x+2)(x^2+1)}=\frac{3}{5(x+2)}+\frac{\left(\dfrac{2}{5}x+\dfrac{1}{5}\right)}{(x^2+1)}
\displaystyle \text{Now, }\int\frac{x^2+x+1}{(x+2)(x^2+1)}\,dx
\displaystyle =\int\frac{3}{5(x+2)}\,dx+\frac{1}{5}\int\frac{2x}{x^2+1}\,dx+\frac{1}{5}\int\frac{dx}{x^2+1}
\displaystyle =\frac{3}{5}\log|x+2|+\frac{1}{5}\log|x^2+1|+\frac{1}{5}\tan^{-1}x+C
\\

\displaystyle \textbf{Question 71. }\text{Evaluate }\int\frac{\cos 2x+2\sin^2x}{\cos^2x}\,dx. \hspace{1.2cm} \text{[CBSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{\cos 2x+2\sin^2x}{\cos^2x}\,dx
\displaystyle =\int\frac{1-2\sin^2x+2\sin^2x}{\cos^2x}\,dx
\displaystyle [\because\cos 2A=1-2\sin^2A]
\displaystyle =\int\frac{1}{\cos^2x}\,dx=\int\sec^2x\,dx=\tan x+C
\\

\displaystyle \textbf{Question 72. }\text{Find }\int\frac{3-5\sin x}{\cos^2x}\,dx. \hspace{1.2cm} \text{[CBSE 2018C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{3-5\sin x}{\cos^2x}\,dx
\displaystyle =\int\left(\frac{3}{\cos^2x}-\frac{5\sin x}{\cos^2x}\right)dx
\displaystyle =3\int\sec^2x\,dx-5\int\sec x\tan x\,dx
\displaystyle =3\tan x-5\sec x+C
\\

\displaystyle \textbf{Question 73. }\text{Find }\int e^x\frac{\sqrt{1+\sin 2x}}{1+\cos 2x}\,dx. \hspace{1.2cm} \text{[CBSE Sample Paper 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int e^x\frac{\sqrt{1+\sin 2x}}{1+\cos 2x}\,dx
\displaystyle =\int e^x\frac{\sqrt{\sin^2x+\cos^2x+2\sin x\cos x}}{1+\cos 2x}\,dx
\displaystyle [\because\sin^2\theta+\cos^2\theta=1\text{ and }\sin 2\theta=2\sin\theta\cos\theta]
\displaystyle =\int e^x\frac{\sqrt{(\sin x+\cos x)^2}}{2\cos^2x}\,dx
\displaystyle [\because 1+\cos 2\theta=2\cos^2\theta]
\displaystyle =\int e^x\frac{(\sin x+\cos x)}{2\cos^2x}\,dx
\displaystyle =\frac{1}{2}\int e^x\left(\frac{\sin x}{\cos^2x}+\frac{\cos x}{\cos^2x}\right)dx
\displaystyle =\frac{1}{2}\int e^x(\sec x\tan x+\sec x)\,dx
\displaystyle \text{Let }f(x)=\sec x\text{, then }f'(x)=\sec x\tan x
\displaystyle =\frac{1}{2}e^x\sec x+C
\displaystyle \left[\because\int e^x(f(x)+f'(x))\,dx=e^x f(x)+C\right]
\\

\displaystyle \textbf{Question 74. }\text{Find }\int \frac{2\cos x}{(1-\sin x)(1+\sin^{2}x)}\,dx. \qquad \text{[CBSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \int \frac{2\cos x}{(1-\sin x)(1+\sin^{2}x)}\,dx
\displaystyle \text{Let }\sin x=t,\text{ so }\cos x\,dx=dt
\displaystyle =\int \frac{2\,dt}{(1-t)(1+t^{2})}
\displaystyle \text{Using partial fractions: }\frac{2}{(1-t)(1+t^{2})}=\frac{A}{1-t}+\frac{Bt+C}{1+t^{2}}
\displaystyle \Rightarrow 2=A(1+t^{2})+(Bt+C)(1-t)
\displaystyle \text{Putting }t=1:\quad 2=2A\Rightarrow A=1
\displaystyle \text{Comparing coefficients of }t^{2}:\quad 0=A-B\Rightarrow B=1
\displaystyle \text{Comparing constant terms:}\quad 2=A+C\Rightarrow C=1
\displaystyle =\int\left(\frac{1}{1-t}+\frac{t+1}{1+t^{2}}\right)dt
\displaystyle =\int\frac{1}{1-t}\,dt+\frac{1}{2}\int\frac{2t}{1+t^{2}}\,dt+\int\frac{1}{1+t^{2}}\,dt
\displaystyle =-\log|1-t|+\frac{1}{2}\log|1+t^{2}|+\tan^{-1}t+C
\displaystyle \therefore \int \frac{2\cos x}{(1-\sin x)(1+\sin^{2}x)}\,dx=-\log|1-\sin x|+\frac{1}{2}\log|1+\sin^{2}x|+\tan^{-1}(\sin x)+C
\\

\displaystyle \textbf{Question 75. }\text{Find }\int\frac{4}{(x-2)(x^2+4)}\,dx. \hspace{1.2cm} \text{[CBSE 2018C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{4}{(x-2)(x^2+4)}\,dx
\displaystyle \text{Again, let }\frac{4}{(x-2)(x^2+4)}=\frac{A}{(x-2)}+\frac{Bx+C}{x^2+4}
\displaystyle \Rightarrow 4=A(x^2+4)+(Bx+C)(x-2)
\displaystyle \Rightarrow 4=x^2(A+B)+x(-2B+C)+4A-2C
\displaystyle \text{On equating the coefficients of }x^2\text{, }x\text{ and constant terms from both sides, we get}
\displaystyle A+B=0\qquad\ldots\text{(i)}
\displaystyle -2B+C=0\qquad\ldots\text{(ii)}
\displaystyle \text{and }4A-2C=4\qquad\ldots\text{(iii)}
\displaystyle \text{On solving Eqs. (i), (ii) and (iii), we get }A=\frac{1}{2},\ B=-\frac{1}{2}\text{ and }C=-1
\displaystyle \therefore\ I=\int\frac{4}{(x-2)(x^2+4)}\,dx
\displaystyle =\int\frac{\dfrac{1}{2}}{(x-2)}\,dx+\int\frac{\dfrac{-x}{2}-1}{x^2+4}\,dx
\displaystyle =\frac{1}{2}\int\frac{dx}{(x-2)}-\int\frac{x+2}{2(x^2+4)}\,dx
\displaystyle =\frac{1}{2}\log|x-2|-\frac{1}{4}\log|x^2+4|-\frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right)+C
\\

\displaystyle \textbf{Question 76. }\text{Find }\int\frac{\sec x}{1+\text{cosec }x}\,dx. \hspace{1.2cm} \text{[CBSE Sample Paper 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{\sec x}{1+\text{cosec }x}\,dx=\int\frac{\dfrac{1}{\cos x}}{1+\dfrac{1}{\sin x}}\,dx
\displaystyle =\int\frac{\sin x\,dx}{\cos x(1+\sin x)}=\int\frac{\sin x\cos x\,dx}{\cos^2x(1+\sin x)}
\displaystyle =\int\frac{\sin x\cos x\,dx}{(1-\sin^2x)(1+\sin x)}
\displaystyle =\int\frac{\sin x\cos x\,dx}{(1-\sin x)(1+\sin x)^2}\qquad[\because a^2-b^2=(a+b)(a-b)]
\displaystyle \text{On putting }\sin x=t\Rightarrow\cos x\,dx=dt\text{, we get}
\displaystyle I=\int\frac{t}{(1+t)^2(1-t)}\,dt
\displaystyle \text{Now, }\frac{t}{(1+t)^2(1-t)}=\frac{A}{1+t}+\frac{B}{(1+t)^2}+\frac{C}{1-t}
\displaystyle \Rightarrow t=A(1+t)(1-t)+B(1-t)+C(1+t)^2
\displaystyle \text{Put }t=-1\Rightarrow-1=2B\Rightarrow B=\frac{-1}{2}
\displaystyle \text{Put }t=1\therefore\ 1=4C\Rightarrow C=\frac{1}{4}
\displaystyle \text{Put }t=0\therefore\ 0=A+B+C
\displaystyle \Rightarrow 0=A+\left(\frac{-1}{2}\right)+\left(\frac{1}{4}\right)
\displaystyle \Rightarrow A=\frac{1}{2}-\frac{1}{4}=\frac{2-1}{4}=\frac{1}{4}
\displaystyle \text{Hence, the required integral}
\displaystyle =\frac{1}{4}\int\frac{1}{1+t}\,dt+\left(\frac{-1}{2}\right)\int\frac{1}{(1+t)^2}\,dt+\frac{1}{4}\int\frac{1}{(1-t)}\,dt
\displaystyle =\frac{1}{4}\log|1+t|+\frac{-1}{2}\times\frac{-1}{1+t}-\frac{1}{4}\log|1-t|+C
\displaystyle =\frac{1}{4}\log|1+\sin x|+\frac{1}{2(1+\sin x)}-\frac{1}{4}\log|1-\sin x|+C\qquad[\because t=\sin x]
\\

\displaystyle \textbf{Question 77. }\text{Find }\int\frac{\sin^2x-\cos^2x}{\sin x\cos x}\,dx. \hspace{1.2cm} \text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{\sin^2x-\cos^2x}{\sin x\cdot\cos x}\,dx
\displaystyle =\int\left[\frac{\sin^2x}{\sin x\cdot\cos x}-\frac{\cos^2x}{\sin x\cdot\cos x}\right]dx
\displaystyle =\int\left[\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}\right]dx
\displaystyle =\int(\tan x-\cot x)\,dx=\int\tan x\,dx-\int\cot x\,dx
\displaystyle =\log|\sec x|-[-\log|\text{cosec }x|]+C
\displaystyle =\log|\sec x|+\log|\text{cosec }x|+C
\displaystyle =\log|\sec x\cdot\text{cosec }x|+C
\\

\displaystyle \textbf{Question 78. }\text{Find }\int \frac{dx}{5-8x-x^{2}}. \qquad \text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \int \frac{dx}{5-8x-x^{2}}
\displaystyle =\int \frac{dx}{-(x^{2}+8x-5)}
\displaystyle =\int \frac{dx}{-(x^{2}+8x+16-16-5)}
\displaystyle =\int \frac{dx}{-((x+4)^{2}-21)}
\displaystyle =\int \frac{dx}{21-(x+4)^{2}}
\displaystyle =\frac{1}{2\sqrt{21}}\log\left|\frac{\sqrt{21}+(x+4)}{\sqrt{21}-(x+4)}\right|+C
\displaystyle \therefore \int \frac{dx}{5-8x-x^{2}}=\frac{1}{2\sqrt{21}}\log\left|\frac{\sqrt{21}+x+4}{\sqrt{21}-x-4}\right|+C
\\

\displaystyle \textbf{Question 79. }\text{Find }\int\frac{2x}{(x^2+1)(x^2+2)^2}\,dx. \hspace{1.2cm} \text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{2x}{(x^2+1)(x^2+2)^2}\,dx
\displaystyle \text{Put }x^2=t\Rightarrow 2x\,dx=dt
\displaystyle \therefore\ I=\int\frac{dt}{(t+1)(t+2)^2}
\displaystyle \text{Let }\frac{1}{(t+1)(t+2)^2}=\frac{A}{t+1}+\frac{B}{t+2}+\frac{C}{(t+2)^2}
\displaystyle \Rightarrow\frac{1}{(t+1)(t+2)^2}=\frac{A(t+2)^2+B(t+1)(t+2)+C(t+1)}{(t+1)(t+2)^2}
\displaystyle \Rightarrow 1=A(t+2)^2+B(t+1)(t+2)+C(t+1)
\displaystyle \Rightarrow 1=A(t^2+4t+4)+B(t^2+2t+t+2)+C(t+1)
\displaystyle \Rightarrow 1=A(t^2+4t+4)+B(t^2+3t+2)+C(t+1)
\displaystyle \Rightarrow 1=t^2(A+B)+t(4A+3B+C)+4A+2B+C
\displaystyle \text{On comparing the coefficients of }t^2\text{, }t\text{ and constant terms from both sides, we get}
\displaystyle A+B=0\qquad\ldots\text{(i)}
\displaystyle 4A+3B+C=0\qquad\ldots\text{(ii)}
\displaystyle 4A+2B+C=1\qquad\ldots\text{(iii)}
\displaystyle \text{From Eq. (i), }A=-B
\displaystyle \text{Put the value of }A\text{ in Eqs. (ii) and (iii), we get}
\displaystyle -4B+3B+C=0\Rightarrow-B+C=0\Rightarrow B-C=0\qquad\ldots\text{(iv)}
\displaystyle \text{and }-4B+2B+C=1\Rightarrow-2B+C=1\qquad\ldots\text{(v)}
\displaystyle \text{Now, from Eqs. (iv) and (v), we get }-B=1\Rightarrow B=-1
\displaystyle \therefore\ A=1\text{ and }C=-1
\displaystyle \therefore\ I=\int\frac{1}{t+1}\,dt+\int\frac{(-1)}{t+2}\,dt+\int\frac{(-1)}{(t+2)^2}\,dt
\displaystyle =\log|t+1|-\log|t+2|-\frac{(t+2)^{-1}}{-1}+C
\displaystyle =\log|t+1|-\log|t+2|+\frac{1}{(t+2)}+C
\displaystyle \Rightarrow I=\log|x^2+1|-\log|x^2+2|+\frac{1}{(x^2+2)}+C\qquad[\because t=x^2]
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\displaystyle \textbf{Question 80. }\text{Find }\int\frac{2x}{(x^2+1)(x^4+4)}\,dx. \hspace{1.2cm} \text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{2x}{(x^2+1)(x^4+4)}\,dx
\displaystyle \text{Put }x^2=t\Rightarrow 2x\,dx=dt
\displaystyle \therefore\ I=\int\frac{dt}{(t+1)(t^2+4)}
\displaystyle \text{Now, }\frac{1}{(t+1)(t^2+4)}=\frac{A}{t+1}+\frac{Bt+C}{t^2+4}
\displaystyle \Rightarrow 1=A(t^2+4)+(Bt+C)(t+1)
\displaystyle \Rightarrow 1=A(t^2+4)+(Bt^2+Bt+Ct+C)
\displaystyle \Rightarrow 1=t^2(A+B)+t(B+C)+(4A+C)
\displaystyle \text{On comparing the coefficients of }t^2\text{, }t\text{ and constant terms from both sides, we get}
\displaystyle A+B=0\qquad\ldots\text{(i)}
\displaystyle B+C=0\qquad\ldots\text{(ii)}
\displaystyle 4A+C=1\qquad\ldots\text{(iii)}
\displaystyle \text{From Eqs. (i) and (ii), we get }A-C=0\qquad\ldots\text{(iv)}
\displaystyle \text{From Eqs. (iii) and (iv), we get }5A=1\Rightarrow A=\frac{1}{5}
\displaystyle \text{Then, }C=\frac{1}{5}\text{ and }B=-\frac{1}{5}
\displaystyle \text{Now, }I=\int\frac{dt}{(t+1)(t^2+4)}=\int\frac{1}{5(t+1)}\,dt+\int\frac{(-1/5)t+(1/5)}{t^2+4}\,dt
\displaystyle =\frac{1}{5}\int\frac{dt}{t+1}-\frac{1}{5}\int\frac{t-1}{t^2+4}\,dt
\displaystyle =\frac{1}{5}\log|t+1|-\frac{1}{5}\left[\int\frac{t}{t^2+4}\,dt-\int\frac{1}{t^2+4}\,dt\right]
\displaystyle =\frac{1}{5}\log|t+1|-\frac{1}{5}\left[\frac{1}{2}\log|t^2+4|-\frac{1}{2}\tan^{-1}\left(\frac{t}{2}\right)\right]+C
\displaystyle =\frac{1}{5}\log|x^2+1|-\frac{1}{5}\left[\frac{1}{2}\log|x^4+4|-\frac{1}{2}\tan^{-1}\left(\frac{x^2}{2}\right)\right]+C\qquad[\because t=x^2]
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\displaystyle \textbf{Question 81. }\text{Find }\int\frac{\cos\theta}{(4+\sin^2\theta)(5-4\cos^2\theta)}\,d\theta. \hspace{1.2cm} \text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{\cos\theta}{(4+\sin^2\theta)(5-4\cos^2\theta)}\,d\theta
\displaystyle =\int\frac{\cos\theta}{(4+\sin^2\theta)[5-4(1-\sin^2\theta)]}\,d\theta
\displaystyle =\int\frac{\cos\theta}{(4+\sin^2\theta)(5-4+4\sin^2\theta)}\,d\theta
\displaystyle =\int\frac{\cos\theta}{(4+\sin^2\theta)(1+4\sin^2\theta)}\,d\theta
\displaystyle \text{Let }\sin\theta=t\Rightarrow\cos\theta\,d\theta=dt
\displaystyle \text{Then, }I=\int\frac{dt}{(4+t^2)(1+4t^2)}\qquad\ldots\text{(i)}
\displaystyle \text{Again, let }\frac{1}{(4+t^2)(1+4t^2)}=\frac{A}{4+t^2}+\frac{B}{1+4t^2}\qquad\ldots\text{(ii)}\qquad\text{[by partial fraction]}
\displaystyle \text{At }t=0,\ \frac{A}{4}+\frac{B}{1}=\frac{1}{4\times 1}\Rightarrow A+4B=1\qquad\ldots\text{(iii)}
\displaystyle \text{At }t=1,\ \frac{A}{5}+\frac{B}{5}=\frac{1}{5\times 5}\Rightarrow 5A+5B=1\qquad\ldots\text{(iv)}
\displaystyle \text{On solving Eqs. (iii) and (iv), we get }A=-\frac{1}{15}\text{ and }B=\frac{4}{15}
\displaystyle \text{On putting }A=-\frac{1}{15}\text{ and }B=\frac{4}{15}\text{ in Eq. (ii), we get}
\displaystyle \frac{1}{(4+t^2)(1+4t^2)}=\frac{-\dfrac{1}{15}}{4+t^2}+\frac{\dfrac{4}{15}}{1+4t^2}
\displaystyle \Rightarrow\frac{1}{(4+t^2)(1+4t^2)}=\frac{-1}{15(4+t^2)}+\frac{4}{15(1+4t^2)}
\displaystyle \text{Now, }I=\int\frac{1}{(4+t^2)(1+4t^2)}\,dt
\displaystyle =\frac{-1}{15}\int\frac{1}{4+t^2}\,dt+\frac{4}{15}\int\frac{1}{1+4t^2}\,dt
\displaystyle =\frac{-1}{15}\int\frac{1}{2^2+t^2}\,dt+\frac{4}{15\times 4}\int\frac{1}{\left(\dfrac{1}{2}\right)^2+t^2}\,dt
\displaystyle =\frac{-1}{15}\cdot\frac{1}{2}\tan^{-1}\frac{t}{2}+\frac{1}{15}\cdot\frac{1}{1/2}\tan^{-1}\frac{t}{1/2}+C
\displaystyle \left[\because\int\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\frac{x}{a}+C\right]
\displaystyle =\frac{-1}{30}\tan^{-1}\frac{\sin\theta}{2}+\frac{2}{15}\tan^{-1}2\sin\theta+C\qquad[\because t=\sin\theta]
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\displaystyle \textbf{Question 82. }\text{Find }\int\frac{\sin x}{\sin^3x+\cos^3x}\,dx. \hspace{1.2cm} \text{[CBSE Sample Paper 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{\sin x}{\sin^3x+\cos^3x}\,dx=\int\frac{\sin x/\cos^3x}{\dfrac{\sin^3x}{\cos^3x}+\dfrac{\cos^3x}{\cos^3x}}\,dx
\displaystyle [\because\text{divide numerator and denominator by }\cos^3x]
\displaystyle =\int\frac{\tan x\sec^2x}{\tan^3x+1}\,dx
\displaystyle \text{On substituting }\tan x=t\text{ and }\sec^2x\,dx=dt\text{, we get}
\displaystyle I=\int\frac{t}{t^3+1}\,dt=\int\frac{t}{(t+1)(t^2-t+1)}\,dt
\displaystyle =-\frac{1}{3}\int\frac{1}{t+1}\,dt+\frac{1}{3}\int\frac{t+1}{t^2-t+1}\,dt
\displaystyle =-\frac{1}{3}\log|t+1|+\frac{1}{6}\int\frac{(2t-1)+3}{t^2-t+1}\,dt
\displaystyle =-\frac{1}{3}\log|t+1|+\frac{1}{6}\int\frac{2t-1}{t^2-t+1}\,dt+\frac{1}{2}\int\frac{1}{t^2-t+1}\,dt
\displaystyle =-\frac{1}{3}\log|t+1|+\frac{1}{6}\log|t^2-t+1|
\displaystyle +\frac{1}{2}\int\frac{1}{\left(t-\dfrac{1}{2}\right)^2+\left(\dfrac{\sqrt{3}}{2}\right)^2}\,dt
\displaystyle \left[\because\int\frac{2t-1}{t^2-t+1}\,dt=\log|t^2-t+1|\right]
\displaystyle =-\frac{1}{3}\log|t+1|+\frac{1}{6}\log|t^2-t+1|+\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2t-1}{\sqrt{3}}\right)+C
\displaystyle \left[\because\int\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)+C\right]
\displaystyle =-\frac{1}{3}\log|\tan x+1|+\frac{1}{6}\log|\tan^2x-\tan x+1|
\displaystyle +\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2\tan x-1}{\sqrt{3}}\right)+C\qquad[\because t=\tan x]
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\displaystyle \textbf{Question 83. }\text{Find }\int\frac{(3\sin\theta-2)\cos\theta}{5-\cos^2\theta-4\sin\theta}\,d\theta. \hspace{1.2cm} \text{[CBSE 2016]}
\displaystyle \text{Or Find }\int\frac{(3\sin x-2)\cos x}{5-\cos^2x-4\sin x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{(3\sin\theta-2)\cos\theta}{5-\cos^2\theta-4\sin\theta}\,d\theta
\displaystyle =\int\frac{(3\sin\theta-2)\cos\theta}{5-(1-\sin^2\theta)-4\sin\theta}\,d\theta
\displaystyle \text{Put }\sin\theta=t\Rightarrow\cos\theta\,d\theta=dt
\displaystyle \therefore\ I=\int\frac{3t-2}{5-(1-t^2)-4t}\,dt
\displaystyle =\int\frac{3t-2}{4+t^2-4t}\,dt=\int\frac{3t-2}{(t-2)^2}\,dt
\displaystyle =\int\frac{3t-6+4}{(t-2)^2}\,dt=\int\frac{3(t-2)+4}{(t-2)^2}\,dt
\displaystyle =\int\frac{3}{(t-2)}\,dt+\int\frac{4}{(t-2)^2}\,dt
\displaystyle =3\log|t-2|+\frac{4(t-2)^{-2+1}}{-2+1}+C
\displaystyle \left[\because\int x^n\,dx=\frac{x^{n+1}}{n+1}+C,\ n\neq-1\right]
\displaystyle =3\log|t-2|-\frac{4}{(t-2)}+C
\displaystyle =3\log|\sin\theta-2|-\frac{4}{(\sin\theta-2)}+C\qquad[\because t=\sin\theta]
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\displaystyle \textbf{Question 84. }\text{Find }\int\frac{\sqrt{x}}{\sqrt{a^3-x^3}}\,dx. \hspace{1.2cm} \text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{\sqrt{x}}{\sqrt{a^3-x^3}}\,dx=\int\frac{\sqrt{x}}{\sqrt{(a^{3/2})^2-(x^{3/2})^2}}\,dx
\displaystyle \text{Put }x^{3/2}=a^{3/2}t
\displaystyle \Rightarrow\frac{3}{2}x^{1/2}\,dx=a^{3/2}\,dt\Rightarrow\sqrt{x}\,dx=\frac{2}{3}a^{3/2}\,dt
\displaystyle \therefore\ I=\int\frac{\dfrac{2}{3}a^{3/2}}{\sqrt{(a^{3/2})^2-(a^{3/2}t)^2}}\,dt
\displaystyle =\frac{2}{3}a^{3/2}\int\frac{dt}{a^{3/2}\sqrt{1-t^2}}
\displaystyle =\frac{2}{3}\int\frac{dt}{\sqrt{1-t^2}}=\frac{2}{3}\sin^{-1}\left(\frac{t}{1}\right)+C
\displaystyle \left[\because\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\left(\frac{x}{a}\right)+C\right]
\displaystyle =\frac{2}{3}\sin^{-1}\left(\frac{x^{3/2}}{a^{3/2}}\right)+C\qquad\left[\because t=\frac{x^{3/2}}{a^{3/2}}\right]
\displaystyle =\frac{2}{3}\sin^{-1}\left(\sqrt{\frac{x^3}{a^3}}\right)+C
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\displaystyle \textbf{Question 85. }\text{Find }\int\frac{(2x-5)e^{2x}}{(2x-3)^3}\,dx. \hspace{1.2cm} \text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{(2x-5)e^{2x}}{(2x-3)^3}\,dx=\int\frac{(2x-3-2)e^{2x}}{(2x-3)^3}\,dx
\displaystyle =\int\frac{e^{2x}}{(2x-3)^2}\,dx-2\int\frac{e^{2x}}{(2x-3)^3}\,dx
\displaystyle =\int e^{2x}(2x-3)^{-2}\,dx-2\int e^{2x}(2x-3)^{-3}\,dx
\displaystyle \left[(2x-3)^{-2}\int e^{2x}\,dx\right.
\displaystyle \left.-\int\left\{\frac{d}{dx}(2x-3)^{-2}\int e^{2x}\,dx\right\}dx\right]
\displaystyle -2\int e^{2x}(2x-3)^{-3}\,dx\qquad\text{[using integration by parts]}
\displaystyle =(2x-3)^{-2}\frac{e^{2x}}{2}-\int-2(2x-3)^{-3}\times 2\times\frac{e^{2x}}{2}\,dx
\displaystyle -2\int e^{2x}(2x-3)^{-3}\,dx
\displaystyle =\frac{e^{2x}(2x-3)^{-2}}{2}+2\int e^{2x}(2x-3)^{-3}\,dx
\displaystyle -2\int e^{2x}(2x-3)^{-3}\,dx
\displaystyle =\frac{e^{2x}(2x-3)^{-2}}{2}+C
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\displaystyle \textbf{Question 86. }\text{Find }\int\frac{(x^2+1)(x^2+4)}{(x^2+3)(x^2-5)}\,dx. \hspace{1.2cm} \text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{(x^2+1)(x^2+4)}{(x^2+3)(x^2-5)}\,dx
\displaystyle \text{Consider, }\frac{(x^2+1)(x^2+4)}{(x^2+3)(x^2-5)}\text{ and put }x^2=y
\displaystyle \text{Then, }\frac{(x^2+1)(x^2+4)}{(x^2+3)(x^2-5)}=\frac{(y+1)(y+4)}{(y+3)(y-5)}
\displaystyle =\frac{y^2+5y+4}{y^2-2y-15}=\frac{(y^2-2y-15)+(7y+19)}{y^2-2y-15}
\displaystyle =1+\frac{7y+19}{y^2-2y-15}=1+\frac{7y+19}{(y+3)(y-5)}\qquad\ldots\text{(i)}
\displaystyle \text{Now, let us write }\frac{7y+19}{(y+3)(y-5)}=\frac{A}{y+3}+\frac{B}{y-5}
\displaystyle \Rightarrow 7y+19=A(y-5)+B(y+3)
\displaystyle \text{On putting }y=5\text{, we get }35+19=8B\Rightarrow B=\frac{54}{8}=\frac{27}{4}
\displaystyle \text{and on putting }y=-3\text{, we get }-21+19=-8A\Rightarrow A=\frac{(-2)}{(-8)}=\frac{1}{4}
\displaystyle \text{Thus, }\frac{7y+19}{(y+3)(y-3)}=\frac{1}{4}\cdot\frac{1}{(y+3)}+\frac{27}{4}\cdot\frac{1}{(y-5)}\qquad\ldots\text{(ii)}
\displaystyle \text{From Eqs. (i) and (ii), we get}
\displaystyle \frac{(x^2+1)(x^2+4)}{(x^2+3)(x^2-5)}=1+\frac{1}{4}\cdot\frac{1}{(x^2+3)}+\frac{27}{4}\cdot\frac{1}{(x^2-5)}
\displaystyle \text{Now, }I=\int\left(1+\frac{1}{4}\cdot\frac{1}{x^2+3}+\frac{27}{4}\cdot\frac{1}{x^2-5}\right)dx
\displaystyle =\int dx+\frac{1}{4}\int\frac{dx}{x^2+3}+\frac{27}{4}\int\frac{dx}{x^2-5}
\displaystyle =x+\frac{1}{4}\int\frac{dx}{x^2+(\sqrt{3})^2}+\frac{27}{4}\int\frac{dx}{x^2-(\sqrt{5})^2}
\displaystyle =x+\frac{1}{4\sqrt{3}}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right)+\frac{27}{4}\cdot\frac{1}{2\sqrt{5}}\log\left|\frac{x-\sqrt{5}}{x+\sqrt{5}}\right|+C
\displaystyle \left[\because\int\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)+C\text{ and }\int\frac{dx}{x^2-a^2}=\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|+C\right]
\displaystyle =x+\frac{1}{4\sqrt{3}}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right)+\frac{27}{8\sqrt{5}}\log\left|\frac{x-\sqrt{5}}{x+\sqrt{5}}\right|+C
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\displaystyle \textbf{Question 87. }\text{Evaluate }\int\frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx. \hspace{1.2cm} \text{[CBSE 2016; CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx
\displaystyle \text{Put }\sin^{-1}x=t\Rightarrow\frac{1}{\sqrt{1-x^2}}\,dx=dt
\displaystyle \therefore\ I=\int t\sin t\,dt
\displaystyle \text{Using integration by parts, taking }t\text{ as the first function and }\sin t\text{ as the second function, we get}
\displaystyle I=t\int\sin t\,dt-\int\left[\frac{d}{dt}(t)\cdot\int\sin t\,dt\right]dt
\displaystyle =-t\cos t-\int(1\times-\cos t)\,dt
\displaystyle =-t\cos t+\int\cos t\,dt
\displaystyle =-t\cos t+\sin t+C
\displaystyle =-t\sqrt{1-\sin^2t}+\sin t+C
\displaystyle [\because\cos^2\theta=1-\sin^2\theta\Rightarrow\cos\theta=\sqrt{1-\sin^2\theta}]
\displaystyle \therefore\ I=-\sin^{-1}x\sqrt{1-x^2}+x+C
\displaystyle [\because t=\sin^{-1}x\Rightarrow x=\sin t]
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\displaystyle \textbf{Question 88. }\text{Find }\int\frac{dx}{\sin x+\sin 2x}. \hspace{1.2cm} \text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{dx}{\sin x+\sin 2x}=\int\frac{dx}{\sin x+2\sin x\cos x}
\displaystyle [\because\sin 2\theta=2\sin\theta\cos\theta]
\displaystyle =\int\frac{dx}{\sin x(1+2\cos x)}=\int\frac{\sin x}{\sin^2x(1+2\cos x)}\,dx
\displaystyle \text{[multiplying numerator and denominator by }\sin x\text{]}
\displaystyle =\int\frac{\sin x}{(1-\cos^2x)(1+2\cos x)}\,dx
\displaystyle \text{Put }\cos x=t\Rightarrow-\sin x\,dx=dt\Rightarrow\sin x\,dx=-dt
\displaystyle \therefore\ I=\int\frac{-dt}{(1-t^2)(1+2t)}
\displaystyle =\int\frac{-dt}{(1-t)(1+t)(1+2t)}\qquad\ldots\text{(i)}
\displaystyle \text{Now, using partial fraction,}
\displaystyle \text{let }\frac{1}{(1-t)(1+t)(1+2t)}=\frac{A}{1-t}+\frac{B}{1+t}+\frac{C}{1+2t}\qquad\ldots\text{(ii)}
\displaystyle \Rightarrow 1=(1+t)(1+2t)A+(1-t)(1+2t)B+(1-t)(1+t)C\qquad\ldots\text{(iii)}
\displaystyle \text{On putting }t=-1\text{ in Eq. (iii), we get }1=(2)(-1)B\Rightarrow B=-\frac{1}{2}
\displaystyle \text{On putting }t=1\text{ in Eq. (iii), we get }1=2\cdot 3\cdot A\Rightarrow A=\frac{1}{6}
\displaystyle \text{On putting }t=-\frac{1}{2}\text{ in Eq. (iii), we get}
\displaystyle 1=\left(1+\frac{1}{2}\right)\left(1-\frac{1}{2}\right)C\Rightarrow 1=\left(\frac{3}{2}\times\frac{1}{2}\right)C\Rightarrow C=\frac{4}{3}
\displaystyle \therefore\ I=-\left[\int\frac{A}{1-t}\,dt+\int\frac{B}{1+t}\,dt+\int\frac{C}{1+2t}\,dt\right]\qquad\text{[using Eqs. (i) and (ii)]}
\displaystyle =-\left[\frac{1}{6}\int\frac{dt}{1-t}+\left(\frac{-1}{2}\right)\int\frac{dt}{1+t}+\frac{4}{3}\int\frac{dt}{1+2t}\right]
\displaystyle =-\left[\frac{1}{6}\cdot\frac{\log|1-t|}{-1}-\frac{1}{2}\log|1+t|+\frac{4}{3}\cdot\frac{\log|1+2t|}{2}\right]+C
\displaystyle =\frac{1}{6}\log|1-t|+\frac{1}{2}\log|1+t|-\frac{2}{3}\log|1+2t|+C
\displaystyle =\frac{1}{6}\log|1-\cos x|+\frac{1}{2}\log|1+\cos x|-\frac{2}{3}\log|1+2\cos x|+C\qquad[\because t=\cos x]
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\displaystyle \textbf{Question 89. }\text{Find }\int\frac{\log|x|}{(x+1)^2}\,dx. \hspace{1.2cm} \text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{\log|x|}{(x+1)^2}\,dx=\int\log|x|\cdot\frac{1}{(x+1)^2}\,dx
\displaystyle \text{On applying integration by parts, we get}
\displaystyle I=\log|x|\cdot\int\frac{dx}{(x+1)^2}-\int\frac{d}{dx}(\log|x|)\cdot\left(\int\frac{dx}{(x+1)^2}\right)dx
\displaystyle \left[\because\int u\cdot v\,dx=u\int v\,dx-\int\left(\frac{d}{dx}(u)\int v\,dx\right)dx\right]
\displaystyle =\log|x|\cdot\frac{(-1)}{x+1}+\int\frac{1}{x(x+1)}\,dx
\displaystyle =\frac{-\log|x|}{x+1}+I_1\text{ (say)}\qquad\ldots\text{(i)}
\displaystyle \text{Consider, }I_1=\int\frac{dx}{x(x+1)}
\displaystyle \text{Now, by using partial fraction method,}
\displaystyle \text{let }\frac{1}{x(x+1)}=\frac{A}{x}+\frac{B}{x+1}
\displaystyle \Rightarrow 1=A(x+1)+Bx
\displaystyle \text{On putting }x=0\text{, we get }A=1
\displaystyle \text{and again putting }x=-1\text{, we get }B=-1
\displaystyle \therefore\ I_1=\int\left(\frac{1}{x}-\frac{1}{x+1}\right)dx=\int\frac{1}{x}\,dx-\int\frac{dx}{x+1}
\displaystyle =\log|x|-\log|x+1|+C\qquad\ldots\text{(ii)}
\displaystyle \text{Now, from Eqs. (i) and (ii), we get}
\displaystyle I=\frac{-\log|x|}{x+1}+\log|x|-\log|x+1|+C
\displaystyle =\frac{-\log|x|}{x+1}+\log\left|\frac{x}{x+1}\right|+C
\displaystyle \left[\because\log m-\log n=\log\frac{m}{n}\right]
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\displaystyle \textbf{Question 90. }\text{Evaluate }\int\frac{\sin(x-a)}{\sin(x+a)}\,dx. \hspace{1.2cm} \text{[CBSE 2015; CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{\sin(x-a)}{\sin(x+a)}\,dx
\displaystyle \text{Put }x+a=t\Rightarrow dx=dt
\displaystyle \therefore\ I=\int\frac{\sin(t-a-a)}{\sin t}\,dt=\int\frac{\sin(t-2a)}{\sin t}\,dt
\displaystyle =\int\frac{\sin t\cos 2a-\cos t\sin 2a}{\sin t}\,dt
\displaystyle [\because\sin(A-B)=\sin A\cos B-\cos A\sin B]
\displaystyle =\int\cos 2a\,dt-\int\sin 2a\cdot\cot t\,dt
\displaystyle =\cos 2a[t]-\sin 2a[\log|\sin t|]+C_1
\displaystyle =(x+a)\cos 2a-\sin 2a\log|\sin(x+a)|+C_1\qquad[\because t=x+a]
\displaystyle =x\cos 2a-\sin 2a\log|\sin(x+a)|+C\text{,}
\displaystyle \text{where }C=a\cos 2a+C_1
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\displaystyle \textbf{Question 91. }\text{Find }\int e^{2x}\cdot\sin(3x+1)\,dx. \hspace{1.2cm} \text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int e^{2x}\sin(3x+1)\,dx\qquad\ldots\text{(i)}
\displaystyle =\sin(3x+1)\int e^{2x}\,dx-\int\left\{\frac{d}{dx}\sin(3x+1)\int e^{2x}\,dx\right\}dx
\displaystyle \qquad\text{[by using integration by parts]}
\displaystyle =\frac{\sin(3x+1)\cdot e^{2x}}{2}-\int 3\cos(3x+1)\cdot\frac{e^{2x}}{2}\,dx
\displaystyle =\frac{e^{2x}\sin(3x+1)}{2}-\frac{3}{2}\int e^{2x}\cos(3x+1)\,dx
\displaystyle =\frac{e^{2x}\sin(3x+1)}{2}-\frac{3}{2}\left[\cos(3x+1)\int e^{2x}\,dx-\int\left\{\frac{d}{dx}\cos(3x+1)\int e^{2x}\,dx\right\}dx\right]
\displaystyle \qquad\text{[again by using integration by parts]}
\displaystyle =\frac{e^{2x}\sin(3x+1)}{2}-\frac{3}{2}\left[\cos(3x+1)\cdot\frac{e^{2x}}{2}-\int-3\sin(3x+1)\cdot\frac{e^{2x}}{2}\,dx\right]+C_1
\displaystyle \Rightarrow I=\frac{e^{2x}\sin(3x+1)}{2}-\frac{3}{4}e^{2x}\cos(3x+1)-\frac{9}{4}\int e^{2x}\sin(3x+1)\,dx+C_1
\displaystyle \Rightarrow I=\frac{e^{2x}\sin(3x+1)}{2}-\frac{3}{4}e^{2x}\cos(3x+1)-\frac{9}{4}I+C_1\qquad\text{[from Eq. (i)]}
\displaystyle \Rightarrow\frac{13}{4}I=\frac{e^{2x}\sin(3x+1)}{2}-\frac{3e^{2x}\cos(3x+1)}{4}+C_1
\displaystyle \therefore\ I=\frac{2e^{2x}\sin(3x+1)}{13}-\frac{3e^{2x}\cos(3x+1)}{13}+C\text{,}
\displaystyle \text{where }C=\frac{4C_1}{13}
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\displaystyle \textbf{Question 92. }\text{Evaluate }\int\frac{x^2}{(x^2+4)(x^2+9)}\,dx. \hspace{1.2cm} \text{[CBSE 2015; CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{x^2}{(x^2+4)(x^2+9)}\,dx
\displaystyle \text{Consider, }\frac{x^2}{(x^2+4)(x^2+9)}\text{ and put }x^2=t
\displaystyle \text{Then, }\frac{x^2}{(x^2+4)(x^2+9)}=\frac{t}{(t+4)(t+9)}
\displaystyle \text{Now, let }\frac{t}{(t+4)(t+9)}=\frac{A}{(t+4)}+\frac{B}{(t+9)}
\displaystyle \Rightarrow t=A(t+9)+B(t+4)
\displaystyle \text{On putting }t=-9\text{, we get }-9=-5B\Rightarrow B=\frac{9}{5}
\displaystyle \text{On putting }t=-4\text{, we get }-4=5A\Rightarrow A=\frac{-4}{5}
\displaystyle \text{Thus, }\frac{t}{(t+4)(t+9)}=\frac{-4}{5(t+4)}+\frac{9}{5(t+9)}
\displaystyle \therefore\ I=\int\frac{x^2}{(x^2+4)(x^2+9)}\,dx
\displaystyle =\int\frac{-4}{5(x^2+4)}\,dx+\int\frac{9}{5(x^2+9)}\,dx
\displaystyle =-\frac{4}{5}\int\frac{dx}{x^2+(2)^2}+\frac{9}{5}\int\frac{dx}{x^2+(3)^2}
\displaystyle =-\frac{4}{5}\cdot\frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right)+\frac{9}{5}\cdot\frac{1}{3}\tan^{-1}\left(\frac{x}{3}\right)+C
\displaystyle \left[\because\int\frac{1}{x^2+a^2}\,dx=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)+C\right]
\displaystyle =\frac{3}{5}\tan^{-1}\left(\frac{x}{3}\right)-\frac{2}{5}\tan^{-1}\left(\frac{x}{2}\right)+C
\\

\displaystyle \textbf{Question 93. }\text{Find }\int\frac{(x^2+1)e^x}{(x+1)^2}\,dx. \hspace{1.2cm} \text{[CBSE 2015C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int e^x\frac{(x^2+1)}{(x+1)^2}\,dx
\displaystyle =\int e^x\frac{(x^2+1+2x-2x)}{(x+1)^2}\,dx
\displaystyle =\int e^x\frac{(x+1)^2-2x}{(x+1)^2}\,dx
\displaystyle =\int e^x\left(1-\frac{2x}{(x+1)^2}\right)dx
\displaystyle =\int e^x\,dx-2\int e^x\cdot\frac{x}{(x+1)^2}\,dx
\displaystyle =e^x-2\int e^x\cdot\frac{x+1-1}{(x+1)^2}\,dx
\displaystyle =e^x-2\int e^x\left(\frac{1}{(x+1)}+\frac{(-1)}{(x+1)^2}\right)dx
\displaystyle \text{Now, consider }f(x)=\frac{1}{x+1}\text{, then }f'(x)=\frac{-1}{(x+1)^2}
\displaystyle \text{Thus, the above integrand is of the form }e^x[f(x)+f'(x)]
\displaystyle \therefore\ I=e^x-2e^x\cdot\frac{1}{(x+1)}+C
\displaystyle \left[\because\int e^x[f(x)+f'(x)]\,dx=e^x f(x)+C\right]
\displaystyle \Rightarrow I=e^x\left(\frac{x+1-2}{x+1}\right)+C
\displaystyle \Rightarrow I=e^x\left(\frac{x-1}{x+1}\right)+C
\\

\displaystyle \textbf{Question 94. }\text{Find }\int \frac{\sin^{2}x-\cos^{2}x}{\sin^{2}x\cos^{2}x}\,dx. \qquad \text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle \int \frac{\sin^{2}x-\cos^{2}x}{\sin^{2}x\cos^{2}x}\,dx
\displaystyle =\int \left(\frac{\sin^{2}x}{\sin^{2}x\cos^{2}x}-\frac{\cos^{2}x}{\sin^{2}x\cos^{2}x}\right)\,dx
\displaystyle =\int \left(\sec^{2}x-\mathrm{cosec}^{2}x\right)\,dx
\displaystyle =\tan x+\cot x+C
\displaystyle \therefore \int \frac{\sin^{2}x-\cos^{2}x}{\sin^{2}x\cos^{2}x}\,dx=\tan x+\cot x+C
\\

\displaystyle \textbf{Question 95. }\text{Find }\int\frac{\sin^6x}{\cos^8x}\,dx. \hspace{1.2cm} \text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{\sin^6x}{\cos^8x}\,dx=\int\tan^6x\cdot\sec^2x\,dx
\displaystyle \text{Put }\tan x=t\Rightarrow\sec^2x\,dx=dt
\displaystyle \therefore\ I=\int t^6\,dt=\frac{t^7}{7}+C=\frac{\tan^7x}{7}+C\qquad[\because t=\tan x]
\\

\displaystyle \textbf{Question 96. }\text{Evaluate }\int\frac{dx}{\sin^2x\cos^2x}. \hspace{1.2cm} \text{[CBSE 2014C; CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{dx}{\sin^2x\cos^2x}
\displaystyle =\int\frac{(\sin^2x+\cos^2x)}{\sin^2x\cdot\cos^2x}\,dx\qquad[\because\sin^2\theta+\cos^2\theta=1]
\displaystyle =\int(\sec^2x+\text{cosec}^2x)\,dx
\displaystyle =\int\sec^2x\,dx+\int\text{cosec}^2x\,dx
\displaystyle =\tan x-\cot x+C
\displaystyle \textbf{Alternate Method}
\displaystyle \text{On dividing the numerator and denominator by }\cos^4x\text{, we get}
\displaystyle I=\int\frac{\sec^2x\cdot\sec^2x}{\tan^2x}\,dx
\displaystyle \Rightarrow I=\int\frac{(1+\tan^2x)\cdot\sec^2x}{\tan^2x}\,dx
\displaystyle \text{Put }\tan x=t\Rightarrow\sec^2x\,dx=dt
\displaystyle \therefore\ I=\int\frac{1+t^2}{t^2}\,dt=\int 1\,dt+\int\frac{1}{t^2}\,dt
\displaystyle \Rightarrow I=t-\frac{1}{t}+C
\displaystyle \Rightarrow I=\tan x-\cot x+C\qquad[\because t=\tan x]
\\

\displaystyle \textbf{Question 97. }\text{Evaluate }\int\cos^{-1}(\sin x)\,dx. \hspace{1.2cm} \text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\cos^{-1}(\sin x)\,dx
\displaystyle =\int\cos^{-1}\left[\cos\left(\frac{\pi}{2}-x\right)\right]dx
\displaystyle =\int\left(\frac{\pi}{2}-x\right)dx\qquad[\because\cos^{-1}(\cos\theta)=\theta]
\displaystyle =\frac{\pi}{2}\int dx-\int x\,dx
\displaystyle =\frac{\pi}{2}x-\frac{x^2}{2}+C
\\

\displaystyle \textbf{Question 98. }\text{Write the anti-derivative of }\left(3\sqrt{x}+\frac{1}{\sqrt{x}}\right). \hspace{1.2cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Anti-derivative of }\left(3\sqrt{x}+\frac{1}{\sqrt{x}}\right)
\displaystyle =\int\left(3\sqrt{x}+\frac{1}{\sqrt{x}}\right)dx=3\int\sqrt{x}\,dx+\int\frac{1}{\sqrt{x}}\,dx
\displaystyle =3\left(\frac{x^{(1/2)+1}}{(1/2)+1}\right)+\left[\frac{x^{(-1/2)+1}}{(-1/2)+1}\right]+C
\displaystyle =2(x^{3/2}+x^{1/2})+C
\\

\displaystyle \textbf{Question 99. }\text{Evaluate }\int\frac{x+2}{\sqrt{x^2+5x+6}}\,dx. \hspace{1.2cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{x+2}{\sqrt{x^2+5x+6}}\,dx
\displaystyle \text{Now, let us write, }(x+2)\text{ as}
\displaystyle x+2=A\frac{d}{dx}(x^2+5x+6)+B
\displaystyle \Rightarrow x+2=A(2x+5)+B
\displaystyle \text{On equating the coefficients of }x\text{ and constant terms from both sides, we get}
\displaystyle 2A=1\text{ and }5A+B=2
\displaystyle \Rightarrow A=\frac{1}{2}\text{ and }B=-\frac{1}{2}
\displaystyle \therefore\ I=\int\frac{\left[\dfrac{1}{2}(2x+5)-\dfrac{1}{2}\right]}{\sqrt{x^2+5x+6}}\,dx
\displaystyle =\frac{1}{2}\int\frac{2x+5}{\sqrt{x^2+5x+6}}\,dx-\frac{1}{2}\int\frac{1}{\sqrt{x^2+5x+6}}\,dx
\displaystyle =\frac{1}{2}I_1-\frac{1}{2}I_2\text{ (say)}\qquad\ldots\text{(i)}
\displaystyle \text{Consider, }I_1=\int\frac{2x+5}{\sqrt{x^2+5x+6}}\,dx
\displaystyle \text{Put }x^2+5x+6=t\Rightarrow(2x+5)\,dx=dt
\displaystyle \therefore\ I_1=\int\frac{dt}{\sqrt{t}}=2\sqrt{t}+C_1=2\sqrt{x^2+5x+6}+C_1\qquad\ldots\text{(ii)}
\displaystyle [\because t=x^2+5x+6]
\displaystyle \text{and }I_2=\int\frac{1}{\sqrt{x^2+5x+6}}\,dx
\displaystyle =\int\frac{1}{\sqrt{x^2+2\times\dfrac{5}{2}\times x+\dfrac{25}{4}-\dfrac{25}{4}+6}}\,dx
\displaystyle =\int\frac{1}{\sqrt{\left(x+\dfrac{5}{2}\right)^2+6-\dfrac{25}{4}}}\,dx
\displaystyle =\int\frac{1}{\sqrt{\left(x+\dfrac{5}{2}\right)^2-\left(\dfrac{1}{2}\right)^2}}\,dx
\displaystyle =\log\left|\left(x+\frac{5}{2}\right)+\sqrt{\left(x+\frac{5}{2}\right)^2-\left(\frac{1}{2}\right)^2}\right|+C_2
\displaystyle \left[\because\int\frac{dx}{\sqrt{x^2-a^2}}=\log|x+\sqrt{x^2-a^2}|+C\right]
\displaystyle \Rightarrow I_2=\log\left|x+\frac{5}{2}+\sqrt{x^2+5x+6}\right|+C_2\qquad\ldots\text{(iii)}
\displaystyle \text{On putting the values of }I_1\text{ and }I_2\text{ from Eqs. (ii) and (iii) in Eq. (i), we get}
\displaystyle I=\frac{1}{2}\left[2\sqrt{x^2+5x+6}+C_1\right]-\frac{1}{2}\left[\log\left|x+\frac{5}{2}+\sqrt{x^2+5x+6}\right|+C_2\right]
\displaystyle \Rightarrow I=\sqrt{x^2+5x+6}-\frac{1}{2}\log\left|x+\frac{5}{2}+\sqrt{x^2+5x+6}\right|+C
\displaystyle \text{where }C=\frac{C_1}{2}-\frac{C_2}{2}.
\\

\displaystyle \textbf{Question 100. }\text{Find }\int\frac{5x-2}{1+2x+3x^2}\,dx. \hspace{1.2cm} \text{[CBSE 2014C; CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{5x-2}{1+2x+3x^2}\,dx\qquad\ldots\text{(i)}
\displaystyle \text{Here, }(5x-2)\text{ can be written as}
\displaystyle 5x-2=A\frac{d}{dx}(1+2x+3x^2)+B
\displaystyle \Rightarrow 5x-2=A(2+6x)+B
\displaystyle \text{On comparing the coefficients of }x\text{ and constant terms, we get}
\displaystyle 5=6A\Rightarrow A=\frac{5}{6}
\displaystyle \text{and }-2=2A+B\Rightarrow B=-2A-2=-\frac{5}{3}-2=-\frac{11}{3}
\displaystyle \text{Then, from Eq. (i), we get}
\displaystyle I=\int\frac{\dfrac{5}{6}(2+6x)-\dfrac{11}{3}}{1+2x+3x^2}\,dx
\displaystyle \Rightarrow I=I_1-I_2\qquad\ldots\text{(ii)}
\displaystyle \text{where }I_1=\frac{5}{6}\int\frac{2+6x}{1+2x+3x^2}\,dx
\displaystyle \text{Put }1+2x+3x^2=t\Rightarrow(2+6x)\,dx=dt
\displaystyle \therefore\ I_1=\frac{5}{6}\int\frac{dt}{t}=\frac{5}{6}\log|t|+C_1=\frac{5}{6}\log|1+2x+3x^2|+C_1
\displaystyle [\because t=1+2x+3x^2]
\displaystyle \text{and }I_2=\frac{11}{3}\int\frac{dx}{3x^2+2x+1}=\frac{11}{9}\int\frac{dx}{x^2+\dfrac{2x}{3}+\dfrac{1}{3}}
\displaystyle =\frac{11}{9}\int\frac{dx}{\left(x+\dfrac{1}{3}\right)^2+\dfrac{2}{9}}
\displaystyle =\frac{11}{9}\cdot\frac{1}{\sqrt{2}/3}\tan^{-1}\left(\frac{x+\dfrac{1}{3}}{\dfrac{\sqrt{2}}{3}}\right)+C_2
\displaystyle \left[\because\int\frac{1}{x^2+a^2}\,dx=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)+C\right]
\displaystyle =\frac{11}{3\sqrt{2}}\tan^{-1}\left(\frac{3x+1}{\sqrt{2}}\right)+C_2
\displaystyle \text{On putting the values of }I_1\text{ and }I_2\text{ in Eq. (ii), we get}
\displaystyle I=\frac{5}{6}\log|1+2x+3x^2|-\frac{11}{3\sqrt{2}}\tan^{-1}\left(\frac{3x+1}{\sqrt{2}}\right)+C\text{,}
\displaystyle \text{where }C=C_1-C_2
\\

\displaystyle \textbf{Question 101. }\text{Find }\int\frac{x^3}{x^4+3x^2+2}\,dx. \hspace{1.2cm} \text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{x^3}{x^4+3x^2+2}\,dx
\displaystyle \text{Put }x^2=t\Rightarrow 2x\,dx=\frac{dt}{2}\Rightarrow x\,dx=\frac{dt}{2}
\displaystyle \therefore\ I=\frac{1}{2}\int\frac{t}{t^2+3t+2}\,dt=\frac{1}{2}\int\frac{t}{(t+2)(t+1)}\,dt
\displaystyle \text{Put }\frac{t}{(t+1)(t+2)}=\frac{A}{t+1}+\frac{B}{t+2}\qquad\text{[by partial fraction]}
\displaystyle t=A(t+2)+B(t+1)
\displaystyle t=At+2A+Bt+B
\displaystyle A+B=1\text{ and }2A+B=0
\displaystyle \Rightarrow A=-1\text{ and }B=2
\displaystyle \therefore\ I=\frac{1}{2}\int\left(\frac{-1}{t+1}+\frac{2}{t+2}\right)dt
\displaystyle =\frac{1}{2}\left[\int\frac{-1}{t+1}\,dt+\int\frac{2}{t+2}\,dt\right]
\displaystyle =\frac{1}{2}[-\log|t+1|+2\log|t+2|]+C
\displaystyle =\log|t+2|-\frac{1}{2}\log|t+1|+C
\displaystyle =\log\left|\frac{t+2}{\sqrt{t+1}}\right|+C=\log\left|\frac{x^2+2}{\sqrt{x^2+1}}\right|+C\qquad[\because t=x^2]
\\

\displaystyle \textbf{Question 102. }\text{Evaluate }\int \frac{x\cos^{-1}x}{\sqrt{1-x^{2}}}\,dx. \qquad \text{[CBSE 2014C; CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \int \frac{x\cos^{-1}x}{\sqrt{1-x^{2}}}\,dx
\displaystyle \text{Let }\cos^{-1}x=t,\text{ so }\frac{-1}{\sqrt{1-x^{2}}}\,dx=dt\Rightarrow \frac{dx}{\sqrt{1-x^{2}}}=-\,dt\text{ and }x=\cos t
\displaystyle =\int \cos t\cdot t\cdot(-dt)
\displaystyle =-\int t\cos t\,dt
\displaystyle \text{Integrating by parts:}
\displaystyle =-\left[t\sin t-\int \sin t\,dt\right]
\displaystyle =-\left[t\sin t+\cos t\right]+C
\displaystyle =-t\sin t-\cos t+C
\displaystyle \text{Since }t=\cos^{-1}x\text{ and }\sin t=\sqrt{1-x^{2}}:
\displaystyle =-\cos^{-1}x\cdot\sqrt{1-x^{2}}-x+C
\displaystyle \therefore \int \frac{x\cos^{-1}x}{\sqrt{1-x^{2}}}\,dx=-\sqrt{1-x^{2}}\cos^{-1}x-x+C
\\

\displaystyle \textbf{Question 103. }\text{Evaluate }\int\frac{1}{\sin^4x+\sin^2x\cos^2x+\cos^4x}\,dx. \hspace{1.2cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{1}{\sin^4x+\sin^2x\cos^2x+\cos^4x}\,dx
\displaystyle \text{On dividing numerator and denominator by }\cos^4x\text{ in RHS, we get}
\displaystyle I=\int\frac{\sec^4x}{\tan^4x+\tan^2x+1}\,dx
\displaystyle =\int\frac{(\sec^2x)(\sec^2x)}{\tan^4x+\tan^2x+1}\,dx
\displaystyle \Rightarrow I=\int\frac{\sec^2x(1+\tan^2x)}{1+(\tan^2x)^2}\,dx
\displaystyle \text{Put }\tan x=t\Rightarrow\sec^2x\,dx=dt
\displaystyle \text{and }\sec^2x=1+\tan^2x=1+t^2
\displaystyle \therefore\ I=\int\frac{1+t^2}{t^4+t^2+1}\,dt
\displaystyle \text{On dividing numerator and denominator by }t^2\text{ in RHS, we get}
\displaystyle I=\int\frac{1+\dfrac{1}{t^2}}{t^2+\dfrac{1}{t^2}-2+2+1}\,dt=\int\frac{1+\dfrac{1}{t^2}}{\left(t-\dfrac{1}{t}\right)^2+3}\,dt
\displaystyle \text{Again, put }u=t-\frac{1}{t}\Rightarrow\left(1+\frac{1}{t^2}\right)dt=du
\displaystyle \therefore\ I=\int\frac{du}{u^2+(\sqrt{3})^2}\Rightarrow I=\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{u}{\sqrt{3}}\right)+C
\displaystyle \left[\because\int\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)+C\right]
\displaystyle =\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{t-\dfrac{1}{t}}{\sqrt{3}}\right)+C\qquad\left[\because u=t-\frac{1}{t}\right]
\displaystyle =\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{t^2-1}{\sqrt{3}t}\right)+C
\displaystyle \therefore\ I=\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{\tan^2x-1}{\sqrt{3}\tan x}\right)+C\qquad[\because t=\tan x]
\\

\displaystyle \textbf{Question 104. }\text{Evaluate }\int(\sqrt{\cot x}+\sqrt{\tan x})\,dx. \hspace{1.2cm} \text{[CBSE 2014; CBSE 2010C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int(\sqrt{\cot x}+\sqrt{\tan x})\,dx
\displaystyle =\int\sqrt{\tan x}(1+\cot x)\,dx
\displaystyle \text{Put }\tan x=t^2\Rightarrow\sec^2x\,dx=2t\,dt
\displaystyle \Rightarrow dx=\frac{2t}{1+t^4}\,dt\qquad[\because 1+\tan^2x=\sec^2x\Rightarrow 1+t^4=\sec^2x]
\displaystyle \therefore\ I=\int t\left(1+\frac{1}{t^2}\right)\cdot\frac{2t}{1+t^4}\,dt\Rightarrow I=2\int\frac{t^2+1}{t^4+1}\,dt
\displaystyle \text{On dividing numerator and denominator by }t^2\text{ in RHS, we get}
\displaystyle I=2\int\frac{1+\dfrac{1}{t^2}}{t^2+\dfrac{1}{t^2}-2+2}\,dt=2\int\frac{1+\dfrac{1}{t^2}}{\left(t-\dfrac{1}{t}\right)^2+2}\,dt
\displaystyle \text{Again, put }t-\frac{1}{t}=y\Rightarrow\left(1+\frac{1}{t^2}\right)dt=dy
\displaystyle \therefore\ I=2\int\frac{dy}{y^2+(\sqrt{2})^2}\Rightarrow I=\frac{2}{\sqrt{2}}\tan^{-1}\left(\frac{y}{\sqrt{2}}\right)+C
\displaystyle =\sqrt{2}\tan^{-1}\left(\frac{t-\dfrac{1}{t}}{\sqrt{2}}\right)+C\qquad\left[\because y=t-\frac{1}{t}\right]
\displaystyle =\sqrt{2}\tan^{-1}\left(\frac{t^2-1}{\sqrt{2}t}\right)+C
\displaystyle =\sqrt{2}\tan^{-1}\left(\frac{\tan x-1}{\sqrt{2\tan x}}\right)+C\qquad[\because t^2=\tan x]
\\

\displaystyle \textbf{Question 105. }\text{Evaluate }\int\frac{1}{\cos^4x+\sin^4x}\,dx. \hspace{1.2cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{1}{\cos^4x+\sin^4x}\,dx
\displaystyle \text{On dividing numerator and denominator by }\cos^4x\text{ in RHS, we get}
\displaystyle I=\int\frac{\sec^4x}{1+\tan^4x}\,dx
\displaystyle \Rightarrow I=\int\frac{(\sec^2x)(\sec^2x)}{1+(\tan^2x)^2}\,dx
\displaystyle \Rightarrow I=\int\frac{\sec^2x(1+\tan^2x)}{1+(\tan^2x)^2}\,dx
\displaystyle \text{Put }\tan x=t\Rightarrow\sec^2x\,dx=dt
\displaystyle \therefore\ I=\int\frac{1+t^2}{1+t^4}\,dt
\displaystyle \text{Again, dividing numerator and denominator by }t^2\text{ in RHS, we get}
\displaystyle I=\int\frac{1+\dfrac{1}{t^2}}{t^2+\dfrac{1}{t^2}-2+2}\,dt=\int\frac{1+\dfrac{1}{t^2}}{\left(t-\dfrac{1}{t}\right)^2+2}\,dt
\displaystyle \text{Again, put }t-\frac{1}{t}=u\Rightarrow\left(1+\frac{1}{t^2}\right)dt=du
\displaystyle \therefore\ I=\int\frac{du}{u^2+(\sqrt{2})^2}\Rightarrow I=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{u}{\sqrt{2}}\right)+C
\displaystyle \left[\because\int\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)+C\right]
\displaystyle \Rightarrow I=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t-\dfrac{1}{t}}{\sqrt{2}}\right)+C\qquad\left[\because u=t-\frac{1}{t}\right]
\displaystyle \Rightarrow I=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t^2-1}{\sqrt{2}t}\right)+C
\displaystyle \therefore\ I=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\tan^2x-1}{\sqrt{2}\tan x}\right)+C\qquad[\because t=\tan x]
\\

\displaystyle \textbf{Question 106. }\text{Find }\int \frac{x^{2}}{(x^{2}+1)(x^{2}+4)}\,dx. \qquad \text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle \int \frac{x^{2}}{(x^{2}+1)(x^{2}+4)}\,dx
\displaystyle \text{Using partial fractions: }\frac{x^{2}}{(x^{2}+1)(x^{2}+4)}=\frac{A}{x^{2}+1}+\frac{B}{x^{2}+4}
\displaystyle \Rightarrow x^{2}=A(x^{2}+4)+B(x^{2}+1)
\displaystyle \text{Putting }x^{2}=-1:\quad -1=3A\Rightarrow A=-\frac{1}{3}
\displaystyle \text{Putting }x^{2}=-4:\quad -4=-3B\Rightarrow B=\frac{4}{3}
\displaystyle =\int\left(\frac{-\frac{1}{3}}{x^{2}+1}+\frac{\frac{4}{3}}{x^{2}+4}\right)dx
\displaystyle =-\frac{1}{3}\int\frac{dx}{x^{2}+1}+\frac{4}{3}\int\frac{dx}{x^{2}+4}
\displaystyle =-\frac{1}{3}\tan^{-1}x+\frac{4}{3}\cdot\frac{1}{2}\tan^{-1}\frac{x}{2}+C
\displaystyle \therefore \int \frac{x^{2}}{(x^{2}+1)(x^{2}+4)}\,dx=-\frac{1}{3}\tan^{-1}x+\frac{2}{3}\tan^{-1}\frac{x}{2}+C
\\

\displaystyle \textbf{Question 107. }\text{Find }\int\frac{\sin^{-1}\sqrt{x}-\cos^{-1}\sqrt{x}}{\sin^{-1}\sqrt{x}+\cos^{-1}\sqrt{x}}\,dx,\ x\in[0,1]. \hspace{1.2cm} \text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle \text{We know that }\sin^{-1}\sqrt{x}+\cos^{-1}\sqrt{x}=\pi/2
\displaystyle \Rightarrow\cos^{-1}\sqrt{x}=\frac{\pi}{2}-\sin^{-1}\sqrt{x}
\displaystyle \therefore\ I=\int\frac{\sin^{-1}\sqrt{x}-\left(\dfrac{\pi}{2}-\sin^{-1}\sqrt{x}\right)}{\pi/2}\,dx
\displaystyle =\int\frac{2\sin^{-1}\sqrt{x}-\dfrac{\pi}{2}}{\dfrac{\pi}{2}}\,dx=\frac{2}{\pi}\int\left(2\sin^{-1}\sqrt{x}-\frac{\pi}{2}\right)dx
\displaystyle =\frac{4}{\pi}\int\sin^{-1}\sqrt{x}\,dx-\int 1\,dx=\frac{4}{\pi}\int\sin^{-1}\sqrt{x}\,dx-x
\displaystyle \Rightarrow I=\frac{4}{\pi}I_1-x\qquad\ldots\text{(i)}
\displaystyle \text{where }I_1=\int\sin^{-1}\sqrt{x}\,dx
\displaystyle \text{Put }\sqrt{x}=t\Rightarrow x=t^2\text{ and }dx=2t\,dt
\displaystyle \therefore\ I_1=\int\sin^{-1}t\cdot 2t\,dt=2\int\sin^{-1}t\cdot t\,dt
\displaystyle =2\left[\sin^{-1}t\int t\,dt-\int\left\{\frac{d}{dt}(\sin^{-1}t)\int t\,dt\right\}dt\right]\qquad\text{[using integration by parts]}
\displaystyle =2\left[\sin^{-1}t\cdot\frac{t^2}{2}-\int\frac{1}{\sqrt{1-t^2}}\cdot\frac{t^2}{2}\,dt\right]
\displaystyle =t^2\sin^{-1}t+\int\frac{(1-t^2)-1}{\sqrt{1-t^2}}\,dt
\displaystyle =t^2\sin^{-1}t+\int\sqrt{1-t^2}\,dt-\int\frac{1}{\sqrt{1-t^2}}\,dt
\displaystyle =t^2\sin^{-1}t+\frac{t\sqrt{1-t^2}}{2}+\frac{1}{2}\sin^{-1}t-\sin^{-1}t+C_1
\displaystyle \left[\because\int\sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C\text{ and }\int\frac{1}{\sqrt{1-x^2}}=\sin^{-1}x+C\right]
\displaystyle =\left(t^2-\frac{1}{2}\right)\sin^{-1}t+\frac{1}{2}t\sqrt{1-t^2}+C_1
\displaystyle =\frac{1}{2}[(2t^2-1)\sin^{-1}\sqrt{x}+\sqrt{x}\sqrt{1-x}]+C_1\qquad[\because t=\sqrt{x}]
\displaystyle I_1=\frac{1}{2}[(2x-1)\sin^{-1}\sqrt{x}+\sqrt{x-x^2}]+C_1
\displaystyle \text{On putting the value of }I_1\text{ in Eq. (i), we get}
\displaystyle I=\frac{2}{\pi}[(2x-1)\sin^{-1}\sqrt{x}+\sqrt{x-x^2}]-x+C\text{,}
\displaystyle \text{where }C=\frac{4}{\pi}C_1
\\

\displaystyle \textbf{Question 108. }\text{Find }\int\frac{\sqrt{x^2+1}[\log(x^2+1)-2\log x]}{x^4}\,dx. \hspace{1.2cm} \text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{\sqrt{x^2+1}[\log(x^2+1)-2\log x]}{x^4}\,dx
\displaystyle =\int\frac{\sqrt{x^2+1}\log\left(\dfrac{x^2+1}{x^2}\right)}{x^4}\,dx\qquad\left[\because\log m-\log n=\log\frac{m}{n}\right]
\displaystyle =\int\frac{x\sqrt{1+\dfrac{1}{x^2}}\cdot\log\left(1+\dfrac{1}{x^2}\right)}{x^4}\,dx
\displaystyle =\int\frac{\sqrt{1+\dfrac{1}{x^2}}\cdot\log\left(1+\dfrac{1}{x^2}\right)}{x^3}\,dx
\displaystyle \text{Put }1+\frac{1}{x^2}=t\Rightarrow\frac{-2}{x^3}\,dx=dt\Rightarrow\frac{dx}{x^3}=-\frac{dt}{2}
\displaystyle \therefore\ I=-\frac{1}{2}\int\sqrt{t}\log t\,dt=-\frac{1}{2}\int t^{1/2}\log t\,dt
\displaystyle =-\frac{1}{2}\left[\log t\int t^{1/2}\,dt-\int\left\{\frac{d}{dt}(\log t)\int t^{1/2}\,dt\right\}dt\right]\qquad\text{[using integration by parts]}
\displaystyle =-\frac{1}{2}\left[\log t\times\frac{t^{3/2}}{3/2}-\int\frac{3/2}{t}\times\frac{t^{3/2}}{3/2}\,dt\right]
\displaystyle =-\frac{1}{2}\left[\frac{t^{3/2}\log t}{3/2}-\int\sqrt{t}\,dt\right]
\displaystyle =-\frac{1}{3}\left[t^{3/2}\log t-\frac{t^{3/2}}{3/2}\right]+C
\displaystyle =-\frac{1}{3}\left[t^{3/2}\log|t|-\frac{t^{3/2}}{3/2}\right]+C
\displaystyle =-\frac{1}{3}t^{3/2}\left[\log|t|-\frac{2}{3}\right]+C
\displaystyle =-\frac{1}{3}\left(1+\frac{1}{x^2}\right)^{3/2}\left[\log\left(1+\frac{1}{x^2}\right)-\frac{2}{3}\right]+C\qquad\left[\because t=1+\frac{1}{x^2}\right]
\\

\displaystyle \textbf{Question 109. }\text{Evaluate }\int e^{2x}\left(\frac{1-\sin 2x}{1-\cos 2x}\right)dx. \hspace{1.2cm} \text{[CBSE 2013C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int e^{2x}\left(\frac{1-\sin 2x}{1-\cos 2x}\right)dx
\displaystyle =\int e^{2x}\left(\frac{1-2\sin x\cos x}{2\sin^2x}\right)dx
\displaystyle \left[\because 1-\cos 2\theta=2\sin^2\theta\text{ and }\sin 2\theta=2\sin\theta\cos\theta\right]
\displaystyle =\frac{1}{2}\int e^{2x}(\text{cosec}^2x-2\cot x)\,dx
\displaystyle =\frac{1}{2}\int e^{2x}\text{cosec}^2x\,dx-\int e^{2x}\cot x\,dx
\displaystyle =\frac{1}{2}\left[e^{2x}\int\text{cosec}^2x\,dx-\int\left\{\frac{d}{dx}(e^{2x})\int\text{cosec}^2x\,dx\right\}dx\right]
\displaystyle -\int e^{2x}\cot x\,dx\qquad\text{[by using integration by parts]}
\displaystyle =\frac{1}{2}\left[-e^{2x}\cot x+\int 2e^{2x}\cot x\,dx\right]+C-\int e^{2x}\cot x\,dx
\displaystyle =-\frac{e^{2x}}{2}\cot x+\int e^{2x}\cot x\,dx-\int e^{2x}\cot x\,dx+C
\displaystyle =-\frac{e^{2x}}{2}\cot x+C
\\

\displaystyle \textbf{Question 110. }\text{Evaluate }\int\frac{3x+1}{(x+1)^2(x+3)}\,dx. \hspace{1.2cm} \text{[CBSE 2013C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{3x+1}{(x+1)^2(x+3)}\,dx
\displaystyle \text{Again, let }\frac{3x+1}{(x+1)^2(x+3)}=\frac{A}{(x+1)}+\frac{B}{(x+1)^2}+\frac{C}{(x+3)}\qquad\ldots\text{(i)}
\displaystyle \Rightarrow 3x+1=A(x+1)(x+3)+B(x+3)+C(x+1)^2
\displaystyle \Rightarrow 3x+1=A(x^2+4x+3)+B(x+3)+C(x^2+1+2x)
\displaystyle \Rightarrow 3x+1=(A+C)x^2+(4A+B+2C)x+3A+3B+C
\displaystyle \text{On comparing like powers of }x\text{ from both sides, we get}
\displaystyle A+C=0
\displaystyle 4A+B+2C=3
\displaystyle \text{and }3A+3B+C=1
\displaystyle \text{On solving, we get }A=2,\ B=-1\text{ and }C=-2
\displaystyle \therefore\text{ Eq. (i) becomes}
\displaystyle \frac{3x+1}{(x+1)^2(x+3)}=\frac{2}{(x+1)}-\frac{1}{(x+1)^2}-\frac{2}{(x+3)}
\displaystyle \text{On integrating both sides, we get}
\displaystyle \int\frac{3x+1}{(x+1)^2(x+3)}\,dx=\int\frac{2}{(x+1)}\,dx-\int\frac{1}{(x+1)^2}\,dx-\int\frac{2}{(x+3)}\,dx
\displaystyle \Rightarrow I=2\log|x+1|-\frac{(x+1)^{-2+1}}{(-2+1)}-2\log|x+3|+C
\displaystyle =2\log\left|\frac{x+1}{x+3}\right|+\frac{1}{(x+1)}+C
\displaystyle \left[\because\log m-\log n=\log\frac{m}{n}\right]
\\

\displaystyle \textbf{Question 111. }\text{Evaluate }\int\frac{2x^2+1}{x^2(x^2+4)}\,dx. \hspace{1.2cm} \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{2x^2+1}{x^2(x^2+4)}\,dx
\displaystyle \text{Consider, }\frac{2x^2+1}{x^2(x^2+4)}\text{ and put }x^2=t\text{ and then}
\displaystyle \frac{2x^2+1}{x^2(x^2+4)}=\frac{2t+1}{t(t+4)}\text{ and by using partial fraction,}
\displaystyle \text{we get }\frac{2t+1}{t(t+4)}=\frac{A}{t}+\frac{B}{t+4}\Rightarrow 2t+1=A(t+4)+Bt
\displaystyle \text{On comparing the coefficients of }t\text{ and constant terms, we get}
\displaystyle 2=A+B\text{ and }1=4A\Rightarrow A=\frac{1}{4}
\displaystyle \therefore\ B=2-A=2-\frac{1}{4}=\frac{7}{4}
\displaystyle \text{Thus, }\frac{2x^2+1}{x^2(x^2+4)}=\frac{1}{4x^2}+\frac{7}{4(x^2+4)}
\displaystyle \therefore\ I=\frac{1}{4}\int\frac{dx}{x^2}+\frac{7}{4}\int\frac{dx}{x^2+4}
\displaystyle =-\frac{1}{4x}+\frac{7}{4}\times\frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right)+C
\displaystyle \left[\because\int\frac{dx}{a^2+x^2}=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)+C\right]
\displaystyle =-\frac{1}{4x}+\frac{7}{8}\tan^{-1}\left(\frac{x}{2}\right)+C
\\

\displaystyle \textbf{Question 112. }\text{Evaluate }\int \frac{x^{2}+1}{(x^{2}+4)(x^{2}+25)}\,dx. \qquad \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \int \frac{x^{2}+1}{(x^{2}+4)(x^{2}+25)}\,dx
\displaystyle \text{Using partial fractions: }\frac{x^{2}+1}{(x^{2}+4)(x^{2}+25)}=\frac{Ax+B}{x^{2}+4}+\frac{Cx+D}{x^{2}+25}
\displaystyle \Rightarrow x^{2}+1=(Ax+B)(x^{2}+25)+(Cx+D)(x^{2}+4)
\displaystyle \text{Comparing coefficients of }x^{3}:\quad 0=A+C
\displaystyle \text{Comparing coefficients of }x^{2}:\quad 1=B+D
\displaystyle \text{Comparing coefficients of }x:\quad 0=25A+4C
\displaystyle \text{Comparing constant terms:}\quad 1=25B+4D
\displaystyle \text{From }A+C=0\text{ and }25A+4C=0\Rightarrow A=0,\ C=0
\displaystyle \text{From }B+D=1\text{ and }25B+4D=1\Rightarrow B=-\frac{1}{7},\ D=\frac{8}{7}
\displaystyle =\int\left(\frac{-\frac{1}{7}}{x^{2}+4}+\frac{\frac{8}{7}}{x^{2}+25}\right)dx
\displaystyle =-\frac{1}{7}\int\frac{dx}{x^{2}+4}+\frac{8}{7}\int\frac{dx}{x^{2}+25}
\displaystyle =-\frac{1}{7}\cdot\frac{1}{2}\tan^{-1}\frac{x}{2}+\frac{8}{7}\cdot\frac{1}{5}\tan^{-1}\frac{x}{5}+C
\displaystyle \therefore \int \frac{x^{2}+1}{(x^{2}+4)(x^{2}+25)}\,dx=-\frac{1}{14}\tan^{-1}\frac{x}{2}+\frac{8}{35}\tan^{-1}\frac{x}{5}+C
\\

\displaystyle \textbf{Question 113. }\text{Evaluate }\int\frac{\cos 2x-\cos 2\alpha}{\cos x-\cos\alpha}\,dx. \hspace{1.2cm} \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{\cos 2x-\cos 2\alpha}{\cos x-\cos\alpha}\,dx
\displaystyle =\int\frac{(2\cos^2x-1)-(2\cos^2\alpha-1)}{\cos x-\cos\alpha}\,dx
\displaystyle [\because\cos 2\theta=2\cos^2\theta-1]
\displaystyle =\int\frac{2(\cos^2x-\cos^2\alpha)}{\cos x-\cos\alpha}\,dx
\displaystyle =\int\frac{2(\cos x-\cos\alpha)(\cos x+\cos\alpha)}{(\cos x-\cos\alpha)}\,dx
\displaystyle [\because a^2-b^2=(a+b)(a-b)]
\displaystyle =\int 2(\cos x+\cos\alpha)\,dx
\displaystyle =2\left[\int\cos x\,dx+\cos\alpha\int 1\,dx\right]
\displaystyle \therefore\ I=2(\sin x+x\cos\alpha)+C
\\

\displaystyle \textbf{Question 114. }\text{Evaluate }\int \frac{x+2}{\sqrt{x^{2}+2x+3}}\,dx. \qquad \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \int \frac{x+2}{\sqrt{x^{2}+2x+3}}\,dx
\displaystyle \text{Let }x+2=A\frac{d}{dx}(x^{2}+2x+3)+B=A(2x+2)+B
\displaystyle \text{Comparing coefficients: }A=\frac{1}{2},\quad B=1
\displaystyle =\frac{1}{2}\int\frac{2x+2}{\sqrt{x^{2}+2x+3}}\,dx+\int\frac{dx}{\sqrt{x^{2}+2x+3}}
\displaystyle =\frac{1}{2}\int\frac{2x+2}{\sqrt{x^{2}+2x+3}}\,dx+\int\frac{dx}{\sqrt{(x+1)^{2}+2}}
\displaystyle =\frac{1}{2}\cdot 2\sqrt{x^{2}+2x+3}+\log\left|(x+1)+\sqrt{(x+1)^{2}+2}\right|+C
\displaystyle \therefore \int \frac{x+2}{\sqrt{x^{2}+2x+3}}\,dx=\sqrt{x^{2}+2x+3}+\log\left|x+1+\sqrt{x^{2}+2x+3}\right|+C
\\

\displaystyle \textbf{Question 115. }\text{Evaluate }\int\frac{dx}{x(x^5+3)}. \hspace{1.2cm} \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{dx}{x(x^5+3)}=\int\frac{x^4}{x^5(x^5+3)}\,dx
\displaystyle [\because\text{multiplying numerator and denominator by }x^4]
\displaystyle \text{Put }t=x^5\Rightarrow dt=5x^4\,dx
\displaystyle \therefore\ I=\int\frac{dt}{5t(t+3)}=\frac{1}{5}\int\left[\frac{1/3}{t}-\frac{1/3}{t+3}\right]dt\qquad\text{[by using partial fraction]}
\displaystyle =\frac{1}{5}\cdot\frac{1}{3}\int\left[\frac{1}{t}-\frac{1}{t+3}\right]dt
\displaystyle =\frac{1}{15}[\log|t|-\log|t+3|]+C
\displaystyle =\frac{1}{15}\log\left|\frac{t}{t+3}\right|+C
\displaystyle =\frac{1}{15}\log\left|\frac{x^5}{x^5+3}\right|+C\qquad[\because t=x^5]
\\

\displaystyle \textbf{Question 116. }\text{Evaluate }\int \frac{dx}{x(x^{3}+1)}. \qquad \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \int \frac{dx}{x(x^{3}+1)}
\displaystyle \text{Multiply numerator and denominator by }x^{2}:
\displaystyle =\int \frac{x^{2}}{x^{3}(x^{3}+1)}\,dx
\displaystyle \text{Let }x^{3}=t,\text{ so }3x^{2}\,dx=dt\Rightarrow x^{2}\,dx=\frac{dt}{3}
\displaystyle =\frac{1}{3}\int \frac{dt}{t(t+1)}
\displaystyle \text{Using partial fractions: }\frac{1}{t(t+1)}=\frac{1}{t}-\frac{1}{t+1}
\displaystyle =\frac{1}{3}\int\left(\frac{1}{t}-\frac{1}{t+1}\right)dt
\displaystyle =\frac{1}{3}\left[\log|t|-\log|t+1|\right]+C
\displaystyle =\frac{1}{3}\log\left|\frac{t}{t+1}\right|+C
\displaystyle \therefore \int \frac{dx}{x(x^{3}+1)}=\frac{1}{3}\log\left|\frac{x^{3}}{x^{3}+1}\right|+C
\\

\displaystyle \textbf{Question 117. }\text{Evaluate }\int \frac{dx}{x(x^{3}+8)}. \qquad \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \int \frac{dx}{x(x^{3}+8)}
\displaystyle \text{Multiply numerator and denominator by }x^{2}:
\displaystyle =\int \frac{x^{2}}{x^{3}(x^{3}+8)}\,dx
\displaystyle \text{Let }x^{3}=t,\text{ so }3x^{2}\,dx=dt\Rightarrow x^{2}\,dx=\frac{dt}{3}
\displaystyle =\frac{1}{3}\int \frac{dt}{t(t+8)}
\displaystyle \text{Using partial fractions: }\frac{1}{t(t+8)}=\frac{1}{8}\left(\frac{1}{t}-\frac{1}{t+8}\right)
\displaystyle =\frac{1}{3}\cdot\frac{1}{8}\int\left(\frac{1}{t}-\frac{1}{t+8}\right)dt
\displaystyle =\frac{1}{24}\left[\log|t|-\log|t+8|\right]+C
\displaystyle =\frac{1}{24}\log\left|\frac{t}{t+8}\right|+C
\displaystyle \therefore \int \frac{dx}{x(x^{3}+8)}=\frac{1}{24}\log\left|\frac{x^{3}}{x^{3}+8}\right|+C
\\

\displaystyle \textbf{Question 118. }\text{Evaluate }\int\frac{\sqrt{1-\sin x}}{1+\cos x}\cdot e^{-x/2}\,dx. \hspace{1.2cm} \text{[CBSE 2013C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{\sqrt{1-\sin x}}{1+\cos x}\cdot e^{-x/2}\,dx
\displaystyle \text{Put }\frac{-x}{2}=t\Rightarrow dx=-2\,dt
\displaystyle \Rightarrow I=\int\frac{\sqrt{1-\sin(-2t)}}{1+\cos(-2t)}\cdot e^t\cdot(-2\,dt)\qquad[\because x=-2t]
\displaystyle =-2\int e^t\frac{\sqrt{1+\sin 2t}}{1+\cos 2t}\,dt
\displaystyle [\because\sin(-\theta)=-\sin\theta\text{ and }\cos(-\theta)=\cos\theta]
\displaystyle =-2\int e^t\left(\frac{\sqrt{(\cos t+\sin t)^2}}{2\cos^2t}\right)dt
\displaystyle =-2\int e^t\left(\frac{\cos t+\sin t}{2\cos^2t}\right)dt
\displaystyle =-\int e^t(\sec t+\tan t\sec t)\,dt
\displaystyle \text{Now, consider }f(t)=\sec t\text{, then }f'(t)=\sec t\tan t
\displaystyle \text{Thus, the above integrand is of the form }\int e^t[f(t)+f'(t)]\,dt
\displaystyle \therefore\ I=-e^t\sec t+C
\displaystyle \left[\because\int e^t[f(t)+f'(t)]\,dt=e^t f(t)+C\right]
\displaystyle =-e^{-x/2}\sec\frac{x}{2}+C
\displaystyle \left[\because t=\frac{-x}{2}\text{ and }\sec(-\theta)=\sec\theta\right]
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\displaystyle \textbf{Question 119. }\text{Evaluate }\int \frac{3x+5}{x^{3}-x^{2}-x+1}\,dx. \qquad \text{[CBSE 2013C]}
\displaystyle \text{Answer:}
\displaystyle \int \frac{3x+5}{x^{3}-x^{2}-x+1}\,dx
\displaystyle \text{Factorising denominator: }x^{3}-x^{2}-x+1=x^{2}(x-1)-(x-1)=(x-1)(x^{2}-1)=(x-1)(x-1)(x+1)=(x-1)^{2}(x+1)
\displaystyle \text{Using partial fractions: }\frac{3x+5}{(x-1)^{2}(x+1)}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{x+1}
\displaystyle \Rightarrow 3x+5=A(x-1)(x+1)+B(x+1)+C(x-1)^{2}
\displaystyle \text{Putting }x=1:\quad 8=2B\Rightarrow B=4
\displaystyle \text{Putting }x=-1:\quad 2=4C\Rightarrow C=\frac{1}{2}
\displaystyle \text{Comparing coefficients of }x^{2}:\quad 0=A+C\Rightarrow A=-\frac{1}{2}
\displaystyle =\int\left(\frac{-\frac{1}{2}}{x-1}+\frac{4}{(x-1)^{2}}+\frac{\frac{1}{2}}{x+1}\right)dx
\displaystyle =-\frac{1}{2}\log|x-1|-\frac{4}{x-1}+\frac{1}{2}\log|x+1|+C
\displaystyle =\frac{1}{2}\log\left|\frac{x+1}{x-1}\right|-\frac{4}{x-1}+C
\displaystyle \therefore \int \frac{3x+5}{x^{3}-x^{2}-x+1}\,dx=\frac{1}{2}\log\left|\frac{x+1}{x-1}\right|-\frac{4}{x-1}+C
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\displaystyle \textbf{Question 120. }\text{Evaluate }\int(1-x)\sqrt{x}\,dx. \hspace{1.2cm} \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int(1-x)\sqrt{x}\,dx=\int(\sqrt{x}-x\sqrt{x})\,dx
\displaystyle =\int(x^{1/2}-x^{3/2})\,dx=\frac{2}{3}x^{3/2}-\frac{2}{5}x^{5/2}+C
\displaystyle \left[\because\int x^n\,dx=\frac{x^{n+1}}{n+1}+C,\ n\neq-1\right]
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\displaystyle \textbf{Question 121. }\text{Given, }\int e^x(\tan x+1)\sec x\,dx=e^x\cdot f(x)+C\text{. Write }f(x)\text{ satisfying above.} \hspace{1.2cm} \text{[CBSE 2012; CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \Rightarrow\int e^x(\sec x+\sec x\tan x)\,dx=e^x f(x)+C
\displaystyle \Rightarrow e^x\cdot\sec x+C=e^x f(x)+C
\displaystyle \left[\because\int e^x\{f(x)+f'(x)\}\,dx=e^x f(x)+C\right.
\displaystyle \left.\text{and here }\frac{d}{dx}(\sec x)=\sec x\tan x\right]
\displaystyle \text{On comparing both sides, we get }f(x)=\sec x
\\

\displaystyle \textbf{Question 122. }\text{Evaluate }\int\frac{2}{1+\cos 2x}\,dx. \hspace{1.2cm} \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{2}{1+\cos 2x}\,dx
\displaystyle =\int\frac{2}{2\cos^2x}\,dx\qquad[\because\cos 2\theta=2\cos^2\theta-1]
\displaystyle =\int\sec^2x\,dx=\tan x+C
\\

\displaystyle \textbf{Question 123. }\text{Write the value of }\int\frac{x+\cos 6x}{3x^2+\sin 6x}\,dx. \hspace{1.2cm} \text{[CBSE 2012C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{x+\cos 6x}{3x^2+\sin 6x}\,dx
\displaystyle \text{Put }3x^2+\sin 6x=t
\displaystyle \Rightarrow 6x+6\cos 6x=\frac{dt}{dx}
\displaystyle \Rightarrow (x+\cos 6x)\,dx=\frac{dt}{6}
\displaystyle \therefore\ I=\int\frac{dt}{6t}=\frac{1}{6}\log|t|+C\qquad\left[\because\int\frac{1}{x}\,dx=\log|x|+C\right]
\displaystyle =\frac{1}{6}[\log|(3x^2+\sin 6x)|]+C\qquad[\because t=3x^2+\sin 6x]
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\displaystyle \textbf{Question 124. }\text{Write the value of }\int\frac{\sec^2x}{\text{cosec}^2x}\,dx. \hspace{1.2cm} \text{[CBSE 2012C, 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{\sec^2x}{\text{cosec}^2x}\,dx=\int\frac{\left(\dfrac{1}{\cos^2x}\right)}{\left(\dfrac{1}{\sin^2x}\right)}\,dx
\displaystyle =\int\frac{\sin^2x}{\cos^2x}\,dx
\displaystyle =\int\tan^2x\,dx=\int(\sec^2x-1)\,dx
\displaystyle [\because\tan^2\theta=\sec^2\theta-1]
\displaystyle =\int\sec^2x\,dx-\int 1\,dx=\tan x-x+C
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\displaystyle \textbf{Question 125. }\text{Evaluate }\int\frac{2}{(1-x)(1+x^2)}\,dx. \hspace{1.2cm} \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{2}{(1-x)(1+x^2)}\,dx
\displaystyle \text{By using partial fraction,}
\displaystyle \text{let }\frac{2}{(1-x)(1+x^2)}=\frac{A}{1-x}+\frac{Bx+C}{1+x^2}\qquad\ldots\text{(i)}
\displaystyle \Rightarrow\frac{2}{(1-x)(1+x^2)}=\frac{A(1+x^2)+(Bx+C)(1-x)}{(1-x)(1+x^2)}
\displaystyle \Rightarrow 2=A(1+x^2)+(Bx+C)(1-x)
\displaystyle \Rightarrow 2=A+Ax^2+Bx+C-Bx^2-Cx
\displaystyle \Rightarrow 2=(A-B)x^2+(B-C)x+(A+C)
\displaystyle \text{On comparing coefficients of }x^2\text{, }x\text{ and constant terms from both sides, we get}
\displaystyle A-B=0\qquad\ldots\text{(ii)}
\displaystyle B-C=0\qquad\ldots\text{(iii)}
\displaystyle \text{and }A+C=2\qquad\ldots\text{(iv)}
\displaystyle \text{On solving Eqs. (ii), (iii) and (iv), we get }A=1,\ B=1\text{ and }C=1
\displaystyle \text{Now, Eq. (i) becomes}
\displaystyle \frac{2}{(1-x)(1+x^2)}=\frac{1}{1-x}+\frac{x+1}{1+x^2}
\displaystyle \therefore\ I=\int\frac{2}{(1-x)(1+x^2)}\,dx=\int\frac{1}{1-x}\,dx+\int\frac{x+1}{1+x^2}\,dx
\displaystyle =-\log|1-x|+\int\frac{x}{1+x^2}\,dx+\int\frac{1}{1+x^2}\,dx
\displaystyle =-\log|1-x|+\frac{1}{2}\int\frac{2x}{1+x^2}\,dx+\int\frac{1}{1+x^2}\,dx
\displaystyle =-\log|1-x|+\frac{1}{2}\log|1+x^2|+\tan^{-1}x+C
\displaystyle \left[\because\int\frac{f'(x)}{f(x)}\,dx=\log|f(x)|+C\right]
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\displaystyle \textbf{Question 126. }\text{Evaluate }\int\left(\frac{1+\sin x}{1+\cos x}\right)e^x\,dx. \hspace{1.2cm} \text{[CBSE 2012C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\left(\frac{1+\sin x}{1+\cos x}\right)e^x\,dx=\int\frac{1+2\sin\dfrac{x}{2}\cos\dfrac{x}{2}}{2\cos^2\dfrac{x}{2}}\cdot e^x\,dx
\displaystyle \left[\because\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\text{ and }1+\cos\theta=2\cos^2\frac{\theta}{2}\right]
\displaystyle =\int\left(\frac{1}{2}\sec^2\frac{x}{2}+\tan\frac{x}{2}\right)e^x\,dx
\displaystyle =\int e^x\left(\tan\frac{x}{2}+\frac{1}{2}\sec^2\frac{x}{2}\right)dx
\displaystyle \text{We know that }\int e^x[f(x)+f'(x)]\,dx=e^x f(x)+C
\displaystyle \text{Here, }f(x)=\tan\frac{x}{2}\Rightarrow f'(x)=\frac{1}{2}\sec^2\frac{x}{2}
\displaystyle \therefore\ I=e^x\tan\frac{x}{2}+C
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\displaystyle \textbf{Question 127. }\text{Evaluate }\int\frac{x^2}{(x\sin x+\cos x)^2}\,dx. \hspace{1.2cm} \text{[CBSE 2012C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{x^2}{(x\sin x+\cos x)^2}\,dx
\displaystyle \Rightarrow I=\int\frac{x\cos x}{(x\sin x+\cos x)^2}\cdot x\sec x\,dx\qquad\ldots\text{(i)}
\displaystyle \text{Put }x\sin x+\cos x=t
\displaystyle \Rightarrow(x\cos x+\sin x-\sin x)\,dx=dt\Rightarrow x\cos x\,dx=dt
\displaystyle \therefore\ I_1=\int\frac{x\cos x}{(x\sin x+\cos x)^2}\,dx\text{ (say)}
\displaystyle =\int\frac{dt}{t^2}=\frac{-1}{t}=\frac{-1}{x\sin x+\cos x}\qquad[\because t=x\sin x+\cos x]
\displaystyle \text{Now, integrating Eq. (i) by parts, we get}
\displaystyle I=\int x\sec x\cdot\frac{x\cos x}{(x\sin x+\cos x)^2}\,dx
\displaystyle =x\sec x\cdot\frac{(-1)}{x\sin x+\cos x}-\int\left(1\cdot\sec x+x\sec x\tan x\right)\cdot\frac{-dx}{x\sin x+\cos x}
\displaystyle =\frac{-x\sec x}{x\sin x+\cos x}+\int\sec x\left(1+\frac{x\sin x}{\cos x}\right)\frac{dx}{x\sin x+\cos x}
\displaystyle =\frac{-x\sec x}{x\sin x+\cos x}+\int\sec^2x\,dx
\displaystyle =\frac{-x\sec x}{x\sin x+\cos x}+\tan x+C
\\

\displaystyle \textbf{Question 128. }\text{Evaluate }\int \frac{x^{2}+1}{(x-1)^{2}(x+3)}\,dx. \qquad \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \int \frac{x^{2}+1}{(x-1)^{2}(x+3)}\,dx
\displaystyle \text{Using partial fractions: }\frac{x^{2}+1}{(x-1)^{2}(x+3)}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{x+3}
\displaystyle \Rightarrow x^{2}+1=A(x-1)(x+3)+B(x+3)+C(x-1)^{2}
\displaystyle \text{Putting }x=1:\quad 2=4B\Rightarrow B=\frac{1}{2}
\displaystyle \text{Putting }x=-3:\quad 10=16C\Rightarrow C=\frac{5}{8}
\displaystyle \text{Comparing coefficients of }x^{2}:\quad 1=A+C\Rightarrow A=1-\frac{5}{8}=\frac{3}{8}
\displaystyle =\int\left(\frac{\frac{3}{8}}{x-1}+\frac{\frac{1}{2}}{(x-1)^{2}}+\frac{\frac{5}{8}}{x+3}\right)dx
\displaystyle =\frac{3}{8}\log|x-1|-\frac{1}{2(x-1)}+\frac{5}{8}\log|x+3|+C
\displaystyle \therefore \int \frac{x^{2}+1}{(x-1)^{2}(x+3)}\,dx=\frac{3}{8}\log|x-1|+\frac{5}{8}\log|x+3|-\frac{1}{2(x-1)}+C
\\

\displaystyle \textbf{Question 129. }\text{Write the value of }\int\frac{dx}{x^2+16}. \hspace{1.2cm} \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{dx}{x^2+16}=\int\frac{dx}{x^2+(4)^2}
\displaystyle =\frac{1}{4}\tan^{-1}\frac{x}{4}+C
\displaystyle \left[\because\int\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\frac{x}{a}+C\right]
\\

\displaystyle \textbf{Question 130. }\text{Write the value of }\int\frac{2-3\sin x}{\cos^2x}\,dx. \hspace{1.2cm} \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{2-3\sin x}{\cos^2x}\,dx=\int\left(\frac{2}{\cos^2x}-\frac{3\sin x}{\cos^2x}\right)dx
\displaystyle =\int(2\sec^2x-3\sec x\tan x)\,dx
\displaystyle =2\int\sec^2x\,dx-3\int\sec x\tan x\,dx
\displaystyle =2\tan x-3\sec x+C
\\

\displaystyle \textbf{Question 131. }\text{Write the value of }\int\sec x(\sec x+\tan x)\,dx. \hspace{1.2cm} \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\sec x(\sec x+\tan x)\,dx
\displaystyle =\int(\sec^2x+\sec x\tan x)\,dx
\displaystyle =\int\sec^2x\,dx+\int\sec x\tan x\,dx
\displaystyle =\tan x+\sec x+C
\\

\displaystyle \textbf{Question 132. }\text{Evaluate }\int\frac{dx}{\sqrt{1-x^2}}. \hspace{1.2cm} \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{dx}{\sqrt{1-x^2}}=\int\frac{dx}{\sqrt{(1)^2-x^2}}
\displaystyle =\sin^{-1}x+C\qquad\left[\because\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\frac{x}{a}+C\right]
\\

\displaystyle \textbf{Question 133. }\text{Evaluate }\int\frac{(\log x)^2}{x}\,dx. \hspace{1.2cm} \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{(\log x)^2}{x}\,dx
\displaystyle \text{Put }\log x=t\Rightarrow\frac{1}{x}\,dx=dt
\displaystyle \therefore\ I=\int\frac{(\log x)^2}{x}\,dx=\int t^2\,dt
\displaystyle =\frac{t^3}{3}+C=\frac{(\log x)^3}{3}+C\qquad[\because t=\log x]
\\

\displaystyle \textbf{Question 134. }\text{Evaluate }\int\frac{e^{\tan^{-1}x}}{1+x^2}\,dx. \hspace{1.2cm} \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{e^{\tan^{-1}x}}{1+x^2}\,dx
\displaystyle \text{Put }\tan^{-1}x=t\Rightarrow\frac{1}{1+x^2}\,dx=dt
\displaystyle \therefore\ I=\int\frac{e^{\tan^{-1}x}}{1+x^2}\,dx=\int e^t\,dt
\displaystyle =e^t+C\qquad\left[\because\int e^x\,dx=e^x+C\right]
\displaystyle =e^{\tan^{-1}x}+C\qquad[\because t=\tan^{-1}x]
\\

\displaystyle \textbf{Question 135. }\text{Evaluate }\int(ax+b)^3\,dx. \hspace{1.2cm} \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int(ax+b)^3\,dx
\displaystyle \text{Put }t=ax+b\Rightarrow\frac{dt}{dx}=a\Rightarrow\frac{dt}{a}=dx
\displaystyle \therefore\ I=\int\frac{t^3}{a}\,dt=\frac{1}{a}\cdot\frac{t^4}{4}+C=\frac{(ax+b)^4}{4a}+C\qquad[\because t=ax+b]
\\

\displaystyle \textbf{Question 136. }\text{Evaluate }\int\frac{(1+\log x)^2}{x}\,dx. \hspace{1.2cm} \text{[CBSE 2011; CBSE 2009]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{(1+\log x)^2}{x}\,dx
\displaystyle \text{Put }1+\log x=t\Rightarrow\frac{1}{x}\,dx=dt
\displaystyle \therefore\ I=\int\frac{(1+\log x)^2}{x}\,dx
\displaystyle =\int t^2\,dt=\frac{t^3}{3}+C
\displaystyle =\frac{(1+\log x)^3}{3}+C\qquad[\because t=1+\log x]
\\

\displaystyle \textbf{Question 137. }\text{Evaluate }\int\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}\,dx. \hspace{1.2cm} \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}\,dx
\displaystyle \text{Put }e^{2x}+e^{-2x}=t
\displaystyle \Rightarrow(2e^{2x}-2e^{-2x})\,dx=dt\qquad\left[\because\frac{d}{dx}(e^{ax})=ae^{ax}\right]
\displaystyle \Rightarrow(e^{2x}-e^{-2x})\,dx=\frac{dt}{2}
\displaystyle \therefore\ I=\int\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}\,dx=\frac{1}{2}\int\frac{dt}{t}=\frac{1}{2}\log|t|+C
\displaystyle =\frac{1}{2}\log|e^{2x}+e^{-2x}|+C\qquad[\because t=e^{2x}+e^{-2x}]
\\

\displaystyle \textbf{Question 138. }\text{Evaluate }\int\frac{\cos\sqrt{x}}{\sqrt{x}}\,dx. \hspace{1.2cm} \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{\cos\sqrt{x}}{\sqrt{x}}\,dx
\displaystyle \text{Put }\sqrt{x}=t\Rightarrow\frac{1}{2\sqrt{x}}\,dx=dt\Rightarrow\frac{1}{\sqrt{x}}\,dx=2\,dt
\displaystyle \therefore\ I=\int\frac{\cos\sqrt{x}}{\sqrt{x}}\,dx=2\int\cos t\,dt
\displaystyle =2\sin t+C=2\sin\sqrt{x}+C\qquad[\because t=\sqrt{x}]
\\

\displaystyle \textbf{Question 139. }\text{Evaluate }\int\frac{2\cos x}{3\sin^2x}\,dx. \hspace{1.2cm} \text{[CBSE 2011C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{2\cos x}{3\sin^2x}\,dx
\displaystyle \text{Put }\sin x=t\Rightarrow\cos x\,dx=dt
\displaystyle \therefore\ I=\int\frac{2}{3t^2}\,dt=\frac{2}{3}\int\frac{dt}{t^2}=\frac{2}{3}\cdot\frac{t^{-1}}{(-1)}+C
\displaystyle =-\frac{2}{3}(\sin x)^{-1}+C\qquad[\because t=\sin x]
\displaystyle =\frac{-2}{3\sin x}+C
\displaystyle \textbf{Alternate Method}
\displaystyle \text{Let }I=\int\frac{2\cos x}{3\sin^2x}\,dx=\frac{2}{3}\int\frac{1}{\sin x}\cdot\frac{\cos x}{\sin x}\,dx
\displaystyle =\frac{2}{3}\int\text{cosec }x\cot x\,dx
\displaystyle =\frac{2}{3}(-\text{cosec }x)+C=\frac{-2}{3\sin x}+C
\\

\displaystyle \textbf{Question 140. }\text{Evaluate }\int\frac{x^3-x^2+x-1}{x-1}\,dx. \hspace{1.2cm} \text{[CBSE 2011C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{x^3-x^2+x-1}{x-1}\,dx
\displaystyle =\int\frac{x^2(x-1)+1(x-1)}{x-1}\,dx
\displaystyle =\int\frac{(x^2+1)(x-1)}{x-1}\,dx=\int(x^2+1)\,dx
\displaystyle =\frac{x^3}{3}+x+C
\\

\displaystyle \textbf{Question 141. }\text{Write the value of }\int\frac{1-\sin x}{\cos^2x}\,dx. \hspace{1.2cm} \text{[CBSE 2011C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{1-\sin x}{\cos^2x}\,dx=\int\left(\frac{1}{\cos^2x}-\frac{\sin x}{\cos^2x}\right)dx
\displaystyle =\int\sec^2x\,dx-\int\sec x\tan x\,dx
\displaystyle =\tan x-\sec x+C
\\

\displaystyle \textbf{Question 142. }\text{Evaluate }\int \frac{2\cos x}{\sin^{2}x}\,dx. \qquad \text{[CBSE 2011C, 2009, 2008]}
\displaystyle \text{Answer:}
\displaystyle \int \frac{2\cos x}{\sin^{2}x}\,dx
\displaystyle =2\int \frac{\cos x}{\sin^{2}x}\,dx
\displaystyle =2\int \cot x\cdot\mathrm{cosec}\, x\,dx
\displaystyle =-2\,\mathrm{cosec}\, x+C
\displaystyle \therefore \int \frac{2\cos x}{\sin^{2}x}\,dx=-2\,\mathrm{cosec}\, x+C
\\

\displaystyle \textbf{Question 143. }\text{Evaluate }\int e^{2x}\sin x\,dx. \qquad \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \int e^{2x}\sin x\,dx
\displaystyle \text{Integrating by parts: taking }\sin x\text{ as first function and }e^{2x}\text{ as second}
\displaystyle =\sin x\cdot\frac{e^{2x}}{2}-\int\cos x\cdot\frac{e^{2x}}{2}\,dx
\displaystyle =\frac{e^{2x}\sin x}{2}-\frac{1}{2}\int e^{2x}\cos x\,dx
\displaystyle \text{Integrating by parts again:}
\displaystyle =\frac{e^{2x}\sin x}{2}-\frac{1}{2}\left[\cos x\cdot\frac{e^{2x}}{2}-\int(-\sin x)\cdot\frac{e^{2x}}{2}\,dx\right]
\displaystyle =\frac{e^{2x}\sin x}{2}-\frac{1}{2}\left[\frac{e^{2x}\cos x}{2}+\frac{1}{2}\int e^{2x}\sin x\,dx\right]
\displaystyle =\frac{e^{2x}\sin x}{2}-\frac{e^{2x}\cos x}{4}-\frac{1}{4}\int e^{2x}\sin x\,dx
\displaystyle \text{Let }I=\int e^{2x}\sin x\,dx
\displaystyle I=\frac{e^{2x}\sin x}{2}-\frac{e^{2x}\cos x}{4}-\frac{I}{4}
\displaystyle I+\frac{I}{4}=\frac{e^{2x}\sin x}{2}-\frac{e^{2x}\cos x}{4}
\displaystyle \frac{5I}{4}=\frac{e^{2x}(2\sin x-\cos x)}{4}
\displaystyle I=\frac{e^{2x}(2\sin x-\cos x)}{5}
\displaystyle \therefore \int e^{2x}\sin x\,dx=\frac{e^{2x}(2\sin x-\cos x)}{5}+C
\\

\displaystyle \textbf{Question 144. }\text{Evaluate }\int \frac{3x+5}{\sqrt{x^{2}-8x+7}}\,dx. \qquad \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \int \frac{3x+5}{\sqrt{x^{2}-8x+7}}\,dx
\displaystyle \text{Let }3x+5=A\frac{d}{dx}(x^{2}-8x+7)+B=A(2x-8)+B
\displaystyle \text{Comparing coefficients: }2A=3\Rightarrow A=\frac{3}{2},\quad -8A+B=5\Rightarrow B=5+12=17
\displaystyle =\frac{3}{2}\int\frac{2x-8}{\sqrt{x^{2}-8x+7}}\,dx+17\int\frac{dx}{\sqrt{x^{2}-8x+7}}
\displaystyle =\frac{3}{2}\cdot 2\sqrt{x^{2}-8x+7}+17\int\frac{dx}{\sqrt{(x-4)^{2}-9}}
\displaystyle =3\sqrt{x^{2}-8x+7}+17\log\left|(x-4)+\sqrt{(x-4)^{2}-9}\right|+C
\displaystyle \therefore \int \frac{3x+5}{\sqrt{x^{2}-8x+7}}\,dx=3\sqrt{x^{2}-8x+7}+17\log\left|x-4+\sqrt{x^{2}-8x+7}\right|+C
\\

\displaystyle \textbf{Question 145. }\text{Evaluate }\int\frac{x^2+4}{x^4+16}\,dx. \hspace{1.2cm} \text{[CBSE 2011C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{x^2+4}{x^4+16}\,dx
\displaystyle \text{On dividing numerator and denominator by }x^2\text{, we get}
\displaystyle I=\int\frac{\left(1+\dfrac{4}{x^2}\right)}{\left(x^2+\dfrac{16}{x^2}\right)}\,dx=\int\frac{\left(1+\dfrac{4}{x^2}\right)}{x^2+\left(\dfrac{4}{x}\right)^2-8+8}\,dx
\displaystyle =\int\frac{\left(1+\dfrac{4}{x^2}\right)}{\left(x-\dfrac{4}{x}\right)^2+8}\,dx
\displaystyle \text{Put }x-\frac{4}{x}=t\Rightarrow\left(1+\frac{4}{x^2}\right)dx=dt
\displaystyle \therefore\ I=\int\frac{dt}{t^2+8}=\int\frac{dt}{t^2+(2\sqrt{2})^2}
\displaystyle =\frac{1}{2\sqrt{2}}\tan^{-1}\left(\frac{t}{2\sqrt{2}}\right)+C
\displaystyle \left[\because\int\frac{dx}{a^2+x^2}=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)+C\right]
\displaystyle =\frac{1}{2\sqrt{2}}\tan^{-1}\left(\frac{x-\dfrac{4}{x}}{2\sqrt{2}}\right)+C\qquad\left[\because t=x-\frac{4}{x}\right]
\displaystyle =\frac{1}{2\sqrt{2}}\tan^{-1}\left(\frac{x^2-4}{2\sqrt{2}x}\right)+C
\\

\displaystyle \textbf{Question 146. }\text{Evaluate }\int \frac{x^{2}+1}{x^{4}+1}\,dx. \qquad \text{[CBSE 2011C]}
\displaystyle \text{Answer:}
\displaystyle \int \frac{x^{2}+1}{x^{4}+1}\,dx
\displaystyle \text{Divide numerator and denominator by }x^{2}:
\displaystyle =\int \frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}}\,dx
\displaystyle =\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+2}\,dx
\displaystyle \text{Let }t=x-\frac{1}{x},\text{ so }dt=\left(1+\frac{1}{x^{2}}\right)dx
\displaystyle =\int \frac{dt}{t^{2}+2}
\displaystyle =\frac{1}{\sqrt{2}}\tan^{-1}\frac{t}{\sqrt{2}}+C
\displaystyle =\frac{1}{\sqrt{2}}\tan^{-1}\frac{x-\frac{1}{x}}{\sqrt{2}}+C
\displaystyle \therefore \int \frac{x^{2}+1}{x^{4}+1}\,dx=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{x^{2}-1}{\sqrt{2}\,x}\right)+C
\\

\displaystyle \textbf{Question 147. }\text{Evaluate }\int\frac{\sin x-\cos x}{\sqrt{\sin 2x}}\,dx. \hspace{1.2cm} \text{[CBSE 2011C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int\frac{\sin x-\cos x}{\sqrt{\sin 2x}}\,dx=\int\frac{\sin x-\cos x}{\sqrt{1+\sin 2x-1}}\,dx
\displaystyle =\int\frac{\sin x-\cos x}{\sqrt{\sin^2x+\cos^2x+2\sin x\cos x-1}}\,dx\qquad[\because\sin^2\theta+\cos^2\theta=1]
\displaystyle =\int\frac{\sin x-\cos x}{\sqrt{(\sin x+\cos x)^2-1}}\,dx
\displaystyle \text{Put }\sin x+\cos x=t\Rightarrow(\cos x-\sin x)\,dx=dt
\displaystyle \therefore\ I=\int\frac{-dt}{\sqrt{t^2-1}}=-\log\left|t+\sqrt{t^2-1}\right|+C
\displaystyle \left[\because\int\frac{dx}{\sqrt{x^2-a^2}}=\log\left|x+\sqrt{x^2-a^2}\right|+C\right]
\displaystyle \Rightarrow I=-\log|(\sin x+\cos x)+\sqrt{(\sin x+\cos x)^2-1}|+C\qquad[\because t=\sin x+\cos x]
\displaystyle =-\log|(\sin x+\cos x)+\sqrt{\sin^2x+\cos^2x+2\sin x\cos x-1}|+C
\displaystyle =-\log|(\sin x+\cos x)+\sqrt{\sin 2x}|+C
\\

\displaystyle \textbf{Question 148. }\text{Evaluate }\int \frac{6x+7}{\sqrt{(x-5)(x-4)}}\,dx. \qquad \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \int \frac{6x+7}{\sqrt{(x-5)(x-4)}}\,dx
\displaystyle =\int \frac{6x+7}{\sqrt{x^{2}-9x+20}}\,dx
\displaystyle \text{Let }6x+7=A\frac{d}{dx}(x^{2}-9x+20)+B=A(2x-9)+B
\displaystyle \text{Comparing coefficients: }2A=6\Rightarrow A=3,\quad -9A+B=7\Rightarrow B=7+27=34
\displaystyle =3\int\frac{2x-9}{\sqrt{x^{2}-9x+20}}\,dx+34\int\frac{dx}{\sqrt{x^{2}-9x+20}}
\displaystyle =3\cdot 2\sqrt{x^{2}-9x+20}+34\int\frac{dx}{\sqrt{\left(x-\frac{9}{2}\right)^{2}-\frac{1}{4}}}
\displaystyle =6\sqrt{x^{2}-9x+20}+34\log\left|\left(x-\frac{9}{2}\right)+\sqrt{\left(x-\frac{9}{2}\right)^{2}-\frac{1}{4}}\right|+C
\displaystyle \therefore \int \frac{6x+7}{\sqrt{(x-5)(x-4)}}\,dx=6\sqrt{x^{2}-9x+20}+34\log\left|x-\frac{9}{2}+\sqrt{x^{2}-9x+20}\right|+C
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