\displaystyle \textbf{Question 1: } \text{If } \int (\cot x - \mathrm{cosec}^{2}x)e^{x}\,dx = e^{x}f(x) + C, \text{ then } f(x) \text{ will be } \quad \text{ISC 2024}
\displaystyle \text{(a) } \cot x + \mathrm{cosec}\,x \quad \text{(b) } \cot^{2}x \quad \text{(c) } \cot x \quad \text{(d) } \mathrm{cosec}\,x
\displaystyle \text{Answer:}
\displaystyle \text{(c) We have, } \int (\cot x-\mathrm{cosec}^{2}x)e^{x}\,dx=e^{x}f(x)+C
\displaystyle \text{Taking LHS } \int (\cot x-\mathrm{cosec}^{2}x)e^{x}\,dx=\int e^{x}\cot x\,dx-\int e^{x}\mathrm{cosec}^{2}x\,dx
\displaystyle \text{Integrating by parts, we get }=e^{x}\cot x+\int \mathrm{cosec}^{2}x\cdot e^{x}\,dx-\int e^{x}\mathrm{cosec}^{2}x\,dx
\displaystyle =e^{x}\cot x+C
\displaystyle \text{On comparing with RHS, we get } f(x)=\cot x
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\displaystyle \textbf{Question 2: } \text{Evaluate } \int x^{2}\cos x\,dx. \quad \text{ISC 2024}
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int x^{2}\cos x\,dx
\displaystyle \text{Integrating by parts, we get }=x^{2}\sin x-\int 2x\sin x\,dx
\displaystyle =x^{2}\sin x-\left[-2x\cos x+\int 2\cos x\,dx\right]
\displaystyle =x^{2}\sin x+2x\cos x-2\sin x+C
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\displaystyle \textbf{Question 3: } \text{Evaluate } \int \frac{x+7}{x^{2}+4x+7}\,dx. \quad \text{ISC 2024}
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{x+7}{x^{2}+4x+7}\,dx
\displaystyle \text{Assume } x+7=A\frac{d}{dx}(x^{2}+4x+7)+B
\displaystyle \text{Comparing coefficients, } 2A=1 \Rightarrow A=\frac{1}{2},\quad 4A+B=7 \Rightarrow B=5
\displaystyle \therefore x+7=\frac{1}{2}(2x+4)+5
\displaystyle \Rightarrow I=\frac{1}{2}\int \frac{2x+4}{x^{2}+4x+7}\,dx+5\int \frac{1}{x^{2}+4x+7}\,dx
\displaystyle =\frac{1}{2}\log|x^{2}+4x+7|+5\int \frac{1}{(x+2)^{2}+3}\,dx
\displaystyle =\frac{1}{2}\log|x^{2}+4x+7|+\frac{5}{\sqrt{3}}\tan^{-1}\left(\frac{x+2}{\sqrt{3}}\right)+C
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\displaystyle \textbf{Question 4: } \text{Evaluate } \int \frac{x}{x^{2}+1}\,dx \quad \text{ISC 2023}
\displaystyle \text{(a) } 2\log(x^{2}+1)+C \quad \text{(b) } \frac{1}{2}\log(x^{2}+1)+C \quad \text{(c) } e^{x^{2}+1}+C \quad \text{(d) } \log x + \frac{x^{2}}{2}+C
\displaystyle \text{Answer:}
\displaystyle \text{(b) Let } I=\int \frac{x}{x^{2}+1}\,dx
\displaystyle \text{Put } x^{2}+1=t \Rightarrow 2x\,dx=dt
\displaystyle \therefore I=\frac{1}{2}\int \frac{dt}{t}=\frac{1}{2}\log|t|+C=\frac{1}{2}\log|x^{2}+1|+C
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\displaystyle \textbf{Question 5: } \text{Evaluate } \int \frac{1+\cos x}{\sin^{2}x}\,dx \quad \text{ISC 2023}
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{1+\cos x}{\sin^{2}x}\,dx
\displaystyle \Rightarrow I=\int (\mathrm{cosec}^{2}x+\mathrm{cosec}x\cot x)\,dx
\displaystyle \Rightarrow I=-\cot x-\mathrm{cosec}x+C
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\displaystyle \textbf{Question 6: } \text{Evaluate } \int \cos^{-1}(\sin x)\,dx. \quad \text{ISC 2023}
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \cos^{-1}(\sin x)\,dx=\int \cos^{-1}\left(\cos\left(\frac{\pi}{2}-x\right)\right)dx
\displaystyle =\int \left(\frac{\pi}{2}-x\right)dx
\displaystyle \text{[}\because \cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta \text{ and } \cos^{-1}(\cos\theta)=\theta,\theta\in[0,\pi]\text{]}
\displaystyle \Rightarrow I=\frac{\pi x}{2}-\frac{x^{2}}{2}+C
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\displaystyle \textbf{Question 7: } \text{If } \int x^{5}\cos(x^{6})\,dx = k\sin(x^{6})+C, \text{ then find the value of } k. \quad \text{ISC 2023}
\displaystyle \text{Answer:}
\displaystyle \text{Given, } \int x^{5}\cos(x^{6})\,dx=k\sin(x^{6})+C \quad (i)
\displaystyle \text{Let } I=\int x^{5}\cos(x^{6})\,dx
\displaystyle \text{On putting } x^{6}=t \Rightarrow 6x^{5}dx=dt
\displaystyle \therefore I=\frac{1}{6}\int \cos t\,dt=\frac{1}{6}\sin t+C=\frac{1}{6}\sin(x^{6})+C
\displaystyle \Rightarrow \int x^{5}\cos(x^{6})\,dx=\frac{1}{6}\sin(x^{6})+C \quad (ii)
\displaystyle \text{On comparing Eqs. (i) and (ii), we get } k=\frac{1}{6}
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\displaystyle \textbf{Question 8: } \text{Evaluate } \int \frac{3e^{2x}-2e^{x}}{e^{2x}+2e^{x}-8}\,dx. \quad \text{ISC 2023}
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{3e^{2x}-2e^{x}}{e^{2x}+2e^{x}-8}\,dx
\displaystyle \text{On putting } e^{x}=t \Rightarrow e^{x}dx=dt
\displaystyle \therefore I=\int \frac{3t^{2}-2t}{t^{2}+2t-8}\cdot \frac{dt}{t}
\displaystyle =\int \frac{3t-2}{t^{2}+2t-8}\,dt
\displaystyle \text{Let } 3t-2=\lambda \frac{d}{dt}(t^{2}+2t-8)+\mu
\displaystyle \Rightarrow 3t-2=\lambda(2t+2)+\mu
\displaystyle \Rightarrow 3t-2=2\lambda t+2\lambda+\mu
\displaystyle \text{Equating coefficients, } 3=2\lambda,\ -2=2\lambda+\mu
\displaystyle \Rightarrow \lambda=\frac{3}{2},\ \mu=-5
\displaystyle \therefore I=\frac{3}{2}\int \frac{2t+2}{t^{2}+2t-8}\,dt-5\int \frac{dt}{t^{2}+2t-8}
\displaystyle =\frac{3}{2}\log|t^{2}+2t-8|-5\int \frac{dt}{(t+1)^{2}-9}
\displaystyle =\frac{3}{2}\log|t^{2}+2t-8|-\frac{5}{6}\log\left|\frac{t-2}{t+4}\right|+C
\displaystyle \Rightarrow I=\frac{3}{2}\log|e^{2x}+2e^{x}-8|-\frac{5}{6}\log\left|\frac{e^{x}-2}{e^{x}+4}\right|+C
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\displaystyle \textbf{Question 9: } \text{Evaluate } \int \frac{2}{(1-x)(1+x^{2})}\,dx. \quad \text{ISC 2023}
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\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 2=A(1+x^{2})+(Bx+C)(1-x)
\displaystyle \Rightarrow 2=A+Ax^{2}+Bx-Cx^{2}+C-Bx^{2}
\displaystyle \Rightarrow 2=(A+C)+(B)x+(A-C-B)x^{2}
\displaystyle \text{Equating coefficients, } A-C-B=0,\ B=0,\ A+C=2
\displaystyle \Rightarrow A=1,\ B=0,\ C=1
\displaystyle \therefore I=\int \frac{dx}{1-x}+\int \frac{dx}{1+x^{2}}
\displaystyle =-\log|1-x|+\tan^{-1}x+C
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\displaystyle \textbf{Question 10: } \text{Evaluate } \int \sin^{3}x\cos^{4}x\,dx. \quad \text{ISC 2020}
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \sin^{3}x\cos^{4}x\,dx
\displaystyle =\int \sin^{2}x\cos^{4}x\sin x\,dx
\displaystyle =\int (1-\cos^{2}x)\cos^{4}x\sin x\,dx
\displaystyle \text{On putting } \cos x=t \Rightarrow -\sin x\,dx=dt
\displaystyle \therefore I=-\int (1-t^{2})t^{4}\,dt
\displaystyle =\int (t^{6}-t^{4})\,dt
\displaystyle =\frac{t^{7}}{7}-\frac{t^{5}}{5}+C
\displaystyle \Rightarrow I=\frac{\cos^{7}x}{7}-\frac{\cos^{5}x}{5}+C
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\displaystyle \textbf{Question 11: } \text{Evaluate } \int \frac{dx}{\sqrt{5x-4x^{2}}}. \quad \text{ISC 2020}
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{dx}{\sqrt{5x-4x^{2}}}
\displaystyle =\frac{1}{2}\int \frac{dx}{\sqrt{\frac{5x}{4}-x^{2}}}
\displaystyle =\frac{1}{2}\int \frac{dx}{\sqrt{\frac{25}{64}-x^{2}+\frac{5x}{4}-\frac{25}{64}}}
\displaystyle =\frac{1}{2}\int \frac{dx}{\sqrt{\left(\frac{5}{8}\right)^{2}-\left(x-\frac{5}{8}\right)^{2}}}
\displaystyle =\frac{1}{2}\sin^{-1}\left(\frac{x-\frac{5}{8}}{\frac{5}{8}}\right)+C
\displaystyle =\frac{1}{2}\sin^{-1}\left(\frac{8x-5}{5}\right)+C
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\displaystyle \textbf{Question 12: } \text{Evaluate } \int \tan^{-1}\sqrt{\frac{1-x}{1+x}}\,dx. \quad \text{ISC 2020}
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \tan^{-1}\sqrt{\frac{1-x}{1+x}}\,dx
\displaystyle \text{On putting } x=\cos 2\theta \Rightarrow dx=-2\sin 2\theta\,d\theta
\displaystyle \therefore I=\int \tan^{-1}\left(\frac{\sqrt{1-\cos 2\theta}}{\sqrt{1+\cos 2\theta}}\right)(-2\sin 2\theta)\,d\theta
\displaystyle =\int \tan^{-1}\left(\frac{\sqrt{2\sin^{2}\theta}}{\sqrt{2\cos^{2}\theta}}\right)(-2\sin 2\theta)\,d\theta
\displaystyle =\int \tan^{-1}(\tan\theta)(-2\sin 2\theta)\,d\theta
\displaystyle =-2\int \theta\sin 2\theta\,d\theta
\displaystyle \text{Using integration by parts, } I=2\theta\cos 2\theta-\int 2\cos 2\theta\,d\theta
\displaystyle =2\theta\cos 2\theta-\sin 2\theta+C
\displaystyle =2\cos^{-1}\sqrt{\frac{1+x}{2}}\cdot x-\sqrt{1-x^{2}}+C
\displaystyle =-2\left[\theta\int \sin 2\theta\,d\theta-\int \left\{\frac{d}{d\theta}\theta\right\}\left(\int \sin 2\theta\,d\theta\right)d\theta\right]+C
\displaystyle =-2\left[-\frac{\theta\cos 2\theta}{2}+\int \frac{\cos 2\theta}{2}\,d\theta\right]+C
\displaystyle =\theta\cos 2\theta-\frac{1}{2}\sin 2\theta+C
\displaystyle \Rightarrow I=\frac{x}{2}\cos^{-1}x-\frac{\sqrt{1-x^{2}}}{2}+C
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\displaystyle \textbf{Question 13: } \text{Evaluate } \int \frac{2x+7}{x^{2}-x+2}\,dx. \quad \text{ISC 2020}
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{2x+7}{x^{2}-x+2}\,dx
\displaystyle =\int \frac{2x-1}{x^{2}-x+2}\,dx+\int \frac{8}{x^{2}-x+2}\,dx
\displaystyle =\int \frac{\frac{d}{dx}(x^{2}-x+2)}{x^{2}-x+2}\,dx+8\int \frac{dx}{x^{2}-x+2}
\displaystyle =\log|x^{2}-x+2|+8\int \frac{dx}{\left(x-\frac{1}{2}\right)^{2}+\frac{7}{4}}
\displaystyle =\log|x^{2}-x+2|+8\cdot \frac{2}{\sqrt{7}}\tan^{-1}\left(\frac{2x-1}{\sqrt{7}}\right)+C
\displaystyle \Rightarrow I=\log|x^{2}-x+2|+\frac{16}{\sqrt{7}}\tan^{-1}\left(\frac{2x-1}{\sqrt{7}}\right)+C
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\displaystyle \textbf{Question 14: } \text{Evaluate } \int \frac{x(1+x^{2})}{1+x^{4}}\,dx. \quad \text{ISC 2019}
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{x(1+x^{2})}{1+x^{4}}\,dx
\displaystyle =\int \frac{x}{1+x^{4}}\,dx+\int \frac{x^{3}}{1+x^{4}}\,dx
\displaystyle \text{On putting } x^{2}=t \text{ in first integral, }2x\,dx=dt
\displaystyle \Rightarrow x\,dx=\frac{dt}{2}
\displaystyle \text{and } 1+x^{4}=u \text{ in second integral, }4x^{3}\,dx=du
\displaystyle \therefore I=\int \frac{1}{2(1+t^{2})}\,dt+\frac{1}{4}\int \frac{du}{u}
\displaystyle =\frac{1}{2}\tan^{-1}t+\frac{1}{4}\log|u|+C
\displaystyle =\frac{1}{2}\tan^{-1}x^{2}+\frac{1}{4}\log|1+x^{4}|+C
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\displaystyle \textbf{Question 15: } \text{Evaluate } \int \frac{x^{3}+5x^{2}+4x+1}{x^{2}}\,dx. \quad \text{ISC 2018}
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{x^{3}+5x^{2}+4x+1}{x^{2}}\,dx
\displaystyle =\int \left(x+5+\frac{4}{x}+\frac{1}{x^{2}}\right)dx
\displaystyle =\frac{x^{2}}{2}+5x+4\log|x|-\frac{1}{x}+C
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\displaystyle \textbf{Question 16: } \text{Evaluate } \int \tan^{-1}\sqrt{x}\,dx. \quad \text{ISC 2018}
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \tan^{-1}\sqrt{x}\,dx
\displaystyle \text{On putting } x=t^{2}\Rightarrow dx=2t\,dt
\displaystyle \therefore I=\int 2t\tan^{-1}t\,dt
\displaystyle \text{Using integration by parts, we get}
\displaystyle I=2\left[\tan^{-1}t\int t\,dt-\int \left(\frac{d}{dt}\tan^{-1}t\right)\left(\int t\,dt\right)dt\right]+C
\displaystyle =2\left[\frac{t^{2}}{2}\tan^{-1}t-\int \frac{1}{1+t^{2}}\cdot\frac{t^{2}}{2}\,dt\right]+C
\displaystyle =t^{2}\tan^{-1}t-\int \frac{t^{2}}{1+t^{2}}\,dt+C
\displaystyle =t^{2}\tan^{-1}t-\int \left(1-\frac{1}{1+t^{2}}\right)dt+C
\displaystyle =t^{2}\tan^{-1}t-t+\tan^{-1}t+C
\displaystyle =x\tan^{-1}\sqrt{x}-\sqrt{x}+\tan^{-1}\sqrt{x}+C
\displaystyle =t^{2}\tan^{-1}t-\int 1\,dt+\int \frac{1}{1+t^{2}}\,dt+C
\displaystyle =t^{2}\tan^{-1}t-t+\tan^{-1}t+C
\displaystyle =(t^{2}+1)\tan^{-1}t-t+C
\displaystyle \Rightarrow I=(x+1)\tan^{-1}\sqrt{x}-\sqrt{x}+C
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\displaystyle \textbf{Question 17: } \text{Evaluate } \int \frac{x-1}{\sqrt{x^{2}-x}}\,dx. \quad \text{ISC 2018}
\displaystyle \text{Answer:}
\displaystyle \text{We have, } I=\int \frac{x-1}{\sqrt{x^{2}-x}}\,dx
\displaystyle =\frac{1}{2}\int \frac{2x-2}{\sqrt{x^{2}-x}}\,dx
\displaystyle =\frac{1}{2}\left[\int \frac{2x-1}{\sqrt{x^{2}-x}}\,dx-\int \frac{dx}{\sqrt{x^{2}-x}}\right]
\displaystyle =\frac{1}{2}\left[\int \frac{2x-1}{\sqrt{x^{2}-x}}\,dx-\int \frac{dx}{\sqrt{\left(x-\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}}\right]
\displaystyle \left[\because \text{putting } x^{2}-x=t \Rightarrow (2x-1)dx=dt\right]
\displaystyle =\frac{1}{2}\left[2\sqrt{t}-\log\left|x-\frac{1}{2}+\sqrt{x^{2}-x}\right|\right]+C
\displaystyle \Rightarrow I=\sqrt{x^{2}-x}-\frac{1}{2}\log\left|x-\frac{1}{2}+\sqrt{x^{2}-x}\right|+C
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\displaystyle \textbf{Question 18: } \text{Evaluate } \int \frac{\sin 2x}{(1+\sin x)(2+\sin x)}\,dx. \quad \text{ISC 2017}
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{\sin 2x}{(1+\sin x)(2+\sin x)}\,dx
\displaystyle =\int \frac{2\sin x\cos x}{(1+\sin x)(2+\sin x)}\,dx
\displaystyle \text{On putting } \sin x=t \Rightarrow \cos x\,dx=dt
\displaystyle \therefore I=2\int \frac{t}{(1+t)(2+t)}\,dt \quad (i)
\displaystyle \text{Let } \frac{t}{(1+t)(2+t)}=\frac{A}{1+t}+\frac{B}{2+t}
\displaystyle \Rightarrow t=A(2+t)+B(1+t)
\displaystyle \text{On putting } t=-1,\text{ we get } A=-1
\displaystyle \text{On putting } t=-2,\text{ we get } B=2
\displaystyle \therefore \frac{t}{(1+t)(2+t)}=\frac{-1}{1+t}+\frac{2}{2+t}
\displaystyle \therefore \text{From Eq. (i), } I=2\int \left(\frac{-1}{1+t}+\frac{2}{2+t}\right)dt
\displaystyle =2[-\log|1+t|+2\log|2+t|]+C
\displaystyle =-2\log|1+\sin x|+4\log|2+\sin x|+C
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\displaystyle \textbf{Question 19: } \text{Evaluate } \int \frac{1}{x^{2}}\sin^{2}\left(\frac{1}{x}\right)\,dx. \quad \text{ISC 2017}
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{1}{x^{2}}\sin^{2}\left(\frac{1}{x}\right)dx
\displaystyle \text{On putting } \frac{1}{x}=t,\text{ and differentiating both sides w.r.t. }x,
\displaystyle -\frac{1}{x^{2}}dx=dt \Rightarrow \frac{1}{x^{2}}dx=-dt
\displaystyle \therefore I=\int \sin^{2}t(-dt)=-\int \sin^{2}t\,dt
\displaystyle =-\int \frac{1-\cos 2t}{2}\,dt
\displaystyle =-\left[\frac{1}{2}t-\frac{1}{2}\cdot\frac{\sin 2t}{2}\right]+C
\displaystyle =\frac{1}{4}\sin 2t-\frac{1}{2}t+C
\displaystyle \text{Now, by putting value of }t,\text{ we get}
\displaystyle I=\frac{1}{4}\sin\left(\frac{2}{x}\right)-\frac{1}{2x}+C
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\displaystyle \textbf{Question 20: } \text{Evaluate } \int \frac{2y^{2}}{y^{2}+4}\,dy. \quad \text{ISC 2017}
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{2y^{2}}{y^{2}+4}\,dy=\int \frac{2y^{2}+8-8}{y^{2}+4}\,dy
\displaystyle \text{[adding and subtracting }8\text{ in numerator of integrand]}
\displaystyle =\int \frac{2(y^{2}+4)-8}{y^{2}+4}\,dy
\displaystyle =\int (2)\,dy-\int \frac{8}{y^{2}+4}\,dy
\displaystyle =2y-8\cdot\frac{1}{2}\tan^{-1}\left(\frac{y}{2}\right)+C
\displaystyle =2y-4\tan^{-1}\left(\frac{y}{2}\right)+C
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\displaystyle \textbf{Question 21: } \text{Evaluate } \int \tan^{3}x\,dx. \quad \text{ISC 2016}
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \tan^{3}x\,dx=\int \tan^{2}x\tan x\,dx
\displaystyle =\int (\sec^{2}x-1)\tan x\,dx
\displaystyle =\int (\tan x\sec^{2}x-\tan x)\,dx
\displaystyle =\int \tan x\sec^{2}x\,dx-\int \tan x\,dx
\displaystyle \text{On putting } \tan x=t \text{ and } \sec^{2}x\,dx=dt \text{ in first integral,}
\displaystyle I=\int t\,dt-\int \tan x\,dx
\displaystyle \Rightarrow I=\frac{t^{2}}{2}+\log|\cos x|+C
\displaystyle \therefore \int \tan^{3}x\,dx=\frac{1}{2}\tan^{2}x+\log|\cos x|+C
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\displaystyle \textbf{Question 22: } \text{Evaluate } \int \frac{\sin x+\cos x}{\sqrt{9+16\sin 2x}}\,dx. \quad \text{ISC 2016}
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{\sin x+\cos x}{\sqrt{9+16\sin 2x}}\,dx
\displaystyle \text{On putting } \sin x-\cos x=t \Rightarrow (\cos x+\sin x)\,dx=dt
\displaystyle \text{Again, }(\sin x-\cos x)^{2}=t^{2}
\displaystyle \Rightarrow \sin^{2}x+\cos^{2}x-2\sin x\cos x=t^{2}
\displaystyle \Rightarrow 1-2\sin x\cos x=t^{2}
\displaystyle \Rightarrow 1-\sin 2x=t^{2}
\displaystyle \Rightarrow \sin 2x=1-t^{2}
\displaystyle \therefore I=\int \frac{dt}{\sqrt{9+16(1-t^{2})}}=\int \frac{dt}{\sqrt{25-16t^{2}}}
\displaystyle =\int \frac{dt}{\sqrt{4^{2}\left[\left(\frac{5}{4}\right)^{2}-t^{2}\right]}}
\displaystyle =\frac{1}{4}\int \frac{dt}{\sqrt{\left(\frac{5}{4}\right)^{2}-t^{2}}}
\displaystyle =\frac{1}{4}\sin^{-1}\left(\frac{4t}{5}\right)+C
\displaystyle \Rightarrow I=\frac{1}{4}\sin^{-1}\left[\frac{4}{5}(\sin x-\cos x)\right]+C
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\displaystyle \textbf{Question 23: } \text{Evaluate } \int \frac{\sec x}{1+\mathrm{cosec}\,x}\,dx. \quad \text{ISC 2015}
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{\sec x}{1+\mathrm{cosec}\,x}\,dx
\displaystyle =\int \frac{\sin x}{\cos x(1+\sin x)}\,dx
\displaystyle =\int \frac{\sin x}{\cos x(1+\sin x)}\times\frac{\cos x}{\cos x}\,dx
\displaystyle =\int \frac{\sin x\cos x}{\cos^{2}x(1+\sin x)}\,dx
\displaystyle =\int \frac{\sin x\cos x}{(1-\sin^{2}x)(1+\sin x)}\,dx
\displaystyle =\int \frac{\sin x\cos x}{(1-\sin x)(1+\sin x)^{2}}\,dx
\displaystyle \text{On putting } \sin x=t \Rightarrow \cos x\,dx=dt
\displaystyle \therefore I=\int \frac{t}{(1+t)^{2}(1-t)}\,dt
\displaystyle \text{By using partial fraction, consider}
\displaystyle \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 \Rightarrow t=A(1-t^{2})+B(1-t)+C(1+t^{2}+2t)
\displaystyle \text{On equating coefficients of }t^{2},t,\text{ and constant, we get}
\displaystyle 0=-A+C,\quad 1=-B+2C,\quad 0=A+B+C
\displaystyle \text{On solving, we get } A=\frac{1}{4},\ B=-\frac{1}{2},\ C=\frac{1}{4}
\displaystyle \therefore \frac{t}{(1+t)^{2}(1-t)}=\frac{1}{4(1+t)}-\frac{1}{2(1+t)^{2}}+\frac{1}{4(1-t)}
\displaystyle \therefore I=\int \left[\frac{1}{4(1+t)}-\frac{1}{2(1+t)^{2}}+\frac{1}{4(1-t)}\right]dt
\displaystyle =\frac{1}{4}\log|1+t|+\frac{1}{2(1+t)}-\frac{1}{4}\log|1-t|+C
\displaystyle =\frac{1}{4}\log\left|\frac{1+t}{1-t}\right|+\frac{1}{2(1+t)}+C
\displaystyle \Rightarrow I=\frac{1}{4}\log\left|\frac{1+\sin x}{1-\sin x}\right|+\frac{1}{2(1+\sin x)}+C
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\displaystyle \textbf{Question 24: } \text{Evaluate } \int \frac{x+\sin x}{1+\cos x}\,dx. \quad \text{ISC 2014}
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{x+\sin x}{1+\cos x}\,dx
\displaystyle \text{Then, } I=\int \frac{x}{1+\cos x}\,dx+\int \frac{\sin x}{1+\cos x}\,dx
\displaystyle \Rightarrow I=I_{1}+I_{2} \quad \text{[say]} \quad (i)
\displaystyle \text{Now, } I_{1}=\int \frac{x}{1+\cos x}\,dx
\displaystyle =\int \frac{x}{2\cos^{2}\frac{x}{2}}\,dx \quad \left[\because \cos x=2\cos^{2}\frac{x}{2}-1\right]
\displaystyle =\frac{1}{2}\int x\sec^{2}\frac{x}{2}\,dx
\displaystyle \text{Using integration by parts, we get}
\displaystyle I_{1}=\frac{1}{2}\left[x\cdot 2\tan\frac{x}{2}-\int 1\cdot 2\tan\frac{x}{2}\,dx\right]
\displaystyle \Rightarrow I_{1}=x\tan\frac{x}{2}-\int \tan\frac{x}{2}\,dx
\displaystyle \text{and } I_{2}=\int \frac{\sin x}{1+\cos x}\,dx
\displaystyle =\int \frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^{2}\frac{x}{2}}\,dx
\displaystyle =\int \tan\frac{x}{2}\,dx
\displaystyle \text{From Eq. (i), } I=I_{1}+I_{2}
\displaystyle \Rightarrow I=x\tan\frac{x}{2}-\int \tan\frac{x}{2}\,dx+\int \tan\frac{x}{2}\,dx
\displaystyle \Rightarrow I=x\tan\frac{x}{2}+C
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\displaystyle \textbf{Question 25: } \text{Evaluate } \int e^{x}\frac{2+\sin 2x}{\cos^{2}x}\,dx. \quad \text{ISC 2014}
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{e^{x}(2+\sin 2x)}{\cos^{2}x}\,dx
\displaystyle =\int \frac{e^{x}(2+2\sin x\cos x)}{\cos^{2}x}\,dx
\displaystyle =2\int e^{x}\left(\frac{1}{\cos^{2}x}+\frac{\sin x\cos x}{\cos^{2}x}\right)dx
\displaystyle =2\int e^{x}(\sec^{2}x+\tan x)\,dx
\displaystyle =2\int e^{x}(\tan x+\sec^{2}x)\,dx
\displaystyle \Rightarrow I=2e^{x}\tan x+C
\displaystyle \left[\because \int e^{x}\{f(x)+f'(x)\}\,dx=e^{x}f(x)+C\right]
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\displaystyle \textbf{Question 26: } \text{Evaluate } \int \frac{1}{x+\sqrt{x}}\,dx. \quad \text{ISC 2013}
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{1}{x+\sqrt{x}}\,dx
\displaystyle \text{On putting } x=t^{2}\Rightarrow dx=2t\,dt
\displaystyle I=\int \frac{2t\,dt}{t^{2}+t}
\displaystyle =2\int \frac{1}{1+t}\,dt
\displaystyle =2\log|1+t|+C
\displaystyle =2\log|1+\sqrt{x}|+C
\\

\displaystyle \textbf{Question 27: } \text{Evaluate } \int \frac{\cos^{-1}x}{x^{2}}\,dx. \quad \text{ISC 2013}
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{\cos^{-1}x}{x^{2}}\,dx
\displaystyle \text{On putting } x=\cos\theta \Rightarrow dx=-\sin\theta\,d\theta \text{ and } \theta=\cos^{-1}x
\displaystyle \therefore I=\int \left[-\theta\cdot\frac{\sin\theta}{\cos^{2}\theta}\right]d\theta
\displaystyle \Rightarrow I=-\int \theta(\tan\theta\sec\theta)\,d\theta
\displaystyle \text{Using integration by parts, we get}
\displaystyle I=-\left[\theta\sec\theta-\int 1\cdot\sec\theta\,d\theta\right]
\displaystyle =-\theta\sec\theta+\int \sec\theta\,d\theta
\displaystyle =-\theta\sec\theta+\log|\sec\theta+\tan\theta|+C
\displaystyle =-\frac{\cos^{-1}x}{x}+\log\left|\frac{1}{x}+\sqrt{\frac{1}{x^{2}}-1}\right|+C
\displaystyle =-\frac{\cos^{-1}x}{x}+\log\left|\frac{1+\sqrt{1-x^{2}}}{x}\right|+C
\\

\displaystyle \textbf{Question 28: } \text{Evaluate } \int \frac{dx}{x\{6(\log x)^{2}+7\log x+2\}}. \quad \text{ISC 2012}
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{dx}{x\{6(\log x)^{2}+7\log x+2\}}
\displaystyle \text{On putting } \log x=t \Rightarrow \frac{1}{x}dx=dt
\displaystyle \therefore I=\int \frac{dt}{6t^{2}+7t+2}
\displaystyle =\int \frac{dt}{(3t+2)(2t+1)} \quad (i)
\displaystyle \text{Let } \frac{1}{(3t+2)(2t+1)}=\frac{A}{3t+2}+\frac{B}{2t+1}
\displaystyle \Rightarrow 1=A(2t+1)+B(3t+2)
\displaystyle \text{On putting } t=-\frac{1}{2},\ B=2;\quad \text{on putting } t=-\frac{2}{3},\ A=-3
\displaystyle \therefore I=\int \left(\frac{-3}{3t+2}+\frac{2}{2t+1}\right)dt
\displaystyle =-3\int \frac{dt}{3t+2}+2\int \frac{dt}{2t+1}
\displaystyle =-\log|3t+2|+\log|2t+1|+C
\displaystyle =\log\left|\frac{2t+1}{3t+2}\right|+C
\displaystyle \Rightarrow I=\log\left|\frac{2\log x+1}{3\log x+2}\right|+C
\\

\displaystyle \textbf{Question 29: } \text{Evaluate } \int e^{x}(\tan x+\log \sec x)\,dx. \quad \text{ISC 2012}
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int e^{x}(\tan x+\log \sec x)\,dx
\displaystyle =\int e^{x}\left[\log \sec x+\frac{d}{dx}(\log \sec x)\right]dx
\displaystyle \left[\because \frac{d}{dx}\log \sec x=\tan x\right]
\displaystyle \therefore I=e^{x}\log \sec x+C
\\

\displaystyle \textbf{Question 30: } \text{Evaluate } \int \frac{\mathrm{cosec}\,x}{\log\tan\left(\frac{x}{2}\right)}\,dx. \quad \text{ISC 2011}
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int \frac{\mathrm{cosec}\,x}{\log \tan\frac{x}{2}}\,dx
\displaystyle \text{On putting } \log \tan\frac{x}{2}=t
\displaystyle \Rightarrow \frac{1}{\tan\frac{x}{2}}\cdot \sec^{2}\frac{x}{2}\cdot \frac{1}{2}dx=dt
\displaystyle \Rightarrow \frac{\sec^{2}\frac{x}{2}}{2\tan\frac{x}{2}}dx=dt
\displaystyle \Rightarrow \frac{1}{2}\cdot \frac{1}{\sin\frac{x}{2}\cos\frac{x}{2}}dx=dt
\displaystyle \Rightarrow \frac{1}{\sin x}dx=dt \quad \left[\because \sin x=2\sin\frac{x}{2}\cos\frac{x}{2}\right]
\displaystyle \therefore I=\int \frac{\mathrm{cosec}\,x\,dx}{\log \tan\frac{x}{2}}=\int \frac{dt}{t}
\displaystyle =\log|t|+C
\displaystyle =\log\left|\log \tan\frac{x}{2}\right|+C
\\

 

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