\displaystyle \text{Evaluate the following integrals:}
\displaystyle \textbf{Question 1: }~\int \frac{1}{\sqrt{1-\cos 2x}}\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \frac{1}{\sqrt{1-\cos 2x}}\,dx

\displaystyle = \int \frac{1}{\sqrt{2\sin^2 x}}\,dx \qquad [\because\ 1-\cos 2x = 2\sin^2 x]

\displaystyle = \frac{1}{\sqrt{2}}\int \mathrm{cosec}\,x\,dx

\displaystyle = \frac{1}{\sqrt{2}}\ln\left|\mathrm{cosec}\,x-\cot x\right|+C

\displaystyle = \frac{1}{\sqrt{2}}\ln\left|\frac{1-\cos x}{\sin x}\right|+C

\displaystyle = \frac{1}{\sqrt{2}}\ln\left|\frac{2\sin^2\frac{x}{2}}{\sin x}\right|+C \qquad [\because\ 1-\cos x = 2\sin^2\frac{x}{2}]

\displaystyle = \frac{1}{\sqrt{2}}\ln\left|\frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}}\right|+C \qquad [\because\ \sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}]

\displaystyle = \frac{1}{\sqrt{2}}\ln\left|\tan\frac{x}{2}\right|+C

\displaystyle \textbf{Question 2: }~\int \frac{1}{\sqrt{1+\cos x}}\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \frac{1}{\sqrt{1+\cos x}}\,dx

\displaystyle = \int \frac{1}{\sqrt{2\cos^2\frac{x}{2}}}\,dx \qquad [\because\ 1+\cos x = 2\cos^2\frac{x}{2}]

\displaystyle = \frac{1}{\sqrt{2}}\int \sec\frac{x}{2}\,dx

\displaystyle = \frac{1}{\sqrt{2}}\cdot 2\log\left|\tan\frac{x}{2}+\sec\frac{x}{2}\right|+C

\displaystyle = \sqrt{2}\log\left|\frac{1+\sin\frac{x}{2}}{\cos\frac{x}{2}}\right|+C

\displaystyle = \sqrt{2}\log\left|\frac{\left(\sin\frac{x}{4}+\cos\frac{x}{4}\right)^2}{\cos^2\frac{x}{4}-\sin^2\frac{x}{4}}\right|+C

\displaystyle = \sqrt{2}\log\left|\frac{\sin\frac{x}{4}+\cos\frac{x}{4}}{\cos\frac{x}{4}-\sin\frac{x}{4}}\right|+C

\displaystyle = \sqrt{2}\log\left|\frac{1+\tan\frac{x}{4}}{1-\tan\frac{x}{4}}\right|+C

\displaystyle = \sqrt{2}\log\left|\tan\left(\frac{\pi}{4}+\frac{x}{4}\right)\right|+C

\displaystyle \textbf{Question 3: }~\int \sqrt{\frac{1+\cos 2x}{1-\cos 2x}}\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \sqrt{\frac{1+\cos 2x}{1-\cos 2x}}\,dx

\displaystyle = \int \sqrt{\frac{2\cos^2 x}{2\sin^2 x}}\,dx \qquad [\because\ 1+\cos 2x = 2\cos^2 x,\; 1-\cos 2x = 2\sin^2 x]

\displaystyle = \int \frac{\cos x}{\sin x}\,dx

\displaystyle = \int \cot x\,dx

\displaystyle = \log|\sin x| + C

\displaystyle \textbf{Question 4: }~\int \sqrt{\frac{1-\cos x}{1+\cos x}}\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \sqrt{\frac{1-\cos x}{1+\cos x}}\,dx

\displaystyle = \int \sqrt{\frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}}}\,dx \qquad [\because\ 1-\cos x = 2\sin^2\frac{x}{2},\; 1+\cos x = 2\cos^2\frac{x}{2}]

\displaystyle = \int \tan\frac{x}{2}\,dx

\displaystyle = -2\log\left|\cos\frac{x}{2}\right|+C

\displaystyle \textbf{Question 5: }~\int \frac{\sec x}{\sec 2x}\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \frac{\sec x}{\sec 2x}\,dx

\displaystyle = \int \frac{\cos 2x}{\cos x}\,dx \qquad [\because\ \sec x/\sec 2x=\cos 2x/\cos x]

\displaystyle = \int \frac{2\cos^2 x-1}{\cos x}\,dx \qquad [\because\ \cos 2x = 2\cos^2 x-1]

\displaystyle = \int \left(2\cos x-\sec x\right)\,dx

\displaystyle = 2\int \cos x\,dx-\int \sec x\,dx

\displaystyle = 2\sin x-\log|\sec x+\tan x|+C

\displaystyle \textbf{Question 6: }~\int \frac{\cos 2x}{(\cos x+\sin x)^2}\,dx.\;[\text{NCERT}]
\displaystyle \text{Answer:}

Let \displaystyle I=\int \frac{\cos 2x}{(\cos x+\sin x)^2}\,dx

\displaystyle = \int \frac{\cos^2 x-\sin^2 x}{(\cos x+\sin x)^2}\,dx \qquad [\because\ \cos 2x=\cos^2 x-\sin^2 x]

\displaystyle = \int \frac{\cos x-\sin x}{\cos x+\sin x}\,dx

Let \displaystyle t=\cos x+\sin x

\displaystyle \Rightarrow \frac{dt}{dx}=-\sin x+\cos x

\displaystyle \Rightarrow (\cos x-\sin x)\,dx=dt

\displaystyle \therefore I=\int \frac{1}{t}\,dt

\displaystyle = \log|t|+C

\displaystyle = \log|\cos x+\sin x|+C

\displaystyle \textbf{Question 7: }~\int \frac{\sin(x-a)}{\sin(x-b)}\,dx.
\displaystyle \text{Answer:}

Let \displaystyle I=\int \frac{\sin(x-a)}{\sin(x-b)}\,dx

Put \displaystyle x-b=t

\displaystyle \Rightarrow x=b+t,\qquad dx=dt

\displaystyle \therefore I=\int \frac{\sin(b+t-a)}{\sin t}\,dt

\displaystyle =\int \frac{\sin\{(b-a)+t\}}{\sin t}\,dt

\displaystyle =\int \frac{\sin(b-a)\cos t+\cos(b-a)\sin t}{\sin t}\,dt

\displaystyle =\sin(b-a)\int \cot t\,dt+\cos(b-a)\int dt

\displaystyle =\sin(b-a)\log|\sin t|+t\cos(b-a)+C

\displaystyle =\sin(b-a)\log|\sin(x-b)|+(x-b)\cos(b-a)+C

\displaystyle \textbf{Question 8: }~\int \frac{\sin(x-c)}{\sin(x+a)}\,dx.\;[\text{CBSE 2006, 2013, 2015}]
\displaystyle \text{Answer:}

Let \displaystyle I=\int \frac{\sin(x-\alpha)}{\sin(x+\alpha)}\,dx

Put \displaystyle x+\alpha=t

\displaystyle \Rightarrow x=t-\alpha,\qquad dx=dt

\displaystyle \therefore I=\int \frac{\sin(t-2\alpha)}{\sin t}\,dt

\displaystyle =\int \frac{\sin t\cos 2\alpha-\cos t\sin 2\alpha}{\sin t}\,dt

\displaystyle =\cos 2\alpha\int dt-\sin 2\alpha\int \cot t\,dt

\displaystyle =t\cos 2\alpha-\sin 2\alpha\log|\sin t|+C

\displaystyle =(x+\alpha)\cos 2\alpha-\sin 2\alpha\log|\sin(x+\alpha)|+C

\displaystyle =x\cos 2\alpha-\sin 2\alpha\log|\sin(x+\alpha)|+C

\displaystyle \textbf{Question 9: }~\int \frac{1+\tan x}{1-\tan x}\,dx.
\displaystyle \text{Answer:}

Let \displaystyle I=\int \frac{1+\tan x}{1-\tan x}\,dx

\displaystyle =\int \frac{1+\frac{\sin x}{\cos x}}{1-\frac{\sin x}{\cos x}}\,dx

\displaystyle =\int \frac{\cos x+\sin x}{\cos x-\sin x}\,dx

Put \displaystyle t=\cos x-\sin x

\displaystyle \Rightarrow dt=(-\sin x-\cos x)\,dx

\displaystyle \Rightarrow (\sin x+\cos x)\,dx=-dt

\displaystyle \therefore I=\int \frac{(\sin x+\cos x)\,dx}{t}=-\int \frac{1}{t}\,dt

\displaystyle =-\log|t|+C

\displaystyle =-\log|\cos x-\sin x|+C

\displaystyle \textbf{Question 10: }~\int \frac{\cos x}{\cos(x-a)}\,dx.
\displaystyle \text{Answer:}

Let \displaystyle I=\int \frac{\cos x}{\cos(x-a)}\,dx

Put \displaystyle x-a=t

\displaystyle \Rightarrow x=a+t,\qquad dx=dt

\displaystyle \therefore I=\int \frac{\cos(a+t)}{\cos t}\,dt

\displaystyle =\int \frac{\cos a\cos t-\sin a\sin t}{\cos t}\,dt

\displaystyle =\int \left(\cos a-\sin a\tan t\right)\,dt

\displaystyle =t\cos a-\sin a\int \tan t\,dt

\displaystyle =t\cos a-\sin a\log|\sec t|+C

\displaystyle =(x-a)\cos a-\sin a\log|\sec(x-a)|+C

\displaystyle \textbf{Question 11: }~\int \sqrt{\frac{1-\sin 2x}{1+\sin 2x}}\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \sqrt{\frac{1-\sin 2x}{1+\sin 2x}}\,dx

\displaystyle =\int \sqrt{\frac{\cos^2x+\sin^2x-2\sin x\cos x}{\cos^2x+\sin^2x+2\sin x\cos x}}\,dx \qquad [\because\ 1=\sin^2x+\cos^2x,\; \sin 2x=2\sin x\cos x]

\displaystyle =\int \sqrt{\frac{(\cos x-\sin x)^2}{(\cos x+\sin x)^2}}\,dx

\displaystyle =\int \frac{\cos x-\sin x}{\cos x+\sin x}\,dx

\displaystyle =\int \frac{1-\tan x}{1+\tan x}\,dx

\displaystyle =\int \tan\left(\frac{\pi}{4}-x\right)\,dx

\displaystyle =-\log\left|\sec\left(\frac{\pi}{4}-x\right)\right|+C

\displaystyle =\log\left|\cos\left(\frac{\pi}{4}-x\right)\right|+C

\displaystyle \textbf{Question 12: }~\int \frac{e^{3x}}{e^{3x}+1}\,dx.
\displaystyle \text{Answer:}

Let \displaystyle I=\int \frac{e^{3x}}{e^{3x}+1}\,dx

Put \displaystyle t=e^{3x}+1

\displaystyle \Rightarrow \frac{dt}{dx}=3e^{3x}

\displaystyle \Rightarrow dx=\frac{dt}{3e^{3x}}

\displaystyle \therefore I=\int \frac{e^{3x}}{t}\cdot\frac{dt}{3e^{3x}}

\displaystyle =\frac{1}{3}\int \frac{1}{t}\,dt

\displaystyle =\frac{1}{3}\log|t|+C

\displaystyle =\frac{1}{3}\log\!\left(e^{3x}+1\right)+C

\displaystyle \textbf{Question 13: }~\int \frac{\sec x\,\tan x}{3\sec x+5}\,dx.
\displaystyle \text{Answer:}

Let \displaystyle I=\int \frac{\sec x\,\tan x}{3\sec x+5}\,dx

Put \displaystyle t=\sec x

\displaystyle \Rightarrow \frac{dt}{dx}=\sec x\,\tan x

\displaystyle \Rightarrow dt=\sec x\,\tan x\,dx

\displaystyle \therefore I=\int \frac{dt}{3t+5}

\displaystyle =\frac{1}{3}\log|3t+5|+C

\displaystyle =\frac{1}{3}\log|3\sec x+5|+C

\displaystyle \textbf{Question 14: }~\int \frac{1-\cot x}{1+\cot x}\,dx.
\displaystyle \text{Answer:}

Let \displaystyle I=\int \frac{1-\cot x}{1+\cot x}\,dx

\displaystyle =\int \frac{1-\frac{\cos x}{\sin x}}{1+\frac{\cos x}{\sin x}}\,dx

\displaystyle =\int \frac{\sin x-\cos x}{\sin x+\cos x}\,dx

Put \displaystyle t=\sin x+\cos x

\displaystyle \Rightarrow dt=(\cos x-\sin x)\,dx

\displaystyle \Rightarrow (\sin x-\cos x)\,dx=-dt

\displaystyle \therefore I=-\int \frac{1}{t}\,dt

\displaystyle =-\log|t|+C

\displaystyle =-\log|\sin x+\cos x|+C

\displaystyle \textbf{Question 15: }~\int \frac{\sec x\,\mathrm{cosec}\,x}{\log(\tan x)}\,dx.
\displaystyle \text{Answer:}

\displaystyle \text{Note: Here, we are considering }\log x\text{ as }\log_e x.

Let \displaystyle I=\int \frac{\sec x\,\mathrm{cosec}\,x}{\log(\tan x)}\,dx

Put \displaystyle t=\log(\tan x)

\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{\tan x}\cdot \sec^2 x

\displaystyle \Rightarrow \frac{dt}{dx}=\sec^2 x\cdot \cot x

\displaystyle \Rightarrow \frac{dt}{dx}=\sec^2 x\cdot \frac{\cos x}{\sin x}

\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{\cos^2 x}\cdot \frac{\cos x}{\sin x}

\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{\sin x\cos x}=\sec x\,\mathrm{cosec}\,x

\displaystyle \Rightarrow dt=\sec x\,\mathrm{cosec}\,x\;dx

\displaystyle \therefore I=\int \frac{1}{t}\,dt

\displaystyle =\log|t|+C

\displaystyle =\log\!\big|\log(\tan x)\big|+C

\displaystyle \textbf{Question 16: }~\int \frac{1}{x(3+\log x)}\,dx.
\displaystyle \text{Answer:}

\displaystyle \text{Note: Here, we are considering }\log x\text{ as }\log_e x.

Let \displaystyle I=\int \frac{1}{x(3+\log x)}\,dx

Put \displaystyle t=\log x

\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{x}

\displaystyle \Rightarrow \frac{dx}{x}=dt

\displaystyle \therefore I=\int \frac{dt}{3+t}

\displaystyle =\log|3+t|+C

\displaystyle =\log|3+\log x|+C

\displaystyle \textbf{Question 17: }~\int \frac{e^x+1}{e^x+x}\,dx.
\displaystyle \text{Answer:}

Let \displaystyle I=\int \frac{e^x+1}{e^x+x}\,dx

Put \displaystyle t=e^x+x

\displaystyle \Rightarrow \frac{dt}{dx}=e^x+1

\displaystyle \Rightarrow (e^x+1)\,dx=dt

\displaystyle \therefore I=\int \frac{1}{t}\,dt

\displaystyle =\log|t|+C

\displaystyle =\log|e^x+x|+C

\displaystyle \textbf{Question 18: }~\int \frac{1}{x\log x}\,dx.
\displaystyle \text{Answer:}

\displaystyle \text{Note: Here, we are considering }\log x\text{ as }\log_e x.

Let \displaystyle I=\int \frac{1}{x\log x}\,dx

Put \displaystyle t=\log x

\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{x}

\displaystyle \Rightarrow \frac{dx}{x}=dt

\displaystyle \therefore I=\int \frac{1}{t}\,dt

\displaystyle =\log|t|+C

\displaystyle =\log|\log x|+C

\displaystyle \textbf{Question 19: }~\int \frac{\sin 2x}{a\cos^2 x+b\sin^2 x}\,dx.
\displaystyle \text{Answer:}

Let \displaystyle I=\int \frac{\sin 2x}{a\cos^2 x+b\sin^2 x}\,dx

\displaystyle =\int \frac{\sin 2x}{a(1-\sin^2 x)+b\sin^2 x}\,dx

\displaystyle =\int \frac{\sin 2x}{(b-a)\sin^2 x+a}\,dx

Put \displaystyle t=\sin^2 x

\displaystyle \Rightarrow \frac{dt}{dx}=2\sin x\cos x=\sin 2x

\displaystyle \Rightarrow \sin 2x\,dx=dt

\displaystyle \therefore I=\int \frac{1}{(b-a)t+a}\,dt

\displaystyle =\frac{1}{b-a}\log|(b-a)t+a|+C

\displaystyle =\frac{1}{b-a}\log|(b-a)\sin^2 x+a|+C

\displaystyle =\frac{1}{b-a}\log|b\sin^2 x+a(1-\sin^2 x)|+C

\displaystyle =\frac{1}{b-a}\log|b\sin^2 x+a\cos^2 x|+C

\displaystyle \textbf{Question 20: }~\int \frac{\cos x}{2+3\sin x}\,dx.
\displaystyle \text{Answer:}

Let \displaystyle I=\int \frac{\cos x}{2+3\sin x}\,dx

Put \displaystyle t=\sin x

\displaystyle \Rightarrow \frac{dt}{dx}=\cos x

\displaystyle \Rightarrow \cos x\,dx=dt

\displaystyle \therefore I=\int \frac{dt}{2+3t}

\displaystyle =\frac{1}{3}\log|2+3t|+C

\displaystyle =\frac{1}{3}\log|2+3\sin x|+C

\displaystyle \textbf{Question 21: }~\int \frac{1-\sin x}{x+\cos x}\,dx.
\displaystyle \text{Answer:}

Let \displaystyle I=\int \frac{1-\sin x}{x+\cos x}\,dx

Put \displaystyle t=x+\cos x

\displaystyle \Rightarrow \frac{dt}{dx}=1-\sin x

\displaystyle \Rightarrow (1-\sin x)\,dx=dt

\displaystyle \therefore I=\int \frac{1}{t}\,dt

\displaystyle =\log|t|+C

\displaystyle =\log|x+\cos x|+C

\displaystyle \textbf{Question 22: }~\int \frac{a}{b+c e^x}\,dx.
\displaystyle \text{Answer:}

Let \displaystyle I=\int \frac{a}{\,b+ce^{x}}\,dx

Divide numerator and denominator by \displaystyle e^{x}

\displaystyle \Rightarrow I=\int \frac{ae^{-x}}{be^{-x}+c}\,dx

Put \displaystyle t=e^{-x}

\displaystyle \Rightarrow \frac{dt}{dx}=-e^{-x}

\displaystyle \Rightarrow e^{-x}dx=-dt

\displaystyle \therefore I=\int \frac{-a}{bt+c}\,dt

\displaystyle =-\frac{a}{b}\log|bt+c|+C

\displaystyle =-\frac{a}{b}\log|be^{-x}+c|+C

\displaystyle \textbf{Question 23: }~\int \frac{1}{e^x+1}\,dx.\;[\text{CBSE 2003}]
\displaystyle \text{Answer:}

Let \displaystyle I=\int \frac{1}{e^{x}+1}\,dx

\displaystyle =\int \frac{e^{-x}}{1+e^{-x}}\,dx

Put \displaystyle t=e^{-x}

\displaystyle \Rightarrow \frac{dt}{dx}=-e^{-x}

\displaystyle \Rightarrow e^{-x}\,dx=-dt

\displaystyle \therefore I=\int \frac{-1}{1+t}\,dt

\displaystyle =-\log|1+t|+C

\displaystyle =-\log|1+e^{-x}|+C

\displaystyle \textbf{Question 24: }~\int \frac{\cot x}{\log(\sin x)}\,dx.
\displaystyle \text{Answer:}

\displaystyle \text{Note: Here, we are considering }\log x\text{ as }\log_e x.

Let \displaystyle I=\int \frac{\cot x}{\log(\sin x)}\,dx

Put \displaystyle t=\log(\sin x)

\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{\sin x}\cdot \cos x=\cot x

\displaystyle \Rightarrow \cot x\,dx=dt

\displaystyle \therefore I=\int \frac{1}{t}\,dt

\displaystyle =\log|t|+C

\displaystyle =\log|\log(\sin x)|+C

\displaystyle \textbf{Question 25: }~\int \frac{e^{2x}}{e^{2x}-2}\,dx.
\displaystyle \text{Answer:}

Let \displaystyle I=\int \frac{e^{2x}}{e^{2x}-2}\,dx

Put \displaystyle t=e^{2x}

\displaystyle \Rightarrow \frac{dt}{dx}=2e^{2x}

\displaystyle \Rightarrow e^{2x}\,dx=\frac{dt}{2}

\displaystyle \therefore I=\frac{1}{2}\int \frac{1}{t-2}\,dt

\displaystyle =\frac{1}{2}\log|t-2|+C

\displaystyle =\frac{1}{2}\log|e^{2x}-2|+C

\displaystyle \textbf{Question 26: }~\int \frac{2\cos x-3\sin x}{6\cos x+4\sin x}\,dx.\;[\text{NCERT}]
\displaystyle \text{Answer:}

\displaystyle \int \frac{2\cos x-3\sin x}{6\cos x+4\sin x}\,dx

\displaystyle =\int \frac{2\cos x-3\sin x}{2(3\cos x+2\sin x)}\,dx

Let \displaystyle t=3\cos x+2\sin x

\displaystyle \Rightarrow \frac{dt}{dx}=-3\sin x+2\cos x=2\cos x-3\sin x

\displaystyle \Rightarrow (2\cos x-3\sin x)\,dx=dt

\displaystyle \therefore \int \frac{2\cos x-3\sin x}{2(3\cos x+2\sin x)}\,dx =\frac{1}{2}\int \frac{1}{t}\,dt

\displaystyle =\frac{1}{2}\log|t|+C

\displaystyle =\frac{1}{2}\log|3\cos x+2\sin x|+C

\displaystyle \textbf{Question 27: }~\int \frac{\cos 2x+x+1}{x^2+\sin 2x+2x}\,dx.
\displaystyle \text{Answer:}

Let \displaystyle I=\int \frac{\cos 2x+x+1}{x^2+\sin 2x+2x}\,dx

Put \displaystyle t=x^2+\sin 2x+2x

\displaystyle \Rightarrow \frac{dt}{dx}=2x+2\cos 2x+2

\displaystyle \Rightarrow (x+\cos 2x+1)\,dx=\frac{dt}{2}

\displaystyle \therefore I=\frac{1}{2}\int \frac{1}{t}\,dt

\displaystyle =\frac{1}{2}\log|t|+C

\displaystyle =\frac{1}{2}\log|x^2+\sin 2x+2x|+C

\displaystyle \textbf{Question 28: }~\int \frac{1}{\cos(x+a)\cos(x+b)}\,dx.\;[\text{NCERT}]
\displaystyle \text{Answer:}

\displaystyle \int \frac{1}{\cos(x+a)\cos(x+b)}\,dx

Multiply and divide by
\displaystyle \sin[(x+b)-(x+a)]

\displaystyle =\int \frac{1}{\sin[(x+b)-(x+a)]}\, \frac{\sin[(x+b)-(x+a)]}{\cos(x+a)\cos(x+b)}\,dx

\displaystyle =\frac{1}{\sin(b-a)} \int \frac{\sin(x+b)\cos(x+a)-\sin(x+a)\cos(x+b)} {\cos(x+a)\cos(x+b)}\,dx

\displaystyle =\frac{1}{\sin(b-a)} \left[\int \frac{\sin(x+b)}{\cos(x+b)}\,dx -\int \frac{\sin(x+a)}{\cos(x+a)}\,dx\right]

\displaystyle =\frac{1}{\sin(b-a)} \left[\int \tan(x+b)\,dx-\int \tan(x+a)\,dx\right]

\displaystyle =\frac{1}{\sin(b-a)} \left[\log\sec(x+b)-\log\sec(x+a)\right]+C

\displaystyle =\frac{1}{\sin(b-a)} \log\!\left(\frac{\sec(x+b)}{\sec(x+a)}\right)+C

\displaystyle \textbf{Question 29: }~\int \frac{-\sin x+2\cos x}{2\sin x+\cos x}\,dx.
\displaystyle \text{Answer:}

Let \displaystyle I=\int \frac{-\sin x+2\cos x}{2\sin x+\cos x}\,dx

Put \displaystyle t=2\sin x+\cos x

\displaystyle \Rightarrow \frac{dt}{dx}=2\cos x-\sin x

\displaystyle \Rightarrow (-\sin x+2\cos x)\,dx=dt

\displaystyle \therefore I=\int \frac{1}{t}\,dt

\displaystyle =\log|t|+C

\displaystyle =\log|2\sin x+\cos x|+C

\displaystyle \textbf{Question 30: }~\int \frac{\cos 4x-\cos 2x}{\sin 4x-\sin 2x}\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \frac{\cos 4x-\cos 2x}{\sin 4x-\sin 2x}\,dx

\displaystyle =\int \frac{-2\sin\!\left(\frac{4x+2x}{2}\right)\sin\!\left(\frac{4x-2x}{2}\right)} {2\cos\!\left(\frac{4x+2x}{2}\right)\sin\!\left(\frac{4x-2x}{2}\right)}\,dx \qquad \left[ \begin{aligned} \cos A-\cos B&=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}\\ \sin A-\sin B&=2\cos\frac{A+B}{2}\sin\frac{A-B}{2} \end{aligned} \right]

\displaystyle =-\int \frac{\sin 3x}{\cos 3x}\,dx

\displaystyle =-\int \tan 3x\,dx

\displaystyle =-\frac{1}{3}\log|\sec 3x|+C

\displaystyle =\frac{1}{3}\log|\cos 3x|+C

\displaystyle \textbf{Question 31: }~\int \frac{\sec x}{\log(\sec x+\tan x)}\,dx.
\displaystyle \text{Answer:}

\displaystyle \text{Note: Here, we are considering }\log x\text{ as }\log_e x.

Let \displaystyle I=\int \frac{\sec x}{\log(\sec x+\tan x)}\,dx

Put \displaystyle t=\log(\sec x+\tan x)

\displaystyle \Rightarrow \frac{dt}{dx} =\frac{\sec x\tan x+\sec^2 x}{\sec x+\tan x}

\displaystyle =\frac{\sec x(\tan x+\sec x)}{\sec x+\tan x}

\displaystyle =\sec x

\displaystyle \Rightarrow \sec x\,dx=dt

\displaystyle \therefore I=\int \frac{1}{t}\,dt

\displaystyle =\log|t|+C

\displaystyle =\log\!\big|\log(\sec x+\tan x)\big|+C

\displaystyle \textbf{Question 32: }~\int \frac{\mathrm{cosec}\,x}{\log\!\left(\tan\frac{x}{2}\right)}\,dx.
\displaystyle \text{Answer:}

\displaystyle \text{Note: Here, we are considering }\log x\text{ as }\log_e x.

Let \displaystyle I=\int \frac{\mathrm{cosec}\,x}{\log\!\left(\tan\frac{x}{2}\right)}\,dx

Put \displaystyle t=\log\!\left(\tan\frac{x}{2}\right)

\displaystyle \Rightarrow \frac{dt}{dx} =\frac{1}{\tan\frac{x}{2}}\cdot\frac{1}{2}\sec^2\frac{x}{2}

\displaystyle =\frac{1}{2\sin\frac{x}{2}\cos\frac{x}{2}}

\displaystyle =\frac{1}{\sin x}

\displaystyle \Rightarrow \frac{dt}{dx}=\mathrm{cosec}\,x

\displaystyle \Rightarrow \mathrm{cosec}\,x\,dx=dt

\displaystyle \therefore I=\int \frac{1}{t}\,dt

\displaystyle =\log|t|+C

\displaystyle =\log\!\left|\log\!\left(\tan\frac{x}{2}\right)\right|+C

\displaystyle \textbf{Question 33: }~\int \frac{1}{x\log x\;\log(\log x)}\,dx.
\displaystyle \text{Answer:}

\displaystyle \text{Note: Here, we are considering }\log x\text{ as }\log_e x.

Let \displaystyle I=\int \frac{1}{x\log x\;\log(\log x)}\,dx

Put \displaystyle t=\log(\log x)

\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{x\log x}

\displaystyle \Rightarrow \frac{dx}{x\log x}=dt

\displaystyle \therefore I=\int \frac{1}{t}\,dt

\displaystyle =\log|t|+C

\displaystyle =\log\!\big|\log(\log x)\big|+C

\displaystyle \textbf{Question 34: }~\int \frac{\mathrm{cosec}^2 x}{1+\cot x}\,dx.
\displaystyle \text{Answer:}

Let \displaystyle I=\int \frac{\mathrm{cosec}^2 x}{1+\cot x}\,dx

Put \displaystyle t=\cot x

\displaystyle \Rightarrow \frac{dt}{dx}=-\mathrm{cosec}^2 x

\displaystyle \Rightarrow \mathrm{cosec}^2 x\,dx=-dt

\displaystyle \therefore I=\int \frac{-dt}{1+t}

\displaystyle =-\log|1+t|+C

\displaystyle =-\log|1+\cot x|+C

\displaystyle \textbf{Question 35: }~\int \frac{10x^9+10^x\log_e 10}{10^x+x^{10}}\,dx.
\displaystyle \text{Answer:}

Let \displaystyle I=\int \frac{10x^9+10^x\log_e 10}{10^x+x^{10}}\,dx

Put \displaystyle t=10^x+x^{10}

\displaystyle \Rightarrow \frac{dt}{dx}=10^x\log_e 10+10x^9

\displaystyle \Rightarrow (10^x\log_e 10+10x^9)\,dx=dt

\displaystyle \therefore I=\int \frac{1}{t}\,dt

\displaystyle =\log|t|+C

\displaystyle =\log|10^x+x^{10}|+C

\displaystyle \textbf{Question 36: }~\int \frac{1-\sin 2x}{x+\cos^2 x}\,dx.\;[\text{CBSE 2000}]
\displaystyle \text{Answer:}

Let \displaystyle I=\int \frac{1-\sin 2x}{x+\cos^2 x}\,dx

Put \displaystyle t=x+\cos^2 x

\displaystyle \Rightarrow \frac{dt}{dx}=1-2\cos x\sin x

\displaystyle \Rightarrow (1-\sin 2x)\,dx=dt

\displaystyle \therefore I=\int \frac{1}{t}\,dt

\displaystyle =\log|t|+C

\displaystyle =\log|x+\cos^2 x|+C

\displaystyle \textbf{Question 37: }~\int \frac{1+\tan x}{x+\log(\sec x)}\,dx.\;[\text{CBSE 2000}]
\displaystyle \text{Answer:}

\displaystyle \text{Note: Here, we are considering }\log x\text{ as }\log_e x.

Let \displaystyle I=\int \frac{1+\tan x}{x+\log(\sec x)}\,dx

Put \displaystyle t=x+\log(\sec x)

\displaystyle \Rightarrow \frac{dt}{dx} =1+\frac{\sec x\tan x}{\sec x} =1+\tan x

\displaystyle \Rightarrow (1+\tan x)\,dx=dt

\displaystyle \therefore I=\int \frac{1}{t}\,dt

\displaystyle =\log|t|+C

\displaystyle =\log|x+\log(\sec x)|+C

\displaystyle \textbf{Question 38: }~\int \frac{\sin 2x}{a^2+b^2\sin^2 x}\,dx.
\displaystyle \text{Answer:}

Let \displaystyle I=\int \frac{\sin 2x}{a^2+b^2\sin^2 x}\,dx

Put \displaystyle t=\sin^2 x

\displaystyle \Rightarrow \frac{dt}{dx}=2\sin x\cos x=\sin 2x

\displaystyle \Rightarrow \sin 2x\,dx=dt

\displaystyle \therefore I=\int \frac{1}{a^2+b^2 t}\,dt

\displaystyle =\frac{1}{b^2}\log|a^2+b^2 t|+C

\displaystyle =\frac{1}{b^2}\log|a^2+b^2\sin^2 x|+C

\displaystyle \textbf{Question 39: }~\int \frac{x+1}{x(x+\log x)}\,dx.
\displaystyle \text{Answer:}

\displaystyle \text{Note: Here, we are considering }\log x\text{ as }\log_e x.

Let \displaystyle I=\int \frac{x+1}{x(x+\log x)}\,dx

Put \displaystyle t=x+\log x

\displaystyle \Rightarrow \frac{dt}{dx}=1+\frac{1}{x}

\displaystyle \Rightarrow \frac{x+1}{x}\,dx=dt

\displaystyle \therefore I=\int \frac{1}{t}\,dt

\displaystyle =\log|t|+C

\displaystyle =\log|x+\log x|+C

\displaystyle \textbf{Question 40: }~\int \frac{1}{\sqrt{1-x^2}\,\left(2+3\sin^{-1}x\right)}\,dx.
\displaystyle \text{Answer:}

Let \displaystyle I=\int \frac{1}{\sqrt{1-x^2}\,(2+3\sin^{-1}x)}\,dx

Put \displaystyle t=\sin^{-1}x

\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{\sqrt{1-x^2}}

\displaystyle \Rightarrow \frac{1}{\sqrt{1-x^2}}\,dx=dt

\displaystyle \therefore I=\int \frac{1}{2+3t}\,dt

\displaystyle =\frac{1}{3}\log|2+3t|+C

\displaystyle =\frac{1}{3}\log|2+3\sin^{-1}x|+C

\displaystyle \textbf{Question 41: }~\int \frac{\sec^2 x}{\tan x+2}\,dx.
\displaystyle \text{Answer:}

Let \displaystyle I=\int \frac{\sec^2 x}{\tan x+2}\,dx

Put \displaystyle t=\tan x

\displaystyle \Rightarrow \frac{dt}{dx}=\sec^2 x

\displaystyle \Rightarrow \sec^2 x\,dx=dt

\displaystyle \therefore I=\int \frac{1}{t+2}\,dt

\displaystyle =\log|t+2|+C

\displaystyle =\log|\tan x+2|+C

\displaystyle \textbf{Question 42: }~\int \frac{2\cos 2x+\sec^2 x}{\sin 2x+\tan x-5}\,dx.
\displaystyle \text{Answer:}

Let \displaystyle I=\int \frac{2\cos 2x+\sec^2 x}{\sin 2x+\tan x-5}\,dx

Put \displaystyle t=\sin 2x+\tan x-5

\displaystyle \Rightarrow \frac{dt}{dx}=2\cos 2x+\sec^2 x

\displaystyle \Rightarrow (2\cos 2x+\sec^2 x)\,dx=dt

\displaystyle \therefore I=\int \frac{1}{t}\,dt

\displaystyle =\log|t|+C

\displaystyle =\log|\sin 2x+\tan x-5|+C

\displaystyle \textbf{Question 43: }~\int \frac{\sin 2x}{\sin 5x\;\sin 3x}\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \frac{\sin 2x}{\sin 5x\,\sin 3x}\,dx

\displaystyle =\int \frac{\sin(5x-3x)}{\sin 5x\,\sin 3x}\,dx

\displaystyle =\int \frac{\sin 5x\cos 3x-\cos 5x\sin 3x}{\sin 5x\,\sin 3x}\,dx

\displaystyle =\int \left(\frac{\sin 5x\cos 3x}{\sin 5x\,\sin 3x} -\frac{\cos 5x\sin 3x}{\sin 5x\,\sin 3x}\right)\,dx

\displaystyle =\int (\cot 3x-\cot 5x)\,dx

\displaystyle =\int \cot 3x\,dx-\int \cot 5x\,dx

\displaystyle =\frac{1}{3}\log|\sin 3x|-\frac{1}{5}\log|\sin 5x|+C

\displaystyle \textbf{Question 44: }~\int \frac{1+\cot x}{x+\log(\sin x)}\,dx.\;[\text{CBSE 2000}]
\displaystyle \text{Answer:}

\displaystyle \text{Note: Here, we are considering }\log x\text{ as }\log_e x.

Let \displaystyle I=\int \frac{1+\cot x}{x+\log(\sin x)}\,dx

Put \displaystyle t=x+\log(\sin x)

\displaystyle \Rightarrow \frac{dt}{dx}=1+\frac{\cos x}{\sin x}=1+\cot x

\displaystyle \Rightarrow (1+\cot x)\,dx=dt

\displaystyle \therefore I=\int \frac{1}{t}\,dt

\displaystyle =\log|t|+C

\displaystyle =\log|x+\log(\sin x)|+C

\displaystyle \textbf{Question 45: }~\int \frac{1}{\sqrt{x}\,(\sqrt{x}+1)}\,dx.
\displaystyle \text{Answer:}

Let \displaystyle I=\int \frac{1}{\sqrt{x}\,(\sqrt{x}+1)}\,dx

Put \displaystyle t=\sqrt{x}+1

\displaystyle \Rightarrow \frac{dt}{dx}=\frac{1}{2\sqrt{x}}

\displaystyle \Rightarrow \frac{1}{\sqrt{x}}\,dx=2\,dt

\displaystyle \therefore I=2\int \frac{1}{t}\,dt

\displaystyle =2\log|t|+C

\displaystyle =2\log|\sqrt{x}+1|+C

\displaystyle \textbf{Question 46: }~\int \tan 2x\,\tan 3x\,\tan 5x\,dx.
\displaystyle \text{Answer:}

We know that,
\displaystyle \tan 5x=\tan(2x+3x)

\displaystyle \Rightarrow \tan 5x=\frac{\tan 2x+\tan 3x}{1-\tan 2x\,\tan 3x}

\displaystyle \Rightarrow \tan 5x-\tan 2x\,\tan 3x\,\tan 5x=\tan 2x+\tan 3x

\displaystyle \Rightarrow \tan 2x\,\tan 3x\,\tan 5x=\tan 5x-\tan 2x-\tan 3x

\displaystyle \therefore \int \tan 2x\,\tan 3x\,\tan 5x\,dx =\int (\tan 5x-\tan 2x-\tan 3x)\,dx

\displaystyle =\int \tan 5x\,dx-\int \tan 2x\,dx-\int \tan 3x\,dx

\displaystyle =\frac{1}{5}\log|\sec 5x| -\frac{1}{2}\log|\sec 2x| -\frac{1}{3}\log|\sec 3x| +C

\displaystyle \textbf{Question 47: }~\int \left(1+\tan x\,\tan(x+\theta)\right)\,dx.
\displaystyle \text{Answer:}

Let \displaystyle I=\int \left(1+\tan x\,\tan(x+\theta)\right)\,dx

\displaystyle =\int \left(1+\tan x\cdot \frac{\tan x+\tan\theta}{1-\tan x\,\tan\theta}\right)\,dx

\displaystyle =\int \frac{1+\tan^2 x}{1-\tan x\,\tan\theta}\,dx

\displaystyle =\int \frac{\sec^2 x}{1-\tan x\,\tan\theta}\,dx

Put \displaystyle t=\tan x

\displaystyle \Rightarrow \frac{dt}{dx}=\sec^2 x

\displaystyle \Rightarrow dx=\frac{dt}{\sec^2 x}

\displaystyle \therefore I=\int \frac{1}{1-t\tan\theta}\,dt

\displaystyle =-\frac{1}{\tan\theta}\log|1-t\tan\theta|+C

\displaystyle =-\cot\theta\log|1-\tan x\,\tan\theta|+C

\displaystyle =\cot\theta\log\left|\frac{1}{1-\tan x\,\tan\theta}\right|+C

\displaystyle =\cot\theta\log\left| \frac{\cos x\cos\theta}{\cos x\cos\theta-\sin x\sin\theta} \right|+C

\displaystyle =\cot\theta\log\left| \frac{\cos x}{\cos(x+\theta)} \right|+C'

\displaystyle \textbf{Question 48: }~\int \frac{\sin 2x}{\sin\!\left(x-\frac{\pi}{6}\right)\sin\!\left(x+\frac{\pi}{6}\right)}\,dx.
\displaystyle \text{Answer:}

Let \displaystyle I=\int \frac{\sin 2x}{\sin\!\left(x-\frac{\pi}{6}\right)\sin\!\left(x+\frac{\pi}{6}\right)}\,dx

\displaystyle =\int \frac{\sin 2x}{\sin^2 x-\sin^2\frac{\pi}{6}}\,dx \qquad [\because\ \sin(A+B)\sin(A-B)=\sin^2 A-\sin^2 B]

\displaystyle =\int \frac{\sin 2x}{\sin^2 x-\frac{1}{4}}\,dx

Put \displaystyle t=\sin^2 x-\frac{1}{4}

\displaystyle \Rightarrow \frac{dt}{dx}=2\sin x\cos x=\sin 2x

\displaystyle \Rightarrow \sin 2x\,dx=dt

\displaystyle \therefore I=\int \frac{1}{t}\,dt

\displaystyle =\log|t|+C

\displaystyle =\log\left|\sin^2 x-\frac{1}{4}\right|+C

\displaystyle \textbf{Question 49: }~\int \frac{e^{x-1}+x^{e-1}}{e^x+x^e}\,dx.
\displaystyle \text{Answer:}

Let \displaystyle I=\int \frac{e^{x-1}+x^{e-1}}{e^x+x^e}\,dx

Put \displaystyle t=e^x+x^e

\displaystyle \Rightarrow \frac{dt}{dx}=e^x+e\,x^{e-1} =e\left(e^{x-1}+x^{e-1}\right)

\displaystyle \Rightarrow \left(e^{x-1}+x^{e-1}\right)dx=\frac{dt}{e}

\displaystyle \therefore I=\frac{1}{e}\int \frac{1}{t}\,dt

\displaystyle =\frac{1}{e}\log|t|+C

\displaystyle =\frac{1}{e}\log|e^x+x^e|+C

\displaystyle \textbf{Question 50: }~\int \frac{1}{\sin x\,\cos^2 x}\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \frac{1}{\sin x\,\cos^2 x}\,dx

\displaystyle =\int \frac{\sin^2 x+\cos^2 x}{\sin x\,\cos^2 x}\,dx

\displaystyle =\int \left(\frac{\sin x}{\cos^2 x} +\frac{1}{\sin x}\right)\,dx

\displaystyle =\int (\tan x\,\sec x+\mathrm{cosec}\,x)\,dx

\displaystyle =\sec x+\log|\mathrm{cosec}\,x-\cot x|+C

\displaystyle =\sec x+\log\left|\tan\frac{x}{2}\right|+C

\displaystyle \textbf{Question 51: }~\int \frac{1}{\cos 3x-\cos x}\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \frac{1}{\cos 3x-\cos x}\,dx

\displaystyle =\int \frac{1}{4\cos^3 x-4\cos x}\,dx \qquad [\because\ \cos 3x=4\cos^3 x-3\cos x]

\displaystyle =\int \frac{1}{4\cos x(\cos^2 x-1)}\,dx

\displaystyle =-\frac{1}{4}\int \frac{1}{\cos x\,\sin^2 x}\,dx

\displaystyle =-\frac{1}{4}\int \frac{\sin^2 x+\cos^2 x}{\cos x\,\sin^2 x}\,dx

\displaystyle =-\frac{1}{4}\int \left(\frac{\cos x}{\sin^2 x}+\frac{1}{\cos x}\right)\,dx

\displaystyle =-\frac{1}{4}\left[\int \cot x\,\mathrm{cosec}\,x\,dx+\int \sec x\,dx\right]

\displaystyle =-\frac{1}{4}\left[-\mathrm{cosec}\,x+\log|\sec x+\tan x|\right]+C

\displaystyle =\frac{1}{4}\left(\mathrm{cosec}\,x-\log|\sec x+\tan x|\right)+C

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