\displaystyle \textbf{Evaluate the following integrals (1-35):}

\displaystyle \textbf{Question 1: }~\int_{0}^{\pi/2}\frac{1}{1+\tan x}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\tan x}\,dx
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\tan\left(\frac{\pi}{2}-x\right)}\,dx
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\cot x}\,dx
\displaystyle \text{Adding the two expressions for }I
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\left(\frac{1}{1+\tan x}+\frac{1}{1+\cot x}\right)dx
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\frac{(1+\cot x)+(1+\tan x)}{(1+\tan x)(1+\cot x)}\,dx
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\frac{2+\tan x+\cot x}{1+\tan x+\cot x+\tan x\cot x}\,dx
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\frac{2+\tan x+\cot x}{2+\tan x+\cot x}\,dx
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}dx
\displaystyle 2I=\left[x\right]_{0}^{\frac{\pi}{2}}
\displaystyle 2I=\frac{\pi}{2}
\displaystyle \therefore I=\frac{\pi}{4}

\displaystyle \textbf{Question 2: }~\int_{0}^{\pi/2}\frac{1}{1+\cot x}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\cot x}\,dx
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\cot\left(\frac{\pi}{2}-x\right)}\,dx
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\tan x}\,dx
\displaystyle \text{Adding (1) and (2)}
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\left(\frac{1}{1+\cot x}+\frac{1}{1+\tan x}\right)dx
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\frac{(1+\tan x)+(1+\cot x)}{(1+\cot x)(1+\tan x)}\,dx
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\frac{2+\tan x+\cot x}{1+\tan x+\cot x+\tan x\cot x}\,dx
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\frac{2+\tan x+\cot x}{2+\tan x+\cot x}\,dx
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}dx
\displaystyle 2I=\left[x\right]_{0}^{\frac{\pi}{2}}
\displaystyle 2I=\frac{\pi}{2}
\displaystyle \therefore I=\frac{\pi}{4}

\displaystyle \textbf{Question 3: }~\int_{0}^{\pi/2}\frac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}}\,dx
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\cot\left(\frac{\pi}{2}-x\right)}}{\sqrt{\cot\left(\frac{\pi}{2}-x\right)}+\sqrt{\tan\left(\frac{\pi}{2}-x\right)}}\,dx
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\tan x}}{\sqrt{\tan x}+\sqrt{\cot x}}\,dx
\displaystyle \text{Adding (1) and (2)}
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\left(\frac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}}+\frac{\sqrt{\tan x}}{\sqrt{\tan x}+\sqrt{\cot x}}\right)dx
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}dx
\displaystyle 2I=\left[x\right]_{0}^{\frac{\pi}{2}}
\displaystyle 2I=\frac{\pi}{2}
\displaystyle \therefore I=\frac{\pi}{4}

\displaystyle \textbf{Question 4: }~\int_{0}^{\pi/2}\frac{\sin^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\sin^{n}x}{\sin^{n}x+\cos^{n}x}\,dx
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\sin^{n}\left(\frac{\pi}{2}-x\right)}{\sin^{n}\left(\frac{\pi}{2}-x\right)+\cos^{n}\left(\frac{\pi}{2}-x\right)}\,dx
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\cos^{n}x}{\cos^{n}x+\sin^{n}x}\,dx
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\cos^{n}x}{\sin^{n}x+\cos^{n}x}\,dx
\displaystyle \text{Adding (1) and (2) we get}
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\left(\frac{\sin^{n}x}{\sin^{n}x+\cos^{n}x}+\frac{\cos^{n}x}{\sin^{n}x+\cos^{n}x}\right)dx
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\frac{\sin^{n}x+\cos^{n}x}{\sin^{n}x+\cos^{n}x}\,dx
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}dx
\displaystyle 2I=\left[x\right]_{0}^{\frac{\pi}{2}}
\displaystyle 2I=\frac{\pi}{2}
\displaystyle \therefore I=\frac{\pi}{4}
\displaystyle \text{i.e., }\int_{0}^{\frac{\pi}{2}}\frac{\sin^{n}x}{\sin^{n}x+\cos^{n}x}\,dx=\frac{\pi}{4}
\displaystyle \therefore \int_{0}^{\frac{\pi}{2}}\frac{\sin^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\,dx=\frac{\pi}{4}

\displaystyle \textbf{Question 5: }~\int_{0}^{\pi/2}\frac{\sin^{n}x}{\sin^{n}x+\cos^{n}x}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\sin^{n}x}{\sin^{n}x+\cos^{n}x}\,dx
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\sin^{n}\left(\frac{\pi}{2}-x\right)}{\sin^{n}\left(\frac{\pi}{2}-x\right)+\cos^{n}\left(\frac{\pi}{2}-x\right)}\,dx
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\cos^{n}x}{\cos^{n}x+\sin^{n}x}\,dx
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\cos^{n}x}{\sin^{n}x+\cos^{n}x}\,dx
\displaystyle \text{Adding (1) and (2) we get}
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\left(\frac{\sin^{n}x}{\sin^{n}x+\cos^{n}x}+\frac{\cos^{n}x}{\sin^{n}x+\cos^{n}x}\right)dx
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\frac{\sin^{n}x+\cos^{n}x}{\sin^{n}x+\cos^{n}x}\,dx
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}1\,dx
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}dx
\displaystyle 2I=\left[x\right]_{0}^{\frac{\pi}{2}}
\displaystyle 2I=\frac{\pi}{2}
\displaystyle \therefore I=\frac{\pi}{4}

\displaystyle \textbf{Question 6: }~\int_{0}^{\pi/2}\frac{1}{1+\sqrt{\tan x}}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\sqrt{\tan x}}\,dx
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\sqrt{\tan\left(\frac{\pi}{2}-x\right)}}\,dx
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\sqrt{\cot x}}\,dx
\displaystyle \text{Adding (1) and (2) we get}
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\left(\frac{1}{1+\sqrt{\tan x}}+\frac{1}{1+\sqrt{\cot x}}\right)dx
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\frac{(1+\sqrt{\cot x})+(1+\sqrt{\tan x})}{(1+\sqrt{\tan x})(1+\sqrt{\cot x})}\,dx
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\frac{2+\sqrt{\cot x}+\sqrt{\tan x}}{1+\sqrt{\cot x}+\sqrt{\tan x}+\sqrt{\tan x\cot x}}\,dx
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\frac{2+\sqrt{\cot x}+\sqrt{\tan x}}{2+\sqrt{\cot x}+\sqrt{\tan x}}\,dx
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}dx
\displaystyle 2I=\left[x\right]_{0}^{\frac{\pi}{2}}
\displaystyle 2I=\frac{\pi}{2}
\displaystyle \therefore I=\frac{\pi}{4}

\displaystyle \textbf{Question 7: }~\int_{0}^{a}\frac{1}{x+\sqrt{a^{2}-x^{2}}}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{a}\frac{1}{x+\sqrt{a^{2}-x^{2}}}\,dx
\displaystyle \text{Putting }x=a\sin\theta
\displaystyle dx=a\cos\theta\,d\theta
\displaystyle \text{When }x\to0,\ \theta\to0
\displaystyle \text{When }x\to a,\ \theta\to\frac{\pi}{2}
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{a\cos\theta}{a\sin\theta+\sqrt{a^{2}-a^{2}\sin^{2}\theta}}\,d\theta
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{a\cos\theta}{a\sin\theta+a\cos\theta}\,d\theta
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\cos\theta}{\sin\theta+\cos\theta}\,d\theta
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\cos\left(\frac{\pi}{2}-\theta\right)}{\sin\left(\frac{\pi}{2}-\theta\right)+\cos\left(\frac{\pi}{2}-\theta\right)}\,d\theta
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\sin\theta}{\cos\theta+\sin\theta}\,d\theta
\displaystyle \text{Adding the two expressions for }I
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\frac{\cos\theta+\sin\theta}{\sin\theta+\cos\theta}\,d\theta
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}d\theta
\displaystyle 2I=\left[\theta\right]_{0}^{\frac{\pi}{2}}
\displaystyle 2I=\frac{\pi}{2}
\displaystyle \therefore I=\frac{\pi}{4}

\displaystyle \textbf{Question 8: }~\int_{0}^{\infty}\frac{\log x}{1+x^{2}}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{\infty}\frac{\log x}{1+x^{2}}\,dx
\displaystyle \text{Putting }x=\tan\theta
\displaystyle dx=\sec^{2}\theta\,d\theta
\displaystyle \text{When }x\to0,\ \theta\to0
\displaystyle \text{When }x\to\infty,\ \theta\to\frac{\pi}{2}
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\log(\tan\theta)}{1+\tan^{2}\theta}\sec^{2}\theta\,d\theta
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\log(\tan\theta)\,d\theta
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\log\!\left(\tan\left(\frac{\pi}{2}-\theta\right)\right)\,d\theta
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\log(\cot\theta)\,d\theta
\displaystyle \text{Adding the two expressions for }I
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\log(\tan\theta)\,d\theta+\int_{0}^{\frac{\pi}{2}}\log(\cot\theta)\,d\theta
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\left[\log(\tan\theta)+\log(\cot\theta)\right]d\theta
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\log(\tan\theta\cdot\cot\theta)\,d\theta
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\log 1\,d\theta
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}0\,d\theta
\displaystyle 2I=0
\displaystyle \therefore I=0
\displaystyle \therefore \int_{0}^{\infty}\frac{\log x}{1+x^{2}}\,dx=0

\displaystyle \textbf{Question 9: }~\int_{0}^{1}\frac{\log(1+x)}{1+x^{2}}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{1}\frac{\log(1+x)}{1+x^{2}}\,dx
\displaystyle \text{Putting }x=\tan\theta
\displaystyle dx=\sec^{2}\theta\,d\theta
\displaystyle \text{When }x\to0,\ \theta\to0
\displaystyle \text{When }x\to1,\ \theta\to\frac{\pi}{4}
\displaystyle I=\int_{0}^{\frac{\pi}{4}}\frac{\log(1+\tan\theta)}{\sec^{2}\theta}\sec^{2}\theta\,d\theta
\displaystyle I=\int_{0}^{\frac{\pi}{4}}\log(1+\tan\theta)\,d\theta
\displaystyle I=\int_{0}^{\frac{\pi}{4}}\log\!\left(1+\tan\left(\frac{\pi}{4}-\theta\right)\right)d\theta
\displaystyle I=\int_{0}^{\frac{\pi}{4}}\log\!\left(1+\frac{\tan\frac{\pi}{4}-\tan\theta}{1+\tan\frac{\pi}{4}\tan\theta}\right)d\theta
\displaystyle I=\int_{0}^{\frac{\pi}{4}}\log\!\left(1+\frac{1-\tan\theta}{1+\tan\theta}\right)d\theta
\displaystyle I=\int_{0}^{\frac{\pi}{4}}\log\!\left(\frac{2}{1+\tan\theta}\right)d\theta
\displaystyle I=\int_{0}^{\frac{\pi}{4}}\left[\log 2-\log(1+\tan\theta)\right]d\theta
\displaystyle \text{Adding (1) and (2) we get}
\displaystyle 2I=\int_{0}^{\frac{\pi}{4}}\log 2\,d\theta
\displaystyle 2I=\log 2\left[\theta\right]_{0}^{\frac{\pi}{4}}
\displaystyle 2I=\frac{\pi}{4}\log 2
\displaystyle \therefore I=\frac{\pi}{8}\log 2
\displaystyle \therefore \int_{0}^{1}\frac{\log(1+x)}{1+x^{2}}\,dx=\frac{\pi}{8}\log 2

\displaystyle \textbf{Question 10: }~\int_{0}^{\infty}\frac{x}{(1+x)(1+x^{2})}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{\infty}\frac{x}{(1+x)(1+x^{2})}\,dx
\displaystyle \text{Putting }x=\tan\theta
\displaystyle dx=\sec^{2}\theta\,d\theta
\displaystyle \text{When }x\to0,\ \theta\to0
\displaystyle \text{When }x\to\infty,\ \theta\to\frac{\pi}{2}
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\tan\theta}{(1+\tan\theta)\sec^{2}\theta}\sec^{2}\theta\,d\theta
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\tan\theta}{1+\tan\theta}\,d\theta
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\sin\theta/\cos\theta}{1+\sin\theta/\cos\theta}\,d\theta
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\sin\theta}{\sin\theta+\cos\theta}\,d\theta
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\sin\left(\frac{\pi}{2}-\theta\right)}{\sin\left(\frac{\pi}{2}-\theta\right)+\cos\left(\frac{\pi}{2}-\theta\right)}\,d\theta
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\cos\theta}{\cos\theta+\sin\theta}\,d\theta
\displaystyle \text{Adding (1) and (2) we get}
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\frac{\sin\theta+\cos\theta}{\sin\theta+\cos\theta}\,d\theta
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}d\theta
\displaystyle 2I=\left[\theta\right]_{0}^{\frac{\pi}{2}}
\displaystyle 2I=\frac{\pi}{2}
\displaystyle \therefore I=\frac{\pi}{4}
\displaystyle \therefore \int_{0}^{\infty}\frac{x}{(1+x)(1+x^{2})}\,dx=\frac{\pi}{4}

\displaystyle \textbf{Question 11: }~\int_{0}^{\pi}\frac{x\tan x}{\sec x\,\mathrm{cosec x}}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{\pi}\frac{x\tan x}{\sec x\,\mathrm{cosec}\,x}\,dx
\displaystyle I=\int_{0}^{\pi}\frac{(\pi-x)\tan(\pi-x)}{\sec(\pi-x)\,\mathrm{cosec}(\pi-x)}\,dx
\displaystyle I=\int_{0}^{\pi}\frac{-(\pi-x)\tan x}{-\sec x\,\mathrm{cosec}\,x}\,dx
\displaystyle I=\int_{0}^{\pi}\frac{(\pi-x)\tan x}{\sec x\,\mathrm{cosec}\,x}\,dx
\displaystyle \text{Adding (1) and (2)}
\displaystyle 2I=\int_{0}^{\pi}\left(\frac{x\tan x}{\sec x\,\mathrm{cosec}\,x}+\frac{(\pi-x)\tan x}{\sec x\,\mathrm{cosec}\,x}\right)dx
\displaystyle 2I=\int_{0}^{\pi}\frac{\pi\tan x}{\sec x\,\mathrm{cosec}\,x}\,dx
\displaystyle 2I=\int_{0}^{\pi}\pi\sin^{2}x\,dx
\displaystyle 2I=\pi\int_{0}^{\pi}\sin^{2}x\,dx
\displaystyle 2I=\pi\int_{0}^{\pi}\left(1-\cos^{2}x\right)dx
\displaystyle 2I=\pi\left[x\right]_{0}^{\pi}-\frac{\pi}{2}\int_{0}^{\pi}(1+\cos2x)\,dx
\displaystyle 2I=\frac{\pi}{2}\left[x\right]_{0}^{\pi}-\frac{\pi}{2}\left[\frac{\sin2x}{2}\right]_{0}^{\pi}
\displaystyle 2I=\frac{\pi^{2}}{2}
\displaystyle \therefore I=\frac{\pi^{2}}{4}

\displaystyle \textbf{Question 12: }~\int_{0}^{\pi}x\sin x\cos^{4}x\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{\pi}x\sin x\cos^{4}x\,dx
\displaystyle I=\int_{0}^{\pi}(\pi-x)\sin(\pi-x)\cos^{4}(\pi-x)\,dx
\displaystyle I=\int_{0}^{\pi}(\pi-x)\sin x\cos^{4}x\,dx
\displaystyle \text{Adding (1) and (2) we get}
\displaystyle 2I=\int_{0}^{\pi}(x+\pi-x)\sin x\cos^{4}x\,dx
\displaystyle 2I=\pi\int_{0}^{\pi}\sin x\cos^{4}x\,dx
\displaystyle \text{Let }\cos x=t
\displaystyle -\sin x\,dx=dt
\displaystyle \text{When }x=0,\ t=1
\displaystyle \text{When }x=\pi,\ t=-1
\displaystyle 2I=-\pi\int_{1}^{-1}t^{4}\,dt
\displaystyle 2I=\pi\int_{-1}^{1}t^{4}\,dt
\displaystyle 2I=\pi\left[\frac{t^{5}}{5}\right]_{-1}^{1}
\displaystyle 2I=\frac{\pi}{5}+\frac{\pi}{5}
\displaystyle 2I=\frac{2\pi}{5}
\displaystyle \therefore I=\frac{\pi}{5}

\displaystyle \textbf{Question 13: }~\int_{0}^{\pi}x\sin^{3}x\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{\pi}x\sin^{3}x\,dx
\displaystyle I=\int_{0}^{\pi}(\pi-x)\sin^{3}(\pi-x)\,dx
\displaystyle I=\int_{0}^{\pi}(\pi-x)\sin^{3}x\,dx
\displaystyle \text{Adding (i) and (ii) we get}
\displaystyle 2I=\int_{0}^{\pi}(x+\pi-x)\sin^{3}x\,dx
\displaystyle 2I=\int_{0}^{\pi}\pi\sin^{3}x\,dx
\displaystyle 2I=\pi\int_{0}^{\pi}\sin^{3}x\,dx
\displaystyle 2I=\pi\int_{0}^{\pi}\frac{3\sin x-\sin3x}{4}\,dx
\displaystyle 2I=\frac{\pi}{4}\int_{0}^{\pi}(3\sin x-\sin3x)\,dx
\displaystyle 2I=\frac{\pi}{4}\left[-3\cos x+\frac{\cos3x}{3}\right]_{0}^{\pi}
\displaystyle 2I=\frac{\pi}{4}\left[-3\cos\pi+3\cos0+\frac{\cos3\pi}{3}-\frac{\cos0}{3}\right]
\displaystyle 2I=\frac{\pi}{4}\left[3+3-\frac{1}{3}-\frac{1}{3}\right]
\displaystyle 2I=\frac{\pi}{2}\left[3-\frac{1}{3}\right]
\displaystyle 2I=\frac{\pi}{2}\cdot\frac{8}{3}
\displaystyle 2I=\frac{4\pi}{3}
\displaystyle \therefore I=\frac{2\pi}{3}

\displaystyle \textbf{Question 14: }~\int_{0}^{\pi}x\log(\sin x)\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{\pi}x\log(\sin x)\,dx
\displaystyle I=\int_{0}^{\pi}(\pi-x)\log(\sin(\pi-x))\,dx
\displaystyle I=\int_{0}^{\pi}(\pi-x)\log(\sin x)\,dx
\displaystyle \text{Adding (i) and (ii)}
\displaystyle 2I=\pi\int_{0}^{\pi}\log(\sin x)\,dx
\displaystyle 2I=2\pi\int_{0}^{\frac{\pi}{2}}\log(\sin x)\,dx
\displaystyle I=\pi\int_{0}^{\frac{\pi}{2}}\log(\sin x)\,dx
\displaystyle \text{Let }\int_{0}^{\frac{\pi}{2}}\log(\sin x)\,dx=I_{2}
\displaystyle I_{2}=\int_{0}^{\frac{\pi}{2}}\log\!\left(\sin\left(\frac{\pi}{2}-x\right)\right)dx
\displaystyle I_{2}=\int_{0}^{\frac{\pi}{2}}\log(\cos x)\,dx
\displaystyle 2I_{2}=\int_{0}^{\frac{\pi}{2}}\left(\log(\sin x)+\log(\cos x)\right)dx
\displaystyle 2I_{2}=\int_{0}^{\frac{\pi}{2}}\log(\sin x\cos x)\,dx
\displaystyle 2I_{2}=\int_{0}^{\frac{\pi}{2}}\log(\sin2x)\,dx-\int_{0}^{\frac{\pi}{2}}\log2\,dx
\displaystyle \text{Let }2x=t
\displaystyle 2dx=dt
\displaystyle \text{When }x=0,\ t=0
\displaystyle \text{When }x=\frac{\pi}{2},\ t=\pi
\displaystyle 2I_{2}=\frac{1}{2}\int_{0}^{\pi}\log(\sin t)\,dt-\frac{\pi}{2}\log2
\displaystyle 2I_{2}=\int_{0}^{\frac{\pi}{2}}\log(\sin t)\,dt-\frac{\pi}{2}\log2
\displaystyle 2I_{2}=I_{2}-\frac{\pi}{2}\log2
\displaystyle I_{2}=-\frac{\pi}{2}\log2
\displaystyle \text{From (iii)}
\displaystyle I=\pi\int_{0}^{\frac{\pi}{2}}\log(\sin x)\,dx=\pi I_{2}
\displaystyle I=-\frac{\pi^{2}}{2}\log2

\displaystyle \textbf{Question 15: }~\int_{0}^{\pi}\frac{x\sin x}{1+\sin x}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{\pi}\frac{x\sin x}{1+\sin x}\,dx
\displaystyle I=\int_{0}^{\pi}\frac{(\pi-x)\sin(\pi-x)}{1+\sin(\pi-x)}\,dx
\displaystyle I=\int_{0}^{\pi}\frac{(\pi-x)\sin x}{1+\sin x}\,dx
\displaystyle \text{Adding (1) and (2) we get}
\displaystyle 2I=\int_{0}^{\pi}(x+\pi-x)\frac{\sin x}{1+\sin x}\,dx
\displaystyle 2I=\pi\int_{0}^{\pi}\frac{\sin x}{1+\sin x}\,dx
\displaystyle 2I=\pi\int_{0}^{\pi}\frac{1+\sin x-1}{1+\sin x}\,dx
\displaystyle 2I=\pi\int_{0}^{\pi}dx-\pi\int_{0}^{\pi}\frac{1}{1+\sin x}\,dx
\displaystyle 2I=\pi\int_{0}^{\pi}dx-\pi\int_{0}^{\pi}\frac{1-\sin x}{(1+\sin x)(1-\sin x)}\,dx
\displaystyle 2I=\pi\int_{0}^{\pi}dx-\pi\int_{0}^{\pi}\frac{1-\sin x}{1-\sin^{2}x}\,dx
\displaystyle 2I=\pi\int_{0}^{\pi}dx-\pi\int_{0}^{\pi}\frac{1-\sin x}{\cos^{2}x}\,dx
\displaystyle 2I=\pi\int_{0}^{\pi}dx-\pi\int_{0}^{\pi}\left(\sec^{2}x-\sec x\tan x\right)dx
\displaystyle 2I=\pi\left[x\right]_{0}^{\pi}-\pi\left[\tan x-\sec x\right]_{0}^{\pi}
\displaystyle 2I=\pi^{2}-\pi(0+1-0+1)
\displaystyle 2I=\pi^{2}-2\pi
\displaystyle \therefore I=\pi\left(\frac{\pi}{2}-1\right)

\displaystyle \textbf{Question 16: }~\int_{0}^{\pi}\frac{x}{1+\cos\alpha\,\sin x}\,dx,\ 0<\alpha<\pi.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{\pi}\frac{x}{1+\cos\alpha\sin x}\,dx
\displaystyle I=\int_{0}^{\pi}\frac{\pi-x}{1+\cos\alpha\sin(\pi-x)}\,dx
\displaystyle I=\int_{0}^{\pi}\frac{\pi-x}{1+\cos\alpha\sin x}\,dx
\displaystyle \text{Adding (1) and (2) we get}
\displaystyle 2I=\int_{0}^{\pi}\frac{x+\pi-x}{1+\cos\alpha\sin x}\,dx
\displaystyle I=\frac{\pi}{2}\int_{0}^{\pi}\frac{1}{1+\cos\alpha\sin x}\,dx
\displaystyle I=\frac{\pi}{2}\int_{0}^{\pi}\frac{1}{1+\cos\alpha\cdot\frac{2\tan\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}}\,dx
\displaystyle I=\frac{\pi}{2}\int_{0}^{\pi}\frac{1+\tan^{2}\frac{x}{2}}{1+\tan^{2}\frac{x}{2}+2\cos\alpha\tan\frac{x}{2}}\,dx
\displaystyle I=\frac{\pi}{2}\int_{0}^{\pi}\frac{\sec^{2}\frac{x}{2}}{1+\tan^{2}\frac{x}{2}+2\cos\alpha\tan\frac{x}{2}}\,dx
\displaystyle \text{Putting }\tan\frac{x}{2}=t
\displaystyle \frac{1}{2}\sec^{2}\frac{x}{2}\,dx=dt
\displaystyle \text{When }x\to0,\ t\to0
\displaystyle \text{When }x\to\pi,\ t\to\infty
\displaystyle I=\frac{\pi}{2}\int_{0}^{\infty}\frac{2}{1+t^{2}+2\cos\alpha\,t}\,dt
\displaystyle I=\frac{\pi}{2}\int_{0}^{\infty}\frac{2}{(t+\cos\alpha)^{2}-\cos^{2}\alpha+1}\,dt
\displaystyle I=\pi\int_{0}^{\infty}\frac{1}{(t+\cos\alpha)^{2}+\sin^{2}\alpha}\,dt
\displaystyle I=\pi\left[\frac{1}{\sin\alpha}\tan^{-1}\left(\frac{t+\cos\alpha}{\sin\alpha}\right)\right]_{0}^{\infty}
\displaystyle I=\frac{\pi}{\sin\alpha}\left[\tan^{-1}(\infty)-\tan^{-1}(\cot\alpha)\right]
\displaystyle I=\frac{\pi}{\sin\alpha}\left[\frac{\pi}{2}-\tan^{-1}\left(\tan\left(\frac{\pi}{2}-\alpha\right)\right)\right]
\displaystyle I=\frac{\pi\alpha}{\sin\alpha}

\displaystyle \textbf{Question 17: }~\int_{0}^{\pi}x\cos^{2}x\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{\pi}x\cos^{2}x\,dx
\displaystyle I=\int_{0}^{\pi}(\pi-x)\cos^{2}(\pi-x)\,dx
\displaystyle I=\int_{0}^{\pi}(\pi-x)\cos^{2}x\,dx
\displaystyle \text{Adding (i) and (ii) we get}
\displaystyle 2I=\int_{0}^{\pi}(x+\pi-x)\cos^{2}x\,dx
\displaystyle 2I=\int_{0}^{\pi}\pi\cos^{2}x\,dx
\displaystyle 2I=\pi\int_{0}^{\pi}\cos^{2}x\,dx
\displaystyle 2I=\pi\int_{0}^{\pi}\frac{1+\cos2x}{2}\,dx
\displaystyle 2I=\frac{\pi}{2}\int_{0}^{\pi}(1+\cos2x)\,dx
\displaystyle 2I=\frac{\pi}{2}\left[x+\frac{\sin2x}{2}\right]_{0}^{\pi}
\displaystyle 2I=\frac{\pi}{2}(\pi-0)
\displaystyle \therefore I=\frac{\pi^{2}}{4}

\displaystyle \textbf{Question 18: }~\int_{\pi/6}^{\pi/3}\frac{1}{1+\cot^{3/2}x}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{1}{1+\cot^{3/2}x}\,dx
\displaystyle I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{1}{1+\cot^{3/2}\left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}\,dx
\displaystyle I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{1}{1+\cot^{3/2}\left(\frac{\pi}{2}-x\right)}\,dx
\displaystyle I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{1}{1+\tan^{3/2}x}\,dx
\displaystyle I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\cot^{3/2}x}{1+\cot^{3/2}x}\,dx
\displaystyle \text{Adding (1) and (2) we get}
\displaystyle 2I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{1+\cot^{3/2}x}{1+\cot^{3/2}x}\,dx
\displaystyle 2I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}dx
\displaystyle 2I=\left[x\right]_{\frac{\pi}{6}}^{\frac{\pi}{3}}
\displaystyle 2I=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}
\displaystyle \therefore I=\frac{\pi}{12}

\displaystyle \textbf{Question 19: }~\int_{0}^{\pi/2}\frac{\tan^{7}x}{\tan^{7}x+\cot^{7}x}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\tan^{7}x}{\tan^{7}x+\cot^{7}x}\,dx
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\tan^{7}\left(\frac{\pi}{2}-x\right)}{\tan^{7}\left(\frac{\pi}{2}-x\right)+\cot^{7}\left(\frac{\pi}{2}-x\right)}\,dx
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\cot^{7}x}{\cot^{7}x+\tan^{7}x}\,dx
\displaystyle \text{Adding (1) and (2) we get}
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\left(\frac{\tan^{7}x}{\tan^{7}x+\cot^{7}x}+\frac{\cot^{7}x}{\tan^{7}x+\cot^{7}x}\right)dx
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\frac{\tan^{7}x+\cot^{7}x}{\tan^{7}x+\cot^{7}x}\,dx
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}dx
\displaystyle 2I=\left[x\right]_{0}^{\frac{\pi}{2}}
\displaystyle 2I=\frac{\pi}{2}
\displaystyle \therefore I=\frac{\pi}{4}

\displaystyle \textbf{Question 20: }~\int_{2}^{8}\frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{2}^{8}\frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}}\,dx
\displaystyle I=\int_{2}^{8}\frac{\sqrt{10-(2+8-x)}}{\sqrt{2+8-x}+\sqrt{10-(2+8-x)}}\,dx
\displaystyle I=\int_{2}^{8}\frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}}\,dx
\displaystyle \text{Adding (1) and (2) we have}
\displaystyle 2I=\int_{2}^{8}\frac{\sqrt{10-x}+\sqrt{x}}{\sqrt{x}+\sqrt{10-x}}\,dx
\displaystyle 2I=\int_{2}^{8}dx
\displaystyle 2I=\left[x\right]_{2}^{8}
\displaystyle 2I=8-2=6
\displaystyle \therefore I=3

\displaystyle \textbf{Question 21: }~\int_{0}^{\pi}x\sin x\cos^{2}x\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{\pi}x\sin x\cos^{2}x\,dx
\displaystyle I=\int_{0}^{\pi}(\pi-x)\sin(\pi-x)\cos^{2}(\pi-x)\,dx
\displaystyle I=\int_{0}^{\pi}(\pi-x)\sin x\cos^{2}x\,dx
\displaystyle \text{Adding (1) and (2) we have}
\displaystyle 2I=\int_{0}^{\pi}(\pi-x+x)\sin x\cos^{2}x\,dx
\displaystyle 2I=\pi\int_{0}^{\pi}\sin x\cos^{2}x\,dx
\displaystyle 2I=-\pi\int_{0}^{\pi}\cos^{2}x(-\sin x)\,dx
\displaystyle 2I=-\pi\left[\frac{\cos^{3}x}{3}\right]_{0}^{\pi}
\displaystyle 2I=-\frac{\pi}{3}\left(\cos^{3}\pi-\cos^{3}0\right)
\displaystyle 2I=-\frac{\pi}{3}(-1-1)
\displaystyle 2I=\frac{2\pi}{3}
\displaystyle \therefore I=\frac{\pi}{3}

\displaystyle \textbf{Question 22: }~\int_{0}^{\pi/2}\frac{x\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{x\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\left(\frac{\pi}{2}-x\right)\sin\left(\frac{\pi}{2}-x\right)\cos\left(\frac{\pi}{2}-x\right)}{\sin^{4}\left(\frac{\pi}{2}-x\right)+\cos^{4}\left(\frac{\pi}{2}-x\right)}\,dx
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\left(\frac{\pi}{2}-x\right)\sin x\cos x}{\cos^{4}x+\sin^{4}x}\,dx
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\left(\frac{\pi}{2}-x\right)\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx
\displaystyle \text{Adding (i) and (ii) we get}
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\left(x+\frac{\pi}{2}-x\right)\frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx
\displaystyle 2I=\frac{\pi}{2}\int_{0}^{\frac{\pi}{2}}\frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx
\displaystyle \text{Let }\sin^{2}x=t
\displaystyle dt=2\sin x\cos x\,dx
\displaystyle \text{When }x=0,\ t=0
\displaystyle \text{When }x=\frac{\pi}{2},\ t=1
\displaystyle 2I=\frac{\pi}{4}\int_{0}^{1}\frac{dt}{t^{2}+(1-t)^{2}}
\displaystyle 2I=\frac{\pi}{8}\int_{0}^{1}\frac{dt}{\left(t-\frac{1}{2}\right)^{2}+\frac{1}{4}}
\displaystyle 2I=\frac{\pi}{8}\cdot2\left[\tan^{-1}(2t-1)\right]_{0}^{1}
\displaystyle 2I=\frac{\pi}{4}\left(\tan^{-1}1-\tan^{-1}(-1)\right)
\displaystyle 2I=\frac{\pi}{4}\left(\frac{\pi}{4}+\frac{\pi}{4}\right)
\displaystyle 2I=\frac{\pi^{2}}{8}
\displaystyle \therefore I=\frac{\pi^{2}}{16}

\displaystyle \textbf{Question 23: }~\int_{-\pi/2}^{\pi/2}\sin^{3}x\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^{3}x\,dx
\displaystyle I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin x(1-\cos^{2}x)\,dx
\displaystyle I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin x\,dx-\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin x\cos^{2}x\,dx
\displaystyle I=\left[-\cos x\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}+\left[\frac{\cos^{3}x}{3}\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}
\displaystyle I=0+0
\displaystyle \therefore I=0

\displaystyle \textbf{Question 24: }~\int_{-\pi/2}^{\pi/2}\sin^{4}x\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^{4}x\,dx
\displaystyle I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\sin^{2}x)^{2}\,dx
\displaystyle I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{1-\cos2x}{2}\right)^{2}dx
\displaystyle I=\frac{1}{4}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(1-2\cos2x+\cos^{2}2x)\,dx
\displaystyle I=\frac{1}{4}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}dx-\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos2x\,dx+\frac{1}{4}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^{2}2x\,dx
\displaystyle I=\frac{1}{4}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}dx-\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos2x\,dx+\frac{1}{8}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(1+\cos4x)\,dx
\displaystyle I=\frac{1}{4}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}dx-\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos2x\,dx+\frac{1}{8}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}dx+\frac{1}{8}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos4x\,dx
\displaystyle I=\frac{3}{8}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}dx-\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos2x\,dx+\frac{1}{8}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos4x\,dx
\displaystyle I=\frac{3}{8}\left[x\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}-\frac{1}{4}\left[\sin2x\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}+\frac{1}{32}\left[\sin4x\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}
\displaystyle I=\frac{3}{8}\left(\frac{\pi}{2}+\frac{\pi}{2}\right)-\frac{1}{4}(0-0)+\frac{1}{32}(0-0)
\displaystyle \therefore I=\frac{3\pi}{8}

\displaystyle \textbf{Question 25: }~\int_{-1}^{1}\log\left(\frac{2-x}{2+x}\right)\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{-1}^{1}\log\left(\frac{2-x}{2+x}\right)dx
\displaystyle \text{Here }f(x)=\log\left(\frac{2-x}{2+x}\right)
\displaystyle f(-x)=\log\left(\frac{2+x}{2-x}\right)
\displaystyle f(-x)=-\log\left(\frac{2-x}{2+x}\right)
\displaystyle f(-x)=-f(x)
\displaystyle \text{Hence }f(x)\text{ is an odd function}
\displaystyle \therefore I=0

\displaystyle \textbf{Question 26: }~\int_{-\pi/4}^{\pi/4}\sin^{2}x\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sin^{2}x\,dx
\displaystyle \text{Here }f(x)=\sin^{2}x
\displaystyle f(-x)=\sin^{2}(-x)=\sin^{2}x=f(x)
\displaystyle \text{Hence }\sin^{2}x\text{ is an even function}
\displaystyle \therefore I=2\int_{0}^{\frac{\pi}{4}}\sin^{2}x\,dx
\displaystyle I=2\int_{0}^{\frac{\pi}{4}}\left(\frac{1-\cos2x}{2}\right)dx
\displaystyle I=\int_{0}^{\frac{\pi}{4}}(1-\cos2x)\,dx
\displaystyle I=\left[x-\frac{\sin2x}{2}\right]_{0}^{\frac{\pi}{4}}
\displaystyle I=\frac{\pi}{4}-\frac{1}{2}

\displaystyle \textbf{Question 27: }~\int_{0}^{\pi}\log(1-\cos x)\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{\pi}\log(1-\cos x)\,dx
\displaystyle I=\int_{0}^{\pi}\log\left(2\sin^{2}\frac{x}{2}\right)dx
\displaystyle I=\int_{0}^{\pi}\log2\,dx+2\int_{0}^{\pi}\log\sin\frac{x}{2}\,dx
\displaystyle \text{Let }t=\frac{x}{2}\text{ in the second integral}
\displaystyle dt=\frac{1}{2}dx
\displaystyle \text{When }x\to0,\ t\to0
\displaystyle \text{When }x\to\pi,\ t\to\frac{\pi}{2}
\displaystyle I=\log2\left[x\right]_{0}^{\pi}+4\int_{0}^{\frac{\pi}{2}}\log(\sin t)\,dt
\displaystyle I=\pi\log2+4\left(-\frac{\pi}{2}\log2\right)
\displaystyle I=-\pi\log2

\displaystyle \textbf{Question 28: }~\int_{-\pi/2}^{\pi/2}\log\left(\frac{2-\sin x}{2+\sin x}\right)\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\log\left(\frac{2-\sin x}{2+\sin x}\right)dx
\displaystyle \text{Here }f(x)=\log\left(\frac{2-\sin x}{2+\sin x}\right)
\displaystyle f(-x)=\log\left(\frac{2-\sin(-x)}{2+\sin(-x)}\right)
\displaystyle f(-x)=\log\left(\frac{2+\sin x}{2-\sin x}\right)
\displaystyle f(-x)=-\log\left(\frac{2-\sin x}{2+\sin x}\right)
\displaystyle f(-x)=-f(x)
\displaystyle \text{Hence }f(x)\text{ is an odd function}
\displaystyle \therefore I=0

\displaystyle \textbf{Question 29: }~\int_{-\pi}^{\pi}\frac{2x(1+\sin x)}{1+\cos^{2}x}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{-\pi}^{\pi}\frac{2x(1+\sin x)}{1+\cos^{2}x}\,dx
\displaystyle I=\int_{-\pi}^{\pi}\frac{2x}{1+\cos^{2}x}\,dx+\int_{-\pi}^{\pi}\frac{2x\sin x}{1+\cos^{2}x}\,dx
\displaystyle I=I_{1}+I_{2}
\displaystyle \text{Consider }f(x)=\frac{2x}{1+\cos^{2}x}
\displaystyle f(-x)=\frac{2(-x)}{1+\cos^{2}(-x)}=-\frac{2x}{1+\cos^{2}x}=-f(x)
\displaystyle \therefore I_{1}=\int_{-\pi}^{\pi}\frac{2x}{1+\cos^{2}x}\,dx=0
\displaystyle \text{Again consider }g(x)=\frac{2x\sin x}{1+\cos^{2}x}
\displaystyle g(-x)=\frac{2(-x)\sin(-x)}{1+\cos^{2}(-x)}=\frac{2x\sin x}{1+\cos^{2}x}=g(x)
\displaystyle I_{2}=\int_{-\pi}^{\pi}\frac{2x\sin x}{1+\cos^{2}x}\,dx
\displaystyle I_{2}=2\int_{0}^{\pi}\frac{2x\sin x}{1+\cos^{2}x}\,dx
\displaystyle I_{2}=4\int_{0}^{\pi}\frac{x\sin x}{1+\cos^{2}x}\,dx
\displaystyle I_{2}=4\int_{0}^{\pi}\frac{(\pi-x)\sin(\pi-x)}{1+\cos^{2}(\pi-x)}\,dx
\displaystyle I_{2}=4\int_{0}^{\pi}\frac{(\pi-x)\sin x}{1+\cos^{2}x}\,dx
\displaystyle \text{Adding the last two expressions for }I_{2}
\displaystyle 2I_{2}=4\int_{0}^{\pi}\frac{\pi\sin x}{1+\cos^{2}x}\,dx
\displaystyle 2I_{2}=4\pi\int_{0}^{\pi}\frac{\sin x}{1+\cos^{2}x}\,dx
\displaystyle \text{Put }\cos x=z
\displaystyle -\sin x\,dx=dz
\displaystyle \text{When }x=0,\ z=1
\displaystyle \text{When }x=\pi,\ z=-1
\displaystyle 2I_{2}=-4\pi\int_{1}^{-1}\frac{dz}{1+z^{2}}
\displaystyle 2I_{2}=4\pi\int_{-1}^{1}\frac{dz}{1+z^{2}}
\displaystyle 2I_{2}=4\pi\left[\tan^{-1}z\right]_{-1}^{1}
\displaystyle 2I_{2}=4\pi\left(\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)\right)=2\pi^{2}
\displaystyle I_{2}=\pi^{2}
\displaystyle \therefore I=I_{1}+I_{2}=0+\pi^{2}=\pi^{2}

\displaystyle \textbf{Question 30: }~\int_{-a}^{a}\log\left(\frac{a-\sin\theta}{a+\sin\theta}\right)\,d\theta,\ a>0.
\displaystyle \text{Answer:}
\displaystyle I=\int_{-a}^{a}\log\left(\frac{a-\sin\theta}{a+\sin\theta}\right)d\theta
\displaystyle \text{Consider }f(\theta)=\log\left(\frac{a-\sin\theta}{a+\sin\theta}\right)
\displaystyle f(-\theta)=\log\left(\frac{a-\sin(-\theta)}{a+\sin(-\theta)}\right)
\displaystyle f(-\theta)=\log\left(\frac{a+\sin\theta}{a-\sin\theta}\right)
\displaystyle f(-\theta)=\log\left(\frac{a-\sin\theta}{a+\sin\theta}\right)^{-1}
\displaystyle f(-\theta)=-\log\left(\frac{a-\sin\theta}{a+\sin\theta}\right)
\displaystyle f(-\theta)=-f(\theta)
\displaystyle \text{Hence }f(\theta)\text{ is an odd function}
\displaystyle \therefore I=\int_{-a}^{a}\log\left(\frac{a-\sin\theta}{a+\sin\theta}\right)d\theta=0

\displaystyle \textbf{Question 31: }~\int_{-2}^{2}\frac{3x^{3}+2\lvert x\rvert+1}{x^{2}+\lvert x\rvert+1}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{-2}^{2}\frac{3x^{3}+2|x|+1}{x^{2}+|x|+1}\,dx
\displaystyle I=\int_{-2}^{2}\frac{3x^{3}}{x^{2}+|x|+1}\,dx+\int_{-2}^{2}\frac{2|x|+1}{x^{2}+|x|+1}\,dx
\displaystyle I=I_{1}+I_{2}
\displaystyle \text{Consider }f(x)=\frac{3x^{3}}{x^{2}+|x|+1}
\displaystyle f(-x)=\frac{3(-x)^{3}}{(-x)^{2}+|-x|+1}=-\frac{3x^{3}}{x^{2}+|x|+1}=-f(x)
\displaystyle \therefore I_{1}=\int_{-2}^{2}\frac{3x^{3}}{x^{2}+|x|+1}\,dx=0
\displaystyle \text{Now consider }g(x)=\frac{2|x|+1}{x^{2}+|x|+1}
\displaystyle g(-x)=\frac{2|-x|+1}{(-x)^{2}+|-x|+1}=\frac{2|x|+1}{x^{2}+|x|+1}=g(x)
\displaystyle I_{2}=\int_{-2}^{2}\frac{2|x|+1}{x^{2}+|x|+1}\,dx
\displaystyle I_{2}=2\int_{0}^{2}\frac{2|x|+1}{x^{2}+|x|+1}\,dx
\displaystyle I_{2}=2\int_{0}^{2}\frac{2x+1}{x^{2}+x+1}\,dx
\displaystyle I_{2}=2\int_{0}^{2}\frac{(x^{2}+x+1)'}{x^{2}+x+1}\,dx
\displaystyle I_{2}=2\left[\log(x^{2}+x+1)\right]_{0}^{2}
\displaystyle I_{2}=2(\log7-\log1)
\displaystyle I_{2}=2\log7
\displaystyle \therefore I=I_{1}+I_{2}=0+2\log7=2\log7

\displaystyle \textbf{Question 32: }~\int_{-3\pi/2}^{-\pi/2}\left\{\sin^{2}(3\pi+x)+(\pi+x)^{3}\right\}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{-\frac{3\pi}{2}}^{-\frac{\pi}{2}}\left\{\sin^{2}(3\pi+x)+(\pi+x)^{3}\right\}dx
\displaystyle \text{Put }\pi+x=z
\displaystyle dx=dz
\displaystyle \text{When }x\to-\frac{3\pi}{2},\ z\to-\frac{\pi}{2}
\displaystyle \text{When }x\to-\frac{\pi}{2},\ z\to\frac{\pi}{2}
\displaystyle I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left\{\sin^{2}(2\pi+z)+z^{3}\right\}dz
\displaystyle I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\sin^{2}z+z^{3}\right)dz
\displaystyle I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{1-\cos2z}{2}\,dz+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}z^{3}\,dz
\displaystyle I=\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}dz-\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos2z\,dz+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}z^{3}\,dz
\displaystyle I=\frac{1}{2}\left[z\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}-\frac{1}{2}\left[\frac{\sin2z}{2}\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}+\left[\frac{z^{4}}{4}\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}
\displaystyle I=\frac{1}{2}\left(\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right)-\frac{1}{4}(\sin\pi-\sin(-\pi))+\frac{1}{4}\left(\frac{\pi^{4}}{16}-\frac{\pi^{4}}{16}\right)
\displaystyle I=\frac{1}{2}\pi-\frac{1}{4}(0+0)+\frac{1}{4}(0)
\displaystyle \therefore I=\frac{\pi}{2}

\displaystyle \textbf{Question 33: }~\int_{0}^{2}x\sqrt{2-x}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{2}x\sqrt{2-x}\,dx
\displaystyle I=\int_{0}^{2}(2-x)\sqrt{2-(2-x)}\,dx
\displaystyle I=\int_{0}^{2}(2-x)\sqrt{x}\,dx
\displaystyle I=\int_{0}^{2}(2\sqrt{x}-x\sqrt{x})\,dx
\displaystyle I=\int_{0}^{2}(2x^{1/2}-x^{3/2})\,dx
\displaystyle I=\left[\frac{4}{3}x^{3/2}-\frac{2}{5}x^{5/2}\right]_{0}^{2}
\displaystyle I=\left(\frac{4}{3}\cdot2^{3/2}-\frac{2}{5}\cdot2^{5/2}\right)
\displaystyle I=\frac{8\sqrt{2}}{3}-\frac{8\sqrt{2}}{5}
\displaystyle I=\frac{40\sqrt{2}-24\sqrt{2}}{15}
\displaystyle \therefore I=\frac{16\sqrt{2}}{15}

\displaystyle \textbf{Question 34: }~\int_{0}^{1}\log\left(\frac{1}{x}-1\right)\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{1}\log\left(\frac{1}{x}-1\right)dx
\displaystyle I=\int_{0}^{1}\log\left(\frac{1}{1-x}-1\right)dx
\displaystyle I=\int_{0}^{1}\log\left(\frac{x}{1-x}\right)dx
\displaystyle \text{Adding (1) and (2)}
\displaystyle 2I=\int_{0}^{1}\log\left(\frac{1-x}{x}\right)dx+\int_{0}^{1}\log\left(\frac{x}{1-x}\right)dx
\displaystyle 2I=\int_{0}^{1}\log\left(\frac{1-x}{x}\cdot\frac{x}{1-x}\right)dx
\displaystyle 2I=\int_{0}^{1}\log 1\,dx
\displaystyle 2I=0
\displaystyle \therefore I=0

\displaystyle \textbf{Question 35: }~\int_{-1}^{1}\lvert x\cos(\pi x)\rvert\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{-1}^{1}|x\cos\pi x|\,dx
\displaystyle \text{Consider }f(x)=|x\cos\pi x|
\displaystyle f(-x)=|(-x)\cos\pi(-x)|=|-x\cos\pi x|=|x\cos\pi x|=f(x)
\displaystyle \therefore f(x)\text{ is an even function}
\displaystyle I=2\int_{0}^{1}|x\cos\pi x|\,dx
\displaystyle |x\cos\pi x|=\begin{cases}x\cos\pi x,&0\le x\le\frac12\\-x\cos\pi x,&\frac12<x\le1\end{cases}
\displaystyle I=2\left[\int_{0}^{\frac12}x\cos\pi x\,dx+\int_{\frac12}^{1}(-x\cos\pi x)\,dx\right]
\displaystyle I=2\left[\left(\frac{x\sin\pi x}{\pi}\right)_{0}^{\frac12}-\frac1\pi\int_{0}^{\frac12}\sin\pi x\,dx-\left(\frac{x\sin\pi x}{\pi}\right)_{\frac12}^{1}+\frac1\pi\int_{\frac12}^{1}\sin\pi x\,dx\right]
\displaystyle I=2\left[\frac1{2\pi}-0-\frac1\pi\left(-\frac{\cos\pi x}{\pi}\right)_{0}^{\frac12}-\left(0-\frac1{2\pi}\right)+\frac1\pi\left(-\frac{\cos\pi x}{\pi}\right)_{\frac12}^{1}\right]
\displaystyle I=2\left[\frac1{2\pi}+\frac1{\pi^{2}}(\cos\frac{\pi}{2}-\cos0)+\frac1{2\pi}+\frac1{\pi^{2}}(\cos\pi-\cos\frac{\pi}{2})\right]
\displaystyle I=2\left[\frac1\pi+\frac1{\pi^{2}}(0-1-1+0)\right]
\displaystyle I=2\left(\frac1\pi-\frac2{\pi^{2}}\right)
\displaystyle \therefore I=\frac{2}{\pi}

\displaystyle \textbf{Question 36: }~\int_{0}^{\pi}\left(\frac{x}{1+\sin^{2}x}+\cos^{7}x\right)\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{\pi}\left(\frac{x}{1+\sin^{2}x}+\cos^{7}x\right)dx
\displaystyle I=\int_{0}^{\pi}\left(\frac{\pi-x}{1+\sin^{2}(\pi-x)}+\cos^{7}(\pi-x)\right)dx
\displaystyle I=\int_{0}^{\pi}\left(\frac{\pi-x}{1+\sin^{2}x}-\cos^{7}x\right)dx
\displaystyle \text{Adding (1) and (2) we get}
\displaystyle 2I=\int_{0}^{\pi}\left(\frac{x}{1+\sin^{2}x}+\cos^{7}x+\frac{\pi-x}{1+\sin^{2}x}-\cos^{7}x\right)dx
\displaystyle 2I=\pi\int_{0}^{\pi}\frac{1}{1+\sin^{2}x}\,dx
\displaystyle \text{Dividing numerator and denominator by }\cos^{2}x
\displaystyle 2I=\pi\int_{0}^{\pi}\frac{\sec^{2}x}{\sec^{2}x+\tan^{2}x}\,dx
\displaystyle 2I=\pi\int_{0}^{\pi}\frac{\sec^{2}x}{1+2\tan^{2}x}\,dx
\displaystyle 2I=2\pi\int_{0}^{\frac{\pi}{2}}\frac{\sec^{2}x}{1+2\tan^{2}x}\,dx
\displaystyle \text{Put }\tan x=z
\displaystyle \sec^{2}x\,dx=dz
\displaystyle \text{When }x=0,\ z=0
\displaystyle \text{When }x=\frac{\pi}{2},\ z=\infty
\displaystyle 2I=2\pi\int_{0}^{\infty}\frac{dz}{1+2z^{2}}
\displaystyle 2I=2\pi\left[\frac{1}{\sqrt{2}}\tan^{-1}(\sqrt{2}z)\right]_{0}^{\infty}
\displaystyle I=\frac{\pi}{\sqrt{2}}\left(\tan^{-1}\infty-\tan^{-1}0\right)
\displaystyle I=\frac{\pi}{\sqrt{2}}\left(\frac{\pi}{2}-0\right)
\displaystyle \therefore I=\frac{\pi^{2}}{2\sqrt{2}}

\displaystyle \textbf{Question 37: }~\int_{0}^{\pi}\frac{x}{1+\sin\alpha\,\sin x}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{\pi}\frac{x}{1+\sin\alpha\sin x}\,dx
\displaystyle I=\int_{0}^{\pi}\frac{\pi-x}{1+\sin\alpha\sin(\pi-x)}\,dx
\displaystyle I=\int_{0}^{\pi}\frac{\pi-x}{1+\sin\alpha\sin x}\,dx
\displaystyle I=\int_{0}^{\pi}\frac{\pi}{1+\sin\alpha\sin x}\,dx-\int_{0}^{\pi}\frac{x}{1+\sin\alpha\sin x}\,dx
\displaystyle I=\int_{0}^{\pi}\frac{\pi}{1+\sin\alpha\sin x}\,dx-I
\displaystyle 2I=\int_{0}^{\pi}\frac{\pi}{1+\sin\alpha\sin x}\,dx
\displaystyle 2I=\pi\int_{0}^{\pi}\frac{1}{1+\sin\alpha\sin x}\,dx
\displaystyle \sin x=\frac{2\tan\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}
\displaystyle 2I=\pi\int_{0}^{\pi}\frac{1+\tan^{2}\frac{x}{2}}{1+\tan^{2}\frac{x}{2}+2\sin\alpha\tan\frac{x}{2}}\,dx
\displaystyle I=\frac{\pi}{2}\int_{0}^{\pi}\frac{\sec^{2}\frac{x}{2}}{1+\tan^{2}\frac{x}{2}+2\sin\alpha\tan\frac{x}{2}}\,dx
\displaystyle \text{Let }\tan\frac{x}{2}=t
\displaystyle \sec^{2}\frac{x}{2}\,dx=2\,dt
\displaystyle \text{When }x=0,\ t=0
\displaystyle \text{When }x=\pi,\ t=\infty
\displaystyle I=\frac{\pi}{2}\int_{0}^{\infty}\frac{2\,dt}{t^{2}+2t\sin\alpha+1}
\displaystyle I=\pi\int_{0}^{\infty}\frac{dt}{(t+\sin\alpha)^{2}+\cos^{2}\alpha}
\displaystyle I=\frac{\pi}{\cos\alpha}\left[\tan^{-1}\left(\frac{t+\sin\alpha}{\cos\alpha}\right)\right]_{0}^{\infty}
\displaystyle I=\frac{\pi}{\cos\alpha}\left(\tan^{-1}\infty-\tan^{-1}(\tan\alpha)\right)
\displaystyle I=\frac{\pi}{\cos\alpha}\left(\frac{\pi}{2}-\alpha\right)

\displaystyle \textbf{Question 38: }~\int_{0}^{2\pi}\sin^{100}x\,\cos^{101}x\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{2\pi}\sin^{100}x\cos^{101}x\,dx
\displaystyle \text{Suppose }f(x)=\sin^{100}x\cos^{101}x
\displaystyle f(2\pi-x)=\sin^{100}(2\pi-x)\cos^{101}(2\pi-x)
\displaystyle f(2\pi-x)=(-\sin x)^{100}(\cos x)^{101}=\sin^{100}x\cos^{101}x=f(x)
\displaystyle \therefore I=2\int_{0}^{\pi}\sin^{100}x\cos^{101}x\,dx
\displaystyle f(\pi-x)=\sin^{100}(\pi-x)\cos^{101}(\pi-x)
\displaystyle f(\pi-x)=(\sin x)^{100}(-\cos x)^{101}=-\sin^{100}x\cos^{101}x=-f(x)
\displaystyle \therefore \int_{0}^{\pi}\sin^{100}x\cos^{101}x\,dx=0
\displaystyle \therefore I=2\times0=0

\displaystyle \textbf{Question 39: }~\int_{0}^{\pi/2}\frac{a\sin x+b\cos x}{\sin x+\cos x}\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{a\sin x+b\cos x}{\sin x+\cos x}\,dx
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{a\sin\left(\frac{\pi}{2}-x\right)+b\cos\left(\frac{\pi}{2}-x\right)}{\sin\left(\frac{\pi}{2}-x\right)+\cos\left(\frac{\pi}{2}-x\right)}\,dx
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{a\cos x+b\sin x}{\cos x+\sin x}\,dx
\displaystyle \text{Adding (1) and (2) we get}
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\left(\frac{a\sin x+b\cos x}{\sin x+\cos x}+\frac{a\cos x+b\sin x}{\sin x+\cos x}\right)dx
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\frac{a\sin x+b\cos x+a\cos x+b\sin x}{\sin x+\cos x}\,dx
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\frac{(a+b)\sin x+(a+b)\cos x}{\sin x+\cos x}\,dx
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}(a+b)\,dx
\displaystyle 2I=(a+b)\left[x\right]_{0}^{\frac{\pi}{2}}
\displaystyle 2I=(a+b)\left(\frac{\pi}{2}-0\right)
\displaystyle 2I=\frac{\pi}{2}(a+b)
\displaystyle \therefore I=\frac{\pi}{4}(a+b)

\displaystyle \textbf{Question 40: }~\int_{0}^{3/2}\lvert x\cos(\pi x)\rvert\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{2a}f(x)\,dx
\displaystyle I=\int_{0}^{a}f(x)\,dx+\int_{a}^{2a}f(x)\,dx
\displaystyle I=\int_{0}^{a}f(x)\,dx+I_{1}
\displaystyle \text{where }I_{1}=\int_{a}^{2a}f(x)\,dx
\displaystyle \text{Let }x=2a-t
\displaystyle dx=-dt
\displaystyle \text{When }t=a,\ x=a
\displaystyle \text{When }t=2a,\ x=0
\displaystyle I_{1}=\int_{a}^{0}f(2a-t)(-dt)
\displaystyle I_{1}=\int_{0}^{a}f(2a-t)\,dt
\displaystyle I_{1}=\int_{0}^{a}f(2a-x)\,dx
\displaystyle I=\int_{0}^{a}f(x)\,dx+\int_{0}^{a}f(2a-x)\,dx
\displaystyle I=\int_{0}^{a}f(x)\,dx+\int_{0}^{a}f(x)\,dx
\displaystyle I=2\int_{0}^{a}f(x)\,dx
\displaystyle \text{Hence proved}

\displaystyle \textbf{Question 41: }~\int_{0}^{1}\lvert x\sin(\pi x)\rvert\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{For }0<x<1,\ x>0\text{ and }\sin\pi x>0
\displaystyle \therefore x\sin\pi x>0
\displaystyle \int_{0}^{1}|x\sin\pi x|\,dx=\int_{0}^{1}x\sin\pi x\,dx
\displaystyle \text{Let }I=\int_{0}^{1}x\sin\pi x\,dx
\displaystyle I=x\int\sin\pi x\,dx-\int\left(\frac{d}{dx}x\right)\left(\int\sin\pi x\,dx\right)dx
\displaystyle I=x\left(-\frac{\cos\pi x}{\pi}\right)-\int\left(-\frac{\cos\pi x}{\pi}\right)dx
\displaystyle I=-\frac{x\cos\pi x}{\pi}+\frac{\sin\pi x}{\pi^{2}}
\displaystyle \text{Applying the limits, we get}
\displaystyle \int_{0}^{1}|x\sin\pi x|\,dx=\left(-\frac{x\cos\pi x}{\pi}+\frac{\sin\pi x}{\pi^{2}}\right)_{0}^{1}
\displaystyle =\left(-\frac{\cos\pi}{\pi}+\frac{\sin\pi}{\pi^{2}}\right)-(0+0)
\displaystyle =\frac{1}{\pi}

\displaystyle \textbf{Question 42: }~\int_{0}^{3/2}\lvert x\sin(\pi x)\rvert\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{For }0<x<1,\ x>0\text{ and }\sin\pi x>0\Rightarrow x\sin\pi x>0
\displaystyle \text{For }1<x<\frac{3}{2},\ x>0\text{ and }\sin\pi x<0\Rightarrow x\sin\pi x<0
\displaystyle \therefore \int_{0}^{\frac{3}{2}}|x\sin\pi x|\,dx=\int_{0}^{1}x\sin\pi x\,dx-\int_{1}^{\frac{3}{2}}x\sin\pi x\,dx
\displaystyle \text{Let }I=\int x\sin\pi x\,dx
\displaystyle I=x\int\sin\pi x\,dx-\int\left(\frac{d}{dx}x\right)\left(\int\sin\pi x\,dx\right)dx
\displaystyle I=x\left(-\frac{\cos\pi x}{\pi}\right)-\int\left(-\frac{\cos\pi x}{\pi}\right)dx
\displaystyle I=-\frac{x\cos\pi x}{\pi}+\frac{\sin\pi x}{\pi^{2}}
\displaystyle \text{Applying the limits, we get}
\displaystyle \int_{0}^{\frac{3}{2}}|x\sin\pi x|\,dx=\left[-\frac{x\cos\pi x}{\pi}+\frac{\sin\pi x}{\pi^{2}}\right]_{0}^{1}-\left[-\frac{x\cos\pi x}{\pi}+\frac{\sin\pi x}{\pi^{2}}\right]_{1}^{\frac{3}{2}}
\displaystyle =\left(-\frac{\cos\pi}{\pi}+\frac{\sin\pi}{\pi^{2}}\right)-(0+0)-\left(\frac{3}{2}\cdot\frac{\cos\frac{3\pi}{2}}{\pi}+\frac{\sin\frac{3\pi}{2}}{\pi^{2}}+\frac{\cos\pi}{\pi}\right)
\displaystyle =\frac{1}{\pi}-\left(0-\frac{1}{\pi^{2}}-\frac{1}{\pi}\right)
\displaystyle =\frac{2}{\pi}+\frac{1}{\pi^{2}}
\displaystyle =\frac{2\pi+1}{\pi^{2}}

\displaystyle \text{Evaluate the following integrals (1-35):}

\displaystyle \textbf{Question 43: }~\text{If }f\text{ is an integrable function such that }f(2a-x)=f(x),\ \text{then prove that } \\ \int_{0}^{2a}f(x)\,dx=2\int_{0}^{a}f(x)\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int_{0}^{2a} f(x)\,dx
\displaystyle \text{By additive property,}
\displaystyle I=\int_{0}^{a} f(x)\,dx+\int_{a}^{2a} f(x)\,dx
\displaystyle \text{Consider the integral }\int_{a}^{2a} f(x)\,dx
\displaystyle \text{Let } x=2a-t,\ \text{then }dx=-dt
\displaystyle \text{When }x=a,\ t=a\ \text{and when }x=2a,\ t=0
\displaystyle \int_{a}^{2a} f(x)\,dx=-\int_{a}^{0} f(2a-t)\,dt
\displaystyle =\int_{0}^{a} f(2a-t)\,dt
\displaystyle =\int_{0}^{a} f(2a-x)\,dx
\displaystyle \text{Therefore,}
\displaystyle I=\int_{0}^{a} f(x)\,dx+\int_{0}^{a} f(2a-x)\,dx
\displaystyle =\int_{0}^{a} f(x)\,dx+\int_{0}^{a} f(x)\,dx
\displaystyle \text{(Given }\int_{0}^{a} f(x)\,dx=\int_{0}^{a} f(2a-x)\,dx\text{)}
\displaystyle =2\int_{0}^{a} f(x)\,dx
\displaystyle \text{Hence proved.}

\displaystyle \textbf{Question 44: }~\text{If }f(2a-x)=-f(x),\ \text{prove that }\int_{0}^{2a}f(x)\,dx=0.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int_{0}^{2a} f(x)\,dx
\displaystyle \text{Using additive property,}
\displaystyle I=\int_{0}^{a} f(x)\,dx+\int_{a}^{2a} f(x)\,dx
\displaystyle \text{Consider the integral }\int_{a}^{2a} f(x)\,dx
\displaystyle \text{Let } x=2a-t,\ \text{then }dx=-dt
\displaystyle \text{When }x=a,\ t=a\ \text{and when }x=2a,\ t=0
\displaystyle \int_{a}^{2a} f(x)\,dx=-\int_{a}^{0} f(2a-t)\,dt
\displaystyle =\int_{0}^{a} f(2a-t)\,dt
\displaystyle =\int_{0}^{a} f(2a-x)\,dx
\displaystyle \text{Given }f(2a-x)=-f(x)
\displaystyle \int_{a}^{2a} f(x)\,dx=-\int_{0}^{a} f(x)\,dx
\displaystyle \text{Therefore,}
\displaystyle I=\int_{0}^{a} f(x)\,dx-\int_{0}^{a} f(x)\,dx
\displaystyle =0

\displaystyle \textbf{Question 45: }~\text{If }f\text{ is an integrable function, show that}
\displaystyle \text{(i): }\int_{-a}^{a}f(x^2)\,dx=2\int_{0}^{a}f(x^2)\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{-a}^{a} f(x^2)\,dx
\displaystyle \text{Here } g(x)=f(x^2)
\displaystyle g(-x)=f((-x)^2)=f(x^2)=g(x)
\displaystyle \text{i.e., } g(x)\text{ is an even function}
\displaystyle \text{Therefore,}
\displaystyle I=2\int_{0}^{a} f(x^2)\,dx

\displaystyle \text{(ii): }\int_{-a}^{a}x f(x^2)\,dx=0.
\displaystyle \text{Answer:}
\displaystyle I=\int_{-a}^{a} x f(x^2)\,dx
\displaystyle \text{Let } g(x)=x f(x^2)
\displaystyle g(-x)=(-x)f((-x)^2)
\displaystyle =-x f(x^2)
\displaystyle =-g(x)
\displaystyle \text{i.e., } g(x)\text{ is an odd function}
\displaystyle \text{Therefore,}
\displaystyle I=0

\displaystyle \textbf{Question 46: }~\text{If }f(x)\text{ is a continuous function defined on }[0,2a].\ \text{Then, prove that } \\ \int_{0}^{2a}f(x)\,dx=\int_{0}^{a}\{f(x)+f(2a-x)\}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int_{0}^{2a} f(x)\,dx
\displaystyle \text{By additive property,}
\displaystyle I=\int_{0}^{a} f(x)\,dx+\int_{a}^{2a} f(x)\,dx
\displaystyle \text{Consider the integral }\int_{a}^{2a} f(x)\,dx
\displaystyle \text{Let } x=2a-t,\ \text{then }dx=-dt
\displaystyle \text{When }x=a,\ t=a\ \text{and when }x=2a,\ t=0
\displaystyle \int_{a}^{2a} f(x)\,dx=-\int_{a}^{0} f(2a-t)\,dt
\displaystyle =\int_{0}^{a} f(2a-t)\,dt
\displaystyle =\int_{0}^{a} f(2a-x)\,dx
\displaystyle \text{Therefore,}
\displaystyle I=\int_{0}^{a} f(x)\,dx+\int_{0}^{a} f(2a-x)\,dx
\displaystyle =\int_{0}^{a}\{f(x)+f(2a-x)\}\,dx

\displaystyle \textbf{Question 47: }~\text{If }f(a+b-x)=f(x),\ \text{then prove that } \\ \int_{a}^{b}x f(x)\,dx=\left(\frac{a+b}{2}\right)\int_{a}^{b}f(x)\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_{a}^{b} x f(x)\,dx=\int_{a}^{b} (a+b-x)\,f(a+b-x)\,dx
\displaystyle \text{Using }\int_{a}^{b} f(x)\,dx=\int_{a}^{b} f(a+b-x)\,dx
\displaystyle \int_{a}^{b} x f(x)\,dx=\int_{a}^{b} (a+b-x)f(x)\,dx
\displaystyle \int_{a}^{b} x f(x)\,dx=\int_{a}^{b} (a+b)f(x)\,dx-\int_{a}^{b} x f(x)\,dx
\displaystyle 2\int_{a}^{b} x f(x)\,dx=(a+b)\int_{a}^{b} f(x)\,dx
\displaystyle \int_{a}^{b} x f(x)\,dx=\frac{a+b}{2}\int_{a}^{b} f(x)\,dx

\displaystyle \textbf{Question 48: }~\text{If }f(x)\text{ is a continuous function defined on }[-a,a],\ \text{then prove that } \\ \int_{-a}^{a}f(x)\,dx=\int_{0}^{a}\{f(x)+f(-x)\}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int_{-a}^{a} f(x)\,dx
\displaystyle \text{By additive property,}
\displaystyle I=\int_{-a}^{0} f(x)\,dx+\int_{0}^{a} f(x)\,dx
\displaystyle \text{Let } x=-t,\ \text{then }dx=-dt
\displaystyle \text{When }x=-a,\ t=a\ \text{and when }x=0,\ t=0
\displaystyle \int_{-a}^{0} f(x)\,dx=-\int_{a}^{0} f(-t)\,dt
\displaystyle =\int_{0}^{a} f(-t)\,dt
\displaystyle =\int_{0}^{a} f(-x)\,dx
\displaystyle \text{Therefore,}
\displaystyle I=\int_{0}^{a} f(-x)\,dx+\int_{0}^{a} f(x)\,dx
\displaystyle =\int_{0}^{a}\{f(x)+f(-x)\}\,dx

\displaystyle \textbf{Question 49: }~\text{Prove that: }\int_{0}^{\pi}x f(\sin x)\,dx=\frac{\pi}{2}\int_{0}^{\pi}f(\sin x)\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_{0}^{\pi} x f(\sin x)\,dx=\int_{0}^{\pi} (\pi-x)\,f(\sin(\pi-x))\,dx
\displaystyle \text{Using }\int_{0}^{a} f(x)\,dx=\int_{0}^{a} f(a-x)\,dx
\displaystyle \int_{0}^{\pi} x f(\sin x)\,dx=\int_{0}^{\pi} (\pi-x)\,f(\sin x)\,dx
\displaystyle \int_{0}^{\pi} x f(\sin x)\,dx=\pi\int_{0}^{\pi} f(\sin x)\,dx-\int_{0}^{\pi} x f(\sin x)\,dx
\displaystyle 2\int_{0}^{\pi} x f(\sin x)\,dx=\pi\int_{0}^{\pi} f(\sin x)\,dx
\displaystyle \int_{0}^{\pi} x f(\sin x)\,dx=\frac{\pi}{2}\int_{0}^{\pi} f(\sin x)\,dx

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