Question

Evaluate: ${\int }_0^{\pi }\frac{x\sin x}{1+{\cos }^{2}x}dx$

Answer 1

$\begin{array}{c}\text{Let}I={\int }_0^{\pi }\frac{x\sin x}{1+{\cos }^{2}x}dx\Rightarrow I={\int }_0^{\pi }\frac{(\pi -x)\sin (\pi -x)dx}{1+{\cos }^2(\pi -x)}\\ ={\int }_0^{\pi }\frac{(\pi -x)\sin xdx}{1+{\cos }^{2}x}=\pi {\int }_0^{\pi }\frac{\sin xdx}{1+{\cos }^{2}x}-I\\ \text{or}2I=\pi {\int }_0^{\pi }\frac{\sin xdx}{1+{\cos }^{2}x}\text{or}I=\frac{\pi }{2}{\int }_0^{\pi }\frac{\sin xdx}{1+{\cos }^{2}x}\end{array}$

Put $\cos x=t$ so that $-\sin xdx=dt$

The limits are, when $x=0,t=1$ and $x=\pi ,t=-1,$ we get

$\begin{array}{c}I=\frac{-\pi }{2}{\int }_1^{-1}\frac{dt}{1+t^2}=\pi {\int }_0^1\frac{dt}{1+t^2}\\ [\because {\int }_a^{-a}f(x)dx=-{\int }_{-a}^{a}f(x)dx\text{and}=-2{\int }_0^{2a}f(x)dx,\text{if f(x) is even function}].\\ I=\pi |{\tan }^{-1}t{|}_0^1=\pi [{\tan }^{-1}1-{\tan }^{-1}0]=\pi [\frac{\pi }{4}-0]=\frac{{\pi }^2}{4}\end{array}$