Question

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

Answer 1

$I={\int }_0^{\pi }\frac{x\sin xdx}{1+{\cos }^{2}x}$---(1)
We know that

${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$

$I={\int }_0^{\pi }\frac{(\pi -x)\sin (\pi -x)dx}{1+{\cos }^2(\pi -x)}={\int }_0^{\pi }\frac{(\pi -x)\sin xdx}{1+{\cos }^{2}x}\Rightarrow {\int }_0^{\pi }\frac{\pi \sin xdx}{1+{\cos }^{2}x}-{\int }_0^{\pi }\frac{x\sin xdx}{1+{\cos }^{2}x}$---- (2)

Adding (1) and (2)
$2I=\pi {\int }_0^{\pi }\frac{\sin xdx}{1+{\cos }^{2}x}$

Put $\cos x=t\Rightarrow \sin xdx=-dt$.

Also when $x=0,t=\cos 0=1$ and $x=\pi \Rightarrow t=\cos \pi =-1$

Therefore, $I={\int }_1^{-1}\frac{-dt}{1+t^2}\Rightarrow \frac{\pi }{2}{\int }_{-1}^1\frac{dt}{1+t^2}$

$=\frac{\pi }{2}[{\tan }^{-1}t{]}_{-1}^1$

$I=\frac{\pi }{2}\left[{\tan }^{-1}\right(1)-{\tan }^{-1}(-1\left)\right]=\frac{\pi }{2}[\frac{\pi }{4}-\frac{-\pi }{4}]$

$I={\left(\frac{\pi }{2}\right)}^2$