The value of the integral  ${\int }_0^{\pi } \frac{x\sin x}{1+{\cos }^{2}x}dx$, is 

Solution:

Let $I={\int }_0^{\pi } \frac{x\sin x}{1+{\cos }^{2}x}dx$

Then,

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

Adding (i) and (ii), we get

$\begin{array}{l}2\pi =\pi {\int }_0^{\pi } \frac{\sin x}{1+{\cos }^{2}x}dx\\ \Rightarrow 2I=-\pi {\int }_1^{-1} \frac{1}{1+t^2}dt,\text{ where }t=\cos x\\ \Rightarrow 2I=-\pi {\left[{\tan }^{-1}t\right]}_1^{-1}=-\pi \left(-\frac{\pi }{4}-\frac{\pi }{4}\right)=\frac{{\pi }^2}{2}\\ \Rightarrow I=\frac{{\pi }^2}{4}\end{array}$