Question

$\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{x\sin x}{1+{\cos }^{2}x}dx$

Answer 1

$$\begin{gathered} \mathrm{Let},I={\int }_0^{\mathrm{\pi }}\frac{x\sin x}{1+{\cos }^{2}x}dx...\left(\mathrm{i}\right) \\ ={\int }_0^{\mathrm{\pi }}\frac{\left(\mathrm{\pi }-x\right)\sin \left(\mathrm{\pi }-x\right)}{1+{\cos }^2\left(\mathrm{\pi }-x\right)}dx \\ ={\int }_0^{\mathrm{\pi }}\frac{\left(\mathrm{\pi }-x\right)\sin x}{1+{\cos }^{2}x}dx...\left(\mathrm{ii}\right) \\ \mathrm{Adding}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right) \\ 2I={\int }_0^{\mathrm{\pi }}\left[\frac{x\sin x}{1+{\cos }^{2}x}+\frac{\left(\mathrm{\pi }-x\right)\sin x}{1+{\cos }^{2}x}\right]dx \\ ={\int }_0^{\mathrm{\pi }}\frac{\mathrm{\pi }\sin x}{1+{\cos }^{2}x}dx \\ =\mathrm{\pi }{\left[-{\tan }^{-1}\left(\mathrm{cosx}\right)\right]}_0^{\mathrm{\pi }} \\ =-\mathrm{\pi }\left[{\tan }^{-1}\left(-1\right)-{\tan }^{-1}\left(1\right)\right] \\ =-\mathrm{\pi }\left(-\frac{\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{4}\right) \\ =\frac{{\mathrm{\pi }}^2}{2} \\ \mathrm{Hence},I=\frac{{\mathrm{\pi }}^2}{4} \end{gathered}$$