$$\begin{gathered} \mathrm{I} = {\int }_0^{\mathrm{\pi }}\frac{\mathrm{xsinx}}{1 + {\cos }^2\mathrm{x}}\mathrm{dx} .............\left(1\right) \\ \mathrm{I} = {\int }_0^{\mathrm{\pi }}\frac{(\mathrm{\pi } - \mathrm{x}) \sin (\mathrm{\pi } - \mathrm{x})}{1 + {\cos }^2 (\mathrm{\pi } - \mathrm{x})}\mathrm{dx} \\ \mathrm{I} = {\int }_0^{\mathrm{\pi }}\frac{(\mathrm{\pi } - \mathrm{x}) \sin \mathrm{x}}{1 + {\cos }^2 \mathrm{x}}\mathrm{dx} \\ \mathrm{I} = {\int }_0^{\mathrm{\pi }}\frac{\mathrm{\pi } \sin \mathrm{x}}{1 + {\cos }^2 \mathrm{x}}\mathrm{dx} - {\int }_0^{\mathrm{\pi }}\frac{\mathrm{xsinx}}{1 + {\cos }^2\mathrm{x}}\mathrm{dx} ..........\left(2\right) \end{gathered}$$

Adding (1) and (2), we get:

$$\begin{gathered} 2\mathrm{I} = {\int }_0^{\mathrm{\pi }}\frac{\mathrm{\pi } \mathrm{sinx}}{1 + {\cos }^2\mathrm{x}} \mathrm{dx} \\ \mathrm{Now}, \mathrm{let} \mathrm{cosx} = \mathrm{t} \Rightarrow -\mathrm{sinx} \mathrm{dx} = \mathrm{dt} \\ \mathrm{When} \mathrm{x} = \mathrm{\pi }, \mathrm{t} = \mathrm{cos\pi } = -1 \\ \mathrm{When} \mathrm{x} = 0, \mathrm{t} = \cos 0 = 1 \\ 2\mathrm{I} = {\int }_1^{-1}\frac{\mathrm{\pi } - \mathrm{dt}}{1 + {\mathrm{t}}^2} \\ 2\mathrm{I} = \mathrm{\pi } {\int }_1^{-1} \left(\frac{1}{1 + {\mathrm{t}}^2}\right) \mathrm{dt} \\ 2\mathrm{I} = \mathrm{\pi } {\left[ {\tan }^{-1}\mathrm{t} \right]}_1^{-1} \\ 2\mathrm{I} = \mathrm{\pi } \left[ {\tan }^{-1} 1 - {\tan }^{-1} -1 \right] \\ 2\mathrm{I} = \mathrm{\pi } \left(\frac{\mathrm{\pi }}{4} - \left(-\frac{\mathrm{\pi }}{4}\right)\right) \\ 2\mathrm{I} = \frac{{\mathrm{\pi }}^2}{2} \\ \therefore \mathrm{I} = \frac{{\mathrm{\pi }}^2}{4} \end{gathered}$$