Mathematics - Definite Integration Question with Solution | TestHub

$$\begin{array}{rl}& {\int }_{-\pi }^{\pi }\frac{2x(1+\sin x)}{1+{\cos }^{2}x}dx={\int }_{-\pi }^{\pi }\frac{2x}{1+{\cos }^{2}x}+2{\int }_{-\pi }^{\pi }\frac{x\sin x}{1+{\cos }^{2}x}\\ & =0+4{\int }_0^{\pi }\frac{x\sin xdx}{1+{\cos }^{2}x}I=4{\int }_0^{\pi }\frac{(\pi -x)\sin (\pi -x)}{1+{\cos }^2(\pi -x)}\\ & I=4{\int }_0^{\pi }\frac{(\pi -x)\sin x}{1+{\cos }^{2}x}\Rightarrow I=4\pi {\int }_0^{\pi }\frac{\sin x}{1+{\cos }^{2}x}-4\pi \int \frac{x\sin x}{1+{\cos }^{2}x}\Rightarrow 2I=4\pi {\int }_0^{\pi }\frac{\sin x}{1+{\cos }^{2}x}dx\end{array}$$ put $\cos x=t$ and solve it.