Question

${\int }_{-\pi }^{\pi }\frac{2x(1+\sin x)}{1+{\cos }^{2}x}dx$.

Answer 1

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

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

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

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

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