Question

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

Answer 1

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

Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

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

Add (1) and (2)

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

$\Rightarrow I=-\pi {\int }_0^{\frac{\pi }{2}}\frac{d\left(\cos x\right)}{1+{\cos }^{2}x}=-\pi {\left[{\tan }^{-1}\right(\cos x\left)\right]}_0^{\pi /2}=-\pi (0-\frac{\pi }{4})=\frac{{\pi }^2}{4}$