Question

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

Answer 1

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

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

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

Adding both $\left(1\right)$ and $\left(2\right)$, we get

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

Let $t=\cos x$ $\cos 0=1$

$dt=-\sin xdx$ $\cos \pi =-1$

So, $2I={\int }_1^{-1}\frac{-\pi dt}{1+t^2}$

$2I=\pi {\int }_{-1}^1\frac{dt}{1+t^2}=\pi {\left[{\tan }^{-1}t\right]}_{-1}^1$

$2I=\pi [\frac{\pi }{4}-(-\frac{\pi }{4})]$

$2I=\frac{{\pi }^2}{2}$

$I=\frac{{\pi }^2}{4}$

$I={\left(\frac{\pi }{2}\right)}^2$.