Question

Show that $\underset{0}{\overset{\pi }{\int }}\frac{x\sin x}{1+{\cos }^{2}x}dx=-\frac{{\pi }^2}{4}$

Answer 1

L.H.S

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

We know that

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

Therefore,

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

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

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

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

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

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

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

Let $t=\cos x$

$\frac{dt}{dx}=-\sin x$

$-dt=\sin xdx$

From equation (2), we get

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

$I=-\frac{\pi }{2}{\left[{\tan }^{-1}\left(t\right)\right]}_1^{-1}$

$I=-\frac{\pi }{2}[{\tan }^{-1}(-1)-{\tan }^{-1}\left(1\right)]$

$I=-\frac{\pi }{2}[\frac{3\pi }{4}-\frac{\pi }{4}]$

$I=-\frac{\pi }{2}\left[\frac{\pi }{2}\right]$

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

Hence, proved.