Question

$\left(Evaluate\right)-\int {\pi }_0\frac{xsinx}{1+co{s}^2}dx$

Answer 1

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

$I=\frac{x\sin x}{1+{\cos }^{2}x}dx\to \left(1\right)$

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

$I=\frac{(\pi -x)\sin x}{1+{\cos }^{2}x}dx\to \left(2\right)$

Adding one and two

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

$=-[\frac{(x+\pi -x)}{1+{\cos }^{2}x}\sin x-dx]$

$2I=-\left[\pi \frac{\sin x}{1+{\cos }^{2}x}dx\right]$

$I=-\left[\frac{\pi }{2}\frac{\sin x}{1+{\cos }^{2}x}dx\right]$

$\cos x=t,-\sin xdx=dt$

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

$=\frac{\pi }{2}[\tan -t]+C$