Question

${\int }_0^{\pi }\frac{xsinx}{1+sinx}dx$

Answer 1

Given,

${\int }_0^{\pi }\frac{x\sin \left(x\right)}{1+\sin \left(x\right)}dx$

$=\int x\sin \left(x\right)\frac{1}{1+\sin \left(x\right)}dx$

integration by parts,

$=x(\frac{2}{\tan \left(\frac{x}{2}\right)+1}+x)-\int \frac{2}{\tan \left(\frac{x}{2}\right)+1}+xdx$

$=\frac{2x}{\tan \left(\frac{x}{2}\right)+1}+x^2-2\arctan \left(\tan \left(\frac{x}{2}\right)\right)-2\ln |1+\tan \left(\frac{x}{2}\right)|+\ln \left({\sec }^2\left(\frac{x}{2}\right)\right)-\frac{x^2}{2}+c$

upon applying the limits, we get,

$=\frac{1}{2}(\pi -2)\pi -0$

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