Question

Prove:
${\int }_0^{\pi }\frac{x\sin x}{1+{\cos }^{2}x}dx=\frac{{\pi }^2}{4}$

Answer 1

$I={\int }_0^{\pi }\frac{xsinx}{1+co{s}^{2}x}dx$...(1)

using property $\infty$ definite integral

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

$I={\int }_a^b\frac{(\pi -x)sin(\pi -x)}{1+co{s}^2(\pi -x)}dx$

$I={\int }_a^b\frac{(\pi -x)sinx}{1+co{s}^{2}x}dx$....(2)

adding $\left(1\right)$& $\left(2\right)$, we get

$I+I={\int }_a^b\frac{(\pi -x)sinx}{1+co{s}^{2}x}dx+{\int }_a^b\frac{(\pi -x)sinx}{1+co{s}^{2}x}$

$2I={\int }_a^b\frac{(\pi -Sinx)sinx}{1+co{s}^{2}x}dx$

$I=\frac{\pi }{2}{\int }_0^{\pi }\frac{Sinx}{1+Co{s}^{2}x}dx$

$=\frac{\pi }{2}\cdot 2{\int }_0^{\pi /2}\frac{Sinx}{1+Co{s}^{2}x}dx$

$=-\pi {\int }_0^{\pi /2}\frac{d\left(cosx\right)}{1+Co{s}^{2}x}dx$

$=-\pi \left[ta{n}^{-1}\right(cosx){]}_0^{\pi /2}$

$-\pi [ta{n}^{-1}cos\frac{\pi }{2}-ta{n}^{-1}(cos0\left)\right]$

$-\pi [0-\frac{\pi }{4}]$

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