Question

If $I={\int }_0^{\pi }\frac{x^2\sin x}{(2x-x)(1+{\cos }^{2}x)}dx$ then the value of $\frac{4}{{\pi }^2}I+198$ is

Answer 1

$I={\int }_0^{\pi }\frac{x^2\sin x}{(2x-x)(1+{\cos }^{2}x)}dx={\int }_0^{\pi }\frac{x\sin x}{(1+{\cos }^{2}x)}dx$
Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$I={\int }_0^{\pi }\frac{(\pi -x)\sin (\pi -x)}{(1+{\cos }^2(\pi -x))}dx={\int }_0^{\pi }\frac{(\pi -x)\sin x}{(1+{\cos }^{2}x)}dx2I={\int }_0^{\pi }\frac{\pi \sin x}{(1+{\cos }^{2}x)}dx$

Substituting $t=\cos x\Rightarrow dt=-\sin xdx$

$2I=-\pi {\int }_1^{-1}\frac{1}{(1+t^2)}dt=-{\pi \left[{\tan }^{-1}t\right]}_1^{-1}=\frac{{\pi }^2}{2}$
Therefore,
$\frac{4}{{\pi }^2}I+198=\frac{4}{{\pi }^2}.\frac{{\pi }^2}{4}+198=199$