Question

The value of integral ${\int }_0^{\pi }\frac{x^2\sin x}{(2x-\pi )(1+{\cos }^{2}x)}dx$ is equal to

• A. $\frac{{\pi }^2}{4}$
• B. $\frac{{\pi }^2}{2}$
• C. $\frac{{\pi }^2}{6}$
• D. none of these

Answer 1

Answer: (A)

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

Let $I={\int }_0^{\pi }\frac{x^2\sin x}{(2x-\pi )(1+{\cos }^{2}x)}dx$ $={\int }_0^{\pi }\frac{{(\pi -x)}^2\sin (\pi -x)}{(2\pi -2x-\pi )(1+{\cos }^2(\pi -x))}dx$
$={\int }_0^{\pi }\frac{({\pi }^2-2\pi x+x^2)\sin x}{(\pi -2x)(1+{\cos }^{2}x)}dx$ $={\int }_0^{\pi }\frac{({\pi }^2-2\pi x)\sin x}{(\pi -2x)(1+{\cos }^{2}x)}dx-I$
$\Rightarrow 2I=\pi {\int }_0^{\pi }\frac{\sin x}{1+{\cos }^{2}x}dx=-\pi {\int }_1^{-1}\frac{dt}{1+t^2}$ [putting $\cos x=t\Rightarrow \sin xdx=-dt$]
$=\pi {\left[{\tan }^{-1}t\right]}_{-1}^1=\pi (\frac{\pi }{4}+\frac{\pi }{4})=\frac{{\pi }^2}{2}$
$\therefore I=\frac{{\pi }^2}{4}$