Answer:
Option B
Explanation:
$\int_{-\pi /2}^{\pi /2} \left(x^{2}+\log\frac{\pi-x}{\pi +x}\right)\cos x dx$
As $\int_{-a}^{a} f(x) dx=0$
when f(-x) =-f(x)
$\therefore$ $I=\int_{-\pi/2}^{\pi/2} x^{2} \cos x dx+0$
= $=2\int_{0}^{\pi/2}( x^{2} \cos x) dx$
=$2\left\{\left(x^{2} \sin x\right)_{0}^{\pi/2}-\int_{0}^{\pi/2} 2x.\sin x dx\right\}$
=$2\left[\frac{\pi^{2}}{4}-2\left\{- x \cos x)_{0}^{\pi/2}-\int_{0}^{\pi/2} 1.(-\cos x)dx\right\}\right]$
=$2\left[\frac{\pi^{2}}{4}-2(\sin x)_{0}^{\pi/2}\right]$
=$2\left[\frac{\pi^{2}}{4}-2\right]=\left(\frac{\pi^{2}}{2}-4\right)$
