The value of the integral $\int\limits_{-\pi / 2}^{\pi / 2}\left\{x^2+\log \left(\frac{\pi+x}{\pi-x}\right)\right\} \cos x d x$, is
Answer & explanation
Correct answer: option 2
Let $I=\int\limits_{-\pi / 2}^{\pi / 2}\left\{x^2+\log _e\left(\frac{\pi+x}{\pi-x}\right)\right\} \cos x d x$ Then,
$I=\int\limits_{-\pi / 2}^{\pi / 2} x^2 \cos x d x+\int\limits_{-\pi / 2}^{\pi / 2} \log \left(\frac{\pi+x}{\pi-x}\right) \cos x d x$
$\Rightarrow I=2 \int\limits^{\pi / 2} x^2 \cos x d x+0$ [∵ $\log \left(\frac{\pi+x}{\pi-x}\right) \cos x$ is an odd function]
$\Rightarrow I=2\left[x^2 \sin x+2 x \cos x-2 \sin x\right]_0^{\pi / 2}$
$\Rightarrow I=2\left(\frac{\pi^2}{4}-2\right)=\frac{\pi^2}{2}-4$