Question

The value of the integral $\int\limits ^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \bigg ( x^2 + \log \frac{\pi - x }{ \pi + x } \bigg ) \, \cos \, x \, dx$

• A. $0$
• B. $\frac{\pi^2 }{2} - 4$
• C. $\frac{\pi^2 }{2} + 4$
• D. $\frac{ \pi ^2 }{ 2 }$

Answer 1

Answer: (B)

$I=\int\limits_{-\pi / 2}^{\pi / 2}\left[x^{2}+\log \left(\frac{\pi-x}{\pi+x}\right)\right] \cos x d x$
As, $\int\limits_{-a}^{a} f(x) d x=0$, when $f(-x)=-f(x)$
$\therefore I =\int\limits_{-\pi / 2}^{\pi / 2} x^{2} \cos x d x+0=2 \int\limits_{0}^{\pi / 2}\left(x^{2} \cos x\right) d x$
$=2\left\{\left(x^{2} \sin x\right)_{0}^{\pi / 2}-\int\limits_{0}^{\pi / 2} 2 x \cdot \sin x d x\right\}$
$=2\left[\frac{\pi^{2}}{4}-2\left\{(-x \cdot \cos x)_{0}^{\pi / 2}-\int\limits_{0}^{\pi / 2} 1 \cdot(-\cos x) d x\right\}\right]$
$=2\left[\frac{\pi^{2}}{4}-2(\sin x)_{0}^{\pi / 2}\right]=2\left[\frac{\pi^{2}}{4}-2\right]=\left(\frac{\pi^{2}}{2}-4\right)$