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Q.
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 $
IIT JEEIIT JEE 2012Integrals
Solution:
$I=\int\limits_{-\pi / 2}^{\pi / 2}\left[x^{2}+\log \left(\frac{\pi-x}{\pi+x}\right)\right] \cos x d x$
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] $
$=2\left[\frac{\pi^{2}}{4}-2(\sin x)_{0}^{\pi / 2}\right]=2\left[\frac{\pi^{2}}{4}-2\right]=\left(\frac{\pi^{2}}{2}-4\right)$