Answer:
Option A
Explanation:
Let $I= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^{2}\cos x}{1+e^{x}}dx$ ......(i)
$[\because \int_{a}^{b} f(x) dx= \int_{a}^{b} f(a+b-x)dx]$
$\Rightarrow I= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^{2}\cos (-x)}{1+e^{-x}}dx$ .................(ii)
On adding Eqs(i) and (ii) , we get
$2 I= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^{2}\cos x[\frac{1}{1+e^{x}}+\frac{1}{1+e^{-x}}] dx$
=$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^{2}\cos x.(1)dx$
[ $\therefore \int_{-a}^{a}f(x) dx $
$=2\int_{0}^{a} f(x) dx.whenf(-x)=f(x)]$
$\Rightarrow 2I=2\int_{0}^{\frac{\pi}{2}} x^{2}\cos x dx$
Using integration by parts , we get
$2I= 2[x^{2}(\sin x)-(2x)(-\cos x)+(2)(-\sin x)]_{0}^{\frac{\pi}{2}}$
$\Rightarrow 2I=2[\frac{\pi^{2}}{4}-2]$
$\therefore I=\frac{\pi^{2}}{4}-2$
