The value of $\int\limits_{-\pi / 2}^{\pi / 2} \frac{x^2 \cos x}{1+e^x} d x$ is equal to

$\frac{\pi^2}{4}-2$

Let $I=\int\limits_{-\pi / 2}^{\pi / 2} \frac{x^2 \cos x}{1+e^x} d x$

We know that $\int\limits_{-a}^a f(x) d x=\int\limits_0^a\{f(x)+f(-x)\} d x$

∴  $I=\int\limits_0^{\pi / 2}\left\{\frac{x^2 \cos x}{1+e^x}+\frac{(-x)^2 \cos (-x)}{1+e^{-x}}\right\} d x$

$\Rightarrow I=\int\limits_0^{\pi / 2}\left\{\frac{x^2 \cos x}{1+e^x}+\frac{x^2 \cos x}{1+e^x} e^x\right\} d x$

$\Rightarrow I=\int\limits_0^{\pi / 2} x^2 \cos x d x=\left[x^2 \sin x-2(-x \cos x+\sin x)\right]_0^{\pi / 2}$

$\Rightarrow I=\frac{\pi^2}{4}-2$