Integrate It Man!

Relevant wiki: Integration Tricks

$\begin{aligned} I & = \int_{-\frac \pi 2}^\frac \pi 2 \frac {x^2\cos x}{1+e^x} dx & \small \color{#3D99F6} \text{Using the identity: } \int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_{-\frac \pi 2}^\frac \pi 2 \frac {x^2\cos x}{1+e^x} + \frac {(-x)^2\cos (-x)}{1+e^{-x}} dx \\ & = \frac 12 \int_{-\frac \pi 2}^\frac \pi 2 \frac {x^2\cos x}{1+e^x} + \frac {e^x x^2\cos x}{e^x+1} dx \\ & = \frac 12 \int_{-\frac \pi 2}^\frac \pi 2 x^2\cos x \ dx & \small \color{#3D99F6} \text{By integration by parts} \\ & = \frac 12 x^2 \sin x \ \bigg|_{-\frac \pi 2}^\frac \pi 2 - \frac 12 \int_{-\frac \pi 2}^\frac \pi 2 2x \sin x \ dx & \small \color{#3D99F6} \text{By integration by parts again} \\ & = \frac {\pi^2}4 + x \cos x \ \bigg|_{-\frac \pi 2}^\frac \pi 2 - \int_{-\frac \pi 2}^\frac \pi 2 \cos x \ dx \\ & = \frac {\pi^2}4 - \sin x \ \bigg|_{-\frac \pi 2}^\frac \pi 2 \\ & = \frac {\pi^2}4 - 2 \end{aligned}$

$\implies a+b+n = 4+2+2 = \boxed{8}$