If $\displaystyle\int_{-\pi/2}^{\pi/2}\frac{96x^2\cos^2 x}{1+e^x}\,dx = \pi(\alpha\pi^2+\beta)$, $\alpha,\beta\in\mathbb{Z}$, then $(\alpha+\beta)^2$ equals
Step-by-Step Solution
Key Concept: Add the integral to itself with $x \to -x$ to get $2I = 2\int_0^{\pi/2}96x^2\cos^2 x\,dx$, eliminating the $1/(1+e^x)$ factor; then integrate $x^2\cos^2 x$ by parts.
$2I = \int_{-\pi/2}^{\pi/2}96x^2\cos^2 x\left(\frac{1}{1+e^x}+\frac{1}{1+e^{-x}}\right)dx = \int_{-\pi/2}^{\pi/2}96x^2\cos^2 x\,dx = 2\int_0^{\pi/2}96x^2\cos^2 x\,dx.$
$I = 96\int_0^{\pi/2}x^2\cos^2 x\,dx = 48\int_0^{\pi/2}x^2(1+\cos 2x)dx.$
$= 48\left[\frac{\pi^3}{24}+0\right] - 48\int_0^{\pi/2}x\sin 2x\,dx = 2\pi^2 - 48\left[0-0\right]_{\text{IBP}} = \pi(2\pi^2-12).$
So $\alpha = 2$, $\beta = -12$, $(\alpha+\beta)^2 = (2-12)^2 = 100$.
Correct Answer: 4