_-/2^/2(96x^2^2x)/(1+e^x),dx = (^2+), ,Z.

Question

$\displaystyle\int_{-\pi/2}^{\pi/2}\frac{96x^2\cos^2x}{1+e^x}\,dx = \pi(\alpha\pi^2+\beta)$, $\alpha,\beta\in\mathbb{Z}$. Then $(\alpha+\beta)^2$ is:

Accepted Answer

By King's rule (using the $\frac{1}{1+e^x}+\frac{1}{1+e^{-x}}=1$ symmetry over $[-\pi/2,\pi/2]$):

$$\int_{-\pi/2}^{\pi/2}\frac{96x^2\cos^2x}{1+e^x}\,dx = \int_0^{\pi/2}96x^2\cos^2x\,dx = 48\int_0^{\pi/2}x^2(1+\cos2x)\,dx$$

$$= 48\left[\frac{x^3}{3}\right]_0^{\pi/2} + 48\int_0^{\pi/2}x^2\cos2x\,dx$$

Working through integration by parts for the second term and simplifying gives the total as $\pi(2\pi^2-12)$, so $\alpha=2,\,\beta=-12$, and $(\alpha+\beta)^2 = 100$.

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