JEE Main 2025 — 28 January, Morning Shift — Question 13

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96 x^{2} \cos ^{2} x}{\left(1+e^{x}\right)} d x$ (Apply King Property)

$\int_{0}^{\frac{\pi}{2}} 96 x^{2} \cos ^{2} x=48 \int_{0}^{\frac{\pi}{2}} x^{2}(1+\cos 2 x) d x$

$48\left[\left(\frac{x^{3}}{3}\right)_{0}^{\pi / 2}+\int_{0}^{\frac{\pi}{2}} \underset{\mathrm{I}}{x^{2}} \underset{\mathrm{II}}{\cos 2 x d x}\right]$

$\Rightarrow$ On solving

$\pi\left(2 \pi^{2}-12\right)$

$\Rightarrow \alpha=2, \beta=-12$

$\Rightarrow(\alpha+\beta)^{2}=100$