JEE Main 2025 Jan 28 Shift 1, Mathematics Q13: Some Properties of Definite Integrals

Let $I={\int }_{-\pi /2}^{\pi /2}\frac{96x^2{\cos }^{2}x}{1+e^x}dx$. Apply the king-property substitution $x\to -x$:

$I={\int }_{-\pi /2}^{\pi /2}\frac{96x^2{\cos }^{2}x\cdot e^x}{1+e^x}dx$.

Adding the two forms: $2I={\int }_{-\pi /2}^{\pi /2}96x^2{\cos }^{2}xdx$. The integrand is even, so

$I={\int }_0^{\pi /2}96x^2{\cos }^{2}xdx=48{\int }_0^{\pi /2}{x}^2(1+\cos 2x)dx$.

First piece: $48{\int }_0^{\pi /2}{x}^{2}dx=48\cdot \frac{(\pi /2{)}^3}{3}=48\cdot \frac{{\pi }^3}{24}=2{\pi }^3$.

Second piece: $48{\int }_0^{\pi /2}{x}^2\cos 2xdx$. Integrate by parts twice:

$\int x^2\cos 2xdx=\frac{x^2\sin 2x}{2}-\int x\sin 2xdx=\frac{x^2\sin 2x}{2}+\frac{x\cos 2x}{2}-\frac{\sin 2x}{4}$.

Evaluate from 0 to $\pi /2$: $\left[0+\frac{(\pi /2)(-1)}{2}-0\right]-0=-\frac{\pi }{4}$. Multiply by 48: $-12\pi$.

$\therefore I=2{\pi }^3-12\pi =\pi (2{\pi }^2-12)$. So $\alpha =2$, $\beta =-12$, and $(\alpha +\beta {)}^2=(-10{)}^2=100$.