JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Definite Integrals

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

Apply King's rule (replace $x\to -x$): $f(-x)=\frac{96x^2{\cos }^{2}x}{1+e^{-x}}=\frac{96x^2{\cos }^{2}x\cdot e^x}{1+e^x}$

Adding, $f(x)+f(-x)=96x^2{\cos }^{2}x$, and since $I=\int f(x)dx=\int f(-x)dx$ over the symmetric interval,

$2I={\int }_{-\pi /2}^{\pi /2}96x^2{\cos }^{2}xdx=2{\int }_0^{\pi /2}96x^2{\cos }^{2}xdx$ (even integrand)

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

$48{\int }_0^{\pi /2}{x}^{2}dx=48\cdot \frac{(\pi /2{)}^3}{3}=2{\pi }^3$

For ${\int }_0^{\pi /2}{x}^2\cos 2xdx$, integrate by parts twice:

$\int x^2\cos 2xdx=\frac{x^2\sin 2x}{2}|-\int x\sin 2xdx$, and $\frac{x^2\sin 2x}{2}$ vanishes at both $0$ and $\pi /2$

${\int }_0^{\pi /2}x\sin 2xdx={\left[-\frac{x\cos 2x}{2}+\frac{\sin 2x}{4}\right]}_0^{\pi /2}=\frac{\pi }{4}$

So ${\int }_0^{\pi /2}{x}^2\cos 2xdx=-\frac{\pi }{4}$, giving $48\left(-\frac{\pi }{4}\right)=-12\pi$

$I=2{\pi }^3-12\pi =\pi (2{\pi }^2-12)$

Matching the given form $\pi (\alpha {\pi }^2+\beta )$: $\alpha =2,\beta =-12$

$(\alpha +\beta {)}^2=(-10{)}^2=100$