JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Definite Integrals

Split $I=K+J$.

For $K={\int }_{-\pi /2}^{\pi /2}\frac{x^2\cos x}{1+{\pi }^x}dx$: since $x^2\cos x$ is even, replacing $x\to -x$ gives $K=\int \frac{x^2\cos x}{1+{\pi }^{-x}}dx$ as well. Adding the two forms and using $\frac{1}{1+{\pi }^x}+\frac{1}{1+{\pi }^{-x}}=1$:


Integrating by parts, ${\int }_0^{\pi /2}{x}^2\cos xdx=\frac{{\pi }^2}{4}-2$, so $K=\frac{{\pi }^2}{4}-2$.

For $J={\int }_{-\pi /2}^{\pi /2}\frac{1+{\sin }^{2}x}{1+e^{\sin (x^{2023})}}dx$: since $x^{2023}$ is odd, $\sin (x^{2023})$ is odd in $x$, while $1+{\sin }^{2}x$ is even. Using the same pairing trick with $\frac{1}{1+e^h}+\frac{1}{1+e^{-h}}=1$:




Matching with $I=\frac{\pi }{4}(\pi +a)-2=\frac{{\pi }^2}{4}+\frac{a\pi }{4}-2$ gives $\frac{a\pi }{4}=\frac{3\pi }{4}⟹a=3$.

Key confirmed: A.