JEE Main 2025 Jan 23 Shift 2, Mathematics Q19: Some Properties of Definite Integrals

First find $I$. Apply the King property $x\mapsto \frac{\pi }{2}-x$ to get $I={\int }_0^{\pi /2}\frac{{\cos }^{3/2}x}{{\sin }^{3/2}x+{\cos }^{3/2}x}dx$. Adding the two expressions:

$2I={\int }_0^{\pi /2}1dx=\frac{\pi }{2}\Rightarrow I=\frac{\pi }{4}$, so $2I=\frac{\pi }{2}$.

Let $J={\int }_0^{\pi /2}\frac{x\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$. Apply King again:

$J={\int }_0^{\pi /2}\frac{\left(\frac{\pi }{2}-x\right)\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx.$

Adding the two: $2J=\frac{\pi }{2}{\int }_0^{\pi /2}\frac{\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$, i.e. $J=\frac{\pi }{4}{\int }_0^{\pi /2}\frac{\tan x{\sec }^{2}x}{{\tan }^{4}x+1}dx$ (after dividing numerator and denominator by ${\cos }^{4}x$).

Substitute $u={\tan }^{2}x$, $du=2\tan x{\sec }^{2}xdx$, so $\tan x{\sec }^{2}xdx=\frac{du}{2}$. As $x:0\to \pi /2$, $u:0\to \infty$.

$J=\frac{\pi }{4}\cdot \frac{1}{2}{\int }_0^{\infty }\frac{du}{u^2+1}=\frac{\pi }{8}\cdot \frac{\pi }{2}=\frac{{\pi }^2}{16}.$