JEE Main 2025 — Definite Integration Question with Solution

$\begin{aligned} & I=\int_0^{\frac{\pi}{2}} \frac{(\sin x)^{\frac{3}{2}} d x}{(\sin x)^{\frac{3}{2}} x+(\cos x)^{\frac{3}{2}}}=\int_0^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}}\left(\frac{\pi}{2}-x\right) d x}{\sin ^{\frac{3}{2}}\left(\frac{\pi}{2}-x\right)+\cos ^{\frac{3}{2}}\left(\frac{\pi}{2}-x\right)} \\ & \Rightarrow \text { Adding } 2 l=\int_0^{\frac{\pi}{2}} \frac{(\sin x)^{\frac{3}{2}}+(\cos x)^{\frac{3}{2}}}{(\sin x)^{\frac{3}{2}}+(\cos x)^{\frac{3}{2}}} d x=\frac{\pi}{2} \\ & I_0=\int_0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x=\int_0^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2}-x\right) \sin x \cos x}{(\sin x)^4+(\cos x)^4} d x \end{aligned}$ Adding, $2I_0={\int }_0^{\frac{\pi }{2}}\frac{\frac{\pi }{2}(\sin x)\cos x}{\left({\sin }^{4}x+{\cos }^{4}x\right)}dx$ $\Rightarrow I_0=\frac{\pi }{4}{\int }_0^{\frac{\pi }{2}}\frac{\tan x\left({\sec }^{2}x\right)dx}{1+{\tan }^{4}x}$ put ${\tan }^{2}x=t\Rightarrow 2\tan x{\sec }^{2}xdx=dt$ $\begin{aligned} & \Rightarrow I_0=\frac{\pi}{4} \int_0^{\infty} \frac{\frac{d t}{2}}{\left(1+t^2\right)}=\left.\frac{\pi}{8}\left(\tan ^{-1} t\right)\right|_0 ^{\infty}=\frac{\pi}{8}\left(\frac{\pi}{2}-0\right) \\ & \Rightarrow I_0=\frac{\pi^2}{16} \end{aligned}$