JEE Main 2024 — Definite Integration Question with Solution

\begin{aligned} & \int_\limits0^\pi \frac{x^2 \sin x \cdot \cos x}{\sin ^4 x+\cos ^4 x} d x \\ & =\int_\limits0^{\frac{\pi}{2}} \frac{\sin x \cdot \cos x}{\sin ^4 x+\cos ^4 x}\left(x^2-(\pi-x)^2\right) d x \\ & =\int_\limits0^{\frac{\pi}{2}} \frac{\sin x \cdot \cos x\left(2 \pi x-\pi^2\right)}{\sin ^4 x+\cos ^4 x} \\ & =2 \pi \int_\limits0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin ^4 x+\cos 4 x} d x-\pi^2 \int_\limits0^{\frac{\pi}{2}} \frac{\sin x \cos x}{\sin ^4 x+\cos 4 x} d x \\ & =2 \pi \cdot \frac{\pi}{4} \int_\limits0^{\frac{\pi}{2}} \frac{\sin x \cos ^4 x}{\sin ^4 x+\cos ^4 x} d x-\pi^2 \int_\limits0^{\frac{\pi}{2}} \frac{\sin x \cos ^4 x}{\sin ^4 x+\cos ^4 x} d x \end{aligned}

\begin{aligned} & =-\frac{\pi^2}{2} \int_\limits0^{\frac{\pi}{2}} \frac{\sin x \cos x}{\sin ^4 x+\cos ^4 x} d x \\ & =-\frac{\pi^2}{2} \int_\limits0^{\frac{\pi}{2}} \frac{\sin x \cos x d x}{1-2 \sin ^2 x \times \cos ^2 x} \\ & =-\frac{\pi^2}{2} \int_\limits0^{\frac{\pi}{2}} \frac{\sin 2 x}{2-\sin ^2 2 x} d x \\ & =-\frac{\pi^2}{2} \int_\limits0^{\frac{\pi}{2}} \frac{\sin 2 x}{1+\cos ^2 2 x} d x \end{aligned}

Let $$\cos 2\mathrm{x}=\mathrm{t}$$