JEE Main 2025 — 23 January, Evening Shift — Question 19

For I

Apply king ( $\mathrm{P}-5$ ) and add

$2 \mathrm{I}=\int_{0}^{\pi / 2} \mathrm{dx}=\frac{\pi}{2} \Rightarrow \mathrm{I}=\frac{\pi}{4}$

$I_{2}=\int_{0}^{\pi / 2} \frac{x \sin x \cos x}{\sin ^{4} x+\cos ^{4} x} d x$

Apply king and add

$\mathrm{I}_{2}=\frac{\pi}{4} \int_{0}^{\pi / 2} \frac{\tan \mathrm{x} \sec ^{2} \mathrm{xdx}}{\tan ^{4} \mathrm{x}+1}$

put $\tan ^{2} x=t$

$$\begin{aligned} & \frac{\pi}{8} \int_{0}^{\infty} \frac{\mathrm{dt}}{\mathrm{t}^{2}+1}\\ = & \frac{\pi}{8} \cdot \frac{\pi}{2}=\frac{\pi^{2}}{16} \end{aligned}$$