JEE Main 2024 — 1 February, Shift 1 — Question 2

$\int_{0}^{\frac{\pi}{4}} \frac{x d x}{\sin ^{4}(2 x)+\cos ^{4}(2 x)}$ Let $2 x=t$ then $d x=\frac{1}{2} d t$

$I=\frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{t d t}{\sin ^{4} t+\cos ^{4} t}$

$I=\frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2}-t\right) d t}{\sin ^{4}\left(\frac{\pi}{2}-t\right)+\cos ^{4}\left(\frac{\pi}{2}-t\right)}$ $I=\frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{\frac{\pi}{2} d t}{\sin ^{4} t+\cos ^{4} t}-I$

$2 I=\frac{\pi}{8} \int_{0}^{\frac{\pi}{2}} \frac{d t}{\sin ^{4} t+\cos ^{4} t}$

$2 I=\frac{\pi}{8} \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{4} t d t}{\tan ^{4} t+1}$

Let tant $=\mathrm{y}$ then $\sec ^{2} \mathrm{t} \mathrm{dt}=\mathrm{dy}$

$2 I=\frac{\pi}{8} \int_{0}^{\infty} \frac{\left(1+y^{2}\right) d y}{1+y^{4}}$ $=\frac{\pi}{16} \int_{0}^{\infty} \frac{1+\frac{1}{y^{2}}}{y^{2}+\frac{1}{y^{2}}} \mathrm{dy}$ Put $y-\frac{1}{y}=p$

$\mathrm{I}=\frac{\pi}{16} \int_{-\infty}^{\infty} \frac{\mathrm{dp}}{\mathrm{p}^{2}+(\sqrt{2})^{2}}$

$=\frac{\pi}{16 \sqrt{2}}\left[\tan ^{-1}\left(\frac{\mathrm{p}}{\sqrt{2}}\right)\right]_{-\infty}^{\infty}$

$I=\frac{\pi^{2}}{16 \sqrt{2}}$

$I=\frac{\sqrt{2} \pi^{2}}{32}$