JEE Main 2024 — 5 April, Shift 2 — Question 27

$\quad f(t)=\int_{0}^{\pi} \frac{2 x}{1-\cos ^{2} t \sin ^{2} x} d x$

$=2 \int_{0}^{\pi} \frac{(\pi-x) d x}{1-\cos ^{2} t \sin ^{2} x}$

$2 f(t)=2 \int_{0}^{\pi} \frac{\pi}{1-\cos ^{2} t \sin ^{2} x} d x$ $f(t)=\int_{0}^{\pi} \frac{\pi}{1-\cos ^{2} t \sin ^{2} x} d x$

divide & by $\cos ^{2} \mathrm{x}$

$$\begin{aligned} & \mathrm{f}(\mathrm{t})=\pi \int_{0}^{\pi} \frac{\sec ^{2} \mathrm{xdx}}{\sec ^{2} \mathrm{x}-\cos ^{2} \mathrm{t} \tan ^{2} x} & \mathrm{f}(\mathrm{t})=2 \pi \int_{0}^{\pi / 2} \frac{\sec ^{2} x d x}{\sec ^{2} x-\cos ^{2} t \tan ^{2} x} \end{aligned}$$

$\tan x=z$

$\sec ^{2} x d x=d z$

$\mathrm{f}(\mathrm{t})=2 \pi \int_{0}^{\infty} \frac{\mathrm{dz}}{1+\sin ^{2} \mathrm{t} \cdot \mathrm{z}^{2}}$ $=\frac{\pi^{2}}{\sin t}$

Then $\int_{0}^{\pi / 2} \frac{\pi^{2}}{\mathrm{f}(\mathrm{t})} \mathrm{dt}$ $=\int_{0}^{\pi / 2} \sin t d t$

$=1$