JEE Main 2024 — Definite Integration Question with Solution

$f(t)={\int }_0^{\pi }\frac{2x}{1-{\cos }^{2}t{\sin }^{2}x}dx$ .....(1) $=2{\int }_0^{\pi }\frac{(\pi -x)dx}{1-{\cos }^2{\sin }^{2}x}$ .....(2) $\begin{aligned} & 2 f(t)=2 \int_0^\pi \frac{\pi}{1-\cos ^2 \sin ^2 x} d x \\ & f(t)=\int_0^\pi \frac{\pi}{1-\cos ^2 t \sin ^2 x} d x \end{aligned}$ divide \& by ${\cos }^2\mathrm{x}$ $\begin{aligned} & f(t)=\pi \int_0^\pi \frac{\sec ^2 x d x}{\sec ^2 x-\cos ^2 t^2 x} \\ & f(t)=2 \pi \int_0^{\pi / 2} \frac{\sec ^2 x d x}{\sec ^2 x-\cos ^2 t^2 \tan ^2 x} \\ & \tan x=z \\ & \sec ^2 x d x=d z \\ & f(t)=2 \pi \int_0^{\infty} \frac{d z}{1+\sin ^2 t \cdot z^2} \\ & =\frac{\pi^2}{\sin t} \end{aligned}$ Then ${\int }_0^{\pi /2}\frac{{\pi }^2}{f(t)}dt$ $\begin{aligned} & =\int_0^{\pi / 2} \sin t d t \\ & =1 \end{aligned}$