JEE Main 2024 — Definite Integration Question with Solution

\begin{aligned} & f(t)=\int_\limits0^\pi \frac{2 x d x}{1-\cos ^2 t \sin ^2 x} \\ & x \rightarrow \pi-x \end{aligned}

\begin{aligned} & f(t)=\int_\limits0^\pi \frac{2(\pi-x) d x}{1-\cos ^2 t \sin ^2 x}=2 \pi \int_\limits0^\pi \frac{d x}{1-\cos ^2 t \sin ^2 x}-f(t) \\ & \Rightarrow f(t)=\pi \int_\limits0^\pi \frac{d x}{1-\cos ^2 t \sin ^2 x} \\ & =2 \pi \int_\limits0^{\frac{\pi}{2}} \frac{d x}{1-\cos ^2 t \sin ^2 x} \\ & f(t)=2 \pi \int_\limits0^{\frac{\pi}{2}} \frac{\sec ^2 x d x}{\sec ^2 x-\cos ^2 t \tan ^2 x} \\ & I_1=\int \frac{\sec ^2 x d x}{\sec ^2 x-\cos ^2 t \tan ^2 x} \\ & \text { Put } \cos t \tan x=\lambda \Rightarrow \cos t \sec ^2 x d x=d \lambda \\ & I_1=\int \frac{d \lambda}{\cos t \cdot\left(1+\lambda^2 \sec ^2 t-\lambda^2\right)}=\int \frac{d \lambda}{\cos t\left(1+\lambda^2 \tan ^2 t\right)} \\ \end{aligned}

$$\begin{array}{rl}=\frac{1}{\cos t\cdot {\tan }^{2}t}\cdot \int \frac{d\lambda }{{\lambda }^2+{\cos }^{2}t}=& \frac{1}{\cos t{\tan }^{2}t}\\ & \times \frac{1}{\cos t}{\tan }^{-1}(\lambda \tan t)\end{array}$$

\begin{aligned} & =\frac{1}{\sin t} \tan ^{-1}(\sin t \tan x) \\ \Rightarrow & \left.f(t)=\frac{2 \pi}{\sin t} \tan ^{-1}(\sin t \tan x)\right]_0^{\frac{\pi}{2}}=\frac{\pi^2}{\sin t} \\ \Rightarrow & \int_\limits0^{\frac{\pi}{2}} \frac{\pi^2 d t}{f(t)}=\int_\limits0^{\frac{\pi}{2}} \sin t d t=1 \end{aligned}