JEE Main 2024 — Definite Integration Question with Solution

\begin{aligned} & \text { Given, } \int_\limits{-1}^1 \frac{\cos \alpha x}{1+3^x} d x=\frac{2}{\pi} \\ & \begin{aligned} I & =\int_\limits{-1}^1 \frac{\cos \alpha x}{1+3^x} d x \\ \Rightarrow I & =\int_\limits0^1\left(\frac{\cos \alpha x}{1+3^x}+\frac{\cos \alpha x}{1+3^{-x}}\right) d x \\ & =\int_\limits0^1 \cos \alpha x d x \\ & =\left(\frac{\sin \alpha x}{\alpha}\right)_0^1 \\ & =\frac{\sin \alpha}{\alpha} \\ \Rightarrow & \frac{\sin \alpha}{\alpha}=\frac{2}{\pi} \\ \Rightarrow & \alpha=\frac{\pi}{2} \end{aligned} \end{aligned}