JEE Main 2024 — Definite Integration Question with Solution

\begin{aligned} & I=\int_\limits{-\pi}^\pi \frac{2 y(1+\sin y)}{1+\cos ^2 y} d y \\ & =\int_\limits0^\pi\left(\frac{2 y(1+\sin y)}{1+\cos ^2 y}+\frac{-2 y(1-\sin y)}{1+\cos ^2 y}\right) d y \\ & =\int_\limits0^\pi\left(\frac{2 y+2 y \sin y-2 y+2 y \sin y}{1+\cos ^2 y}\right) d y \end{aligned}

$$I=4{\int }_0^{\pi }\left(\frac{y\sin }{1+{\cos }^{2}y}\right)dy\text{... (1)}$$

\begin{aligned} & I=4 \int_\limits0^\pi\left(\frac{(\pi-y) \sin (\pi-)}{1+\cos ^2(\pi-y)}\right) d y \\ & I=4\left\lfloor\int_\limits0\left(\frac{\pi \sin y}{1+\cos ^2 y}\right) d y-\int_\limits0^\pi \frac{y \sin y}{1+\cos ^2 y} d y\right\rfloor \quad \text{... (2)} \end{aligned}

Adding equation (1) and (2)

\begin{aligned} & 2 I=4 \int_\limits0^\pi\left(\frac{\pi \sin y}{1+\cos ^2 y}\right) d y \\ & I=2 \pi \int_\limits0^\pi \frac{\sin y}{1+\cos ^2 y} d y \\ & =2 \pi \times \frac{\pi}{2} \\ & =\pi^2 \end{aligned}