JEE Main 2024 — 5 April, Shift 1 — Question 19

$\int_{-\pi}^{\pi} \frac{2 y(1+\sin y)}{1+\cos ^{2} y} d y$

$=\int_{-\pi}^{\pi} \frac{2 y}{1+\cos ^{2} y} d y+\int_{-\pi}^{\pi} \frac{2 y \sin y}{1+\cos ^{2} y} d y$

                (Odd)                                            (Even)

$=0+2.2 \int_{0}^{\pi} y\left(\frac{\sin y}{1+\cos ^{2} y}\right) d y$

$I=4 \int_{0}^{\pi} \frac{y \sin y}{1+\cos ^{2} y} d y$

$I=4 \int_{0}^{\pi} \frac{(\pi-y) \sin y}{1+\cos ^{2} y} d y$

$2 I=4 \int_{0}^{\pi} \frac{\pi \sin y}{1+\cos ^{2} y} d y$

$I=2 \pi \int_{0}^{\pi} \frac{\sin y}{1+\cos ^{2} y} d y$

$=2 \pi\left(-\tan ^{-1}(\cos y)\right)_{0}^{\pi}$

$=-2 \pi\left[\left(-\frac{\pi}{4}\right)-\left(\frac{\pi}{4}\right)\right]$

$=-2 \pi\left[-\frac{2 \pi}{4}\right]=\pi^{2}$