JEE Main 2024 — Mathematics Calculus

$\displaystyle \int_{-\pi}^\pi \frac{2 y(1+\sin y)}{1+\cos ^2 y} d y$∫−ππ​1+cos2y2y(1+siny)​dy $\displaystyle =\underset{\text{(Odd)}}{\int_{-\pi}^\pi \frac{2 y}{1+\cos ^2 y} d y}+\underset{\text{(Even)}}{\int_{-\pi}^\pi \frac{2 y \sin y}{1+\cos ^2 y} d y}$=(Odd)∫−ππ​1+cos2y2y​dy​+(Even)∫−ππ​1+cos2y2ysiny​dy​

$\displaystyle =0+2.2 \int_0^\pi y\left(\frac{\sin y}{1+\cos ^2 y}\right) d y$=0+2.2∫0π​y(1+cos2ysiny​)dy $\displaystyle \begin{aligned} & I=4 \int_0^\pi \frac{y \sin y}{1+\cos ^2 y} d y \\ & I=4 \int_0^\pi \frac{(\pi-y) \sin y}{1+\cos ^2 y} d y \\ & 2 I=4 \int_0^\pi \frac{\pi \sin y}{1+\cos ^2 y} d y \\ & I=2 \pi \int_0^\pi \frac{\sin y}{1+\cos ^2 y} d y \\ & =2 \pi\left(-\tan ^{-1}(\cos y)\right)_0^\pi \\ & =-2 \pi\left[\left(-\frac{\pi}{4}\right)-\left(\frac{\pi}{4}\right)\right] \\ & =-2 \pi\left[-\frac{2 \pi}{4}\right]=\pi^2\end{aligned}$​I=4∫0π​1+cos2yysiny​dyI=4∫0π​1+cos2y(π−y)siny​dy2I=4∫0π​1+cos2yπsiny​dyI=2π∫0π​1+cos2ysiny​dy=2π(−tan−1(cosy))0π​=−2π[(−4π​)−(4π​)]=−2π[−42π​]=π2​