The value of $\int_{-\pi}^\pi \frac{2 y(1+\sin y)}{1+\cos ^2 y} d y$ is :

The value of $\int_{-\pi}^\pi \frac{2 y(1+\sin y)}{1+\cos ^2 y} d y$ is :
  1. $2 \pi^2$
  2. $\frac{\pi^2}{2}$
  3. $\frac{\pi}{2}$
  4. $\pi^2$

Solution

$\int_{-\pi}^\pi \frac{2 y(1+\sin y)}{1+\cos ^2 y} d y$ \(=\underset{\text{(Odd)}}{\int_{-\pi}^\pi \frac{2 y}{1+\cos ^2 y} d y}+\underset{\text{(Even)}}{\int_{-\pi}^\pi \frac{2 y \sin y}{1+\cos ^2 y} d y}\)
$=0+2.2 \int_0^\pi y\left(\frac{\sin y}{1+\cos ^2 y}\right) d y$ $\begin{aligned} & I=4 \int_0^\pi \frac{y \sin y}{1+\cos ^2 y} d y \\ & I=4 \int_0^\pi \frac{(\pi-y) \sin y}{1+\cos ^2 y} d y \\ & 2 I=4 \int_0^\pi \frac{\pi \sin y}{1+\cos ^2 y} d y \\ & I=2 \pi \int_0^\pi \frac{\sin y}{1+\cos ^2 y} d y \\ & =2 \pi\left(-\tan ^{-1}(\cos y)\right)_0^\pi \\ & =-2 \pi\left[\left(-\frac{\pi}{4}\right)-\left(\frac{\pi}{4}\right)\right] \\ & =-2 \pi\left[-\frac{2 \pi}{4}\right]=\pi^2\end{aligned}$

Asked in: JEE Main 2024 (05 Apr Shift 1)

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