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JEE Main 20245 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
5 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2024 · 5 April · Morning Shift · Mathematics · Definite Integration

The value of ∫_ -π ^π (2 y(1+ y))/(1+ ^2 y) d y is :

The value of ∫−ππ2y(1+sin⁡y)1+cos⁡2ydy\int_{-\pi}^\pi \frac{2 y(1+\sin y)}{1+\cos ^2 y} d y is :

Options

  1. A2π22 \pi^2
  2. Bπ22\frac{\pi^2}{2}
  3. Cπ2\frac{\pi}{2}
  4. Dπ2\pi^2

Correct answer

D. π2\pi^2

Step-by-step solution

∫−ππ2y(1+sin⁡y)1+cos⁡2ydy\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∫0πy(sin⁡y1+cos⁡2y)dy=0+2.2 \int_0^\pi y\left(\frac{\sin y}{1+\cos ^2 y}\right) d yI=4∫0πysin⁡y1+cos⁡2ydyI=4∫0π(π−y)sin⁡y1+cos⁡2ydy2I=4∫0ππsin⁡y1+cos⁡2ydyI=2π∫0πsin⁡y1+cos⁡2ydy=2π(−tan⁡−1(cos⁡y))0π=−2π[(−π4)−(π4)]=−2π[−2π4]=π2\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}$$
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