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JEE Main 20253 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
3 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2025 · 3 April · Evening Shift · Mathematics · Definite Integration

The integral ∫_0^π (8 x d x)/(4 ^2 x+ ^2 x) is equal to

The integral ∫0π8xdx4cos⁡2x+sin⁡2x\int_0^\pi \frac{8 x d x}{4 \cos ^2 x+\sin ^2 x} is equal to

Options

  1. A2π22 \pi^2
  2. B4π24 \pi^2
  3. Cπ2\pi^2
  4. D3π22\frac{3 \pi^2}{2}

Correct answer

A. 2π22 \pi^2

Step-by-step solution

<br/>I=∫0π8xdx4cos⁡2x+sin⁡2xI=∫0π8(π−x)dx4cos⁡2x+sin⁡2x<br/>\begin{aligned}<br/>& I=\int_0^\pi \frac{8 x d x}{4 \cos ^2 x+\sin ^2 x} \\ & I=\int_0^\pi \frac{8(\pi-x) d x}{4 \cos ^2 x+\sin ^2 x}<br/>\end{aligned}
22\mathrm{I}=8π∫0πdx4cos⁡2x+sin⁡2x=8 \pi \int_0^\pi \frac{d x}{4 \cos ^2 x+\sin ^2 x}
22\mathrm{I}=8π×2∫0π/2sec⁡2x4+tan⁡2xdx=8 \pi \times 2 \int_0^{\pi / 2} \frac{\sec ^2 x}{4+\tan ^2 \mathrm{x}} \mathrm{dx}
<br/>I=8π∫0∞dt4+t2=8π×12(tan⁡−1t2]0∞=4π×π2=2π2<br/>\begin{aligned}<br/>& \mathrm{I}=8 \pi \int_0^{\infty} \frac{\mathrm{dt}}{4+\mathrm{t}^2}=8 \pi \times \frac{1}{2}\left(\tan ^{-1} \frac{\mathrm{t}}{2}\right]_0^{\infty} \\ & =4 \pi \times \frac{\pi}{2}=2 \pi^2<br/>\end{aligned}
option (1)
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