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JEE Main 202621 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2026 · 21 January · Morning Shift · Mathematics · Definite Integration

The value of ∫_ -π / 6 ^ π / 6 ( π+4 x^ 11 1- (|x|+π / 6) ) d x is...

The value of ∫−π/6π/6(π+4x111−sin⁡(∣x∣+π/6))dx\int_{-\pi / 6}^{\pi / 6}\left(\frac{\pi+4 x^{11}}{1-\sin (|x|+\pi / 6)}\right) d x is equal to:

Options

  1. A8π8 \pi
  2. B6π6 \pi
  3. C2π2 \pi
  4. D4π4 \pi

Correct answer

D. 4π4 \pi

Step-by-step solution

Split: ∫−π/6π/6π1−sin⁡(∣x∣+π/6)dx+∫−π/6π/64x111−sin⁡(∣x∣+π/6)dx\int_{-\pi/6}^{\pi/6} \frac{\pi}{1 - \sin(|x| + \pi/6)} dx + \int_{-\pi/6}^{\pi/6} \frac{4x^{11}}{1 - \sin(|x| + \pi/6)} dx.
Second integral = 0 (odd function over symmetric interval).
First integral: 2π∫0π/611−sin⁡(x+π/6)dx2\pi \int_{0}^{\pi/6} \frac{1}{1 - \sin(x + \pi/6)} dx.
Let u=x+π/6u = x + \pi/6: =2π∫π/6π/311−sin⁡udu= 2\pi \int_{\pi/6}^{\pi/3} \frac{1}{1 - \sin u} du.
11−sin⁡u=sec⁡2u+sec⁡utan⁡u\frac{1}{1-\sin u} = \sec^2 u + \sec u \tan u.
=2π[tan⁡u+sec⁡u]π/6π/3=2π[(3+2)−3]=4π= 2\pi [\tan u + \sec u]_{\pi/6}^{\pi/3} = 2\pi[(\sqrt{3}+2) - \sqrt{3}] = 4\pi.
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