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JEE Main 20199 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
9 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Indefinite Integration

JEE Main 2019 · 9 January · Morning Shift · Mathematics · Indefinite Integration

For, x ^ 2 ≠ nπ + 1, n ∈ N (the set of natural numbers), the...

For, x2≠nπ+1,n∈N{x}^{2} \ne n\pi + 1, n \in N (the set of natural numbers), the integral ∫x2sin⁡(x2−1)−sin⁡2(x2−1)2sin⁡(x2−1)+sin⁡2(x2−1)dx∫ x\sqrt{\dfrac{2\sin \left({x}^{2} - 1\right) - \sin 2\left({x}^{2} - 1\right)}{2\sin \left({x}^{2} - 1\right) + \sin 2\left({x}^{2} - 1\right)}}dx, is equal to
(where cc is a constant of integration).

Options

  1. Alog⁡e∣sec⁡(x2−14)∣+c\log_{e}\left|\sec\left(\dfrac{{x}^{2} - 1}{4}\right)\right| + c
  2. Blog⁡e∣12sec⁡2(x2−1)∣+c\log_{e}\left|\dfrac{1}{2}\sec^{2}\left({x}^{2} - 1\right)\right| + c
  3. C12log⁡e∣sec⁡(x2−1)∣+c\dfrac{1}{2}\log_{e}\left|\sec\left({x}^{2} - 1\right)\right| + c
  4. Dlog⁡e∣sec⁡2(x2−12)∣+c\log_{e}\left|\sec^{2}\left(\dfrac{{x}^{2} - 1}{2}\right)\right| + c

Correct answer

D. log⁡e∣sec⁡2(x2−12)∣+c\log_{e}\left|\sec^{2}\left(\dfrac{{x}^{2} - 1}{2}\right)\right| + c

Step-by-step solution

⇒I=∫x2sin⁡(x2−1)−sin⁡2(x2−1)2sin⁡(x2−1)+sin⁡2(x2−1)dx⇒ I = ∫ x\sqrt{\dfrac{2\sin \left({x}^{2} - 1\right) - \sin 2\left({x}^{2} - 1\right)}{2\sin \left({x}^{2} - 1\right) + \sin 2\left({x}^{2} - 1\right)}}dx

Let, x2−1=θ⇒xdx=12dθ{x}^{2} - 1 = \theta ⇒ xdx = \dfrac{1}{2}d\theta

⇒I=12∫2sinθ−sin⁡2θ2sinθ+sin⁡2θdθ⇒ I = \dfrac{1}{2} ∫ \sqrt{\dfrac{2sin\theta - \sin 2\theta}{2sin\theta + \sin 2\theta}}d\theta

=12∫2sinθ−2sinθcosθ2sinθ+2sinθcosθdθ= \dfrac{1}{2} ∫ \sqrt{\dfrac{2sin\theta - 2sin\theta cos\theta}{2sin\theta + 2sin\theta cos\theta}}d\theta

=12∫1−cosθ1+cosθdθ= \dfrac{1}{2} ∫ \sqrt{\dfrac{1 - cos\theta}{1 + cos\theta}}d\theta

=12∫tanθ2dθ= \dfrac{1}{2} ∫ tan\dfrac{\theta}{2}d\theta

=12log⁡e(secθ2)(12)+c= \dfrac{1}{2}\dfrac{\log_{e}\left(sec\dfrac{\theta}{2}\right)}{\left(\dfrac{1}{2}\right)} + c, where cc is the constant of integration.

=log⁡e∣sec⁡2(x2−12)∣+c= \log_{e}\left|\sec^{2}\left(\dfrac{{x}^{2} - 1}{2}\right)\right| + c
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