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JEE Main 202124 FebruaryMorning Shift
Exam
JEE Main
Year
2021
Date
24 February
Shift
Morning Shift
Subject
Mathematics
Chapter
Indefinite Integration

JEE Main 2021 · 24 February · Morning Shift · Mathematics · Indefinite Integration

If ∫ x - x 8 - 2x dx = a ^ - 1 (( x + x)/(b)) + c, where c is a...

If ∫cos⁡x−sin⁡x8−sin⁡2xdx=asin⁡−1(sin⁡x+cos⁡xb)+c,∫ \dfrac{\cos x - \sin x}{\sqrt{8 - \sin 2x}}dx = a\sin^{- 1}\left(\dfrac{\sin x + \cos x}{b}\right) + c, where cc is a constant of integration, then the ordered pair \left(a, b\right) is equal to:

Options

  1. A(1,−3)\left(1, - 3\right)
  2. B(3,1)\left(3, 1\right)
  3. C(−1,3)\left(- 1, 3\right)
  4. D(1,3)\left(1, 3\right)

Correct answer

D. (1,3)\left(1, 3\right)

Step-by-step solution

∫cos⁡x−sin⁡x8−sin⁡2xdx∫ \dfrac{\cos x - \sin x}{\sqrt{8 - \sin 2x}}dx

=∫cos⁡x−sin⁡x9−(sin⁡x+cos⁡x)2dx= ∫ \dfrac{\cos x - \sin x}{\sqrt{9 - {\left(\sin x + \cos x\right)}^{2}}}dx

Let sin⁡x+cos⁡x=t\sin x + \cos x = t

∫dt9−t2=sin⁡−1t3+c∫ \dfrac{dt}{\sqrt{9 - {t}^{2}}} = \sin^{- 1}\dfrac{t}{3} + c

=sin⁡−1(sin⁡x+cos⁡x3)+c= \sin^{- 1}\left(\dfrac{\sin x + \cos x}{3}\right) + c

So a=1,b=3a = 1, b = 3

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