Verified Solution • 2025–26 Session moderate

QJEEM21Mae87b: JEE Main 2021 Mathematics - Class 12 JEEM21Mae87b Solution: If , where c is a constant of integration, then the ordered…

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

Multiple Choice Options

  • A (-1, 3)
  • B (1, 3)
  • C (1, -3)
  • D (3, 1)

Step-by-Step Verified Solution

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

Write sin2x = 1 + sin2x - 1

= ∫cos⁡x−sin⁡x8−[1+sin⁡2x−1]dx\int {{{\cos x - \sin x} \over {\sqrt {8 - \left[ {1 + \sin 2x - 1} \right]} }}} dx

= ∫cos⁡x−sin⁡x8−[sin⁡2x+cos⁡2x+2sin⁡xcos⁡x−1]dx\int {{{\cos x - \sin x} \over {\sqrt {8 - \left[ {{{\sin }^2}x + {{\cos }^2}x + 2\sin x\cos x - 1} \right]} }}} dx

= ∫cos⁡x−sin⁡x8−[(sin⁡x+cos⁡x)2−1]dx\int {{{\cos x - \sin x} \over {\sqrt {8 - \left[ {{{\left( {\sin x + \cos x} \right)}^2} - 1} \right]} }}} dx

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

put sin x + cos x = t
⇒\Rightarrow (cos x – sin x) dx = dt

= ∫dt9−(t)2\int {{{dt} \over {\sqrt {9 - {{\left( t \right)}^2}} }}}

= sin⁡−1(t3){\sin ^{ - 1}}\left( {{t \over 3}} \right) + C

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

∴\therefore a = 1 and b = 3

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring