If ∫ (cos x - sin x)/(√(8 - sin 2x)) dx = a sin ⁻¹ ((sin x + cos x)/b) + c, where c is a constant of integration, then the ordered pair (a, b) is equal to:

Integral Calculus: JEE Main 2021 Mathematics question

JEE Main 2021 · 24 Feb, Morning shift · Q4028

JEE Main 202124 Feb, Morning shiftMathematicsMediumIntegral Calculus369× since 2002

\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 :

JEE Main 2021 · 24 Feb, Morning shift · Q4028

JEE Main 202124 Feb, Morning shiftMathematicsMediumIntegral Calculus369× since 2002

\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 :