NTA Abhyas JEE Main2020MathematicsIndefinite IntegrationPractice

∫ sin 8 x − cos 8 x 1 − 2 sin 2 x cos 2 x dx is equal to (where C is an arbitrary constant)

Options

A1 2 sin2x + C
B− 1 2 sin2x + C
C− 1 2 sinx + C
D− sin 2 x + C

Correct answer

B. − 1 2 sin2x + C

Step-by-step solution

∫ sin 8 x - cos 8 x 1 - 2sin 2 xcos 2 x dx = ∫ sin 4 x - cos 4 x sin 4 x + cos 4 x 1 - 2sin 2 xcos 2 x dx = ∫ sin 2 x - cos 2 x sin 4 x + cos 4 x 1 - 2 sin 2 xcos 2 x dx = ∫ - cos2x dx = - 1 2 sin2x + C

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