∫ sin³ x dx (using sin³ x = sin x (1 - cos² x)):
- -cos x + cos³ x/3 + C
- cos x - cos³ x/3 + C
- sin³ x/3 + C
- x sin³ x + C
Show answer and solution
Correct answer: A. -cos x + cos³ x/3 + C
sin³ x = sin x (1 - cos² x).
Let u = cos x, du = -sin x dx.
Integral = -∫ (1 - u²) du = -(u - u³/3) + C = -cos x + cos³ x/3 + C.