∫ sin³ x dx (using sin³ x = sin x (1 - cos² x)):
A-cos x + cos³ x/3 + C
Bcos x - cos³ x/3 + C
Csin³ x/3 + C
Dx sin³ x + C
Answer & 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.
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