Evaluate the indefinite integral? Integral (4sin^(2)x cos x dx)

Recognize that the derivative of sine is cosine, so let u = sinx and the cosine will cancel out because du = cosx dx. That means dx = du/cosx

Int[4sin²x cosx dx]
Int[4 u² cosx du/cosx]
Int[4 u² du]
4/3 u³ + c
4/3 sin³x + c

Take the 4 out of the integral, so we have 4 ∫sin²x cos x dx

Then differentiate sin³x by the chain rule. You get 3 sin² x cos x
So the remaining integral is ⅓ sin³ x.
Multiply that by the 4 we took out and the answer is 4/3 sin³ x + c.

u= sinx

du= cosx dx

I= int 4 u^2 du = (4/3) u^3 + C = (4/3) (sin(x))^3 +C