integral of sinx*(cosx)^2?

Answer 1

I = ∫ (sin x) (cos ² x) dx
Let u = cos x
du = - sin x dx
I = - ∫ u ² du
I = - u ³ / 3 + C
I = - cos ³ x / 3 + C

Answer 2

∫sinx * (cosx)^2 dx

let cosx = y

-sinx(dx) = dy

-∫y^2 dy = -y^3/3 + c

= -(1/3)(cosx)^3 + c

Answer 3

(-1/3)[cosx]^3 + C
recognize it as a function to a power multiplied by its derivative