What's the integral of sin(x)cos^2(x)?

The integral of sin(x)cos^2(x) is (cos^3(x))/3 - (cos(x)sin(x))/2 + C.

To solve this integral, we can use the substitution u = cos(x). Then, du/dx = -sin(x) and dx = -du/sin(x). Substituting these into the integral, we get:

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

Substituting back u = cos(x), we get:

-(cos^3(x))/3 + C

However, we're not done yet. We need to use the identity cos^2(x) + sin^2(x) = 1 to simplify the answer. Rearranging, we get sin^2(x) = 1 - cos^2(x). Substituting this into the original integral, we get:

∫sin(x)cos^2(x) dx = ∫sin(x)(1 - sin^2(x)) dx

Using the substitution u = sin(x), we get:

∫u(1 - u^2) du = (u^2)/2 - (u^4)/4 + C

Substituting back u = sin(x), we get:

(sin^2(x))/2 - (sin^4(x))/4 + C

Using the identity cos^2(x) = 1 - sin^2(x), we can simplify this to:

(cos^3(x))/3 - (cos(x)sin(x))/2 + C

Therefore, the integral of sin(x)cos^2(x) is (cos^3(x))/3 - (cos(x)sin(x))/2 + C.