What's the integral of cos^3(x)sin^2(x)?

The integral of cos^3(x)sin^2(x) is -1/3cos^4(x) + 1/5cos^2(x) + C.

To solve this integral, we can use the substitution u = cos(x). Then, du/dx = -sin(x) and dx = -du/sin(x). We can rewrite the integral as:

∫cos^3(x)sin^2(x) dx = ∫cos^3(x)(1-cos^2(x)) dx
= ∫u^3(1-u^2) (-du/sqrt(1-u^2)) (using the identity sin^2(x) = 1-cos^2(x))
= -∫u^3(1-u^2) du/sqrt(1-u^2)

We can now use partial fractions to simplify the integrand:

-∫u^3(1-u^2) du/sqrt(1-u^2) = -∫(u^2-1) du/sqrt(1-u^2) + ∫u^2 du/sqrt(1-u^2)
= -1/2∫d(1-u^2)^1/2 + 1/2∫d(1-u^2)^1/2 - 1/2∫d(1-u^2)^1/2

Simplifying further, we get:

= -1/2(1-u^2)^1/2 - 1/6(1-u^2)^3/2 + 1/2(1-u^2)^1/2 + C
= -1/3(1-u^2)^3/2 + 1/5(1-u^2)^1/2 + C

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

∫cos^3(x)sin^2(x) dx = -1/3cos^4(x) + 1/5cos^2(x) + C.