AP Calculus AB and BC

Integral of sin^2 x cos x: Substitution

The integral of sin squared x times cos x is sin cubed x over 3, plus C. The cosine factor is exactly the derivative of sin x, so substituting u equals sin x turns the integrand into u squared.

∫sin⁡2xcos⁡x dx=sin⁡3x3+C\int \sin^{2}x\cos x\,dx = \frac{\sin^{3}x}{3} + C

The odd power is what makes it work

u=sin⁡x,du=cos⁡x dx  ⟹  ∫u2du=sin⁡3x3+Cu = \sin x, \quad du = \cos x\,dx \implies \int u^{2}du = \frac{\sin^{3}x}{3} + C

The strategy for products of sines and cosines is decided by parity: with an ODD power available, peel one factor off to serve as dudu. Only when both powers are even do you need the power reducing identities.

Contrast with the even case

∫sin⁡2x dx\int \sin^{2}x\,dx has no cosine to absorb, so it needs sin⁡2x=1−cos⁡2x2\sin^{2}x = \frac{1-\cos 2x}{2} and gives x2−sin⁡2x4+C\frac{x}{2} - \frac{\sin 2x}{4} + C. Same-looking integrand, completely different method.

Common mistakes

  • Answering sin⁡3xsin⁡x3\frac{\sin^{3}x\sin x}{3} or multiplying antiderivatives.
  • Reaching for the power reducing identity when a spare cosine is available.
  • Substituting u=cos⁡xu = \cos x, which does not match what is present.

Every answer on this page is machine checked

An automated test differentiates the antiderivative above and confirms it returns the integrand. A wrong sign or a missing factor fails the build, so it cannot reach you.

Frequently asked questions

What is the integral of sin^2 x cos x?

It is sin⁡3x3+C\frac{\sin^{3}x}{3} + C.

When do I need the power reducing identity instead?

When both powers are even, so there is no spare factor to serve as dudu.