AP Calculus AB and BC

Integral of cos^2 x sin x: Substitution

The integral of cos squared x times sin x is negative cos cubed x over 3, plus C. Substituting u equal to cos x gives du equal to negative sin x dx, so the sine factor is consumed and the minus sign survives.

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

The odd power supplies du

u=cos⁡x,du=−sin⁡x dx  ⟹  ∫u2(−du)=−cos⁡3x3+Cu = \cos x, \quad du = -\sin x\,dx \implies \int u^{2}\left(-du\right) = -\frac{\cos^{3}x}{3} + C

The single sine factor is exactly what the substitution needs, which is why an odd power somewhere in the product is the signal for this method.

Compare with the sine version

∫sin⁡2xcos⁡x dx=sin⁡3x3+C\int \sin^{2}x\cos x\,dx = \frac{\sin^{3}x}{3} + C has no minus sign, because there the substitution is on sine, whose derivative is positive. Which function you substitute decides the sign.

Common mistakes

  • Losing the minus sign from du=−sin⁡x dxdu = -\sin x\,dx.
  • Reaching for a power reducing identity when a spare factor is available.

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 cos^2 x sin x?

It is −cos⁡3x3+C-\frac{\cos^{3}x}{3} + C.

Why is it negative?

Because u=cos⁡xu = \cos x gives du=−sin⁡x dxdu = -\sin x\,dx, and that minus carries through.