AP Calculus AB and BC

Integral of sin^3 x cos x: Substitution

The integral of sin cubed x times cos x is sin to the fourth x over 4, plus C. Substituting u equal to sin x gives du equal to cos x dx, so the integrand becomes u cubed and the power rule finishes it.

∫sin⁡3xcos⁡x dx=sin⁡4x4+C\int \sin^{3}x\cos x\,dx = \frac{\sin^{4}x}{4} + C

The u du pattern

u=sin⁡x,du=cos⁡x dx  ⟹  ∫u3du=sin⁡4x4+Cu = \sin x, \quad du = \cos x\,dx \implies \int u^{3}du = \frac{\sin^{4}x}{4} + C

The general form ∫sin⁡nxcos⁡x dx=sin⁡n+1xn+1+C\int \sin^{n}x\cos x\,dx = \frac{\sin^{n+1}x}{n+1} + C covers every case with a single spare cosine, for any n≠−1n \neq -1.

Parity picks the method

An odd power of cosine leaves a spare factor for dudu, so substitution works. When BOTH powers are even there is no spare factor and the power reducing identities are needed instead.

Common mistakes

  • Answering sin⁡4xcos⁡2x4\frac{\sin^{4}x\cos^{2}x}{4} or multiplying antiderivatives.
  • Substituting u=cos⁡xu = \cos x, which does not match the spare factor present.

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Frequently asked questions

What is the integral of sin^3 x cos x?

It is sin⁡4x4+C\frac{\sin^{4}x}{4} + C.

What is the general rule?

∫sin⁡nxcos⁡x dx=sin⁡n+1xn+1+C\int \sin^{n}x\cos x\,dx = \frac{\sin^{n+1}x}{n+1} + C for n≠−1n \neq -1.