AP Calculus AB and BC

Integral of cos^3 x: Answer, Proof, and Steps

The integral of cos^3(x) is sin(x) - (sin^3(x))/3 + C. To get it, write cos^3(x) as cos(x)(1 - sin^2(x)) and substitute u = sin(x), which turns the integral into a simple polynomial in u.

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

Splitting off one cosine factor

An odd power of cosine reduces to one substitution. Set a single cos⁡x\cos x aside and rewrite the rest with cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 x, so the remaining factors are all sines.

cos⁡3x=cos⁡x (1−sin⁡2x)\cos^3 x = \cos x\,(1-\sin^2 x)

Let u=sin⁡xu = \sin x, so du=cos⁡x dxdu = \cos x\,dx. The saved cos⁡x dx\cos x\,dx is exactly dudu, with no sign to fix.

∫cos⁡3x dx=∫(1−u2) du\int \cos^3 x\,dx = \int (1-u^2)\,du

Finishing the integral

Integrate term by term and restore u=sin⁡xu = \sin x.

∫(1−u2) du=u−u33+C=sin⁡x−sin⁡3x3+C\int (1-u^2)\,du = u - \frac{u^3}{3} + C = \sin x - \frac{\sin^3 x}{3} + C

Check by differentiating: ddx[sin⁡x−sin⁡3x3]=cos⁡x−sin⁡2xcos⁡x=cos⁡x (1−sin⁡2x)=cos⁡3x\frac{d}{dx}\left[\sin x - \frac{\sin^3 x}{3}\right] = \cos x - \sin^2 x\cos x = \cos x\,(1-\sin^2 x) = \cos^3 x.

The mistake students make

Reaching for the power rule and writing cos⁡4x4\frac{\cos^4 x}{4} is the common error. The power rule raises the exponent on the variable, not on a function of the variable, and ddx[cos⁡4x4]=−cos⁡3xsin⁡x\frac{d}{dx}\left[\frac{\cos^4 x}{4}\right] = -\cos^3 x\sin x, which is not cos⁡3x\cos^3 x.

Odd power? Save one factor

For an odd power of cosine, keep one cos⁡x\cos x for the dudu and convert the even remainder with cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 x. Then u=sin⁡xu = \sin x finishes the job.

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

Why does u = sin x work so cleanly here?

Because the saved factor is cos⁡x dx\cos x\,dx, which is precisely dudu for u=sin⁡xu = \sin x. No sign correction is needed, unlike the sine case where du=−sin⁡x dxdu = -\sin x\,dx introduces a minus.

How is this different from the integral of cos^2 x?

An even power has no factor to save, so substitution stalls. For cos⁡2x\cos^2 x you use the half-angle identity cos⁡2x=1+cos⁡2x2\cos^2 x = \frac{1+\cos 2x}{2} instead. Odd powers use the split-and-substitute method shown here.

Can the answer be written with a common denominator?

Yes. sin⁡x−sin⁡3x3=3sin⁡x−sin⁡3x3\sin x - \frac{\sin^3 x}{3} = \frac{3\sin x - \sin^3 x}{3}. Both forms are correct; the split form is the one substitution produces directly.