AP Calculus AB and BC

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

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

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

Splitting off one sine factor

An odd power of sine always yields to a single substitution. Peel one sin⁡x\sin x aside and rewrite the rest with the Pythagorean identity sin⁡2x=1−cos⁡2x\sin^2 x = 1 - \cos^2 x, so every remaining factor is a cosine.

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

Now let u=cos⁡xu = \cos x, so du=−sin⁡x dxdu = -\sin x\,dx. The stray sin⁡x dx\sin x\,dx becomes −du-du, and the integral turns into a polynomial in uu.

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

Finishing the integral

Integrate the polynomial term by term, then put u=cos⁡xu = \cos x back.

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

A quick check: differentiating cos⁡3x3−cos⁡x\frac{\cos^3 x}{3} - \cos x gives cos⁡2x⋅(−sin⁡x)+sin⁡x=sin⁡x (1−cos⁡2x)=sin⁡3x\cos^2 x\cdot(-\sin x) + \sin x = \sin x\,(1-\cos^2 x) = \sin^3 x, the function you started with.

The mistake students make

The tempting shortcut is to treat sin⁡3x\sin^3 x like a power of the variable and write sin⁡4x4\frac{\sin^4 x}{4}, copying the power rule for x3x^3. That rule applies to powers of xx, not powers of a function, and differentiating sin⁡4x4\frac{\sin^4 x}{4} returns sin⁡3xcos⁡x\sin^3 x\cos x, not sin⁡3x\sin^3 x.

Odd power? Save one factor

When sine or cosine appears to an odd power, split off a single factor and substitute for the other function. The leftover even power always converts cleanly through sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1.

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Frequently asked questions

Why substitute u = cos x instead of u = sin x?

Because the saved factor is sin⁡x dx\sin x\,dx, which is exactly −du-du when u=cos⁡xu = \cos x. Choosing u=sin⁡xu = \sin x would need a cos⁡x dx\cos x\,dx to match dudu, and none is available for an odd power of sine.

Does this method work for sin^5 x?

Yes. Save one sin⁡x\sin x and write sin⁡4x=(1−cos⁡2x)2\sin^4 x = (1-\cos^2 x)^2, then substitute u=cos⁡xu = \cos x. Every odd power of sine reduces to a polynomial in cos⁡x\cos x the same way.

Is -cos x + (cos^3 x)/3 a different answer?

No. cos⁡3x3−cos⁡x\frac{\cos^3 x}{3} - \cos x and −cos⁡x+cos⁡3x3-\cos x + \frac{\cos^3 x}{3} are identical, since the order of addition does not matter.