AP Calculus AB and BC

Integral of sin(x)cos^2(x): Answer and Proof

Q: Why is u = cos x the right choice and not u = sin x?

Because sine appears to an odd power. Peeling off one \sin x for du leaves only even powers of cosine, and every remaining factor becomes u. With u = \sin x an unmatched \cos x survives.

Q: How does this extend to other powers of sine and cosine?

When sine has an odd power, save one \sin x for du and set u = \cos x. When cosine has an odd power, save one \cos x and set u = \sin x. When both powers are even, use the half-angle identities.

The integral of sin(x)cos^2(x) is -cos^3(x)/3 + C. Substituting u = cos x gives du = -sin x dx, so the integral becomes the negative of the integral of u^2 du, which is -u^3/3. Differentiating -cos^3(x)/3 returns sin(x)cos^2(x).

\int \sin x\cos^2 x\,dx = -\frac{\cos^3 x}{3} + C

Substituting u = cos x

One factor of $\sin x$ sits beside a power of $\cos x$ , and $\sin x$ is what the derivative of $\cos x$ produces up to sign. Let the cosine be the new variable.

u = \cos x, \qquad du = -\sin x\,dx, \qquad \sin x\,dx = -du

The minus sign travels with the substitution and stays in the final answer.

\int \sin x\cos^2 x\,dx = -\int u^2\,du = -\frac{u^3}{3} + C = -\frac{\cos^3 x}{3} + C

Track the sign at the substitution step

Writing $\sin x\,dx = -du$ once, in a line of its own, is what keeps the minus sign from disappearing later.

Checking by differentiating

Differentiate with the chain rule. Two negatives appear and cancel.

\frac{d}{dx}\left(-\frac{\cos^3 x}{3}\right) = -\frac{1}{3}\cdot 3\cos^2 x\cdot(-\sin x) = \sin x\cos^2 x

One minus sign comes from the antiderivative and the other from $\frac{d}{dx}\cos x = -\sin x$ , so the integrand comes back positive.

The mistakes students make

Dropping the minus sign and answering $\frac{\cos^3 x}{3}$ . Differentiating that gives $-\sin x\cos^2 x$ , the negative of the integrand.
Choosing $u = \sin x$ . Then $du = \cos x\,dx$ consumes only one cosine and leaves a stray $\cos x$ that cannot be written in terms of $u$ .
Answering $\frac{\sin^2 x}{2}\cdot\frac{\cos^3 x}{3}$ by splitting the product and then applying the power rule to each trig factor. Two rules break at once: $\int fg$ is not $\left(\int f\right)\left(\int g\right)$ , and the reverse power rule needs the variable itself raised to a power, so $\int\sin x\,dx = -\cos x$ , not $\frac{\sin^2 x}{2}$ .
Using the identity $\cos^2 x = \frac{1+\cos 2x}{2}$ first. It is legal but creates extra work that substitution avoids.

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 is u = cos x the right choice and not u = sin x?

Because sine appears to an odd power. Peeling off one $\sin x$ for $du$ leaves only even powers of cosine, and every remaining factor becomes $u$ . With $u = \sin x$ an unmatched $\cos x$ survives.

What is the integral of sin(x)cos^2(x) from 0 to pi/2?

$\int_0^{\pi/2}\sin x\cos^2 x\,dx = \left[-\frac{\cos^3 x}{3}\right]_0^{\pi/2} = 0 - \left(-\frac{1}{3}\right) = \frac{1}{3}$ .

How does this extend to other powers of sine and cosine?

When sine has an odd power, save one $\sin x$ for $du$ and set $u = \cos x$ . When cosine has an odd power, save one $\cos x$ and set $u = \sin x$ . When both powers are even, use the half-angle identities.