AP Calculus AB and BC

Integral of cos(3x): Answer, Proof, and Steps

Q: How does the integral of cos(3x) relate to the integral of cos(2x)?

Both divide by their inner coefficient: \int \cos 3x\,dx = \frac{\sin 3x}{3} + C and \int \cos 2x\,dx = \frac{\sin 2x}{2} + C. A larger coefficient gives a smaller factor out front.

Q: What is the general integral of cos(kx)?

\int \cos kx\,dx = \frac{1}{k}\sin kx + C for any nonzero constant k, with no negative sign since cosine integrates to sine.

The integral of cos(3x) is sin(3x)/3 + C. The antiderivative of cosine is sine, and because the inside is 3x you divide by the inner coefficient 3. Differentiating sin(3x)/3 gives back cos(3x).

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

Setting up the antiderivative

The antiderivative of cosine is sine, and the inside function $3x$ calls for a u-substitution that divides by the inner coefficient $3$ .

\int \cos x\,dx = \sin x + C

Let $u = 3x$ , so $du = 3\,dx$ and $dx = \frac{du}{3}$ .

\int \cos 3x\,dx = \int \cos u \cdot \frac{du}{3} = \frac{1}{3}\int \cos u\,du = \frac{1}{3}\sin u + C

Substituting $u = 3x$ back finishes the problem.

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

Verifying the 1/3

A quick derivative confirms both the coefficient and the absence of a sign flip.

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

The chain rule's factor of $3$ cancels the $\frac{1}{3}$ , and the derivative of sine is cosine, so the integrand comes back untouched with no leading minus.

Cosine stays positive

Integrating cosine never introduces a negative, so keep $\frac{\sin 3x}{3}$ positive. The sign only appears for sine.

Common mistakes

Forgetting the $\frac{1}{3}$ , which leaves the derivative as $3\cos 3x$ instead of $\cos 3x$ .
Inserting a negative sign. Integrating cosine gives positive sine.
Dividing by $2$ out of habit. The inside is $3x$ , so the divisor is $3$ .

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(3x) from 0 to pi/6?

$\int_0^{\frac{\pi}{6}} \cos 3x\,dx = \left[\frac{\sin 3x}{3}\right]_0^{\frac{\pi}{6}} = \frac{\sin\left(\frac{\pi}{2}\right)}{3} - 0 = \frac{1}{3}$ .

How does the integral of cos(3x) relate to the integral of cos(2x)?

Both divide by their inner coefficient: $\int \cos 3x\,dx = \frac{\sin 3x}{3} + C$ and $\int \cos 2x\,dx = \frac{\sin 2x}{2} + C$ . A larger coefficient gives a smaller factor out front.

What is the general integral of cos(kx)?

$\int \cos kx\,dx = \frac{1}{k}\sin kx + C$ for any nonzero constant $k$ , with no negative sign since cosine integrates to sine.