AP Calculus AB and BC

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

Q: Why divide by 3 when integrating sin(3x)?

Because the substitution u = 3x gives dx = \frac{du}{3}, pulling out a \frac{1}{3}. Equivalently, differentiating \cos 3x multiplies by 3, so integrating must divide by 3.

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

\int \sin kx\,dx = -\frac{1}{k}\cos kx + C for any nonzero k. The negative comes from integrating sine and the \frac{1}{k} from the inner coefficient.

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

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

Setting up the antiderivative

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

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

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

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

Substituting $u = 3x$ back gives the antiderivative.

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

Verifying the sign and the 1/3

Differentiate the answer and confirm the integrand returns.

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

The inner factor $3$ from the chain rule cancels the $\frac{1}{3}$ , and the two negatives, one from integrating sine and one from differentiating cosine, cancel to leave $+\sin 3x$ .

Bigger inside, smaller factor

The inner coefficient $3$ becomes a $\frac{1}{3}$ . Every linear inside divides by its own slope.

Common mistakes

Losing the negative sign. The antiderivative of sine is negative cosine, so the answer is $-\frac{\cos 3x}{3}$ .
Dividing by the wrong number. The inside is $3x$ , so the divisor is $3$ , giving $\frac{1}{3}$ , not $\frac{1}{2}$ .
Multiplying by $3$ instead of dividing, which reverses the chain-rule adjustment.

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 sin(3x) from 0 to pi/3?

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

Why divide by 3 when integrating sin(3x)?

Because the substitution $u = 3x$ gives $dx = \frac{du}{3}$ , pulling out a $\frac{1}{3}$ . Equivalently, differentiating $\cos 3x$ multiplies by $3$ , so integrating must divide by $3$ .

What is the general integral of sin(kx)?

$\int \sin kx\,dx = -\frac{1}{k}\cos kx + C$ for any nonzero $k$ . The negative comes from integrating sine and the $\frac{1}{k}$ from the inner coefficient.