AP Calculus AB and BC

Integral of sin(4x): Answer, Proof, and Steps

Q: Why is the answer negative?

Because \frac{d}{dx}\cos x = -\sin x. To land on +\sin 4x after differentiating, the antiderivative has to start with a minus sign, giving -\frac{\cos 4x}{4} + C.

Q: Does the same rule apply to sin(kx) for any k?

Yes, for any nonzero constant k: \int\sin kx\,dx = -\frac{\cos kx}{k} + C. It fails only when the inside is not linear, as in \sin(x^2), where no elementary antiderivative exists.

The integral of sin(4x) is -cos(4x)/4 + C. Integrating sine produces negative cosine, and dividing by the inner coefficient 4 cancels the factor the chain rule would create. Differentiating -cos(4x)/4 gives 4 sin(4x) divided by 4, which is sin(4x).

\int \sin 4x\,dx = -\frac{\cos 4x}{4} + C

Two adjustments: the sign and the coefficient

Integrating $\sin x$ gives $-\cos x$ , so the minus sign is built in. The inner coefficient $4$ adds a second adjustment, because differentiating anything with $4x$ inside multiplies by $4$ .

\int \sin kx\,dx = -\frac{\cos kx}{k} + C

\int \sin 4x\,dx = -\frac{\cos 4x}{4} + C

Substitution makes the division explicit. Let $u = 4x$ , so $du = 4\,dx$ and $dx = \frac{du}{4}$ .

\int \sin 4x\,dx = \frac{1}{4}\int \sin u\,du = -\frac{1}{4}\cos u + C = -\frac{\cos 4x}{4} + C

Checking by differentiating

Differentiate the answer and watch both adjustments reverse.

\frac{d}{dx}\left(-\frac{\cos 4x}{4}\right) = -\frac{1}{4}\cdot(-\sin 4x)\cdot 4 = \sin 4x

The chain rule brings out the $4$ , which cancels the $\frac{1}{4}$ , and the derivative of cosine supplies the second minus sign that turns the answer positive.

The mistakes students make

Losing the minus sign and answering $\frac{\cos 4x}{4}$ . Differentiating that gives $-\sin 4x$ .
Answering $-\cos 4x$ with no division. Its derivative is $4\sin 4x$ , four times too large.
Multiplying by $4$ instead of dividing, giving $-4\cos 4x$ , whose derivative is $16\sin 4x$ .
Mixing up the sign rules. Sine integrates to negative cosine; cosine integrates to positive sine.

Divide by the inner coefficient

For any constant $k \neq 0$ , both $\int\sin kx\,dx$ and $\int\cos kx\,dx$ carry a factor of $\frac{1}{k}$ , which cancels the $k$ the chain rule would introduce.

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 the answer negative?

Because $\frac{d}{dx}\cos x = -\sin x$ . To land on $+\sin 4x$ after differentiating, the antiderivative has to start with a minus sign, giving $-\frac{\cos 4x}{4} + C$ .

What is the integral of sin(4x) from 0 to pi/4?

$\int_0^{\pi/4}\sin 4x\,dx = \left[-\frac{\cos 4x}{4}\right]_0^{\pi/4} = -\frac{\cos\pi}{4} + \frac{\cos 0}{4} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$ .

Does the same rule apply to sin(kx) for any k?

Yes, for any nonzero constant $k$ : $\int\sin kx\,dx = -\frac{\cos kx}{k} + C$ . It fails only when the inside is not linear, as in $\sin(x^2)$ , where no elementary antiderivative exists.