AP Calculus AB and BC

Integral of cos 4x: Answer, Proof, and Steps

Q: Why divide by 4 instead of multiplying by 4?

Because \frac{d}{dx}[\sin 4x] = 4\cos 4x, so the raw guess already overshoots by a factor of 4. Dividing cancels it exactly. Multiplying would leave a derivative of 16\cos 4x.

Q: Does the same rule work for any coefficient?

Yes. For any nonzero constant k, \int \cos kx\,dx = \frac{\sin kx}{k} + C. It stops working the moment the inside is nonlinear, since then the inside derivative is not a constant.

Q: What is the definite integral of cos 4x from 0 to pi/8?

Evaluate \frac{\sin 4x}{4} at the endpoints: \frac{\sin\frac{\pi}{2}}{4} - \frac{\sin 0}{4} = \frac{1}{4}.

The integral of cos(4x) is sin(4x)/4 + C. The 4 inside the cosine reappears as a factor when you differentiate sin(4x), so the antiderivative divides by 4 to cancel it. The substitution u = 4x makes the same point, since dx = du/4.

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

Undoing the chain rule on the inside

The antiderivative of $\cos x$ is $\sin x$ , so the shape of the answer is a sine. What changes is the 4 sitting in front of $x$ . Differentiating $\sin 4x$ triggers the chain rule, and the inside derivative shows up as a stray factor.

\frac{d}{dx}\left[\sin 4x\right] = \cos 4x \cdot 4 = 4\cos 4x

That first guess is four times too big, so scale it down by 4.

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

The substitution that makes it automatic

If guessing feels risky, run the substitution. Let $u = 4x$ , so $du = 4\,dx$ and $dx = \frac{du}{4}$ . The constant leaves the integral, and what remains is the basic cosine rule.

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

Putting $u = 4x$ back gives $\frac{\sin 4x}{4} + C$ . Differentiating it returns $\frac{4\cos 4x}{4} = \cos 4x$ , the function you started with.

The mistake students make

The frequent error is multiplying by 4 instead of dividing, writing $4\sin 4x + C$ . Differentiating that gives $16\cos 4x$ , off by a factor of 16. Derivatives of composite functions multiply by the inside derivative, so antiderivatives have to divide by it.

Only a linear inside gets this shortcut

Dividing by the inside coefficient works because the derivative of $4x$ is the constant 4, which can be pulled out front. For an inside like $x^2$ the derivative still contains $x$ , no constant cancels it, and $\int \cos(x^2)\,dx$ has no elementary answer at all.

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 divide by 4 instead of multiplying by 4?

Because $\frac{d}{dx}[\sin 4x] = 4\cos 4x$ , so the raw guess already overshoots by a factor of 4. Dividing cancels it exactly. Multiplying would leave a derivative of $16\cos 4x$ .

Does the same rule work for any coefficient?

Yes. For any nonzero constant $k$ , $\int \cos kx\,dx = \frac{\sin kx}{k} + C$ . It stops working the moment the inside is nonlinear, since then the inside derivative is not a constant.

What is the definite integral of cos 4x from 0 to pi/8?

Evaluate $\frac{\sin 4x}{4}$ at the endpoints: $\frac{\sin\frac{\pi}{2}}{4} - \frac{\sin 0}{4} = \frac{1}{4}$ .