AP Calculus AB and BC

Integral of cos(2x): Answer, Proof, and Steps

Q: Why is there no negative sign in the integral of cos(2x)?

Because the antiderivative of cosine is +\sin. The negative sign only shows up when integrating sine, where \int \sin x\,dx = -\cos x + C.

Q: How does the integral of cos(2x) differ from the integral of cos(x)?

Only by the factor of \frac{1}{2}. \int \cos x\,dx = \sin x + C, while \int \cos 2x\,dx = \frac{\sin 2x}{2} + C. The inner coefficient 2 becomes a \frac{1}{2} out front.

The integral of cos(2x) is sin(2x)/2 + C. The antiderivative of cosine is sine, and because the inside is 2x you divide by the inner coefficient 2. Differentiating sin(2x)/2 gives (1/2) times cos(2x) times 2, which is cos(2x).

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

Setting up the antiderivative

The antiderivative of cosine is sine, and the inside function $2x$ calls for a u-substitution that divides by the inner coefficient.

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

Let $u = 2x$ , so $du = 2\,dx$ and $dx = \frac{du}{2}$ .

\int \cos 2x\,dx = \int \cos u \cdot \frac{du}{2} = \frac{1}{2}\int \cos u\,du = \frac{1}{2}\sin u + C

Return to $x$ by substituting $u = 2x$ .

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

Verifying the 1/2

Differentiating confirms the coefficient. The chain rule brings out the inner factor $2$ .

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

The $2$ from the chain rule cancels the $\frac{1}{2}$ , and the derivative of sine is cosine, so the result is exactly $\cos 2x$ . Unlike the sine case, no sign flip is involved.

No leading minus

Cosine integrates without a sign change: $\int \cos u\,du = \sin u + C$ . The negative only appears when you integrate sine.

Common mistakes

Forgetting the $\frac{1}{2}$ . Without it the derivative comes out as $2\cos 2x$ , twice the integrand.
Adding a stray negative sign. Integrating cosine gives positive sine; the minus belongs to sine's integral, not cosine's.
Multiplying by $2$ rather than dividing. The inner coefficient enters as a reciprocal under integration.

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(2x) from 0 to pi/4?

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

Why is there no negative sign in the integral of cos(2x)?

Because the antiderivative of cosine is $+\sin$ . The negative sign only shows up when integrating sine, where $\int \sin x\,dx = -\cos x + C$ .

How does the integral of cos(2x) differ from the integral of cos(x)?

Only by the factor of $\frac{1}{2}$ . $\int \cos x\,dx = \sin x + C$ , while $\int \cos 2x\,dx = \frac{\sin 2x}{2} + C$ . The inner coefficient $2$ becomes a $\frac{1}{2}$ out front.