AP Calculus AB and BC

Integral of cos x: Answer, Proof, and Mistakes

Q: What is the integral of cos x?

\int \cos x\,dx = \sin x + C. Sine is the antiderivative because \frac{d}{dx}[\sin x] = \cos x, and the constant C covers every function in that family. The rule is valid for all real x.

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

\int \cos(kx)\,dx = \frac{1}{k}\sin(kx) + C for any nonzero constant k. Substituting u = kx gives du = k\,dx, which is where the \frac{1}{k} comes from. For example, \int \cos(5x)\,dx = \frac{1}{5}\sin(5x) + C.

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

It is 0. By the Fundamental Theorem, \int_0^{\pi} \cos x\,dx = \sin \pi - \sin 0 = 0. The positive area on \left(0, \frac{\pi}{2}\right) exactly cancels the negative area on \left(\frac{\pi}{2}, \pi\right). The total unsigned area is 2.

The integral of cos x is sin x + C. Because the derivative of sin x is cos x, sin x is an antiderivative of cos x, and adding an arbitrary constant C gives every antiderivative. For a definite integral the constant cancels, so the integral of cos x from a to b equals sin(b) minus sin(a).

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

Why the integral of cos x is sin x + C

Antidifferentiation is differentiation run backwards. A function $F$ is an antiderivative of $f$ when $F'(x) = f(x)$ , so to integrate $\cos x$ you only need to name the function whose derivative is $\cos x$ . That function is $\sin x$ , one of the basic derivatives memorized in Unit 2, Topic 2.7.

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

Reading that statement right to left gives the indefinite integral. Note the differential $dx$ : it is part of the notation, not decoration, and AP readers expect it.

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

Why the + C is not optional

Constants differentiate to zero, so $\sin x$ , $\sin x + 4$ , and $\sin x - 100$ all have derivative $\cos x$ . The symbol $C$ stands for that whole family. On an indefinite integral, leaving it off is a lost point; on a definite integral it cancels and does not appear.

There is no domain restriction to state. Both $\cos x$ and $\sin x$ are defined and continuous for every real $x$ , so the rule holds on all of $\mathbb{R}$ , unlike $\int \frac{1}{x}\,dx = \ln|x| + C$ , which breaks at $x = 0$ .

Definite integrals of cos x

Once you have an antiderivative, the Fundamental Theorem of Calculus (Topic 6.7) turns any definite integral of $\cos x$ into one subtraction.

\int_a^b \cos x\,dx = \sin b - \sin a

Over a quarter period starting at zero, the area under one arch of the cosine curve comes out to exactly 1.

\int_0^{\pi/2} \cos x\,dx = \sin\frac{\pi}{2} - \sin 0 = 1 - 0 = 1

Extend the interval to $\pi$ and the answer collapses to zero, because the curve sits above the axis on $\left(0, \frac{\pi}{2}\right)$ and an equal amount below it on $\left(\frac{\pi}{2}, \pi\right)$ .

\int_0^{\pi} \cos x\,dx = \sin \pi - \sin 0 = 0

Signed area against total area

A definite integral reports signed area, so the zero above is correct and is not a mistake. If a question asks for the total area between $y = \cos x$ and the $x$ -axis on $[0, \pi]$ , split at $x = \frac{\pi}{2}$ and add the absolute values, which gives $1 + 1 = 2$ .

Integrating composites such as cos(u)

Bare $\cos x$ is rare on the free-response section. What actually appears is $\cos$ of something, which calls for u-substitution (Topic 6.9), the reverse of the chain rule.

\int \cos(u)\,du = \sin(u) + C

Example 1. For $\int \cos(5x)\,dx$ , let $u = 5x$ , so $du = 5\,dx$ and $dx = \frac{1}{5}\,du$ . The reciprocal of the inside coefficient comes out front.

\int \cos(5x)\,dx = \frac{1}{5}\sin(5x) + C

Example 2. For $\int x\cos(x^2)\,dx$ , let $u = x^2$ , so $du = 2x\,dx$ . The stray $x$ outside is exactly what makes the substitution work, up to the factor of $\frac{1}{2}$ .

\int x\cos(x^2)\,dx = \frac{1}{2}\sin(x^2) + C

Example 3. $\int \cos^2 x\,dx$ is a different problem entirely, since no substitution clears the square. Use the power-reduction identity $\cos^2 x = \frac{1 + \cos(2x)}{2}$ first, then integrate term by term.

\int \cos^2 x\,dx = \frac{x}{2} + \frac{\sin(2x)}{4} + C

Check any antiderivative by differentiating

Every result on this page can be confirmed in one line. Differentiate $\frac{x}{2} + \frac{\sin(2x)}{4}$ and you get $\frac{1}{2} + \frac{\cos(2x)}{2} = \frac{1 + \cos(2x)}{2} = \cos^2 x$ . If differentiating your answer does not return the integrand, the answer is wrong, and this costs about ten seconds.

Common mistakes with the integral of cos x

Importing a minus sign. Differentiating cosine produces one, $\frac{d}{dx}[\cos x] = -\sin x$ , but integrating it does not: $\int \cos x\,dx = \sin x + C$ , positive. The negative belongs to the other pair, $\int \sin x\,dx = -\cos x + C$ .
Dropping the constant of integration. An indefinite integral names a family of functions, so $\sin x$ alone is an incomplete answer.
Forgetting the $\frac{1}{k}$ on $\cos(kx)$ . The answer to $\int \cos(3x)\,dx$ is $\frac{1}{3}\sin(3x) + C$ ; differentiate $\sin(3x)$ and the chain rule hands back a stray 3.
Working in degrees. The rule depends on $\frac{d}{dx}[\sin x] = \cos x$ , which only holds in radians, so keep a graphing calculator in radian mode for every integral on the AP exam.
Confusing $\cos^2 x$ with $\cos(x^2)$ . The first needs power reduction; the second has no elementary antiderivative at all and will never be asked for as an indefinite integral.
Carrying limits through a substitution unchanged. If you switch to $u$ in a definite integral, either convert the limits to $u$ values or convert back to $x$ before evaluating.

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 x?

$\int \cos x\,dx = \sin x + C$ . Sine is the antiderivative because $\frac{d}{dx}[\sin x] = \cos x$ , and the constant $C$ covers every function in that family. The rule is valid for all real $x$ .

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

The minus sign appears when you differentiate cosine, since $\frac{d}{dx}[\cos x] = -\sin x$ . Integration reverses the derivative of sine instead, and that derivative is $+\cos x$ , so $\int \cos x\,dx = \sin x + C$ . The companion rule with the negative is $\int \sin x\,dx = -\cos x + C$ .

What is the integral of cos(kx)?

$\int \cos(kx)\,dx = \frac{1}{k}\sin(kx) + C$ for any nonzero constant $k$ . Substituting $u = kx$ gives $du = k\,dx$ , which is where the $\frac{1}{k}$ comes from. For example, $\int \cos(5x)\,dx = \frac{1}{5}\sin(5x) + C$ .

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

It is 0. By the Fundamental Theorem, $\int_0^{\pi} \cos x\,dx = \sin \pi - \sin 0 = 0$ . The positive area on $\left(0, \frac{\pi}{2}\right)$ exactly cancels the negative area on $\left(\frac{\pi}{2}, \pi\right)$ . The total unsigned area is 2.

Is the integral of cos squared x just sine squared?

No. There is no product rule for integrals, so you cannot square the answer. Apply the identity $\cos^2 x = \frac{1 + \cos(2x)}{2}$ first, which gives $\int \cos^2 x\,dx = \frac{x}{2} + \frac{\sin(2x)}{4} + C$ .