AP Calculus AB and BC

Integral of cos^2 x: Answer and Power Reduction

Q: What is the integral of cos^2 x?

It is \frac{x}{2} + \frac{\sin 2x}{4} + C, equivalently \frac{x}{2} + \frac{\sin x\cos x}{2} + C. Use \cos^{2} x = \frac{1 + \cos 2x}{2} first, then integrate term by term.

Q: Why can I not just use the power rule on cos^2 x?

The power rule for antiderivatives applies to x^{n}, where the base is the variable itself. Here the base is \cos x, a function of x, so differentiating \frac{\cos^{3} x}{3} triggers the chain rule and produces an extra -\sin x that is not in the integrand.

Q: What is the average value of cos^2 x?

Over any whole number of periods it is exactly \frac{1}{2}. The \frac{\sin 2x}{4} term returns to where it started, so only \frac{x}{2} contributes, and dividing by the interval length leaves \frac{1}{2}.

Q: How does the integral of sin^2 x compare?

It is \frac{x}{2} - \frac{\sin 2x}{4} + C, identical except for the sign, because \sin^{2} x = \frac{1 - \cos 2x}{2}. Adding the two antiderivatives gives x + C, matching \sin^{2} x + \cos^{2} x = 1.

The integral of cos^2 x with respect to x is x/2 + sin(2x)/4 + C. You cannot integrate cos^2 x directly, so you first replace it with the power reducing identity cos^2 x = (1 + cos 2x)/2, which turns the problem into two basic antiderivatives. An equivalent form is x/2 + (sin x cos x)/2 + C.

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

How to integrate cos^2 x with power reduction

There is no antiderivative rule for a squared trig function, and substitution fails because $\cos^{2} x\,dx$ contains no $\sin x$ to absorb a $du$ . The move is to change the integrand before integrating, using the power reducing identity that trades the square for a doubled angle.

\cos^{2} x = \frac{1 + \cos 2x}{2}

That identity comes straight from the double angle formula $\cos 2x = 2\cos^{2} x - 1$ , rearranged. Once the square is gone, both pieces are basic antiderivatives.

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

The $\frac{1}{2}$ inside comes from the chain rule running backwards: the antiderivative of $\cos 2x$ is $\frac{\sin 2x}{2}$ , not $\sin 2x$ . Distributing the outer $\frac{1}{2}$ gives the standard form.

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

Check it by differentiating

Differentiate $\frac{x}{2} + \frac{\sin 2x}{4}$ to get $\frac{1}{2} + \frac{2\cos 2x}{4} = \frac{1 + \cos 2x}{2}$ , which is exactly the identity you started from, so it equals $\cos^{2} x$ .

Equivalent forms of the answer

Graders accept several forms, and a calculator or textbook may hand you a different one. They differ only by algebra, not by a constant, so none is more correct than another.

\frac{x}{2} + \frac{\sin 2x}{4} + C = \frac{x}{2} + \frac{\sin x\cos x}{2} + C = \frac{x + \sin x\cos x}{2} + C

The middle form follows from the double angle identity $\sin 2x = 2\sin x\cos x$ . If your answer and the answer key disagree, try that substitution before assuming you made an error.

The companion result for $\sin^{2} x$ has the same shape with one sign flipped, because its power reducing identity is $\sin^{2} x = \frac{1 - \cos 2x}{2}$ .

\int \sin^{2} x\,dx = \frac{x}{2} - \frac{\sin 2x}{4} + C

Adding the two results gives $x + C$ , which is the check that they are consistent: $\sin^{2} x + \cos^{2} x = 1$ , and the antiderivative of $1$ is $x$ .

Where the integral of cos^2 x shows up on the AP exam

Basic antiderivatives are Topic 6.8 and substitution is Topic 6.9, both on AB and BC, and Topic 6.14 (Selecting Techniques for Antidifferentiation) is where you have to notice that an identity comes before any integration technique. Unit 6 carries a weighting of 15 to 20 percent on both exams.

The most quotable consequence is the average value of $\cos^{2} x$ over a full period, which is $\frac{1}{2}$ . This is why the result appears in physics contexts such as average power.

\frac{1}{2\pi}\int_{0}^{2\pi} \cos^{2} x\,dx = \frac{1}{2\pi}\left[\frac{x}{2} + \frac{\sin 2x}{4}\right]_{0}^{2\pi} = \frac{1}{2\pi}\cdot \pi = \frac{1}{2}

The $\sin 2x$ term contributes nothing over a full period because $\sin 4\pi = \sin 0 = 0$ , so only the $\frac{x}{2}$ part survives. The same argument gives average value $\frac{1}{2}$ for $\sin^{2} x$ .

On a volume of revolution problem, $\cos^{2} x$ appears the moment you square a $\cos x$ radius, so this antiderivative is often the last step of a disk method question rather than the question itself.

Common mistakes with the integral of cos^2 x

Writing $\frac{\cos^{3} x}{3}$ . The power rule antidifferentiates $x^{n}$ , not $\cos^{n} x$ . Differentiating $\frac{\cos^{3} x}{3}$ gives $-\cos^{2} x\sin x$ , which is not the integrand.
Forgetting the inner $\frac{1}{2}$ on $\int \cos 2x\,dx$ and writing $\frac{\sin 2x}{2}$ as the final term instead of $\frac{\sin 2x}{4}$ .
Using the wrong sign in the identity. $\cos^{2} x = \frac{1 + \cos 2x}{2}$ takes a plus; the minus belongs to $\sin^{2} x$ . Test at $x = 0$ : $\cos^{2} 0 = 1$ , and $\frac{1 + 1}{2} = 1$ confirms the plus.
Trying $u = \cos x$ . That needs $du = -\sin x\,dx$ , and there is no $\sin x$ anywhere in $\cos^{2} x\,dx$ to supply it.
Assuming the answer is periodic. The $\frac{x}{2}$ term grows without bound, which is correct: $\cos^{2} x$ is never negative, so its accumulated area must keep increasing.

Odd powers are a different problem

Power reduction is for EVEN powers. For $\int \cos^{3} x\,dx$ , peel off one factor and use $\cos^{2} x = 1 - \sin^{2} x$ , then substitute $u = \sin x$ . Recognizing which parity you are looking at picks the method.

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^2 x?

It is $\frac{x}{2} + \frac{\sin 2x}{4} + C$ , equivalently $\frac{x}{2} + \frac{\sin x\cos x}{2} + C$ . Use $\cos^{2} x = \frac{1 + \cos 2x}{2}$ first, then integrate term by term.

Why can I not just use the power rule on cos^2 x?

The power rule for antiderivatives applies to $x^{n}$ , where the base is the variable itself. Here the base is $\cos x$ , a function of $x$ , so differentiating $\frac{\cos^{3} x}{3}$ triggers the chain rule and produces an extra $-\sin x$ that is not in the integrand.

Is the integral of cos^2 x on the AP Calculus AB exam?

Yes. It uses only the power reducing identity from precalculus plus basic antiderivatives and substitution, all of which are on AB. It shows up most often as the final step of a disk method volume problem.

What is the average value of cos^2 x?

Over any whole number of periods it is exactly $\frac{1}{2}$ . The $\frac{\sin 2x}{4}$ term returns to where it started, so only $\frac{x}{2}$ contributes, and dividing by the interval length leaves $\frac{1}{2}$ .

How does the integral of sin^2 x compare?

It is $\frac{x}{2} - \frac{\sin 2x}{4} + C$ , identical except for the sign, because $\sin^{2} x = \frac{1 - \cos 2x}{2}$ . Adding the two antiderivatives gives $x + C$ , matching $\sin^{2} x + \cos^{2} x = 1$ .