AP Calculus AB and BC

Integral of cos^2(2x): Double Angle Twice

Q: What does cos^2(2x) integrate to?

The double angle identity turns it into \frac{x}{2} + \frac{\sin 4x}{8} + C.

Q: Which identity do I use for cos squared?

\cos^{2}\theta = \frac{1 + \cos 2\theta}{2}. Substitute the actual argument for \theta, which here is 2x, so the right hand side carries \cos 4x.

Because an even power of cosine cannot be reduced by substitution alone, some trig identity has to enter the problem, and the double angle identity is the standard route. It gives x over 2 plus sin 4x over 8, plus C, with the inner angle of 2x doubling to 4x.

\int \cos^{2}\left(2x\right)\,dx = \frac{x}{2} + \frac{\sin 4x}{8} + C

An even power forces an identity

Substitution needs a leftover derivative to absorb, and $\cos^{2}\left(2x\right)$ has none. Even setting $u = 2x$ only moves the problem: you get $\frac{1}{2}\int \cos^{2}u\,du$ , still an even power. Some identity has to enter somewhere, and the double angle identity is the direct way in.

\cos^{2}\theta = \frac{1 + \cos 2\theta}{2}, \qquad \theta = 2x \;\Rightarrow\; \cos^{2}\left(2x\right) = \frac{1 + \cos 4x}{2}

What is left is a constant plus a cosine, and both integrate on sight.

\int \cos^{2}\left(2x\right)dx = \frac{1}{2}\int \left(1 + \cos 4x\right)dx = \frac{x}{2} + \frac{\sin 4x}{8} + C

It is not the only route. Parts with $u = \cos 2x$ and $dv = \cos 2x\,dx$ , followed by $\sin^{2}\left(2x\right) = 1 - \cos^{2}\left(2x\right)$ , brings the original integral $I$ back on the right hand side: $2I = \frac{\sin 2x\cos 2x}{2} + x$ , so $I = \frac{x}{2} + \frac{\sin 2x\cos 2x}{4}$ . Since $\sin 4x = 2\sin 2x\cos 2x$ , that is the same answer written differently, which makes it a useful check.

Where the 8 comes from

Two separate halvings build that denominator. The identity contributes the $\frac{1}{2}$ out front, and integrating $\cos 4x$ contributes $\frac{1}{4}$ because the inner angle is $4x$ . Multiply them and you get $\frac{1}{8}$ .

Track the angle as it moves: the argument starts at $2x$ and the identity doubles it to $4x$ . Students who apply the identity by rote with $\cos 2x$ on the right hand side are answering a different question.

A sanity check on the shape

Over a full period, cos^2 averages one half, so any antiderivative must grow roughly like x/2 with a small ripple on top. The x/2 term is that growth and the sine term is the ripple. An answer with no linear term cannot be right.

The mistakes students make

Each of these produces a specific wrong answer that a quick derivative check would catch.

Using the power rule on the whole function and writing $\frac{\cos^{3}\left(2x\right)}{6}$ . Differentiating that gives a sine, and the original integrand has no sine in it.
Applying the identity with the wrong inner angle: $\frac{1 + \cos 2x}{2}$ gives $\frac{x}{2} + \frac{\sin 2x}{4}$ , which is the antiderivative of $\cos^{2}x$ , not $\cos^{2}\left(2x\right)$ .
Integrating $\cos 4x$ as $\sin 4x$ with no division by $4$ , which leaves $\frac{x}{2} + \frac{\sin 4x}{2}$ .

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 does cos^2(2x) integrate to?

The double angle identity turns it into $\frac{x}{2} + \frac{\sin 4x}{8} + C$ .

Why can't I use u-substitution on cos^2(2x)?

Substitution needs the derivative of the inner function present in the integrand. There is no $\sin 2x$ factor here, so nothing cancels. Rewriting with an identity is the standard route for any even power of sine or cosine.

Which identity do I use for cos squared?

$\cos^{2}\theta = \frac{1 + \cos 2\theta}{2}$ . Substitute the actual argument for $\theta$ , which here is $2x$ , so the right hand side carries $\cos 4x$ .