Integral of cos²x

Q: How can I check the answer?

Differentiate it: the derivative of \frac{1}{2}x + \frac{1}{4}\sin\left(2 x\right) is \cos^{2}\left(x\right). CalcViz checks every antiderivative this way before showing it.

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

Use cos²x = (1 + cos 2x)/2. Over a full period, sin² and cos² each have average value ½.

Step by step

Trig identity
Rewrite powers/products of sin and cos with the power-reduction, double-angle or Pythagorean identities (sin²u = (1 − cos 2u)/2, cos²u = (1 + cos 2u)/2, sin²u + cos²u = 1).
$\cos^{2}\left(x\right) = \frac{1}{2}\cos\left(2 x\right) + \frac{1}{2}$
Sum rule
Integrate term by term.
$\int \left(\frac{1}{2}\cos\left(2 x\right) + \frac{1}{2}\right)\,dx = \int \frac{1}{2}\cos\left(2 x\right)\,dx + \int \frac{1}{2}\,dx$
Constant multiple rule
Move the constant outside the integral.
$\int \frac{1}{2}\cos\left(2 x\right)\,dx = \frac{1}{2} \int \cos\left(2 x\right)\,dx$
Trig rule
∫cos u du = sin u. Divide by the inner slope 2 (reverse chain rule).
$\int \cos\left(2 x\right)\,dx = \frac{1}{2}\sin\left(2 x\right)$
Constant rule
The integral of a constant k is k·x.
$\int \frac{1}{2}\,dx = \frac{1}{2}x$

f(x) = cos²x
F(x) + C

Constant C = 0.0

Every choice of C gives a valid antiderivative, since shifting up or down doesn’t change the slope.

Open in the integral calculator

Questions students ask

What is the integral of cos²x?

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

How do you integrate cos²x?

Use cos²x = (1 + cos 2x)/2. Over a full period, sin² and cos² each have average value ½.

How can I check the answer?

Differentiate it: the derivative of $\frac{1}{2}x + \frac{1}{4}\sin\left(2 x\right)$ is $\cos^{2}\left(x\right)$ . CalcViz checks every antiderivative this way before showing it.