# How do you find the integral of cos^2(2x)dx?

Aug 28, 2016

$\setminus \frac{1}{2} \setminus \left(x + \setminus \frac{1}{4} \setminus \sin \setminus \left(4 x\right)\right) + C$

#### Explanation:

$\int {\cos}^{2} \left(2 x\right) \mathrm{dx}$

using the following identity,

${\cos}^{2} \left(x\right) = \frac{1 + \cos \left(2 x\right)}{2}$

$= \int \frac{1 + \cos \left(2 \cdot 2 x\right)}{2} \mathrm{dx}$

taking the constant out,

$\int a \cdot f \left(x\right) \mathrm{dx} = a \cdot \int f \left(x\right) \mathrm{dx}$

so,$= \frac{1}{2} \int 1 + \cos \left(2 \cdot 2 x\right) \mathrm{dx}$

applying the sum rule,

$\int f \left(x\right) \pm g \left(x\right) \mathrm{dx} = \int f \left(x\right) \mathrm{dx} \pm \int g \left(x\right) \mathrm{dx}$

we have,$\int 1 \mathrm{dx}$ $= x$
and,
$\int \cos \left(2 \cdot 2 x\right) \mathrm{dx} = \frac{1}{4} \sin \left(4 x\right)$

finally,
$= \frac{1}{2} \left(x + \frac{1}{4} \sin \left(4 x\right)\right)$

$\frac{1}{2} \left(x + \frac{1}{4} \sin \left(4 x\right)\right) + C$