# How do you integrate int sin^2(2x)dx?

Mar 15, 2017

$\frac{1}{8} \left(4 - \sin \left(4 x\right)\right) + C$

#### Explanation:

Use the cosine double-angle identity to rewrite the function:

$\cos \left(2 \alpha\right) = 1 - 2 {\sin}^{2} \left(\alpha\right) \text{ "=>" } {\sin}^{2} \left(\alpha\right) = \frac{1 - \cos \left(2 \alpha\right)}{2}$

Then:

${\sin}^{2} \left(2 x\right) = \frac{1 - \cos \left(4 x\right)}{2}$

So:

$\int {\sin}^{2} \left(2 x\right) \mathrm{dx} = \frac{1}{2} \int \left(1 - \cos \left(4 x\right)\right) \mathrm{dx} = \frac{1}{2} \int \mathrm{dx} - \frac{1}{2} \int \cos \left(4 x\right) \mathrm{dx}$

The second can be solved with the substitution $u = 4 x \implies \mathrm{du} = 4 \mathrm{dx}$:

$\int {\sin}^{2} \left(2 x\right) \mathrm{dx} = \frac{1}{2} x - \frac{1}{8} \int 4 \cos \left(4 x\right) \mathrm{dx} = \frac{1}{2} x - \frac{1}{8} \int \cos \left(u\right) \mathrm{du}$

The integral of cosine is sine:

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