# What is the integral of int sin^2(2x) dx?

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May 13, 2017

We have that

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

Hence we have that

$\int {\sin}^{2} \left(2 x\right) \mathrm{dx} = \int \left[\frac{1}{2} \cdot \left(1 - \cos \left(4 x\right)\right)\right] \mathrm{dx} = \frac{x}{2} - \sin \frac{4 x}{8} + c$

Footnote

We used the following trig identities

1) $\cos \left(2 \cdot a\right) = {\cos}^{2} a - {\sin}^{2} a$

2) ${\cos}^{2} a = 1 - {\sin}^{2} a$

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