# How do you integrate int sin^5(2x)cos2xdx?

Dec 8, 2016

$u$-substitution!!

I'm so happy I just learned about these.

Just looking at this beautifully set up equation, we can see that we want to substitute $\sin \left(2 x\right)$ as $u$ so as to have a $\cos \left(2 x\right) \mathrm{dx}$ in the equation.

Thus:

Let $\sin \left(2 x\right) = u$

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

So:

$\int {\sin}^{5} \left(2 x\right) \cos \left(2 x\right) \mathrm{dx}$
$= \frac{1}{2} \int {u}^{5} \mathrm{du}$

$= \frac{1}{2} \cdot {u}^{6} / 6 + c = {u}^{6} / 12 + c$

Plug $u$ back in:

$= {\sin}^{6} \frac{2 x}{12} + c$