How do you integrate $\int {sin}^{5} (2 x) cos 2 x d x$ $int sin^5(2x)cos2xdx$ ?

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How do you integrate $\int {sin}^{5} (2 x) cos 2 x d x$ $int sin^5(2x)cos2xdx$ ?

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Answer 1 · 2016-12-08T07:52:47

$u$ $u$ -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 $sin (2 x)$ $sin(2x)$ as $u$ $u$ so as to have a $cos (2 x) d x$ $cos(2x)dx$ in the equation.

Thus:

Let $sin (2 x) = u$ $sin(2x)=u$

$d u = 2 cos (2 x) d x$ $du=2cos(2x)dx$
$\frac{1}{2} d u = cos (2 x) d x$ $1/2du=cos(2x)dx$

So:

$\int {sin}^{5} (2 x) cos (2 x) d x$ $intsin^5(2x)cos(2x)dx$
$= \frac{1}{2} \int u^{5} d u$ $=1/2intu^5du$

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

Plug $u$ $u$ back in:

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

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