# stuck with sin^4(x)

• Mar 24th 2009, 04:47 PM
TYTY
stuck with sin^4(x)
Here is what I am trying to do:

$

\int{\sin^4{x}} = \int{\sin^3{x}\sin{x}}
$

now use product rule
$
\int{udv} = uv - \int{vdu}

$

notably

$

dv = \sin^3{x}
v = \frac{\cos^3{x}}{3} - \cos{x}
$

I derived this earlier and just copied the result. Now am I supposed to do :

$
\int{vdu}
$

Isn't that just

$
(\frac{\cos^3{x}}{3} - \cos{x})\cos{x}
$

I feel like I am going backwards...
• Mar 24th 2009, 04:56 PM
skeeter
$\sin^4{x} =$

$(\sin^2{x})^2 =$

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

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

$\frac{1}{4}\left[1 - 2\cos(2x) + \frac{1 + \cos(4x)}{2}\right] =
$

$\frac{3}{8} - \frac{\cos(2x)}{2} + \frac{\cos(4x)}{8}$

now integrate