# Thread: [SOLVED] Trig substitution

1. ## [SOLVED] Trig substitution

$
\int_0^\frac {\pi}{5} sin^4(5x)~dx
$

I know how to solve this problem if it was only a second power, but since it is to the fourth power, I am a little confused.

2. Use half angle identity to break it down

3. Originally Posted by redman223
$
\int_0^\frac {\pi}{5} sin^4(5x)~dx
$

I know how to solve this problem if it was only a second power, but since it is to the fourth power, I am a little confused.
$
\int_0^\frac {\pi}{5} sin^4(5x)~dx
$

$= \int_0^\frac {\pi}{5} \left[ sin^2(5x) \right]^2~dx$

$= \int_0^\frac {\pi}{5} \left[ \frac{1- cos(10x)}{2} \right]^2~dx$

$= \int_0^\frac {\pi}{5} \left[ \frac{1- 2cos(10x) + cos^2(10x)}{4} \right]~dx$

$= \frac {1}{4} \int_0^\frac {\pi}{5} \left[1- 2cos(10x) \right]~dx + \frac {1}{4} \int_0^\frac {\pi}{5} cos^2(10x) ~dx$

$= \frac {1}{4} \int_0^\frac {\pi}{5} \left[1- 2cos(10x) \right]~dx + \frac {1}{8} \int_0^\frac {\pi}{5} \left[1+cos(20x) \right] ~dx$

4. So is

$
sin^4(5x) = sin^2(5x)sin^2(5x)
$

Edit: Looks like we posted at the same time, I guess it is equal, thanks.