1. ## A Trig Integral

Hi everyone. I did a trig integral problem. Can someone please check my work for me? Thanks.

$\displaystyle \int sin(3x)^3cos(3x)$

$\displaystyle u=sin(3x) du=cos3x dx$

$\displaystyle \int u^3 du$

$\displaystyle u^4/4+c$

$\displaystyle sin(4x)^4/4+c$

Thanks a lot, everyone!

2. Originally Posted by zhangxupage2
[FONT=Verdana]Hi everyone. I did a trig integral problem. Can someone please check my work for me? Thanks.

$\displaystyle \int sin(3x)^3cos(3x)$

$\displaystyle u=sin(3x) du=cos3x dx \int u^3 du u^4/4+c sin(4x)^4/4+c$

Thanks a lot, everyone!
$\displaystyle du=3\cos{(3x)}dx$

3. Thanks a lot. I thought I made a mistake.

$\displaystyle u=sin(3x) du=3cos(3x)dx$

$\displaystyle \int u^3 du/3$

$\displaystyle u^4/4*1/3$

$\displaystyle sin(3x)^4/12+c$

Is this correct?

5. Originally Posted by zhangxupage2

$\displaystyle u=sin(3x) du=3cos(3x)dx$

$\displaystyle \int u^3 du/3$

$\displaystyle u^4/4*1/3$

$\displaystyle sin(3x)^4/12+c$

Is this correct?
Correct

6. Thanks a lot.