I have the integral from 0 to pi of sin^4(3x)dx.
This should be really easy, and I can see that I probably need to use the half angle trig identity. However, I'm not seeing how the 3x might mess that identity up. How does this work?
I have the integral from 0 to pi of sin^4(3x)dx.
This should be really easy, and I can see that I probably need to use the half angle trig identity. However, I'm not seeing how the 3x might mess that identity up. How does this work?
$\displaystyle \sin^4(3x) =$
$\displaystyle \left[\sin^2(3x)\right]^2 =$
$\displaystyle \left[\frac{1-\cos(6x)}{2}\right]^2 =
$
$\displaystyle \frac{1}{4}\left[1 - 2\cos(6x) + \cos^2(6x)\right] =$
$\displaystyle \frac{1}{4}\left[1 - 2\cos(6x) + \frac{1+\cos(12x)}{2}\right] =
$
$\displaystyle \frac{3}{8} - \frac{\cos(6x)}{2} + \frac{\cos(12x)}{8}$