Having an issue with a seemingly simple integration with trig.

**charleschafsky** · May 31st 2010, 05:06 PM

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?

**pickslides** · May 31st 2010, 05:19 PM

Here's a start

$\int \sin^4(3x)~dx = \int \sin^2(3x)\sin^2(3x)~dx= \int \frac{1-\cos(3x)}{2}\times \frac{1-\cos(3x)}{2}~dx$

Expand and then reduce $\cos^2(3x)$ using $\cos^2(u)=\frac{1+\cos(u)}{2}$

**skeeter** · May 31st 2010, 05:26 PM

Originally Posted by charleschafsky

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?

$\sin^4(3x) =$

$\left[\sin^2(3x)\right]^2 =$

$\left[\frac{1-\cos(6x)}{2}\right]^2 =<br />$

$\frac{1}{4}\left[1 - 2\cos(6x) + \cos^2(6x)\right] =$

$\frac{1}{4}\left[1 - 2\cos(6x) + \frac{1+\cos(12x)}{2}\right] =<br />$

$\frac{3}{8} - \frac{\cos(6x)}{2} + \frac{\cos(12x)}{8}$

**AllanCuz** · May 31st 2010, 06:15 PM

Could we not take advantage of the reduction formula for sin here? It's already computed for us

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