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Math Help - Having an issue with a seemingly simple integration with trig.

  1. #1
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    Having an issue with a seemingly simple integration with trig.

    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?
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  2. #2
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    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}
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  3. #3
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    Quote Originally Posted by charleschafsky View Post
    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}
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  4. #4
    Senior Member AllanCuz's Avatar
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    Could we not take advantage of the reduction formula for sin here? It's already computed for us
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