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Math Help - Tricky integral

  1. #1
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    Tricky integral

    \int_0^\pi \! 29sin^4(3t) \, dt.

    Not sure if I have to substitute just once or twice. Any suggestions?
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  2. #2
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    Quote Originally Posted by mmattson07 View Post
    \int_0^\pi \! 29sin^4(3t) \, dt.

    Not sure if I have to substitute just once or twice. Any suggestions?
    \sin^4{u} =

    (\sin^2{u})^2 =

    \left[\frac{1 - \cos(2u)}{2}\right]^2 =

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

    \frac{1}{4}\left[1 - 2\cos(2u) + \frac{1 + \cos(4u)}{2}\right] =

    \frac{3}{8} - \frac{\cos(2u)}{2} + \frac{\cos(4u)}{8}
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  3. #3
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    I see how you got this and i know sin^2x=1-cos^2x but how did you get 1-cos(2u)/2
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    Half-Angle formula.
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  5. #5
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    Quote Originally Posted by mmattson07 View Post


    I see how you got this and i know sin^2x=1-cos^2x but how did you get 1-cos(2u)/2
    Before attempting integrals like the one you posted you should ensure that you are thoroughly familiar with all required trigonometric identities. Otherwise you have no chance.

    For example, you should know that \cos (2u) = \cos^2 u - \sin^2 u = 1 - 2 \sin^2 u ....
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  6. #6
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    Yeah maybe that would help. He skipped a few algebraic steps so i was lost for a second.
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