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Math Help - Integration of trig functions

  1. #1
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    Integration of trig functions

    \int sin^2xcos^4xdx


    Attempt to solution:

    sin^2x=1-cos^2x

    \int(1-cos^2x)(cos^4x)dx

    \int(cos^4x-cos^6x)dx ( I am stuck here, how l do evaluate this integral)
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by nyasha View Post
    \int sin^2xcos^4xdx
    recall that \sin^2 x = \frac {1 - \cos 2x}2 and \cos^2 x = \frac {1 + \cos 2x}2
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    recall that \sin^2 x = \frac {1 - \cos 2x}2 and \cos^2 x = \frac {1 + \cos 2x}2




    \frac{1}{2}\int(1-cos^2x)(cos^4x)dx<br />

    \frac{1}{2}\int(cos^4x-cos^6x)dx

    \frac{1}{2}\int cos^4xdx-\frac{1}{2}\int cos^6xdx

    \frac{1}{8}\int (1+cos2x)^2-\frac{1}{16}\int(1+cos2x)^3

    I'm l now on the right path
    Last edited by nyasha; March 15th 2009 at 12:46 AM.
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