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Thread: Integration of trig functions

  1. #1
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    Integration of trig functions

    $\displaystyle \int sin^2xcos^4xdx$


    Attempt to solution:

    $\displaystyle sin^2x=1-cos^2x$

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

    $\displaystyle \int(cos^4x-cos^6x)dx$ ( 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
    $\displaystyle \int sin^2xcos^4xdx$
    recall that $\displaystyle \sin^2 x = \frac {1 - \cos 2x}2$ and $\displaystyle \cos^2 x = \frac {1 + \cos 2x}2$
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    recall that $\displaystyle \sin^2 x = \frac {1 - \cos 2x}2$ and $\displaystyle \cos^2 x = \frac {1 + \cos 2x}2$




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

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

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

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

    I'm l now on the right path
    Last edited by nyasha; Mar 14th 2009 at 11:46 PM.
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