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Math Help - Difficult integral

  1. #1
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    Difficult integral

    I have been asked to integrate the folowing function:

    sin(x)/(1+9cos^2(x))

    and am having considerable difficulties.

    Could anyone please tell me how to progress - I don't want the answer, just a method for solving this indefinate integral.

    Thanks in advance for any help.

    Regards,

    Kirstie
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  2. #2
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    Let u=3cos(x), \;\ \frac{-du}{3}=sin(x)dx.

    Then you should see an arctan somewhere.
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by kirstie View Post
    I have been asked to integrate the folowing function:

    sin(x)/(1+9cos^2(x))

    and am having considerable difficulties.

    Could anyone please tell me how to progress - I don't want the answer, just a method for solving this indefinate integral.

    Thanks in advance for any help.

    Regards,

    Kirstie
    let u=3\cos(x) \to du=-3\sin(x)dx

    \int \frac{\sin(u)}{1+9\cos(u)}dx=-\frac{1}{3}\int\frac{du}{1+u^2}

    This is in the form of the arctangent so we get

    -\frac{1}{3}\int\frac{du}{1+u^2}=-\frac{1}{3}\tan^{-1}(u)+C=-\frac{1}{3}\tan^{-1}\left( 3\cos(x)\right)+C
    Last edited by TheEmptySet; July 30th 2008 at 10:27 AM.
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  4. #4
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    Thanks for your prompt replies.

    However shouldn't the final answer be

    1/3tan^-1(3cos(x))+c

    since u = 3cos(x) not cos(x)/3

    Kirstie
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  5. #5
    Behold, the power of SARDINES!
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    Yes it should. I fixed the above post.
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