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Math Help - Basic (I think) integral involving sin

  1. #1
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    Basic (I think) integral involving sin

    How would I go about solving the following integral?

    <br />
\int {{1}\over{5-2\sin{x}}} dx<br />
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  2. #2
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    Make the u-substitution: u=\tan\left(\frac{x}{2}\right)
    Then dx=\frac{2\,du}{u^2+1}, and \sin{x}=\frac{2u}{u^2+1}. Thus
    \int\frac{1}{5-2\sin{x}}\,dx=\int\frac{2}{5u^2-4u+5}\,du=\frac{2}{5}\int\frac{du}{u^2-\frac{4}{5}u+1}
    Completing the square will allow you to find the answer.

    --Kevin C.
    Last edited by TwistedOne151; August 11th 2009 at 06:47 PM. Reason: Typo
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    Wow. That was much more complicated than I expected. Not quite sure how you made the step to \sin{x}=\frac{2u}{u^2+1}, but I'll digest everything and try and re-attempt it. Thanks for the reply.
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  4. #4
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    Quote Originally Posted by drew.walker View Post
    Wow. That was much more complicated than I expected. Not quite sure how you made the step to \sin{x}=\frac{2u}{u^2+1}, but I'll digest everything and try and re-attempt it. Thanks for the reply.
    Google Weierstrass substitution.
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  5. #5
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    I was only able to make one further step and then I hit a roadblock. After completing the square I have:

    \int\frac{1}{(u-\frac{2}{5})^2+\frac{21}{25}} du

    However, I don't know which rule to then use to calculate the integral. Any guidance would be greatly appreciated.
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  6. #6
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    Quote Originally Posted by drew.walker View Post
    I was only able to make one further step and then I hit a roadblock. After completing the square I have:

    \int\frac{1}{(u-\frac{2}{5})^2+\frac{21}{25}} du

    However, I don't know which rule to then use to calculate the integral. Any guidance would be greatly appreciated.
    The integrand can be put into the form \frac{a}{w^2 + a^2} and the integral of this (with respect to w) is \tan^{-1} \left( \frac{w}{a}\right) + C.
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