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Math Help - integral help

  1. #1
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    integral help

    The problem is:

    int (1+sinx)/(1-sin)dx

    I tried doing this:
    Let u=sinx, x=sin^(-1)u, so dx= du/sqrt(1-u^2)
    Then I got
    integral of (1+u)/(1-u)*(du)/(sqrt(1-u^2))
    which is equal to
    integral of ((1+u)*sqrt(1-u^2))/((1-u)^2(1+u)) du

    I tried to figure it out using partial fractions from there, but that didn't work out. Any suggestions? Thanks!



    When I multiply by (1-sinx)/(1-sinx) I just end up with
    integral of cosx/(2-2sinx-cos^2x)dx
    which I still don't know how to integrate.
    Last edited by ilovecalculus; October 26th 2009 at 05:48 PM.
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  2. #2
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    Try multiplying either by (1-sinx)/(1-sinx).
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  3. #3
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    When I multiply by (1-sinx)/(1-sinx) I just end up with
    integral of cosx/(2-2sinx-cos^2x)dx
    which I still don't know how to integrate.
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  4. #4
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    Sorry meant (1+sinx)/(1+sinx). That will give you 1-sin^2(x)=cos^2(x) in the denominator and sin^2(x)+2sinx+1 in the numerator. Break it up into 3 fractions.
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  5. #5
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    Just to verify-- did you solve it all the way through for an answer? I just want to make sure I did the rest correctly. I ended up with:

    2tanx+2secx-x+C

    Let me know if that looks right.
    Thanks a lot!!
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  6. #6
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     \frac{ 1 + \sin(x)}{1 - \sin(x)} = \frac{ ( \sin(x/2) + \cos(x/2) )^2 }{ ( \sin(x/2) - \cos(x/2))^2}

     = \tan^2(x/2 + \pi/4)
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