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Reciprocal of Trig - Integral

  1. Feb 21, 2006 #1
    Hi I have a proof i'm doing

    [tex]
    \int \frac{1}{1+\sin(x)}dx
    [/tex]

    I know that the answer i'm looking for is

    [tex]
    \frac{\sin(x) - 1}{\cos(x)}
    [/tex]
    and then
    [tex]
    \tan(x) - \sec(x)
    [/tex]

    I have tried integration by parts making
    [tex]
    u = (1+\sin(x))^{-1}[/tex] and [tex]dv = dx
    [/tex]

    Eventually I get an answer that contains an ln and an unsolvable integral. I have been at this for 2 hours, can anyone give me a hint or a push in the right direction?
     
    Last edited: Feb 21, 2006
  2. jcsd
  3. Feb 21, 2006 #2
    And I will be fixing my latex right now...
     
  4. Feb 21, 2006 #3

    TD

    User Avatar
    Homework Helper

    It's possible to do a t-substitution (t = tan(x/2)) but you can also use a little trick.

    [tex]
    \int {\frac{1}
    {{1 + \sin x}}dx} = \int {\frac{{1 - \sin x}}
    {{\left( {1 + \sin x} \right)\left( {1 - \sin x} \right)}}} dx = \int {\frac{{1 - \sin x}}
    {{1 - \sin ^2 x}}} dx = \int {\frac{{1 - \sin x}}
    {{\cos ^2 x}}} dx
    [/tex]

    Now split the integral (so the fraction...) in two: the first part should be a standard integral and the second will go easily using an obvious substitution.
     
  5. Feb 21, 2006 #4
    I tried that t thing and it seemed too complicated for the level we are currently at. The trick worked, and I didn't have to use any substitution - just rearrange the sinx/(cosx)^2 to secxtanx which is an easy integral.

    Thanks for the fast response.
     
  6. Feb 21, 2006 #5

    TD

    User Avatar
    Homework Helper

    You're welcome.

    If you'd like to try it using the t-substitution too, post your work or ask for help :smile:
     
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