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Integral of 1 - 2*sinx in square root?

  1. Mar 31, 2013 #1
    Hello PF members,

    I want to solve this integral but I cannot find a method.

    [itex]\int\sqrt{1 - 2sin(x)}dx[/itex] for 0 < x < ∏/6

    Or more generally [itex]\int\sqrt{a - bsin(x)}dx[/itex] for a > b

    How can I solve this?

    Thanks in advance
     
    Last edited: Mar 31, 2013
  2. jcsd
  3. Mar 31, 2013 #2
  4. Apr 1, 2013 #3
    There is no imaginary part in the solution.
    The imaginary terms cancel each other. Finally, there is no simpler closed form than Re(-2*i*E(1/4 , 4) ) where E is the incomplete elliptic function of the second kind and Re is the real part. It is the simplest formal expression.
    http://www.wolframalpha.com/input/?i=Re(-2*i*E(pi/4,4))
    Alternatively, you might express the solution on the form of infinite series (rather complicated), or on the form of a real number (approximative solution thanks to direct numerical integration)
     
  5. Apr 2, 2013 #4
    Thank you for your response. But in the definition of the elliptic integral, we have the square of sine in the square root. In my case it is not square of the k*sinx.
     
  6. Apr 2, 2013 #5
    Of course some transformations are necessary to bring back the integral to a standard form of elliptic function.
    Be carreful with confusing concurent notations.
     

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    Last edited: Apr 2, 2013
  7. Apr 2, 2013 #6
    Thank you very much for your help :)
     
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