Integral of 1 - 2*sinx in square root?

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Discussion Overview

The discussion revolves around the integral of the function \(\sqrt{1 - 2\sin(x)}\) over the interval \(0 < x < \frac{\pi}{6}\). Participants explore methods for solving this integral, including the use of elliptic integrals and potential transformations to standard forms.

Discussion Character

  • Mathematical reasoning
  • Exploratory
  • Technical explanation

Main Points Raised

  • One participant seeks a method to solve the integral \(\int\sqrt{1 - 2\sin(x)}dx\) and its general form \(\int\sqrt{a - b\sin(x)}dx\) for \(a > b\).
  • Another participant mentions that Mathematica provides a solution involving an elliptic integral, but expresses a need for a simpler solution without imaginary parts.
  • A different participant asserts that the solution does not contain imaginary parts, stating that the imaginary terms cancel out, and presents a formal expression involving the incomplete elliptic function.
  • One participant notes the necessity of transformations to convert the integral into a standard form of elliptic function, cautioning against confusion with notation.

Areas of Agreement / Disagreement

Participants express differing views on the nature of the solution, particularly regarding the presence of imaginary parts and the need for transformations. There is no consensus on a simpler method or form for the integral.

Contextual Notes

The discussion highlights the complexity of the integral and the potential need for transformations to relate it to known forms of elliptic integrals. Some assumptions about the nature of the functions involved may not be fully explored.

coki2000
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Hello PF members,

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

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

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

How can I solve this?

Thanks in advance
 
Last edited:
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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)
 
JJacquelin said:
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)

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.
 
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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Thank you very much for your help :)
 

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