Integral of 1 - 2*sinx in square root?

[coki2000]

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

 

[coki2000]

Mathematica gives a solution with an elliptic integral but since my operation interval does not require any imaginary part, it is not useful for me.

http://www.wolframalpha.com/input/?i=int[sqrt{1+-+2sin(x)}dx]

So I need another and simpler solution.

 

[JJacquelin]

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)

 

[coki2000]

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.

 

[JJacquelin]

Of course some transformations are necessary to bring back the integral to a standard form of elliptic function.
Be carreful with confusing concurent notations.

 

[coki2000]

Thank you very much for your help :)