# Integral of 1 - 2*sinx in square root?

1. Mar 31, 2013

### 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?

Last edited: Mar 31, 2013
2. Mar 31, 2013

3. Apr 1, 2013

### 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)

4. Apr 2, 2013

### coki2000

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.

5. Apr 2, 2013

### 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.

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Last edited: Apr 2, 2013
6. Apr 2, 2013

### coki2000

Thank you very much for your help :)