How Can You Integrate (1-a-cos x)^(-1/2) in Terms of Elliptic Integrals?

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Homework Statement


how does one integrate (1-a-cos x)-1/2, where a is an arbitrary constant?


Homework Equations


as above


The Attempt at a Solution


Thought of writing cos x as 1-2*(sin(\frac{x}{2}))2
then the integrand simplifies to [2*(sin(\frac{x}{2}))2 - a]-1/2...
But then...? Is there a more elegant way of integrating this?
 
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c299792458 said:

Homework Statement


how does one integrate (1-a-cos x)-1/2, where a is an arbitrary constant?


Homework Equations


as above


The Attempt at a Solution


Thought of writing cos x as 1-2*(sin(\frac{x}{2}))2
then the integrand simplifies to [2*(sin(\frac{x}{2}))2 - a]-1/2...
But then...? Is there a more elegant way of integrating this?

It is a non-elementary integral. Maple 9.5 gets:
J:=int(1/sqrt(b-cos(x)),x):simplify(J,symbolic);

- 2*EllipticF(cos(x/2),sqrt(2/(b+1)) )/sqrt(b+1),

where EllipticF is the incomplete elliptic integral of the first kind.

RGV
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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