Turns out that you don't need to evaluate the integral of sqrt(cos(x)).
I've had a quick chat with a mate of mine (transcipt below) who has a solution and gets the same answer
Hope this helps.
WB: Actually, that integral comes out quite nice. The indefinite integral turns out to be 2*Sqrt(1+Cos(x))*Tan(x/2)
portej05: How come integrals.wolfram.com throughs out something hideous?
WB: Probably because you didn't enter exactly what I was evaluating. Note that in the previous expression, I had a y'(x) ... where y(x) is the integral from the forum.
portej05:
http://integrals.wolfram.com/index.jsp?expr=Sqrt(Cos(x))
WB: Exactly, and then you take the derivative of that (Sqrt(Cos[x))), square it (Cos(x)) and you're left with Integrate(Sqrt(1+Cos(x)),{x,x_1,x_2}), which evaluates nicely.
portej05:ah, I see
WB: Actually, I may have just figured out the last bit of that question. The integral is complex if x isn't in (-pi/2,pi/2)...