u = 1 + cos(x)
-du/dx=sin(x) (Numerator once the square root is simplified)
Or does this not work because you need to consider both + and - sin(x)
I played around with this a bit more, and yes, your method can be made to work but you're right, you need to consider the fact that sin(x) changes from positive to negative in the interval [2pi/3, 2pi]. This is important because when we make the substitution u = 1 + cos(x) and end up with sqrt(sin^2(x)) / sqrt(1 + cos(x)), that integrand is always positive, so we can't just simplify sqrt(sin^2(x)) to sin(x).
So, using your method, you'd need to break it up into the sum of two definite integrals: one where you integrate sin(x) / sqrt(1 + cos(x)) over [2pi/3, pi] and another where you integrate sin(x) / sqrt(1 + cos(x)) over [pi, 2pi].