x(t) = 2t - 2sin(t)

y(t) = 2 - 2cos(t)

t is from 0 to 2Pi

Okay, I am able to set up the problem correctly:

x'(t) = 2 - 2cos(t)

y'(t) = 2sin(t)

Length = Integral from 0 to 2Pi of (Sqrt[[2 - 2cos(t)]^2 + [2sin(t)]^2])

= Integral from 0 to 2Pi of (Sqrt[4[1-cos(t)]^2 + [2sin(t)]^2])

I know I can pull the 4 out (making it a 2), but I'm stuck after that. What good does making a substitution of 1 - cos(t) = 2sin(t/2)^2 do?

Help would be greatly appreciated!