next step: bring out 4 from root
next step: break down the bracket under the root and simplify
next step: you can bring out again 2 from root
Now one trick:
Now it's not so wild, isn't it?
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!