Question is: Find the length of the curve x = t-sint, y = 1-cost for 0<=t<=2pi

I've started the question using the formula $\displaystyle \int_{0}^{2\pi}\sqrt{{(\frac{dx}{dt}})^2 + (\frac{dy}{dt})^2}} dt$

differentiated and subbed back into the formula: $\displaystyle \int_{0}^{2\pi}\sqrt{{(1-cost)^2 + (sint)^2}} dt$

expanded simplified using trig identity: $\displaystyle \int_{0}^{2\pi}\sqrt{{2-2cost}} dt$

did some u substitution. u = 2-2cost du/dt=2sint du=2sint dt

$\displaystyle \int_{0}^{2\pi}\sqrt{{u}} dt$

this is where im getting stuck. I'm trying to change to dt to a du and i get:

$\displaystyle \int_{0}^{2\pi}\frac{\sqrt{u}}{sint} du$

looking at wolframalpha I should be getting

integrate sqrt(2-2cost) dt - Wolfram|Alpha

$\displaystyle \int_{0}^{2\pi}- \frac{1}{\sqrt{4-u}} du$

can anyone help me see how they completed this step please?

Interestingly the final solution to the question is zero(i think), is there any insight that would mean you wouldnt need to complete the integration but come to the solution earlier.. ?

again, any help would be great(man latex code takes forever to write your first time) thanks