I've parametrized the cycloid with the function $$\gamma(t)=(\cos(\frac{3}{2}\pi-t)+t,\sin(\frac{3}{2}\pi-t)+1) ; \space t \in [0,+\infty)$$
I am asked to find the arc lenght of the curve which corresponds to one complete rotation of the disk.
So, the arc lenght is calculated by the integral $$\int_0^{2\pi} ||\gamma'(t)||dt=\int_0^{2\pi} \sqrt{(\sin(\frac{3}{2}\pi-t)+1)^2+(-\cos(\frac{3}{2}\pi-t))^2}dt.$$
By the change of variables $s=\frac{3}{2}\pi-t$, $ds=-dt$, the integral reduces to $$\int_{-\frac{\pi}{2}}^{\frac{3}{2}\pi} \sqrt{(\sin(s)+1)^2+(-\cos(s))^2}ds.$$
And by some simple calculations, this integral is the same as $$\int_{-\frac{\pi}{2}}^{\frac{3}{2}\pi} \sqrt{2+2\sin(s)}ds.$$
I got stuck trying to solve this integral, I can't think of any trig identity which could be useful. I would appreciate any help.
Note that $\sin\left(\frac{3\pi}{2}-t\right)=-\cos t$, so we are integrating $\sqrt{2-2\cos t}$.
For the integration, use the fact that $1-\cos t=2\sin^2(t/2)$. This is a version of the more familiar $\cos 2x=1-2\sin^2 x$.