How to evaluate $\int_0^{2\pi}\sqrt{2-2\cos t}$

Question

I having some problems with the integration of $\sqrt{2-2\cos t}$. I got it by calculating the length of the arc

\begin{align} \gamma (t) &= (t - \sin t,1 - \cos t), \quad t\in[0,2\pi] \\ \implies \dot{\gamma(t)} &= (1 - \cos t, \sin t) \\ \implies \vert\vert\dot{\gamma(t)}\vert\vert &= \sqrt{(1 - \cos t)^2+(\sin t)^2} \\ &= \sqrt{2-2\cos t} \end{align}

How can I do?

Answer 1

We have,

$\sqrt2 \sqrt{1 - \cos t }$

= $\sqrt2 \sqrt{2 \sin^2 \frac{t}{2}}$

= $ 2 \sin \frac{t}{2}$

Now integrate it.

= 2 $\int_0^{2\pi} \sin \frac{t}{2}$