Simple way to do an integral

I am trying to do ...

$\displaystyle \int_0^{2\pi}\sqrt{1-cos(t)}dt$

Here is what I did. Can you either comment on my work or post a better solution?

*Assume the domain of all functions is *$\displaystyle (0,\pi)$. *Let *$\displaystyle u=1-cos(t)$

$\displaystyle \displaystyle\int\sqrt{1-cos(x)}dx=\int\frac{sin(x)}{\sqrt{1+cos(x)}}dx=\in t\frac{1}{\sqrt{2-u}}du=-2\sqrt{1+cos(x)}$

*Therefore, *$\displaystyle \int_0^\pi\sqrt{1-cos(x)}dx=2\sqrt{2}$

*Since the integrand is symmetric about *$\displaystyle y=\pi$*, *

$\displaystyle \int_0^{2\pi}\sqrt{1-cos(t)}dt=2*2\sqrt{2}$