Ta có:
$\int\limits_{0}^{2\pi}\sqrt{1-\cos2x}dx$
$=\int\limits_0^{2\pi}\sqrt2|\sin x|dx$
$=\int\limits_0^{\pi}\sqrt2\sin xdx-\int\limits_{\pi}^{2\pi}\sqrt2\sin xdx$
$=-\sqrt2\cos x\left|\begin{array}{l}\pi\\0\end{array}\right.+\sqrt2\cos x\left|\begin{array}{l}2\pi\\\pi\end{array}\right.=4\sqrt2$