Ta có:
$\int\limits_0^{2\pi}\sqrt{1+\sin2x}dx$
$=\int\limits_0^{2\pi}\sqrt{(\sin x+\cos x)^2}dx$
$=\int\limits_0^{2\pi}|\sin x+\cos x|dx$
$=\int\limits_0^{\frac{3\pi}{4}}(\sin x+\cos x)dx-\int\limits_{\frac{3\pi}{4}}^{\frac{7\pi}{4}}(\sin x+\cos x)dx+\int\limits_{\frac{7\pi}{4}}^{2\pi}(\sin x+\cos x)dx$
$=(\sin x-\cos x)\left|\begin{array}{l}\dfrac{3\pi}{4}\\0\end{array}\right.-(\sin x-\cos x)\left|\begin{array}{l}\dfrac{7\pi}{4}\\\dfrac{3\pi}{4}\end{array}\right.+(\sin x-\cos x)\left|\begin{array}{l}2\pi\\\dfrac{7\pi}{4}\end{array}\right.=4\sqrt2$