$I=\int\limits_{0}^{2\pi }\sqrt{2cos^2x}dx$$=\sqrt{2}\int\limits_{0}^{\pi /2}|cosx|dx+\sqrt{2}\int\limits_{\pi /2}^{\pi }|cosx|dx+\sqrt{2}\int\limits_{\pi }^{3\pi /2}|cosx|dx+\sqrt{2}\int\limits_{3\pi /2}^{2\pi }|cosx|dx$
$=\sqrt{2}\int\limits_{0}^{\pi /2}cosxdx-\sqrt{2}\int\limits_{\pi /2}^{\pi }cosxdx-\sqrt{2}\int\limits_{\pi }^{3\pi/2}cosxdx+\sqrt{2}\int\limits_{3\pi /2}^{2\pi }cosxdx$
tự làm nốt nha