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$I=\int\limits_{0}^{2\pi }\sqrt{2cos^22x}dx$$=\sqrt{2}\int\limits_{0}^{\pi /2}|cos2x|dx+\sqrt{2}\int\limits_{\pi /2}^{\pi }|cos2x|dx+\sqrt{2}\int\limits_{\pi }^{3\pi /2}|cos2x|dx+\sqrt{2}\int\limits_{3\pi /2}^{2\pi }|cos2x|dx$$=\sqrt{2}\int\limits_{0}^{\pi /2}cos2xdx-\sqrt{2}\int\limits_{\pi /2}^{\pi }cos2xdx-\sqrt{2}\int\limits_{\pi }^{3\pi/2}cos2xdx+\sqrt{2}\int\limits_{3\pi /2}^{2\pi }cos2xdx$tự làm nốt nha

$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