are these integrals same $$\int_{0}^{2\pi}|\sin x - \cos x|\,dx.$$
$$\int_{0}^{2\pi} \sqrt{1-\sin 2x}\,dx$$
If in the second integration I write $(\sin x - \cos x)^2$, after removing root it will become $|\sin x - \cos x|$, but online calculator calculation gives $\sqrt{32}$ as answer of first one and second integration answer is $0$? Why? Please explain. I am new to integration.
$$|\sin x - \cos x|=\sqrt{(\sin x-\cos x)^2}=\sqrt{\sin^2x+\cos^2x-2\sin x\cos x}=\sqrt{1-\sin 2x}$$