The integral ${\int }_0^{\pi /2}|\sin x-\cos x|dx$ is equal to:

Let $I={\int }_0^{\pi /2}|\sin x-\cos x|dx$.The expression $\sin x-\cos x$ changes sign at $x=\pi /4$ in the interval $[0,\pi /2]$.For $0\le x<\pi /4$, $\cos x>\sin x⟹|\sin x-\cos x|=\cos x-\sin x$.For $\pi /4\le x\le \pi /2$, $\sin x\ge \cos x⟹|\sin x-\cos x|=\sin x-\cos x$.$I={\int }_0^{\pi /4}(\cos x-\sin x)dx+{\int }_{\pi /4}^{\pi /2}(\sin x-\cos x)dx$$I=[\sin x+\cos x{]}_0^{\pi /4}+[-\cos x-\sin x{]}_{\pi /4}^{\pi /2}$$I=[(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})-(0+1)]+[(-0-1)-(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}})]$$I=[\sqrt{2}-1]+[-1+\sqrt{2}]$$I=2\sqrt{2}-2=2(\sqrt{2}-1)$