$\begin{aligned} & \int_{0}^{\pi / 2}|\sin x-\cos x| d x \\ &=\int_{0}^{\pi / 4}(\sin x-\cos x) \cdot d x+\int_{\pi / 4}^{\pi / 2}(\sin x-\cos x) d x \\ &=\int_{0}^{\pi / 4}(\cos x-\sin x) \cdot d x+\int_{\pi / 4}^{\pi / 2}(\sin x-\cos x) \cdot d x \\ &=(\sin x+\cos x)_{0}^{\pi / 4}+(-\cos x-\sin x)_{\pi / 4}^{\pi / 2} \\ &=2 \sqrt{2}-2=2(\sqrt{2}-1) \end{aligned}$