Question

$\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\left|\sin x-\cos x\right|dx$

Answer 1

$$\begin{gathered} {\int }_0^{\frac{\mathrm{\pi }}{2}}\left|\sin x-\cos x\right|dx \\ =\sqrt{2}{\int }_0^{\frac{\mathrm{\pi }}{2}}\left|\sin x\frac{1}{\sqrt{2}}-\cos x\frac{1}{\sqrt{2}}\right|dx \\ =\sqrt{2}{\int }_0^{\frac{\mathrm{\pi }}{2}}\left|\sin x\cos \frac{\mathrm{\pi }}{4}-\cos x\sin \frac{\mathrm{\pi }}{4}\right|dx \\ =\sqrt{2}{\int }_0^{\frac{\mathrm{\pi }}{2}}\left|\sin \left(x-\frac{\mathrm{\pi }}{4}\right)\right|dx \\ \mathrm{We}\mathrm{have}, \\ \left|\sin \left(x-\frac{\mathrm{\pi }}{4}\right)\right|=\left\{\begin{array}{l}-\sin \left(x-\frac{\mathrm{\pi }}{4}\right),0\le x\le \frac{\mathrm{\pi }}{4}\\ \sin \left(x-\frac{\mathrm{\pi }}{4}\right),\frac{\mathrm{\pi }}{4}\le x\le \frac{\mathrm{\pi }}{2}\end{array}\right. \\ \therefore {\int }_0^{\frac{\mathrm{\pi }}{2}}\left|\sin x-\cos x\right|dx=\sqrt{2}{\int }_0^{\frac{\mathrm{\pi }}{4}}-\sin \left(x-\frac{\mathrm{\pi }}{4}\right)dx+\sqrt{2}{\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}\sin \left(x-\frac{\mathrm{\pi }}{4}\right)dx \\ =\sqrt{2}{\left[\cos \left(x-\frac{\mathrm{\pi }}{4}\right)\right]}_0^{\frac{\mathrm{\pi }}{4}}-\sqrt{2}{\left[\cos \left(x-\frac{\mathrm{\pi }}{4}\right)\right]}_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}} \\ =\sqrt{2}\left[\cos \left(0\right)-\cos \left(-\frac{\mathrm{\pi }}{4}\right)\right]-\sqrt{2}\left[\cos \left(\frac{\mathrm{\pi }}{4}\right)-\cos \left(0\right)\right] \\ =\sqrt{2}\left(1-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}+1\right) \\ =\sqrt{2}\left(2-\frac{2}{\sqrt{2}}\right) \\ =2\sqrt{2}-2 \\ =2\left(\sqrt{2}-1\right) \end{gathered}$$
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