Question

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}|\sin x-\cos x|dx$ is equal to

• A. $0$
• B. $2(\sqrt{2}-1)$
• C. $\sqrt{2}-1$
• D. $2(\sqrt{2}+1)$

Answer 1

The correct option is B $2(\sqrt{2}-1)$

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}|\sin x-\cos x|dx=\underset{0}{\overset{\frac{\pi }{4}}{\int }}-(\sin x-\cos x)dx+\underset{\frac{\pi }{4}}{\overset{\frac{\pi }{2}}{\int }}(\sin x-\cos x)dx={[\cos x+\sin x]}_0^{\frac{\pi }{4}}-{[\cos x+\sin x]}_{\frac{\pi }{4}}^{\frac{\pi }{2}}=[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1-0]-[0+1-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}]=2\sqrt{2}-2=2(\sqrt{2}-1)$