Question

The value of the integral ${\int }_0^{\pi /2}|\sin x-\cos x|dx$ is

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

Answer 1

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

${\int }_0^{\pi /2}|\sin x-\cos x|dx={\int }_0^{\pi /4}-(\sin x-\cos x)dx+{\int }_{\pi /4}^{\pi /2}(\sin x-\cos x)dx$

$=(\cos x+\sin x{)}_0^{\pi /4}+(-\cos x-\sin x{)}_{\pi /4}^{\pi /2}$

$=(\frac{2}{\sqrt{2}}-1)+(-1+\frac{2}{\sqrt{2}})$

$=2(\frac{2}{\sqrt{2}}-1)=2(\sqrt{2}-1)$