Question

${\int }_0^{\pi /2}|sinx-cosx|dx$ is equal to

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

Answer 1

Answer: (C)

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

Consider, $I={\int }_0^{\pi /2}|\sin x-\cos x|dx$

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

$I={[\sin x+\cos x]}_0^{\pi /4}+{[-\cos x-\sin x]}_{\pi /4}^{\pi /2}$

$I=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-0-1-0-1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}$

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