Question

Evaluate the integral

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

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

Answer 1

Answer: (A)

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

${\int }_0^{\frac{\pi }{2}}|cosx-sinx|dx$

$\sqrt{2}{\int }_0^{\frac{\pi }{2}}|sin(x-\frac{\pi }{4})|dx$

$=-2{\int }_0^{\frac{\pi }{4}}sin(x-\frac{\pi }{4})dx+\sqrt{2}{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}sin(x-\frac{\pi }{4})dx$

$=\sqrt{2}cos{(x-\frac{\pi }{4})}_0^{\frac{\pi }{4}}-\sqrt{2}cos{(x-\frac{\pi }{4})}_{\frac{\pi }{4}}^{\frac{\pi }{2}}$

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

$=\sqrt{2}\left(2\right)[1-\frac{1}{\sqrt{2}}]$
$=2(\sqrt{2}-1)$