Question

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

Answer 1

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

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

$=\int (\cos -\sin x)dx$

$=\sin x+\cos {]}^{\pi /2}$

$=\sin \frac{\pi }{2}+\cos \frac{\pi }{2}-[\sin o+\cos x]$

$=1+0-[0+1]$

$=1-1=0$