Question

The integral ${\int }_0^{\pi /2}|\sin x-\cos x|dx$ is equal $to:$

Answer 1

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

First look at the graph of $\sin x$ & $\cot x$

Both the curve $\sin x$ & $\cos x$ are positive in $I^{st}$ quadrant.

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

$={\int }_0^{\pi /2}\sin xdx-{\int }_0^{\pi /2}\cos xdx$

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

$=-[\cos \frac{\pi }{2}-\cos 0]-[\sin \frac{\pi }{2}-\sin 0]$

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

$=1-1=0$.