Answer
$$\sqrt{2}-1$$
Work Step by Step
Our aim is to integrate the integral as follows:
$ Area =\iint_D dA $
$\int^{\pi/4}_{0} \int^{\cos (x)}_{\sin (x)} \space dy \space dx $
=$\int^{\pi/4}_{0} [\cos (x)-\sin (x)] \space dx $
or, $=[\sin (x)+\cos (x) ]^{\pi/4}_0$
or, $=(\dfrac{\sqrt{2}}{2}+ \dfrac{\sqrt{2}}{2})-(0+1)$
and $ Area=\sqrt{2}-1$
