Answer
$=\frac{\pi-1}{2}$
Work Step by Step
$r=2 sin \theta$ and $r=sin \theta+cos \theta$
$Area,A=
\frac{1}{2}\int_{0}^{\pi/4}2sin\theta)^2d\theta+\frac{1}{2}\int_{\pi/4}^{3\pi/4}(\sqrt 2sin( \theta+ \pi/4))^2d\theta$
$=[\theta-\frac{1}{2}sin 2\theta]_{0}^{\pi/4}+[\frac{1}{2} \theta -\frac{1}{4} sin (2 \theta+ \pi/2)]_{\pi/4}^{3\pi/4}$
$=\frac{\pi-1}{2}$
