Answer
$\frac{\pi}{2}-1$
Work Step by Step
$A=\frac{1}{2}\int_{0}^{\pi/8}(sin2\theta)^{2}d \theta+\frac{1}{2}\int_{\pi/8}^{\pi/4}(cos2\theta)^{2}d \theta$
$=2.\frac{1}{2}\int_{0}^{\pi/2}(sin2\theta)^{2}d \theta$
$=\frac{1}{2}[\theta-\frac{1}{4}sin4\theta]_{0}^{\pi/8}$
$=\frac{\pi}{16}-\frac{1}{8}$
Therefore the area of the region is,
$A=8(\frac{\pi}{16}-\frac{1}{8})=\frac{\pi}{2}-1$
