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