Chapter 10 - Parametric Equations and Polar Coordinates - 10.4 Areas and Lengths in Polar Coordinates - 10.4 Exercises - Page 713: 22

$$A=(2-\frac{\pi}{2})$$

Given: $r=2cos\theta-sec\theta$ It can be rewritten as: $r^{2}=2cos(2\theta)-2+sec^{2}\theta$ $$A=2.\frac{1}{2}\int_{0}^{\pi/4}[2cos(2\theta)-2+sec^{2}\theta]d \theta=[sin(2\theta)-2\theta+tan\theta]_{0}^{\pi/4}$$ Hence, $$A=(2-\frac{\pi}{2})$$