Answer
Area of the shaded region: $\frac{3}{8}\pi - 1$.
Work Step by Step
The shaded region corresponds to the interval: $0 \le \theta \le \frac{\pi }{2}$. So, the area of the shaded region is
$area = \frac{1}{2}\cdot\mathop \smallint \limits_0^{\pi /2} {\left( {1 - \cos \theta } \right)^2}{\rm{d}}\theta = \frac{1}{2}\cdot\mathop \smallint \limits_0^{\pi /2} \left( {1 - 2\cos \theta + {{\cos }^2}\theta } \right){\rm{d}}\theta $
Since ${\cos ^2}\theta = \frac{1}{2}\left( {1 + \cos 2\theta } \right)$, the integral becomes
$area = \frac{1}{2}\cdot\mathop \smallint \limits_0^{\pi /2} \left( {1 - 2\cos \theta + \frac{1}{2}\left( {1 + \cos 2\theta } \right)} \right){\rm{d}}\theta $
$area = \frac{1}{2}\left( {\theta - 2\sin \theta + \frac{1}{2}\theta + \frac{1}{4}\sin 2\theta } \right)|_0^{\pi /2}$
$area = \frac{1}{2}\left( {\frac{\pi }{2} - 2 + \frac{\pi }{4}} \right) = \frac{3}{8}\pi - 1$.
