Answer
shaded area $ = \frac{{5\pi }}{8} - 1$
Work Step by Step
First, we find the intersection of the unit circle $r=1$ and the curve $r = 1 + \cos 2\theta $ by solving the equation
$r = 1 = 1 + \cos 2\theta $
$\cos 2\theta = 0$
The solutions are $\theta = \pm \frac{\pi }{4} + \pi n$, for $n = 0, \pm 1, \pm 2, \pm 3,...$
For our purpose we choose the solution in the first quadrant, that is, $\theta = \frac{\pi }{4}$. Let $A$ denote the shaded area in the first quadrant. From Figure 3 we see that for the first quadrant, the area $A$ is the area of the curve $r = 1 + \cos 2\theta $ minus the area between the unit circle and the curve $r = 1 + \cos 2\theta $. So,
$A = \frac{1}{2}\cdot\mathop \smallint \limits_0^{\pi /2} {\left( {1 + \cos 2\theta } \right)^2}{\rm{d}}\theta - \frac{1}{2}\cdot\mathop \smallint \limits_0^{\pi /4} \left( {{{\left( {1 + \cos 2\theta } \right)}^2} - {1^2}} \right){\rm{d}}\theta $
$A = \frac{1}{2}\cdot\mathop \smallint \limits_0^{\pi /2} {\left( {1 + \cos 2\theta } \right)^2}{\rm{d}}\theta + \frac{1}{2}\cdot\mathop \smallint \limits_{\pi /4}^0 {\left( {1 + \cos 2\theta } \right)^2}{\rm{d}}\theta + \frac{1}{2}\cdot\mathop \smallint \limits_0^{\pi /4} {\rm{d}}\theta $
$A = \frac{1}{2}\cdot\mathop \smallint \limits_{\pi /4}^{\pi /2} {\left( {1 + \cos 2\theta } \right)^2}{\rm{d}}\theta + \frac{1}{2}\cdot\mathop \smallint \limits_0^{\pi /4} {\rm{d}}\theta $
$A = \frac{1}{2}\cdot\mathop \smallint \limits_{\pi /4}^{\pi /2} \left( {1 + 2\cos 2\theta + {{\cos }^2}2\theta } \right){\rm{d}}\theta + \frac{\pi }{8}$
Since ${\cos ^2}2\theta = \frac{1}{2}\left( {1 + \cos 4\theta } \right)$, the integral becomes
$A = \frac{1}{2}\cdot\mathop \smallint \limits_{\pi /4}^{\pi /2} \left( {1 + 2\cos 2\theta + \frac{1}{2}\left( {1 + \cos 4\theta } \right)} \right){\rm{d}}\theta + \frac{\pi }{8}$
$A = \frac{1}{2}\left( {\frac{3}{2}\theta + \sin 2\theta + \frac{1}{8}\sin 4\theta } \right)|_{\pi /4}^{\pi /2} + \frac{\pi }{8}$
$A = \frac{1}{2}\left( {\frac{{3\pi }}{4} - \frac{{3\pi }}{8} - 1} \right) + \frac{\pi }{8} = \frac{{5\pi }}{{16}} - \frac{1}{2}$
By symmetry, the total shaded area is twice the area $A$. So,
shaded area $ = \frac{{5\pi }}{8} - 1$
