Answer
The area inside both curves:
$area = \frac{{9\pi }}{2} - 4\sqrt 2 $
Work Step by Step
First, we find the intersection points between the two curves $r = 2 + \sin 2\theta $ and $r = 2 + \cos 2\theta $ by solving the equation
$r = 2 + \sin 2\theta = 2 + \cos 2\theta $
$\sin 2\theta = \cos 2\theta $
The solution is in the form $2\theta = \frac{\pi }{4} + 2\pi n$ or $2\theta = - \frac{{3\pi }}{4} + 2\pi n$ for $n = 1,2,3,...$
For the interval $0 \le \theta \le 2\pi $, the solutions are $\theta = \frac{\pi }{8},\frac{{5\pi }}{8},\frac{{9\pi }}{8},\frac{{13\pi }}{8}$.
Next, we find part of the area of the curve $r = 2 + \cos 2\theta $ in the interval $\frac{\pi }{8} \le \theta \le \frac{{5\pi }}{8}$. Let $A$ denote this area. So,
$A = \frac{1}{2}\cdot\mathop \smallint \limits_{\pi /8}^{5\pi /8} {\left( {2 + \cos 2\theta } \right)^2}{\rm{d}}\theta $
$A = \frac{1}{2}\cdot\mathop \smallint \limits_{\pi /8}^{5\pi /8} \left( {4 + 4\cos 2\theta + {{\cos }^2}2\theta } \right){\rm{d}}\theta $
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 /8}^{5\pi /8} \left( {4 + 4\cos 2\theta + \frac{1}{2}\left( {1 + \cos 4\theta } \right)} \right){\rm{d}}\theta $
$A = \frac{1}{2}\left( {\frac{9}{2}\theta + 2\sin 2\theta + \frac{1}{8}\sin 4\theta } \right)|_{\pi /8}^{5\pi /8}$
$A = \frac{1}{2}\left( {\frac{{9\pi }}{4} + 2\sin \frac{{5\pi }}{4} - 2\sin \frac{\pi }{4} + \frac{1}{8}\sin \frac{{5\pi }}{2} - \frac{1}{8}\sin \frac{\pi }{2}} \right)$
$A = \frac{1}{2}\left( {\frac{{9\pi }}{4} - 2\sqrt 2 } \right)$
By symmetry, the area inside both curves are four times the area $A$. So,
$area = 4\cdot\frac{1}{2}\left( {\frac{{9\pi }}{4} - 2\sqrt 2 } \right) = \frac{{9\pi }}{2} - 4\sqrt 2 $
