Answer
$area = \frac{\pi }{2}$
Work Step by Step
Write
$r = 2\sin \left( {\theta + \frac{\pi }{4}} \right) = 2\left( {\sin \theta \cos \frac{\pi }{4} + \cos \theta \sin \frac{\pi }{4}} \right) = \sqrt 2 \left( {\sin \theta + \cos \theta } \right)$
and
$r = \sec \left( {\theta - \frac{\pi }{4}} \right) = \frac{1}{{\cos \left( {\theta - \frac{\pi }{4}} \right)}} = \frac{1}{{\cos \theta \cos \frac{\pi }{4} + \sin \theta \sin \frac{\pi }{4}}} = \frac{2}{{\sqrt 2 \left( {\cos \theta + \sin \theta } \right)}}$
We find the points of intersection between the two polar curves by solving the equation
$r = 2\sin \left( {\theta + \frac{\pi }{4}} \right) = \sec \left( {\theta - \frac{\pi }{4}} \right)$
So,
$\sqrt 2 \left( {\sin \theta + \cos \theta } \right) = \frac{2}{{\sqrt 2 \left( {\cos \theta + \sin \theta } \right)}}$
${\left( {\sin \theta + \cos \theta } \right)^2} = 1$
${\sin ^2}\theta + 2\sin \theta \cos \theta + {\cos ^2}\theta = 1$
$2\sin \theta \cos \theta = 0$
The solutions are $\theta = 0,\frac{\pi }{2}$.
Using Eq. (2) of Theorem 1, the area of the region inside the circle $r = 2\sin \left( {\theta + \frac{\pi }{4}} \right)$ and above the line $r = \sec \left( {\theta - \frac{\pi }{4}} \right)$ is
$area = \frac{1}{2}\cdot\mathop \smallint \limits_0^{\pi /2} \left( {{{\left( {2\sin \left( {\theta + \frac{\pi }{4}} \right)} \right)}^2} - {{\left( {\sec \left( {\theta - \frac{\pi }{4}} \right)} \right)}^2}} \right){\rm{d}}\theta $
$area = \frac{1}{2}\cdot\mathop \smallint \limits_0^{\pi /2} \left( {4{{\sin }^2}\left( {\theta + \frac{\pi }{4}} \right) - {{\sec }^2}\left( {\theta - \frac{\pi }{4}} \right)} \right){\rm{d}}\theta $
$area = 2\cdot\mathop \smallint \limits_0^{\pi /2} {\sin ^2}\left( {\theta + \frac{\pi }{4}} \right){\rm{d}}\theta - \frac{1}{2}\cdot\mathop \smallint \limits_0^{\pi /2} {\sec ^2}\left( {\theta - \frac{\pi }{4}} \right){\rm{d}}\theta $
Write $\alpha = \theta + \frac{\pi }{4}$ and $\beta = \theta - \frac{\pi }{4}$. So, $d\alpha = d\theta $ and $d\beta = d\theta $. The integral becomes
$area = 2\cdot\mathop \smallint \limits_{\pi /4}^{3\pi /4} {\sin ^2}\alpha {\rm{d}}\alpha - \frac{1}{2}\cdot\mathop \smallint \limits_{ - \pi /4}^{\pi /4} {\sec ^2}\beta {\rm{d}}\beta $
From the Table of Trigonometric Integrals of Section 8.2 on page 402, we have
$\smallint {\sin ^2}x{\rm{d}}x = \frac{x}{2} - \frac{{\sin 2x}}{4} + C$
and
$\smallint {\sec ^m}x{\rm{d}}x = \frac{{\tan x{{\sec }^{m - 2}}x}}{{m - 1}} + \frac{{m - 2}}{{m - 1}}\smallint {\sec ^{m - 2}}x{\rm{d}}x$
Thus,
$area = 2\left( {\frac{\alpha }{2} - \frac{{\sin 2\alpha }}{4}} \right)|_{\pi /4}^{3\pi /4} - \frac{1}{2}\tan \beta |_{ - \pi /4}^{\pi /4}$
$area = 2\left( {\frac{{3\pi }}{8} - \frac{\pi }{8} + \frac{1}{4} + \frac{1}{4}} \right) - \frac{1}{2}\left( 2 \right) = \frac{\pi }{2} + 1 - 1 = \frac{\pi }{2}$
