Answer
total area $ = {\rm{e}} - {{\rm{e}}^{ - 1}}$
Work Step by Step
Notice from Figure 2, the right-hand side of the curve from the $y$-axis resides in the first and and the third quadrant. So, the limits of integration is $ - \frac{\pi }{2}$ and $\frac{\pi }{2}$. Let $A$ denote the region enclosed by this part of the curve. Using Eq. (2) of Theorem 1 of Section 12.4, the area of $A$ is
$area = \frac{1}{2}\cdot\mathop \smallint \limits_{ - \pi /2}^{\pi /2} \cos \theta {{\rm{e}}^{\sin \theta }}{\rm{d}}\theta $
Let $t=\sin \theta$. So, $dt = \cos \theta d\theta $. The integral becomes
$area = \frac{1}{2}\cdot\mathop \smallint \limits_{ - 1}^1 {{\rm{e}}^t}{\rm{d}}t$
$area = \frac{1}{2}{{\rm{e}}^t}|_{ - 1}^1 = \frac{1}{2}\left( {{\rm{e}} - {{\rm{e}}^{ - 1}}} \right)$
By symmetry, total area enclosed by the curve is twice the area of $A$. So,
total area $ = {\rm{e}} - {{\rm{e}}^{ - 1}}$
