Looks like a bird...

Finding the area nearest to the origin

hi

$\textbf{Finding intersection between the 2 functions}$

(We are only interested in the positive area)

Solving $sin(x)+cos(x)=sin(x)-cos(x)$ gives $x=\frac { \pi }{ 2 }$ There is more than 1 solution but we are only taking this solution as it is the only one useful. $.$ $.$ $\textbf{Finding the intersect between the x-axis and the two functions}$

We are doing so to be able to integrate with the proper limits later.

So, using some geometry, we are again able to find the intersections. $sin(x)+cos(x)=0$ We are only interested in the smallest negative solution. $sin(x)+cos(x)=0, opposite=-adjacent$ This is possible when ( $x=-\frac{\pi}{4}$ ). $.$ Again, using some geometry, we can solve $sin(x)-cos(x)=0$ Here, we are only interested in the smallest positive solution. $sin(x)-cos(x)=0, opposite=adjacent$ This is possible when $x=\frac{\pi}{4}$ . $.$ $.$ $\textbf{Integrating the area}$

So, using the solutions above, putting in the proper limits, the area of the shaded figure above is $\left[ \int _{ -\frac { \pi }{ 4 } }^{ \frac { \pi }{ 2 } }{ sin(x)+cos(x)dx } \right] -\left[ \int _{ \frac { \pi }{ 4 } }^{ \frac { \pi }{ 2 } }{ sin(x)+cos(x)dx } \right] =2$ However, this is only half of the area we want. So, $A=2\times2=4$ And since the question requires ( ${A}^{2}$ ), ${A}^{2}={4}^{2}=\boxed{16}$