Answer
$\frac{\pi}{2}$
Work Step by Step
First let's graph our function and see our orientation. We draw the graph of cosx from $(\frac{-\pi}{2},0) to (\frac{\pi}{2},0)$ then the line segment from $(\frac{\pi}{2},0) to (\frac{-\pi}{2},0)$
When we graph this, we realize that our orientation is going clockwise meaning our orientation is negative, so before we set up our integral, there will be a negative sign in front of the integral.
Now for the values, x is trapped from $-\pi/2$ to $\pi/2$.
y is trapped from $0$ to $\cos x$.
That is if we were to do dydx
$∫cPdx+Qdy=∫∫D(dQdx−dPdy)DA$
We were given $F(x,y)=\lt e^- (^{x})+y^2 , e^{-y}+x^2 >$
$dQ/dx = 2x $
$dP/dy= 2y$
Our integral will be
$-\int _{\frac{-\pi}{2}}^{\frac{\pi}{2}}\int ^{\cos x}_{0} (2x-2y) dydx $
$-\int _{\frac{-\pi}{2}}^{\frac{\pi}{2}} |2xy-y^2|^{\cos x}_{0}dx $
$-\int _{\frac{-\pi}{2}}^{\frac{\pi}{2}} 2x\cos x-\cos^2xdx $
For $2x\cos x$, we do integration by part $\int udv=uv-\int vdu$
$u=2x ,dV=\cos x $
$du=2 , V=\sin x$
We will get
$-[|2x\sin x| _{\frac{-\pi}{2}}^{\frac{\pi}{2}} -\int _{\frac{-\pi}{2}}^{\frac{\pi}{2}} 2\sin x dx -\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos^2xdx]$
$-[\pi-(\pi)(-1)+|2\cos x|_{\frac{-\pi}{2}}^{\frac{\pi}{2}}-\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}1/2+\cos 2x/2]$
$-[-\frac{x}{2}+\frac{\sin 2x}{4}]_{\frac{-\pi}{2}}^{\frac{\pi}{2}}$=
$-[-\frac{\pi}{4}-\frac{\pi}{4}]$=
$-[\frac{-\pi}{2}]$=
$\frac{\pi}{2}$
