Notice that [tex]C[/tex] is traversed clockwise. Green's theorem applies to curves with a counterclockwise orientation, so we'll have to multiply the area integral by -1.
By Green's theorem, with the vector field [tex]\vec F(x,y)=P(x,y)\,\vec\imath+Q(x,y)\,\vec\jmath[/tex],
[tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec r=-\iint_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,\mathrm dA[/tex]
where [tex]D[/tex] is the region with boundary [tex]C[/tex]. The partial derivatives are
[tex]\dfrac{\partial(e^{-y}+x^2)}{\partial x}=2x[/tex]
[tex]\dfrac{\partial(e^{-x}+y^2)}{\partial y}=2y[/tex]
so that the double integral is
[tex]\displaystyle\iint_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,\mathrm dA=\int_{-\pi/2}^{\pi/2}\int_0^{\cos x}2(y-x)\,\mathrm dy\,\mathrm dx=\boxed{\frac\pi2}[/tex]