Given that y = cos(x) makes up part of the boundary of C, I suspect you mean the given points to be (-π/2, 0) and (π/2, 0).
I also assume the given vector field is
[tex]\vec F(x,y) = \left\langle e^{-x} + y^2, e^{-y} + x^2 \right\rangle[/tex]
Since [tex]\vec F[/tex] has no singularities on C or in its interior, Green's theorem applies:
[tex]\displaystyle \int_C \vec F(x,y) \cdot d\vec r = \iiint_D \frac{\partial(e^{-y}+x^2)}{\partial x} - \frac{\partial(e^{-x}+y^2)}{\partial y} \, dA = 2 \iiint_D (x + y) \, dA[/tex]
where D is the interior of C, the region
[tex]D = \left\{ (x, y) : -\dfrac\pi2 \le x \le \dfrac\pi2 \text{ and } 0 \le y \le \cos(x) \right\}[/tex]
The integral then reduces to
[tex]\displaystyle 2 \iint_D (x + y) \, dA = 2 \int_{-\frac\pi2}^{\frac\pi2} \int_0^{\cos(x)} (x + y) \, dy \, dx[/tex]
[tex]\displaystyle 2 \iint_D (x + y) \, dA = 2 \int_{-\frac\pi2}^{\frac\pi2} \left( x\cos(x) + \frac12 \cos^2(x) \right) \, dx[/tex]
[tex]\displaystyle 2 \iint_D (x + y) \, dA = 2 \int_{-\frac\pi2}^{\frac\pi2} \left( x\cos(x) + \frac{1 + \cos(2x)}4 \right) \, dx[/tex]
[tex]\displaystyle 2 \iint_D (x + y) \, dA = \frac12 \int_{-\frac\pi2}^{\frac\pi2} \left( 4x\cos(x) + 1 + \cos(2x) \right) \, dx[/tex]
Since 4x cos(x) is an odd function over the symmetric interval [-π/2, π/2], its contribution to the integral is 0, and the remaining integral is trivial.
[tex]\displaystyle 2 \iint_D (x + y) \, dA = \frac12 \int_{-\frac\pi2}^{\frac\pi2} \left( 1 + \cos(2x) \right) \, dx = \boxed{\frac\pi2} [/tex]
