No, the integrand is not even odd.

In order to solve the problem, we will focus us on a specific theorem: Let $a \in \mathbb{R}$ and let $f, g$ be functions such that $f(-x) = f(x)$ and $g(-x)=-g(x)$ for all $x \in [-a, a].$ Then for all $b > 0$ : $\int_{-a}^a \frac{f(x)}{1+b^{\,g(x)}}\mathrm{d}x = \int_0^a f(x)\mathrm{d}x.$ In the description of the problem a link can be found to a proof of this theorem. However, I will get over the proof quickly. Notice that $\int_{-a}^a \frac{f(x)}{1+b^{\,g(x)}}\mathrm{d}x = \int_{-a}^0 \frac{f(x)}{1+b^{\,g(x)}}\mathrm{d}x + \int_{0}^a \frac{f(x)}{1+b^{\,g(x)}}\mathrm{d}x. \hskip{2em}(1)$ Using the substitution $u = -x$ on the first integral at the right hand side gives us the information that $\int_{-a}^0 \frac{f(x)}{1+b^{\,g(x)}}\mathrm{d}x = -\int_{a}^0 \frac{f(-u)}{1+b^{\,g(-u)}}\mathrm{d}u = \int_{0}^a \frac{f(u)}{1+b^{-g(u)}}\mathrm{d}u = \int_{0}^a \frac{f(u)b^{\,g(u)}}{1+b^{\,g(u)}}\mathrm{d}u = \int_{0}^a \frac{f(x)b^{\,g(x)}}{1+b^{\,g(x)}}\mathrm{d}x.$ We can thus combine both integrals in (1), which results in $\int_{-a}^a \frac{f(x)}{1+b^{\,g(x)}}\mathrm{d}x = \int_0^a \frac{f(x) + f(x)b^{\,g(x)}}{1+b^{\,g(x)}}\mathrm{d}x = \int_0^a \frac{f(x)(1+b^{\,g(x)})}{1+b^{\,g(x)}}\mathrm{d}x = \int_0^a f(x)\mathrm{d}x. \hskip{2em} \blacksquare$

Now, note that we can make use of this theorem to see that $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2\cos(x)}{1+e^{x^2\sin(x)}}\mathrm{d}x = \int_0^\frac{\pi}{2} x^2\cos(x)\mathrm{d}x,$ as the functions $x^2\cos(x)$ and $x^2\sin(x)$ are even and odd, respectively. Using repeated integration by parts we can show that $\int_0^\frac{\pi}{2} x^2\cos(x)\mathrm{d}x = \left[ x^2\sin(x) + 2x\cos(x) - 2\sin(x) \right]\bigg|_0^{\frac{\pi}{2}} = \frac{\pi^2}{4}-2 = \frac{\pi^2-8}{4},$ from which follows that $a+b+c= 2 + 8 + 4 = \boxed{14}.$

Let $f(x) = \frac{x^2 \cos(x)}{1+e^{x^2 \sin(x)}}.$ Then, we can write $f(x)$ as the sum of an even and odd function: $f(x) = \frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2}.$ The latter term vanishes due the symmetry of the bounds. We calculate $\frac{f(x)+f(-x)}{2} = \frac{x^2\cos(x)}{2}$ and so we can use integration by parts to obtain $\frac{\pi^2 - 8}{4}$ as explained Joel's solution.