A power, a cosine, an exponential, oh my!

First we should kknow that $\frac{1}{e^x+1}+\frac{1}{e^{-x}+1}=1$ Assume that $I =\displaystyle\int \limits^{\frac{\pi }{2} }_{\frac{-\pi }{2}}\frac{x^2 \cos(x) }{e^x+1} dx$ Letting $u=-x$ , you can show that $\displaystyle I=\int_{-\pi/2}^{\pi/2}\frac{x^2\cos x}{e^x+1}\,dx=\int_{-\pi/2}^{\pi/2}\frac{x^2\cos x}{e^{-x}+1}dx$ , hence $\displaystyle 2I=\int_{-\pi/2}^{\pi/2} x^2\cos x\,dx$ . Since $x^2\cos x$ is even, you have $\displaystyle I=\int_0^{\pi/2}x^2\cos xdx$ $I=\displaystyle\int_0^{\pi/2}x^2\cos xdx$ $I=\displaystyle\int_0^{\pi/2}x^2\cos xdx =\left.2x\cos x+(x^2-2)\sin x\,\right|_{0}^{\pi/2}\\ = \frac{\pi^2-8}{4}$ Hence $A +B +C =2+4+2=\boxed{8}$