Let's do some calculus! (62)

The integral is $\begin{aligned}\int_{-\frac12\pi}^{\frac12\pi} \frac{e^x \cos(\tan^2x)}{(1 + \cos2x)(e^x+1)}\,dx &= \; \tfrac12\int_{-\frac12\pi}^{\frac12\pi} \frac{\cos(\tan^2x)}{1+ \cos2x}\,dx \\ & = \; \tfrac14\int_{-\frac12\pi}^{\frac12\pi}\cos(\tan^2x)\,\sec^2x\,dx \\ & = \; \tfrac14\int_{-\infty}^\infty \cos(x^2)\,dx \; = \; \tfrac{\sqrt{\pi}}{4\sqrt{2}} \end{aligned}$ for a solution of $1 + 4 + 2 = \boxed{7}$ .

We have

$I = \int_{\frac {-\pi}{2}}^{\frac {\pi}{2}} \frac {e^{x}\cos(\tan^{2}x)}{(1+\cos2x)(1+e^{x})} dx = \frac {1}{2}\int_{\frac {-\pi}{2}}^{\frac {\pi}{2}} \frac{\cos(\tan^{2}x)}{\cos^{2}x(1+e^{-x})} dx$

Now, using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+ b-x) dx$ ,

$I = \frac {1}{2} \int_{\frac {-\pi}{2}}^{\frac {\pi}{2}} \frac {\cos(\tan^{2}x)}{\cos^{2}x(1 + e^{x})} dx$

as $\tan^{2}x$ and $\cos^{2}x$ are both even functions.

$\therefore 2I = \frac {1}{2} \int_{\frac {-\pi}{2}}^{\frac {\pi}{2}} \frac {(e^{x}+1)\cos(\tan^{2}x)}{\cos^{2}x(1+e^{x})} = \int_{0}^{\frac {\pi}{2}} \frac {\cos(\tan^{2}x)}{(\cos^{2}x)}$

Substituting $\tan x = u$ , we get the Fresnel integral as $x \to \infty$

$2I = \int_{0}^{\infty} \cos(u^{2}) du = \sqrt {\frac {\pi}{8}} = \frac {\sqrt {\pi}}{2\sqrt{2}} \Rightarrow I = \frac {\sqrt {\pi}}{4\sqrt {2}}$

$\boxed{\therefore \Large {a + b + c = 7}}$