Why all integration problems have integrand?

$\displaystyle\int\limits_{-\pi/3}^{\pi/3} \frac{\pi + 4x^3}{2 - \cos^2(|x| + \frac{\pi}{3})} dx$ $=\pi\int\limits_{-\pi/3}^{\pi/3} \frac{1}{1 + 2sin^2(\frac{|x|}{2} + \frac{\pi}{6})} dx + \int\limits_{-\pi/3}^{\pi/3} \frac{4x^3}{1 + 2sin^2(\frac{|x|}{2} + \frac{\pi}{6})} dx$ The second integral is $0$ (zero), since it is an odd function.

We are left with first integral which is splitted into two parts....

$\displaystyle = \pi[\int\limits_{-\pi/3}^0 \frac{1}{1 + 2sin^2(\frac{-x}{2} + \frac{\pi}{6})} dx + \int\limits_0^{\pi/3} \frac{1}{1 + 2sin^2(\frac{x}{2} + \frac{\pi}{6})} dx]$

Substituting $\frac{-x}{2} + \frac{\pi}{6} = u$ and $\frac{x}{2} + \frac{\pi}{6} = v$ , we get

$=2\pi {\int\limits_{\pi/6}^{\pi/3} \frac{2}{1 + \sin^2u} du}$

because $\int_a^b f(u)du = \int_a^b f(v)dv$

Solving the integral , we easily get the result $=4\pi(\frac{4\tan^{-1}{3} - \pi}{4\sqrt{3}})$

which can be written in the form of $=\frac{4\pi}{\sqrt{3}} (\tan^{-1}{3} - \frac{\pi}{4})$

which on further simplification yeilds,

$=\frac{4\pi}{\sqrt{3}} \tan^{-1}{\frac{1}{2}}$

Therefore, $a+b+c = 4+3+2 = 9$