Periodic

If $\omega = -\tfrac12 + \tfrac12i\sqrt{3}$ is a primitive cube root of unity, then $\begin{aligned} \int_0^{2\pi} \frac{dx}{\sin x - \omega} & = \; \int_{|z|=1} \frac{dz}{iz\left[\big(\frac{z-z^{-1}}{2i}\big) - \omega\right]} \; = \; 2\int_{|z|=1} \frac{dz}{z^2 - 2i\omega z - 1} \\ & = \; 2\int_{|z|=1} \frac{dz}{(z-\alpha_+)(z-\alpha_-)} \; = \; 4\pi i\,\mathrm{Res}_{z=\alpha_+} \frac{1}{(z-\alpha_+)(z-\alpha_-)} \; = \; \frac{4\pi i}{\alpha_+ - \alpha_-} \end{aligned}$ where $\alpha_{\pm} \; = \; i\omega \pm \sqrt{1-\omega^2} \; = \; \frac{\pm\sqrt{3+2\sqrt{3}}-\sqrt{3}}{2} + i\frac{\pm\sqrt{2\sqrt{3}-3} - 1}{2}$ noting that $|\alpha_+| < 1 < |\alpha_-|$ . Thus $\begin{aligned} \int_0^{2\pi} \frac{dx}{\sin x - \omega} & = \; \frac{4\pi i}{\sqrt{3+2\sqrt{3}} + i\sqrt{2\sqrt{3}-3}} \; = \; \frac{\pi i}{\sqrt{3}}\Big[\sqrt{3+2\sqrt{3}} - i\sqrt{2\sqrt{3}-3}\Big] \end{aligned}$ Thus $\begin{aligned} \int_0^{2\pi} \frac{dx}{\sin^2x + \sin x + 1} & = \; \int_0^{2\pi} \frac{dx}{(\sin x - \omega)(\sin x - \omega^2)} \; = \; \frac{1}{\omega - \omega^2}\int_0^{2\pi} \left(\frac{1}{\sin x - \omega} - \frac{1}{\sin x - \omega^2}\right) \\ & = \; \frac{1}{i\sqrt{3}} \times 2i \,\mathrm{Im}\left(\int_0^{2\pi} \frac{dx}{\sin x - \omega}\right) \; = \; \frac{2}{\sqrt{3}} \times \frac{\pi}{\sqrt{3}}\sqrt{3 + 2\sqrt{3}} \\ & = \; \tfrac23\pi\sqrt{3+ 2\sqrt{3}} \end{aligned}$ so that $p=2$ and $q=3$ .

Since the arc-length is the integral of $g(x) = \sqrt{1 + f'(x)^2}$ with respect to $x$ , where $f(x) = \frac{1}{\sin^2x + \sin x + 1}$ is a periodic function of $x$ of period $2\pi$ , the function $g(x)$ is also periodic of period $2\pi$ , and hence the arc-length over an interval of length $2\pi$ is independent of that interval's position. This makes the answer $2 + 3 + 1 = \boxed{6}$ .