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This type of integral can be handled easily using contour integration. The substitution $z = e^{ix}$ converts the integral into one of a rational function of $z$ around the unit circle, centre $0$ , which is then equal to $2\pi i$ times the sum of the residues of the integrand at singularities of modulus less than $1$ . Thus $\begin{array}{rcl} \displaystyle \int_0^{2\pi} \frac{1}{6\cos x - 10}\,dx & = & \displaystyle \int_{|z|=1} \frac{1}{3z + 3z^{-1} - 10} \,\frac{dz}{iz} \; = \; \frac{1}{i} \int_{|z|=1} \frac{1}{3z^2 - 10z + 3}\,dz \\ & = & \displaystyle \frac{1}{i}\int_{|z|=1} \frac{dz}{(3z-1)(z-3)} \; = \; 2\pi \,\mathrm{Res}_{z=\frac13} \frac{1}{(3z-1)(z-3)} \\ & = & \displaystyle \frac{2\pi}{3(\frac13-3)} \; =\; -\tfrac14\pi \end{array}$ making the answer $\boxed{4}$ .

$\begin{aligned} I & = \int_0^{2 \pi} \frac{dx}{6\cos x - 10} \quad \quad \small \color{#3D99F6}{\text{Since the integrand is even}} \\ & = \color{#3D99F6}{2} \int_0^{\color{#3D99F6}{\pi}} \frac{dx}{6\color{#D61F06}{\cos x} - 10} \quad \quad \small \color{#D61F06}{\text{Let }t = \tan \left(\frac{x}{2}\right) \implies dt = \frac{1+t^2}{2}dx \implies \cos x = \frac{1-t^2}{1+t^2}} \\ & = 2 \int_0^{\pi} \frac{dx}{6\left(\color{#D61F06}{\frac{1-t^2}{1+t^2}}\right) - 10} \\ & = 2 \int_0^{\pi} \frac {\color{#D61F06}{(1+ t^2) dx}}{\color{#D61F06}{2}\left(3-3t^2-5-5t^2\right)} \\ & = 2 \int_0^{\color{#D61F06}\infty} \frac {-\color{#D61F06}{dt}}{2+8t^2} \\ & = - \int_0^{\infty} \frac{d\color{#3D99F6}{t}}{1+\color{#3D99F6}{4t^2}} \quad \quad \small \color{#3D99F6}{\text{Let } u = 2t \implies du = 2 dt } \\ & = - \color{#3D99F6}{\frac{1}{2}} \int_0^{\infty} \frac{d\color{#3D99F6}{u}}{1+\color{#3D99F6}{u^2}} \\ & = - \frac{\tan^{-1} u}{2} \bigg|_0^\infty = - \frac{1}{2} \times \frac{\pi}{2} = - \frac{\pi}{4} \end{aligned}$

$\implies -A = \boxed{4}$