Let's do some calculus! (43)

$\int_{-\pi}^{\pi} \frac{1}{3\cos^{2}x+1}dx=4\int_{0}^{\frac{\pi}{2}}\frac{1}{3\cos^{2}x+1}dx=4\int_{0}^{\frac{\pi}{2}}\frac{\sec^{2}x}{4+\tan^{2}x}dx=4\int_{0}^{\infty}\frac{dt}{4+t^{2}}= \pi$

The integral is $I \; = \; \int_{-\pi}^{\pi}\frac{dx}{3\cos^2x + 1} \; = \; \int_{-\pi}^{\pi} \frac{2\,dx}{3\cos2x + 5} \; = \; \int_{-2\pi}^{2\pi}\frac{dx}{3\cos x + 5} \; = \; 2\int_{-\pi}^\pi \frac{dx}{3\cos x + 5}$ The substitution $t = \tan\tfrac12x$ yields $I \; = \; 2\int_{-\infty}^\infty \frac{2\,dt}{3(1-t^2) + 5(1+t^2)} \; = \; 2\int_{-\infty}^\infty \frac{dt}{4+t^2} \; = \; 2\Big[\tfrac12\tan^{-1}\tfrac12t\Big]_{-\infty}^\infty \; = \; \pi$ making the answer $\boxed{3.141592654}$ to 9 DP.