Contorno of the day

$\sin^2{x}+\cos^2{t}=1 \Rightarrow 1+\tan^2{t}=\sec^2{t}$

By symettry of $\cos^2{x}$ :

$\int_{-\pi}^{\pi} \dfrac{1}{1+3 cos^2{t}} \; dt=4 \int_{\frac{\pi}{2}}^{\pi} \dfrac{1}{1+3 cos^2{t}}=4 \int_{\frac{\pi}{2}}^{\pi} \sec^2{t} \dfrac{1}{3+\sec^2{t}} \; dt=4 \int_{\frac{\pi}{2}}^{\pi} \sec^2{t} \dfrac{1}{4+\tan^2{t}} \; dt$

Use the substitution $u= \dfrac{\tan{t}}{2} \Rightarrow 2 du=\sec^2{t} \; dt$ :

$\cdots=\int_{\frac{\pi}{2}}^{\pi} \sec^2{t} \dfrac{1}{1+\left (\dfrac{\tan{t}}{2} \right)^2} \; dt=2 \int_{0}^{\infty} \dfrac{1}{1+u^2} \; du$

$\cdots= 2 \left [\arctan{u} \right]_{0}^{\infty}=2 \left (\dfrac{\pi}{2}-0 \right) =\boxed{\pi}$