= Solution
On an <interval> where $\tan x$ is defined, use the <substitution rule> $t=\tan x$. Then $dx=dt/(1+t^2)$ and $\cos^2x=1/(1+t^2)$, giving
$$
\int\frac{dx}{1+3\cos^2x}=\int\frac{dt}{t^2+4}=\boxed{\frac12\arctan\left(\frac{\tan x}{2}\right)+C.}
$$
The <inverse tangent> expression is valid on each <interval> $(j\pi-\pi/2,j\pi+\pi/2)$ separately. The original integrand is continuous for all real $x$, since its denominator is at least $1$. To obtain one continuous <antiderivative> on the whole real axis, add $j\pi/2$ on the $j$th such <interval> and use the continuous limits at their endpoints.
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