 Solution

On an interval where $\tan x$ is defined, use the substitution rule $t=\tan x$. Then $dx=dt/(1+t^2)$ and ${\cos }^{2}x=1/(1+t^2)$, givingThe 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.