Answer
$\frac{tan^{3}x}{3}+C$
Work Step by Step
Plug in: t=tan x so that $dx=\frac{dt}{sec^{2}x}$ in the given integral. Then
$\int sec^{2}x tan^{2}xdx= \int sec^{2}x t^{2}\frac{dt}{sec^{2}x}$
$=\int t^{2}dt= \frac{t^{3}}{3}+C$
Undoing substitution, we have
$\int sec^{2}x tan^{2}xdx=\frac{tan^{3}x}{3}+C$
