## University Calculus: Early Transcendentals (3rd Edition)

$\frac{tan^{3}x}{3}+C$
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 $\int sec^{2}x tan^{2}xdx=\frac{tan^{3}x}{3}+C$