Answer
$$\frac{1}{3}{\left( {\tan x} \right)^3} + C $$
Work Step by Step
$$\eqalign{
& \int {{{\sec }^2}x} {\tan ^2}xdx \cr
& = \int {{{\tan }^2}x{{\sec }^2}x} dx \cr
& {\text{Integrate by substitution method}} \cr
& {\text{Let }}u = \tan x,\,\,\,\,du = {\sec ^2}xdx,\,\,\,\,dx = \frac{{du}}{{{{\sec }^2}x}} \cr
& {\text{Then}}{\text{,}} \cr
& \int {{{\tan }^2}x{{\sec }^2}xdx} = \int {{u^2}{{\sec }^2}x\left( {\frac{{du}}{{{{\sec }^2}x}}} \right)} \cr
& {\text{Cancel common factor }}{\sec ^2}x \cr
& = \int {{u^2}} du \cr
& {\text{integrating}} \cr
& = \frac{1}{3}{u^3} + C \cr
& {\text{write in terms of }}x \cr
& = \frac{1}{3}{\left( {\tan x} \right)^3} + C \cr} $$
