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