Answer
$$\frac{1}{2}\sec x\tan x - \frac{1}{2}\ln \left| {\sec x + \tan x} \right| + C $$
Work Step by Step
$$\eqalign{
& \int {\sec x} {\tan ^2}xdx \cr
& {\text{Use the fundamental identiy ta}}{{\text{n}}^2}x = {\sec ^2}x - 1 \cr
& \int {\sec x} {\tan ^2}xdx = \int {\sec x} \left( {{{\sec }^2}x - 1} \right)dx \cr
& {\text{Use distributive property}} \cr
& = \int {\left( {{{\sec }^3}x - \sec x} \right)dx} \cr
& {\text{Sum rule for integration}} \cr
& = \int {{{\sec }^3}xdx} - \int {\sec x} dx \cr
& \cr
& {\text{From example 6, we obtain: }} \cr
& \int {{{\sec }^3}xdx} = \frac{1}{2}\sec x\tan x + \frac{1}{2}\ln \left| {\sec x + \tan x} \right| + C \cr
& {\text{and }}\int {\sec x} dx = \ln \left| {\sec x + \tan x} \right| + C \cr
& {\text{Then}}{\text{,}} \cr
& \cr
& = \frac{1}{2}\sec x\tan x + \frac{1}{2}\ln \left| {\sec x + \tan x} \right| - \ln \left| {\sec x + \tan x} \right| + C \cr
& {\text{simplifying}} \cr
& = \frac{1}{2}\sec x\tan x - \frac{1}{2}\ln \left| {\sec x + \tan x} \right| + C \cr} $$
