Answer
$$\frac{1}{2}\sec x\tan x + \frac{1}{2}\ln \left| {\sec x + \tan x} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sec }^3}x} dx \cr
& {\text{Rewrite}} \cr
& = \int {\sec x{{\sec }^2}x} dx \cr
& {\text{Integrate by parts }} \cr
& {\text{Let }}u = \sec x,{\text{ }} \to {\text{ }}du = \sec x\tan xdx \cr
& dv = {\sec ^2}xdx \to v = \tan x \cr
& {\text{Use integration by parts formula}} \cr
& \int {udv} = uv - \int {vdu} \cr
& \int {{{\sec }^3}x} dx = \sec x\tan x - \int {\tan x\left( {\sec x\tan x} \right)} dx \cr
& \int {{{\sec }^3}x} dx = \sec x\tan x - \int {{{\tan }^2}x\sec x} dx \cr
& {\text{Use pythagorean identity}} \cr
& \int {{{\sec }^3}x} dx = \sec x\tan x - \int {\left( {{{\sec }^2}x - 1} \right)\sec x} dx \cr
& \int {{{\sec }^3}x} dx = \sec x\tan x - \int {\left( {{{\sec }^3}x - \sec x} \right)} dx \cr
& \int {{{\sec }^3}x} dx = \sec x\tan x - \int {{{\sec }^3}x} dx + \int {\sec x} dx \cr
& {\text{Combine }}\int {{{\sec }^3}x} dx \cr
& 2\int {{{\sec }^3}x} dx = \sec x\tan x + \int {\sec x} dx \cr
& \int {{{\sec }^3}x} dx = \frac{1}{2}\sec x\tan x + \frac{1}{2}\int {\sec x} dx \cr
& {\text{Integrate }}\int {\sec x} dx \cr
& \int {{{\sec }^3}x} dx = \frac{1}{2}\sec x\tan x + \frac{1}{2}\ln \left| {\sec x + \tan x} \right| + C \cr} $$
