Answer
$$\tan x + \frac{{{{\tan }^3}x}}{3} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sec }^4}x} dx \cr
& {\text{split exponent se}}{{\text{c}}^4}x \cr
& = \int {{{\sec }^2}x} {\sec ^2}xdx \cr
& {\text{pythagorean identity 1 + ta}}{{\text{n}}^2}x = {\sec ^2}x \cr
& = \int {{{\sec }^2}x} \left( {{\text{1 + ta}}{{\text{n}}^2}x} \right)dx \cr
& = \int {{{\sec }^2}x} \left( {{\text{1 + ta}}{{\text{n}}^2}x} \right)dx \cr
& {\text{substitute }}u = \tan x,{\text{ }}du = {\sec ^2}xdx \cr
& = \int {\left( {{\text{1 + }}{u^2}} \right)} du \cr
& {\text{find the antiderivative by the power rule}} \cr
& = u + \frac{{{u^3}}}{3} + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = \tan x \cr
& = \tan x + \frac{{{{\tan }^3}x}}{3} + C \cr} $$
