Answer
$$\frac{{{{\sec }^5}x}}{5} + C$$
Work Step by Step
$$\eqalign{
& \int {\tan x{{\sec }^5}x} dx \cr
& {\text{split exponent se}}{{\text{c}}^3}x \cr
& \int {\tan x{{\sec }^4}x} \sec xdx \cr
& \int {{{\sec }^4}x} \sec x\tan xdx \cr
& {\text{substitute }}u = \sec x,{\text{ }}du = \sec x\tan xdx \cr
& = \int {{u^4}} du \cr
& {\text{find the antiderivative by the power rule}} \cr
& = \frac{{{u^5}}}{5} + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = \sec x \cr
& = \frac{{{{\sec }^5}x}}{5} + C \cr} $$
