Answer
$$\frac{{{{\sec }^3}x\tan x}}{4} + \frac{{{{\sec }^2}x\tan x}}{4} + \frac{2}{3}\tan x + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sec }^5}x} dx \cr
& {\text{A matching integral in a table of integrals at the end of the book is the }} \cr
& {\text{formula }}55 \cr
& \int {{{\sec }^n}x} dx = \frac{{{{\sec }^{n - 2}}x\tan x}}{{n - 1}} + \frac{{n - 2}}{{n - 1}}\int {{{\sec }^{n - 2}}} xdx \cr
& {\text{ }}n = 5 \cr
& then \cr
& \int {{{\sec }^5}x} dx = \frac{{{{\sec }^{5 - 2}}x\tan x}}{{5 - 1}} + \frac{{5 - 2}}{{5 - 1}}\int {{{\sec }^{5 - 2}}} xdx \cr
& = \frac{{{{\sec }^3}x\tan x}}{4} + \frac{3}{4}\int {{{\sec }^3}} xdx \cr
& n = 4 \cr
& = \frac{{{{\sec }^3}x\tan x}}{4} + \frac{3}{4}\left( {\frac{{{{\sec }^{4 - 2}}x\tan x}}{{4 - 1}} + \frac{{4 - 2}}{{4 - 1}}\int {{{\sec }^{4 - 2}}} xdx} \right) \cr
& = \frac{{{{\sec }^3}x\tan x}}{4} + \frac{3}{4}\left( {\frac{{{{\sec }^2}x\tan x}}{3} + \frac{2}{3}\int {{{\sec }^2}} xdx} \right) \cr
& = \frac{{{{\sec }^3}x\tan x}}{4} + \frac{{{{\sec }^2}x\tan x}}{4} + \frac{2}{3}\int {{{\sec }^2}} xdx + C \cr
& = \frac{{{{\sec }^3}x\tan x}}{4} + \frac{{{{\sec }^2}x\tan x}}{4} + \frac{2}{3}\tan x + C \cr} $$
