Answer
$$\frac{1}{5}{\left( {\tan x} \right)^5} + \frac{2}{3}{\left( {\tan x} \right)^3} + \tan x + C $$
Work Step by Step
$$\eqalign{
& \int {{{\sec }^6}x} dx \cr
& {\text{We write }}{\sec ^6}x{\text{ as }}{\sec ^4}x{\sec ^2}x{\text{ }} \cr
& = \int {{{\sec }^4}x{{\sec }^2}x} dx \cr
& = \int {{{\left( {{{\sec }^2}x} \right)}^2}{{\sec }^2}x} dx \cr
& \cr
& {\text{Use the fundamental identiy }}{\sec ^2}\theta = {\tan ^2}\theta + 1 \cr
& = \int {{{\left( {{{\tan }^2}x + 1} \right)}^2}{{\sec }^2}x} dx \cr
& {\text{Integrate by substitution method}} \cr
& {\text{Let }}u = \tan x,\,\,\,\,du = {\sec ^2}xdx,\,\,\,\,dx = \frac{{du}}{{{{\sec }^2}x}} \cr
& {\text{Then}}{\text{,}} \cr
& = \int {{{\left( {{u^2} + 1} \right)}^2}} {\sec ^2}x\left( {\frac{{du}}{{{{\sec }^2}x}}} \right) \cr
& {\text{Cancel common factor }}{\sec ^2}x \cr
& = \int {{{\left( {{u^2} + 1} \right)}^2}} du \cr
& {\text{expanding}} \cr
& = \int {\left( {{u^4} + 2{u^2} + 1} \right)} du \cr
& {\text{integrating}} \cr
& = \frac{1}{5}{u^5} + \frac{2}{3}{u^3} + u + C \cr
& {\text{write in terms of }}x,\,\,\,u = \tan x \cr
& = \frac{1}{5}{\left( {\tan x} \right)^5} + \frac{2}{3}{\left( {\tan x} \right)^3} + \tan x + C \cr} $$
