Answer
$$\int {{{\sec }^n}x} dx = \frac{{{{\sec }^{n - 2}}x\tan x}}{{n - 1}} + \frac{{n - 2}}{{n - 1}}\int {{{\sec }^{n - 2}}x} dx$$
Work Step by Step
$$\eqalign{
& \int {{{\sec }^n}x} dx \cr
& {\text{*Rewrite the integrand}} \cr
& \int {{{\sec }^n}x} dx = \int {{{\sec }^{n - 2}}x{{\sec }^2}x} dx \cr
& {\text{Integrate by parts }} \cr
& {\text{Let }}u = {\sec ^{n - 2}}x \to du = \left( {n - 2} \right){\sec ^{n - 2}}x\tan xdx \cr
& dv = {\sec ^2}xdx \to v = \tan x \cr
& {\text{*Use integration by parts formula}} \cr
& \int {udv} = uv - \int {vdu} \cr
& \int {{{\sec }^n}x} dx = {\sec ^{n - 2}}x\tan x - \int {\tan x} \left( {n - 2} \right){\sec ^{n - 3}}xdx \cr
& \int {{{\sec }^n}x} dx = {\sec ^{n - 2}}x\tan x - \left( {n - 2} \right)\int {{{\sec }^{n - 2}}x{{\tan }^2}x} dx \cr
& {\text{*Use the pythagorean identity }}{\tan ^2}x = {\sec ^2}x - 1 \cr
& \int {{{\sec }^n}x} dx = {\sec ^{n - 2}}x\tan x - \left( {n - 2} \right)\int {{{\sec }^{n - 2}}x\left( {{{\sec }^2}x - 1} \right)} dx \cr
& \int {{{\sec }^n}x} dx = {\sec ^{n - 2}}x\tan x - \left( {n - 2} \right)\int {\left( {{{\sec }^n}x - {{\sec }^{n - 2}}x} \right)} dx \cr
& \int {{{\sec }^n}x} dx = {\sec ^{n - 2}}x\tan x - \left( {n - 2} \right)\int {{{\sec }^n}x} dx \cr
& + \left( {n - 2} \right)\int {{{\sec }^{n - 2}}x} dx \cr
& {\text{Combine like terms}} \cr
& \int {{{\sec }^n}x} dx + \left( {n - 2} \right)\int {{{\sec }^n}x} dx = {\sec ^{n - 2}}x\tan x \cr
& + \left( {n - 2} \right)\int {{{\sec }^{n - 2}}x} dx \cr
& \left( {n - 1} \right)\int {{{\sec }^n}x} dx = {\sec ^{n - 2}}x\tan x + \left( {n - 2} \right)\int {{{\sec }^{n - 2}}x} dx \cr
& {\text{Divide both sides by }}n - 1 \cr
& \int {{{\sec }^n}x} dx = \frac{{{{\sec }^{n - 2}}x\tan x}}{{n - 1}} + \frac{{n - 2}}{{n - 1}}\int {{{\sec }^{n - 2}}x} dx \cr} $$
