Evaluate the integral
\[ \int \sec^4 x \; dx \]
Write the integrand \( \sec^4 x \) as the product \( \sec^2 x \sec^2 x \)
\[ \int \sec^4 x \; dx = \int \sec^2 x \; \sec^2 x \; dx \]
Use the trigonometric identity \( \sec^2 x = \tan^2 x + 1 \) to write the integral as follows
\[ \int \sec^4 x \; dx = \int (\tan^2 x + 1) \sec^2 x \; dx \]
Expand the integrand and rewrite the integral as a sum of integrals
\[ \int \sec^4 x \; dx = \int \tan^2 x \sec^2 x \; dx + \int \sec^2 x \; dx \]
Use Integration by Substitution: Let \( u = \tan x \) and hence \( \dfrac{du}{dx} = \sec^2 x \) or \( dx = \dfrac{1}{\sec^2 x} du \) to write
\[ \int \sec^4 x \; dx = \int u^2 \sec^2 x \dfrac{1}{\sec^2 x} du + \int \sec^2 x \; dx \]
Simplify
\[ \int \sec^4 x \; dx = \int u^2 du + \int \sec^2 x \; dx \]
Evaluate using the integral formulas \( \displaystyle \int u^2 du = (1/3) u^3 \) and the common integral \( \displaystyle \int \sec^2 x \; dx = \tan x\) to write
\[ \int \sec^4 x \; dx = \dfrac{1}{3} u^3 + \tan x + c \]
Substitute back \( \displaystyle u = \tan x \) to obtain the final answer
\[ \boxed { \int \sec^4 x \; dx = \dfrac{1}{3} \tan^3 x + \tan x + c } \]
