Integral of \( \sec^4(x) \)

Derivation and Evaluation

Evaluate the integral:

\[ \int \sec^4 x \, dx \]

Write the integrand \( \sec^4 x \) as the product \( \sec^2 x \sec^2 x \):

\[ \int \sec^4 x \, dx = \int \sec^2 x \cdot \sec^2 x \, dx \]

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 \( u = \tan x \), which gives \( \dfrac{du}{dx} = \sec^2 x \) or \( dx = \dfrac{1}{\sec^2 x} \, du \). Substituting this yields:

\[ \int \sec^4 x \, dx = \int u^2 \sec^2 x \left(\dfrac{1}{\sec^2 x}\right) du + \int \sec^2 x \, dx \]

Simplify the expression:

\[ \int \sec^4 x \, dx = \int u^2 \, du + \int \sec^2 x \, dx \]

Evaluate using standard integral formulas \( \displaystyle \int u^2 \, du = \dfrac{1}{3} u^3 \) and the common integral \( \displaystyle \int \sec^2 x \, dx = \tan x \):

\[ \int \sec^4 x \, dx = \dfrac{1}{3} u^3 + \tan x + c \]

where \( c \) is the constant of integration.

Substitute back \( u = \tan x \) to obtain the final answer: