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

 

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 $\boxed { \int \sec^4 x \; dx = \dfrac{1}{3} \tan^3 x + \tan x + c }$