# Integral of $\tan^3(x)$

 

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