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
\[ \boxed { \int \tan^3(x) \; dx = \dfrac{1}{2} \tan^2 x - \ln |\sec x| + c } \]