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

Derivation and Evaluation

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

\[ \int \tan^3(x) \, dx = \int u \sec^2 x \left(\dfrac{1}{\sec^2 x}\right) du - \int \tan x \, dx \]

Simplify the expression:

\[ \int \tan^3(x) \, dx = \int u \, du - \int \tan x \, dx \]

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

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

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

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