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