# Reduction Formula for Integral of Power of Tangent

## Theorem

For all $n \in \Z_{> 1}$:

$\ds \int \map {\tan^n} x \rd x = \frac {\map {\tan^{n - 1} } x} {n - 1} - \int \map {\tan^{n - 2} } x \rd x$

## Proof

 $\ds \int \map {\tan^n} x \rd x$ $=$ $\ds \int \map {\tan^{n - 2} } x \map {\tan^2} x \rd x$ $\ds$ $=$ $\ds \int \map {\tan^{n - 2} } x \paren {\map {\sec^2} x - 1} \rd x$ Sum of Squares of Sine and Cosine $\ds$ $=$ $\ds \int \map {\tan^{n - 2} } x \map {\sec^2} x \rd x - \int \map {\tan^{n - 2} } x \rd x$ Linear Combination of Primitives

Let:

 $\ds t$ $=$ $\ds \map \tan x$ $\ds \leadsto \ \$ $\ds \frac {\d t} {\d x}$ $=$ $\ds \map {\sec^2} x$ Derivative of Tangent Function

Then:

 $\ds \int \map {\tan^{n - 2} } x \map {\sec^2} x \rd x - \int \map {\tan^{n - 2} } x \rd x$ $=$ $\ds \int t^{n - 2} \rd t - \int \map {\tan^{n - 2} } x \rd x$ Integration by Substitution $\ds$ $=$ $\ds \frac {t^{n - 1} } {n - 1} - \int \map {\tan^{n - 2} } x \rd x$ Primitive of Power $\ds$ $=$ $\ds \frac {\map {\tan^{n - 1} } x} {n - 1} - \int \map {\tan^{n - 2} } x \rd x$ substituting back $t \to \map \tan x$

$\blacksquare$