# Reduction Formula for Integral of Power of Tangent

## Theorem

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

Let:

$I_n := \ds \int \tan^n x \rd x$

Then:

$I_n = \dfrac {\tan^{n - 1} x} {n - 1} - I_{n - 2}$

is a reduction formula for $\ds \int \tan^n x \rd x$.

## Proof

 $\ds I_n$ $=$ $\ds \int \tan^n x \rd x$ by definition $\ds$ $=$ $\ds \int \tan^{n - 2} x \tan^2 x \rd x$ $\ds$ $=$ $\ds \int \tan^{n - 2} x \paren {\sec^2 x - 1} \rd x$ Sum of Squares of Sine and Cosine $\ds$ $=$ $\ds \int \tan^{n - 2} x \sec^2 x \rd x - \int \tan^{n - 2} x \rd x$ Linear Combination of Primitives $\text {(1)}: \quad$ $\ds$ $=$ $\ds \int \tan^{n - 2} x \sec^2 x \rd x - I_{n - 2}$ by definition

Let:

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

Then:

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

Hence the result.

$\blacksquare$