Reduction Formula for Integral of Power of Tangent

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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.

$\blacksquare$


Also see


Sources

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