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
- 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): Appendix $8$: Integrals: Reduction Formulae