Hyperbolic Tangent Half-Angle Substitution
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Theorem
- $\ds \int \map F {\sinh x, \cosh x} \rd x = 2 \int \map F {\frac {2 u} {1 - u^2}, \frac {1 + u^2} {1 - u^2} } \frac {\d u} {1 - u^2}$
where $u = \tanh \dfrac x 2$.
Proof
| \(\ds u\) | \(=\) | \(\ds \tanh \dfrac x 2\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds 2 \tanh^{-1} u\) | Definition 1 of Inverse Hyperbolic Tangent | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \dfrac {\d x} {\d u}\) | \(=\) | \(\ds \dfrac 2 {1 - u^2}\) | Derivative of Inverse Hyperbolic Tangent and Derivative of Constant Multiple | ||||||||||
| \(\ds \sinh x\) | \(=\) | \(\ds \dfrac {2 u} {1 - u^2}\) | Hyperbolic Tangent Half-Angle Substitution for Sine | |||||||||||
| \(\ds \cosh x\) | \(=\) | \(\ds \dfrac {1 + u^2} {1 - u^2}\) | Hyperbolic Tangent Half-Angle Substitution for Cosine |
The result follows from Integration by Substitution.
$\blacksquare$