4.9: Hyperbolic Functions

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Definition of the Hyperbolic Functions

We define the hyperbolic functions as follows:

\[ \sinh x = \dfrac{e^x - e^{-x}}{2}, \]

\[ \cosh x = \dfrac{e^x + e^{-x}}{2},\]

\[ \tanh x = \dfrac{\sinh x}{\cosh x}.\]

$hyperb8a.gif$ $hyperb8b.gif$

Properties of Hyperbolic Functions

$ \cosh^2 x - \sinh^2 x = 1 $,
$ \dfrac{d}{dx} \sinh x = \cosh x$,
$\dfrac{d}{dx} \cosh x = \sinh x$.

Proof of Property A

We find

\[\begin{align*} \cosh^2 x - \sinh^2 x &= \left( \dfrac{e^x+e^{-x}}{2} \right)^2 - \left( \dfrac{e^x-e^{-x}}{2} \right)^2 \\[4pt] &= \dfrac{e^{2x}+2+e^{-2x}}{4} - \dfrac{e^{2x}-2+e^{-2x}}{4} \\[4pt] &= \dfrac{4}{4} =1. \end{align*} \]

$\square$

The Derivative of the Inverse Hyperbolic Trig Functions

\[ \dfrac{d}{dx} \sinh^{-1} x = \dfrac{1}{\sqrt{1+x^2}},\]

\[ \dfrac{d}{dx} \cosh^{-1} x = \dfrac{1}{\sqrt{x^2-1}},\]

\[ \dfrac{d}{dx} \tanh^{-1} x = \dfrac{d}{dx} \coth^{-1} x = \dfrac{1}{1-x^2},\]

\[ \dfrac{d}{dx} \text{sech}^{-1} x = \dfrac{1}{x\sqrt{1-x^2}},\]

\[ \dfrac{d}{dx} \text{csch}^{-1} x = \dfrac{1}{x\sqrt{1+x^2}}.\]

Proof of the third identity

We have

\[ \tanh (\tanh^{-1} x) = x. \nonumber\]

Taking derivatives implicitly, we have

\[ \dfrac{d}{dx} \text{sech}^2 (\tanh^{-1} x = \tanh^{-1} x = 1. \nonumber\]

Dividing gives

\[ \dfrac{d}{dx} \tanh^{-1} x = \dfrac{1}{\text{sech}^2 (\tan^{-1} x)}. \nonumber\]

Since

\[ \cosh^2(x) - \sinh^2(x) = 1, \nonumber\]

dividing by $\cosh^2(x)$, we get

\[1 - \tanh^2(x) = \text{sech}^2(x) \nonumber\]

so that

\[\begin{align*} \dfrac{d}{dx} \tan^{-1} x &= \dfrac{1}{1-\tanh^2 (\tanh^{-1} x)} \\[4pt] &= \dfrac{1}{1-x^2}. \end{align*}

$\square$

For the derivative of the $\text{sech}^{-1} (x)$ click here.

Integration and Hyperbolic Functions

Now we are ready to use the arc hyperbolic functions for integration.

Example $\PageIndex{1}$

Evaluate

\[ \int \dfrac{dx}{4-x^2} \nonumber\]

Solution

\[ \int \dfrac{dx}{4-x^2} = \int \dfrac{1}{4} \int \dfrac{dx}{1-(2/3)^2}\nonumber\]

let $ u = \dfrac{x}{2}$, then $du = \dfrac{1}{2}dx$

\[ \dfrac{1}{2} \int \dfrac{du}{1-u^2}= \dfrac{1}{2}\tanh^{-1} u +C = \dfrac{1}{2} \tanh^{-1} \left(\dfrac{x}{2}\right) + C.\nonumber\]

Example $\PageIndex{2}$

Evaluate

\[ \int \dfrac{x}{1-x^4} dx. \nonumber\]

Solution

Although this is not directly a derivative of a hyperbolic trig function, we can use the substitution $ u = x^2 $ and $du = 2x\, dx$.

To change the integral to

\[\begin{align*} \dfrac{1}{2} \int \dfrac{du}{1-u^2} &= \dfrac{1}{2} \tanh^{-1} u + C \\[4pt] &= \dfrac{1}{2} \tanh^{-1} (x^2) + C. \end{align*}\]

Contributors and Attributions

Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.