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Mathematics LibreTexts

4.9: Hyperbolic Functions

  • Page ID
    525
  • [ "article:topic", "hyperbolic functions", "authorname:green" ]

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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

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

    Proof of Property A

    We find

    \[\begin{align*} \cosh^2 x - \sinh^2 x &= \left( \dfrac{e^x+e^{-x}}{2} \right)^2 - \left( \dfrac{e^x-e^{-x}}{2} \right)^2 \\[5pt] &= \dfrac{e^{2x}+2+e^{-2x}}{4} - \dfrac{e^{2x}-2+e^{-2x}}{4} \\[5pt] &= \dfrac{4}{4} =1. \end{align*} \]

    \(\square\)

    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)} \\[5pt] &= \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

    \[\begin{align*} \dfrac{1}{2} \int \dfrac{du}{1-u^2} &= \dfrac{1}{2} \tanh^{-1} u + C \\[5pt] &= \dfrac{1}{2} \tanh^{-1} (x^2) + C. \end{align*}\]

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