4.9: Hyperbolic Functions

Definition of the Hyperbolic Functions

We define the hyperbolic functions as follows:

$$\sinh x = \dfrac{e^x - e^{-x}}{2}, \nonumber$$

$$\cosh x = \dfrac{e^x + e^{-x}}{2}, \nonumber$$

$$\tanh x = \dfrac{\sinh x}{\cosh x}. \nonumber$$

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

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

$\square$

The Derivative of the Inverse Hyperbolic Trig Functions

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

$$\dfrac{d}{dx} \cosh^{-1} x = \dfrac{1}{\sqrt{x^2-1}}, \nonumber$$

$$\dfrac{d}{dx} \tanh^{-1} x = \dfrac{d}{dx} \coth^{-1} x = \dfrac{1}{1-x^2}, \nonumber$$

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

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

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.