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4.9: Hyperbolic Functions

Definition of the Hyperbolic Functions

We define the hyperbolic functions as follows:

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

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

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

Note: Properties

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

$\cosh^2 x - \sinh^2 x = \left( \dfrac{e^x+e^{-x}}{2} \right)^2 - \left( \dfrac{e^x-e^{-x}}{2} \right)^2$

$= \dfrac{e^{2x}+2+e^{-2x}}{4} - \dfrac{e^{2x}-2+e^{-2x}}{4}$

$= \dfrac{4}{4} =1.$

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

$\tanh (\tanh^{-1} x) = x.$

Taking derivatives implicitly, we have

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

Dividing gives

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

Since

$\cosh^2(x) - \sinh^2(x) = 1,$

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

$1 - \tanh^2(x) = \text{sech}^2(x)$

so that

$\dfrac{d}{dx} \tan^{-1} x = \dfrac{1}{1-\tanh^2 (\tanh^{-1} x)} = \dfrac{1}{1-x^2}.$

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 1

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

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

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

Example 2

Evaluate

$\int \dfrac{x}{1-x^4} dx.$

Solution

Although this is not directly a derivative of a hyperbolic trig function, we can use the substitution

$u = x^2 \;\;\; \text{and}\;\;\; du = 2x\, dx$

To change the integral to

$\dfrac{1}{2} \int \dfrac{du}{1-u^2} = \dfrac{1}{2} \tanh^{-1} u + C$

$= \dfrac{1}{2} \tanh^{-1} (x^2) + C.$

Contributors

• Integrated by Justin Marshall.