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Calculus of Inverse Hyperbolic Functions

Looking at the graphs of the hyperbolic functions, we see that with appropriate range restrictions, they all have inverses. Most of the necessary range restrictions can be discerned by close examination of the graphs. The domains and ranges of the inverse hyperbolic functions are summarized in the following table.

Function Domain Range
\(\text{sinh}^{−1} x\) \(\left(\right. − \infty , \infty \left.\right)\) \(\left(\right. − \infty , \infty \left.\right)\)
\(\text{cosh}^{−1} x\) \(\left[\right. 1 , \infty \left.\right)\) \(\left[\right. 0 , \infty \left.\right)\)
\(\text{tanh}^{−1} x\) \(\left(\right. −1 , 1 \left.\right)\) \(\left(\right. − \infty , \infty \left.\right)\)
\(\text{coth}^{−1} x\) \(\left(\right. − \infty , −1 \left.\right) \cup \left(\right. 1 , \infty \left.\right)\) \(\left(\right. − \infty , 0 \left.\right) \cup \left(\right. 0 , \infty \left.\right)\)
\(\text{sech}^{−1} x\) \(\left(\right. 0 ,\text{ 1} \left]\right.\) \(\left[\right. 0 , \infty \left.\right)\)
\(\text{csch}^{−1} x\) \(\left(\right. − \infty , 0 \left.\right) \cup \left(\right. 0 , \infty \left.\right)\) \(\left(\right. − \infty , 0 \left.\right) \cup \left(\right. 0 , \infty \left.\right)\)
Table 6.3 Domains and Ranges of the Inverse Hyperbolic Functions

The graphs of the inverse hyperbolic functions are shown in the following figure.

This figure has six graphs. The first graph labeled “a” is of the function y=sinh^-1(x). It is an increasing function from the 3rd quadrant, through the origin to the first quadrant. The second graph is labeled “b” and is of the function y=cosh^-1(x). It is in the first quadrant, beginning on the x-axis at 2 and increasing. The third graph labeled “c” is of the function y=tanh^-1(x). It is an increasing function from the third quadrant, through the origin, to the first quadrant. The fourth graph is labeled “d” and is of the function y=coth^-1(x). It has two pieces, one in the third quadrant and one in the first quadrant with a vertical asymptote at the y-axis. The fifth graph is labeled “e” and is of the function y=sech^-1(x). It is a curve decreasing in the first quadrant and stopping on the x-axis at x=1. The sixth graph is labeled “f” and is of the function y=csch^-1(x). It has two pieces, one in the third quadrant and one in the first quadrant with a vertical asymptote at the y-axis.
Figure 6.82 Graphs of the inverse hyperbolic functions.

To find the derivatives of the inverse functions, we use implicit differentiation. We have

\[\begin{aligned} y & = & \text{sinh}^{−1} x \\ \text{sinh} y & = & x \\ \frac{d}{d x} \text{sinh} y & = & \frac{d}{d x} x \\ \text{cosh} y \frac{d y}{d x} & = & 1 . \end{aligned}\]

Recall that \(\text{cosh}^{2} y - \text{sinh}^{2} y = 1 ,\) so \(\text{cosh} y = \sqrt{1 + \text{sinh}^{2} y} .\) Then,

\[\frac{d y}{d x} = \frac{1}{\text{cosh} y} = \frac{1}{\sqrt{1 + \text{sinh}^{2} y}} = \frac{1}{\sqrt{1 + x^{2}}} .\]

We can derive differentiation formulas for the other inverse hyperbolic functions in a similar fashion. These differentiation formulas are summarized in the following table.

\(f \left(\right. x \left.\right)\) \(\frac{d}{d x} f \left(\right. x \left.\right)\)
\(\text{sinh}^{−1} x\) \(\frac{1}{\sqrt{1 + x^{2}}}\)
\(\text{cosh}^{−1} x\) \(\frac{1}{\sqrt{x^{2} - 1}}\)
\(\text{tanh}^{−1} x\) \(\frac{1}{1 - x^{2}}\)
\(\text{coth}^{−1} x\) \(\frac{1}{1 - x^{2}}\)
\(\text{sech}^{−1} x\) \(\frac{−1}{x \sqrt{1 - x^{2}}}\)
\(\text{csch}^{−1} x\) \(\frac{−1}{\left|\right. x \left|\right. \sqrt{1 + x^{2}}}\)
Table 6.4 Derivatives of the Inverse Hyperbolic Functions

Note that the derivatives of \(\text{tanh}^{−1} x\) and \(\text{coth}^{−1} x\) are the same. Thus, when we integrate \(1 / \left(\right. 1 - x^{2} \left.\right) ,\) we need to select the proper antiderivative based on the domain of the functions and the values of \(x .\) Integration formulas involving the inverse hyperbolic functions are summarized as follows.

\[\int \frac{1}{\sqrt{1 + u^{2}}} d u & = & \text{sinh}^{−1} u + C & & & \int \frac{1}{u \sqrt{1 - u^{2}}} d u & = & − \text{sech}^{−1} \left|\right. u \left|\right. + C \\ \int \frac{1}{\sqrt{u^{2} - 1}} d u & = & \text{cosh}^{−1} u + C & & & \int \frac{1}{u \sqrt{1 + u^{2}}} d u & = & − \text{csch}^{−1} \left|\right. u \left|\right. + C \\ \int \frac{1}{1 - u^{2}} d u & = & \left{\right. \begin{matrix}\text{tanh}^{−1} u + C \text{if} \left|\right. u \left|\right. < 1 \\ \text{coth}^{−1} u + C \text{if} \left|\right. u \left|\right. > 1\end{matrix} & & & & &\]

Example 6.49

Differentiating Inverse Hyperbolic Functions

Evaluate the following derivatives:

  1. \(\frac{d}{d x} \left(\right. \text{sinh}^{−1} \left(\right. \frac{x}{3} \left.\right) \left.\right)\)
  2. \(\frac{d}{d x} \left(\right. \text{tanh}^{−1} x \left.\right)^{2}\)

Solution

Using the formulas in Table 6.4 and the chain rule, we obtain the following results:

  1. \(\frac{d}{d x} \left(\right. \text{sinh}^{−1} \left(\right. \frac{x}{3} \left.\right) \left.\right) = \frac{1}{3 \sqrt{1 + \frac{x^{2}}{9}}} = \frac{1}{\sqrt{9 + x^{2}}}\)
  2. \(\frac{d}{d x} \left(\right. \text{tanh}^{−1} x \left.\right)^{2} = \frac{2 \left(\right. \text{tanh}^{−1} x \left.\right)}{1 - x^{2}}\)

Checkpoint 6.49

Evaluate the following derivatives:

  1. \(\frac{d}{d x} \left(\right. \text{cosh}^{−1} \left(\right. 3 x \left.\right) \left.\right)\)
  2. \(\frac{d}{d x} \left(\right. \text{coth}^{−1} x \left.\right)^{3}\)

Example 6.50

Integrals Involving Inverse Hyperbolic Functions

Evaluate the following integrals:

  1. \(\int \frac{1}{\sqrt{4 x^{2} - 1}} d x\)
  2. \(\int \frac{1}{2 x \sqrt{1 - 9 x^{2}}} d x\)

Solution

We can use \(u -\text{substitution}\) in both cases.

  1. Let \(u = 2 x .\) Then, \(d u = 2 d x\) and we have
    \[\int \frac{1}{\sqrt{4 x^{2} - 1}} d x = \int \frac{1}{2 \sqrt{u^{2} - 1}} d u = \frac{1}{2} \text{cosh}^{−1} u + C = \frac{1}{2} \text{cosh}^{−1} \left(\right. 2 x \left.\right) + C .\]
  2. Let \(u = 3 x .\) Then, \(d u = 3 d x\) and we obtain
    \[\int \frac{1}{2 x \sqrt{1 - 9 x^{2}}} d x = \frac{1}{2} \int \frac{1}{u \sqrt{1 - u^{2}}} d u = - \frac{1}{2} \text{sech}^{−1} \left|\right. u \left|\right. + C = - \frac{1}{2} \text{sech}^{−1} \left|\right. 3 x \left|\right. + C .\]

Checkpoint 6.50

Evaluate the following integrals:

  1. \(\int \frac{1}{\sqrt{x^{2} - 4}} d x , x > 2\)
  2. \(\int \frac{1}{\sqrt{1 - e^{2 x}}} d x\)

This lesson is part of:

Applications of Integration

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