5.7: Integration of Hyperbolic Functions

Learning Objectives

• Apply the formulas for integration of the hyperbolic functions.
• Apply the formulas for integration of the inverse hyperbolic functions.
• Describe the common applied conditions of a catenary curve.

We were introduced to hyperbolic functions previously, along with some of their properties and derivatives. In this section, we look at integration formulas for the hyperbolic functions and their inverses.

Integrals of the Hyperbolic Functions

Recall that the hyperbolic sine and hyperbolic cosine are defined as

\[\sinh x=\dfrac{e^x−e^{−x}}{2} \nonumber \]

and

\[\cosh x=\dfrac{e^x+e^{−x}}{2}. \nonumber \]

The other hyperbolic functions are then defined in terms of \(\sinh x\) and \(\cosh x\).

As we continue our examination of the hyperbolic functions, we must be mindful of their similarities and differences to the standard trigonometric functions. The differentiation formulas for the hyperbolic functions lead directly to the following integral formulas.

\[ \begin{align} \int \sinh u \,du &=\cosh u+C \\[4pt] \int \text{csch}^2 u \, du &=−\coth u+C \\[4pt] \int \cosh u \,du &=\sinh u+C \\[4pt] \int \text{sech} \,u \tanh u \,du &=−\text{sech } \,u+C−\text{csch} \,u+C \\[4pt] \int \text{sech }^2u \,du &=\tanh u+C \\[4pt] \int \text{csch} \,u \coth u \,du &=−\text{csch} \,u+C \end{align} \nonumber \]

Example \(\PageIndex{1}\): Integrals Involving Hyperbolic Functions

Evaluate the following integrals:

1. \( \displaystyle \int x\cosh(x^2)dx\)
2. \( \displaystyle \int \tanh x\,dx\)

Solution

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

a. Let \(u=x^2\). Then, \(du=2x\,dx\) and

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

b. Let \(u=\cosh x\). Then, \(du=\sinh x\,dx\) and

\[\begin{align*} \int \tanh x \,dx &=\int \dfrac{\sinh x}{\cosh x}\,dx \\[4pt] &=\int \dfrac{1}{u}du \\[4pt] &=\ln|u|+C \\[4pt] &= \ln|\cosh x|+C.\end{align*}\]

Note that \(\cosh x>0\) for all \(x\), so we can eliminate the absolute value signs and obtain

\[\int \tanh x \,dx=\ln(\cosh x)+C. \nonumber \]

Exercise \(\PageIndex{1}\)

Evaluate the following integrals:

1. \(\displaystyle \int \sinh^3x \cosh x \,dx\)
2. \(\displaystyle \int \text{sech }^2(3x)\, dx\)

Hint

Use the formulas above and apply \(u\)-substitution as necessary.

Answer a

\(\displaystyle \int \sinh^3x \cosh x \,dx=\dfrac{\sinh^4x}{4}+C\)

Answer b

\(\displaystyle \int \text{sech }^2(3x) \, dx=\dfrac{\tanh(3x)}{3}+C\)

Integrals of Inverse Hyperbolic Functions

Integration formulas involving the inverse hyperbolic functions are summarized as follows.

\[\int \dfrac{1}{\sqrt{1+u^2}}du=\sinh^{−1}u+C \nonumber \]

\[\int \dfrac{1}{u\sqrt{1−u^2}}du=−\text{sech}^{−1}|u|+C \nonumber \]

\[\int \dfrac{1}{\sqrt{u^2−1}}du=\cosh^{−1}u+C \nonumber \]

\[\int \dfrac{1}{u\sqrt{1+u^2}}du=−\text{csch}^{−1}|u|+C \nonumber \]

\[\int \dfrac{1}{1−u^2}du=\begin{cases}\tanh^{−1}u+C & \text{if }|u|<1\\ \coth^{−1}u+C & \text{if }|u|>1\end{cases} \nonumber \]

Example \(\PageIndex{2}\): Integrals Involving Inverse Hyperbolic Functions

Evaluate the following integrals:

1. \(\displaystyle \int \dfrac{1}{\sqrt{4x^2−1}}dx\)
2. \(\displaystyle \int \dfrac{1}{2x\sqrt{1−9x^2}}dx\)

Solution

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

Let \(u=2x\). Then, \(du=2\,dx\) and we have

\[\begin{align*} \int \dfrac{1}{\sqrt{4x^2−1}}\,dx &=\int \dfrac{1}{2\sqrt{u^2−1}}\,du \\[4pt] &=\dfrac{1}{2}\cosh^{−1}u+C \\[4pt] &=\dfrac{1}{2}\cosh^{−1}(2x)+C. \end{align*} \nonumber \]

Let \(u=3x.\) Then, \(du=3\,dx\) and we obtain

\[\begin{align*} \int \dfrac{1}{2x\sqrt{1−9x^2}}dx &=\dfrac{1}{2}\int \dfrac{1}{u\sqrt{1−u^2}}du \\[4pt] &=−\dfrac{1}{2}\text{sech}^{−1}|u|+C \\[4pt] &=−\dfrac{1}{2}\text{sech}^{−1}|3x|+C \end{align*}\]

Exercise \(\PageIndex{2}\)

Evaluate the following integrals:

1. \(\displaystyle \int \dfrac{1}{\sqrt{x^2−4}}dx,x>2\)
2. \(\displaystyle \int \dfrac{1}{\sqrt{1−e^{2x}}}dx\)

Hint

Use the formulas above and apply \(u\)-substitution as necessary.

Answer a

\(\displaystyle \int \dfrac{1}{\sqrt{x^2−4}}dx=\cosh^{−1}(\dfrac{x}{2})+C\)

Answer b

\( \displaystyle \int \dfrac{1}{\sqrt{1−e^{2x}}}dx=−\text{sech}^{−1}(e^x)+C\)

Applications

One physical application of hyperbolic functions involves hanging cables. If a cable of uniform density is suspended between two supports without any load other than its own weight, the cable forms a curve called a catenary. High-voltage power lines, chains hanging between two posts, and strands of a spider’s web all form catenaries.

Hyperbolic functions can be used to model catenaries. Specifically, functions of the form \(y=a\cdot \cosh(x/a)\) are catenaries.

Example \(\PageIndex{3}\): Using a Catenary to Find the Length of a Cable

Assume a hanging cable has the shape \(10\cosh(x/10)\) for \(−15≤x≤15\), where \(x\) is measured in feet. Determine the length of the cable (in feet).

Solution

The formula for arc length is

\[\underbrace{\int ^b_a\sqrt{1+[f′(x)]^2}dx}_{\text{Arc Length}}. \nonumber \]

We have \(f(x)=10 \cosh(x/10)\), so \(f′(x)=\sinh(x/10)\). Then the arc length is

\[\int ^b_a\sqrt{1+[f′(x)]^2}dx=\int ^{15}_{−15}\sqrt{1+\sinh^2 \left(\dfrac{x}{10}\right)}dx. \nonumber \]

Now recall that

\[1+\sinh^2x=\cosh^2x, \nonumber \]

so we have

\[\begin{align*} \text{Arc Length} &= \int ^{15}_{−15}\sqrt{1+\sinh^2 \left(\dfrac{x}{10}\right)}dx \\[4pt] &=\int ^{15}_{−15}\cosh \left(\dfrac{x}{10}\right)dx \\[4pt] &= \left.10\sinh \left(\dfrac{x}{10}\right)\right|^{15}_{−15}\\[4pt] &=10\left[\sinh\left(\dfrac{3}{2}\right)−\sinh\left(−\dfrac{3}{2}\right)\right]\\[4pt] &=20\sinh \left(\dfrac{3}{2}\right) \\[4pt] &≈42.586\,\text{ft.} \end{align*}\]

Exercise \(\PageIndex{3}\):

Assume a hanging cable has the shape \(15 \cosh (x/15)\) for \(−20≤x≤20\). Determine the length of the cable (in feet).

Answer

\(52.95\) ft

Key Concepts

• Hyperbolic functions are defined in terms of exponential functions.
• Term-by-term differentiation yields differentiation formulas for the hyperbolic functions. These differentiation formulas give rise, in turn, to integration formulas.
• With appropriate range restrictions, the hyperbolic functions all have inverses.
• Implicit differentiation yields differentiation formulas for the inverse hyperbolic functions, which in turn give rise to integration formulas.
• The most common physical applications of hyperbolic functions are calculations involving catenaries.

Glossary

catenary

a curve in the shape of the function \(y=a\cdot\cosh(x/a)\) is a catenary; a cable of uniform density suspended between two supports assumes the shape of a catenary