7.4 Integrals Involving Hyperbolic and Inverse Hyperbolic Functions

In Section 1.6, we noted the similarities between hyperbolic and trigonometric functions. We also saw in Section 3.9 that the formulas for their derivatives resemble each other, differing in at most a sign. The derivative formulas for the hyperbolic functions are equivalent to the following integral formulas.

Reminder

\[ \begin{alignat*}{2} &\sinh x=\frac{e^x-e^{-x}}2&\quad&\cosh x =\frac{e^x+e^{-x}}2\\ &\frac{d}{dx}\sinh x = \cosh x&\quad &\frac{d}{dx}\cosh x = \sinh x \end{alignat*} \] \[ \begin{alignat*}{2} &\frac{d}{dx}\tanh x = \textrm{sech}^2 x&&\\ &\frac{d}{dx}\coth x = - \textrm{csch}^2 x&&\\ &\frac{d}{dx}\textrm{sech } x = - \textrm{sech } x~\tanh x &&\\ &\frac{d}{dx}\textrm{csch } x = - \textrm{sech } x \coth x&& \end{alignat*} \]

Hyperbolic Integral Formulas

\[ \begin{alignat*}{2} &\int \sinh x \,dx = \cosh x + C,& \qquad &\int~\cosh x\,dx = \sinh x + C \\ &\int \textrm{sech}^2 x \,dx = \tanh x + C,& \qquad &\int \textrm{csch}^2 x \,dx = -\coth x + C \\ &\int \textrm{sech } x \tanh x\,dx = -\textrm{sech } x + C,& \qquad &\int \textrm{csch } x \coth x\,dx = -\textrm{csch } x + C \end{alignat*} \]

EXAMPLE 1

Calculate \(\displaystyle \int x~\cosh (x^2)\,dx\).

Solution The substitution \(u = x^2, du = 2x\,dx\) yields \[ \int x~\cosh (x^2)\,dx = \frac{1}{2}\int~\cosh u\,du = \frac{1}{2} \sinh u+C =\frac{1}{2}\sinh (x^2) + C \]

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The techniques for computing trigonometric integrals discussed in Section 7.2 apply with little change to hyperbolic integrals. In place of trigonometric identities, we use the corresponding hyperbolic identities (see margin).

Hyperbolic Identities \[ \begin{align*} \cosh^2x&-\sinh^2x=1,\\ \cosh^2x&=1+\sinh^2x\\ \cosh^2x &=\tfrac12(\cosh 2x+1),\\ \sinh^2x &=\tfrac12(\cosh 2x-1)\\ \sinh 2x &= 2\sinh x~\cosh x,\\ \cosh 2x &= \cosh^2 x +\sinh^2x \end{align*} \]

EXAMPLE 2 Powers of \(\sinh x\) and \(\cosh x\)

Calculate:\( \quad \) (a) \(\displaystyle \int \sinh^4x~\cosh^5x\,dx\)\(\quad\) and\(\quad\) (b) \(\displaystyle \int \cosh^2x\,dx\).

Solution

  1. Since \(\cosh x\) appears to an odd power, use \(\cosh^2x=1+\sinh^2x\) to write \[ \begin{align*} \cosh^5x &=\cosh^4x\cdot \cosh x=(\sinh^2x+1)^2\cosh x\\ \end{align*} \]

    Then use the substitution \(u=\sinh x\), \(du=\cosh x\,dx\): \[ \begin{align*} \int \sinh^4x~\cosh^5x\,dx&=\int \underbrace{\sinh^4x}_{u^4}(\underbrace{\sinh^2x+1)^2}_{(u^2+1)^2}\underbrace{\cosh x\,dx}_{du}\\ &= \int u^4(u^2+1)^2\,du= \int(u^8+2u^6+u^4)\,du\\ &=\frac{u^9}9+\frac{2u^7}7+\frac{u^5}5+C= \frac{\sinh^9x}9+\frac{2\sinh^7x}7+\frac{\sinh^5x}5+C \end{align*} \]

  2. Use the identity \(\cosh^2x =\frac12(\cosh 2x+1)\): \[ \begin{align*} \int \cosh^2x\,dx &= \frac12\int(\cosh 2x+1)\,dx=\frac12\left(\frac{\sinh 2x}2+x\right)+C \\ &=\frac14\sinh 2x+\frac12x+C \end{align*} \]

Hyperbolic substitution may be used as an alternative to trigonometric substitution to integrate functions involving the following square root expressions:

Square root form Hyperbolic substitution
\(\sqrt{x^2+a^2}\) \(x=a~\sinh u\), \(dx = a~\cosh u\), \(\sqrt{x^2+a^2} = a~\cosh u\)
\(\sqrt{x^2-a^2}\) \(x=a~\cosh u\), \(dx = a~\sinh u\), \(\sqrt{x^2-a^2} = a~\sinh u\)

In trigonometric substitution, we treat \(\sqrt{x^2+a^2}\) using the substitution \(x=a~\tan \theta\) and \(\sqrt{x^2-a^2}\) using \(x=a~\sec\theta\). Identities can be used to show that the results coincide with those obtained from hyperbolic substitution (see Exercises 31–35).

EXAMPLE 3 Hyperbolic Substitution

Calculate \(\int\sqrt{x^2+16}\,dx\).

Solution

Step 1. Substitute to eliminate the square root.

Use the hyperbolic substitution \(x=4\sinh u\), \(dx = 4\cosh u\,du\). Then \[ x^2+16 = 16(\sinh^2u+1) =(4\cosh u)^2\]

Furthermore, \(4~\cosh u>0\), so \(\sqrt{x^2 +16}=4~\cosh u\) and thus, \[\int\sqrt{x^2+16}\,dx = \int (4\cosh u)4\cosh u\,du = 16\int\cosh^2u\,du\]

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Step 2. Evaluate the hyperbolic integral.

We evaluated the integral of \(\cosh^2u\) in Example 2(b): \[ \begin{align*} \int\sqrt{x^2+16}\,dx &= 16\int\cosh^2u\,du =16\left(\frac14\sinh 2u+\frac12u+C\right)\nonumber\\ &= 4\sinh 2u +8u +C\tag{1} \end{align*} \]

Step 3. Convert back to original variable.

To write the answer in terms of the original variable \(x\), we note that \[ \sinh u = \frac{x}4,\qquad u=\sinh^{-1}\frac{x}4 \]

Use the identities recalled in the margin to write \[ \begin{align*} 4\sinh 2u &= 4(2\sinh u\cosh u) = 8\sinh u\sqrt{\sinh^2u+1}\\ &=8\left(\frac{x}4\right)\sqrt{\left(\frac{x}4\right)^2+1} = 2x\sqrt{\frac{x^2}{16} +1} = \frac12x\sqrt{x^2+16} \end{align*} \]

Reminder

\[ \begin{align*} \sinh 2u &= 2\sinh u\cosh u\\ \cosh u&=\sqrt{\sinh^2u+1} \end{align*} \]

Then Eq. (1) becomes \[ \int\sqrt{x^2+16}\,dx =4\sinh 2u +8u +C=\frac12x\sqrt{x^2+16}+8\sinh^{-1}\frac{x}4+C \]

The next theorem states the integral formulas corresponding to the derivative formulas for the inverse hyperbolic functions recorded in Section 3.9. Each formula is valid on the domain where the integrand and inverse hyperbolic function are defined.

THEOREM 1 Integrals Involving Inverse Hyperbolic Functions

\[ \begin{eqnarray*} &&\int \frac{dx}{\sqrt{x^2+1}} = \sinh^{-1}x+C&\\ &&\int \frac{dx}{\sqrt{x^2-1}} = \cosh^{-1}x+C &&& \hbox{(for \(x> 1\))}\\ &&\int \frac{dx}{1-x^2} = \tanh^{-1}x+C &&& \hbox{(for \(|x|<1\))}\\ &&\int \frac{dx}{1-x^2} = \coth^{-1}x+C &&& \hbox{(for \(|x|>1\))}\\ &&\int \frac{dx}{x\sqrt{1-x^2}} = -\textrm{sech}^{-1}x+C &&& \hbox{(for \(0<x <1\))}\\ &&\int \frac{dx}{|x|\sqrt{1+x^2}} = -\textrm{csch}^{-1}x+C &&& \hbox{(for \(x\ne 0\))} \end{eqnarray*} \]

If your calculator does not provide values of inverse hyperbolic functions, you can use an online resource such as https://www.wolframalpha.com/.

EXAMPLE 4

Evaluate:\(\quad\)(a) \(\int_2^4\frac{dx}{\sqrt{x^2-1}}\)\(\quad\) and\(\quad\) (b) \(\int_{0.2}^{0.6}\frac{x\,dx}{1-x^4}\).

Solution

  1. By Theorem 1, \[ \int_2^4\frac{dx}{\sqrt{x^2-1}} =\cosh^{-1}x\Big|_2^4=\cosh^{-1}4-\cosh^{-1}2 \approx 0.75 \]

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  2. First use the substitution \(u=x^2\), \(du=2x\,du\). The new limits of integration become \(u=(0.2)^2=0.04\) and \(u=(0.6)^2=0.36\), so \[ \int_{0.2}^{0.6} \frac{x\,dx}{1-x^4} =\int_{0.04}^{0.36} \frac{\frac12 du}{1-u^2} =\frac12 \int_{0.04}^{0.36} \frac{du}{1-u^2} \]

    By Theorem 1, both \(\tanh^{-1}u\) and \(\coth^{-1}u\) are antiderivatives of \(f(u)=(1-u^2)^{-1}\). We use \(\tanh^{-1}u\) because the interval of integration \([0.04,0.36]\) is contained in the domain \((-1,1)\) of \(\tanh^{-1}u\). If the limits of integration were contained in \((1, \infty)\) or \((-\infty,-1)\), we would use \(\coth^{-1}u\). The result is \[ \frac12 \int_{0.04}^{0.36} \frac{du}{1-u^2} = \frac12\bigl(\tanh^{-1}(0.36)- \tanh^{-1}(0.04)\bigr) \approx 0.1684 \]

Excursion: A Leap of Imagination

The terms “hyperbolic sine” and “hyperbolic cosine” suggest a connection between the hyperbolic and trigonometric functions. This excursion explores the source of this connection, which leads us to complex numbers and a famous formula of Euler (Figure 7.6).

Recall that \(y = e^{t}\) satisfies the differential equation \(y' = y\). In fact, we know that every solution is of the form \(y = Ce^t\) for some constant \(C\). Observe that both \(y = e^t\) and \(y = e^{-t}\) satisfy the second-order differential equation \[ \boxed{\bbox[#FAF8ED,5pt]{{y'' = y}}}\tag{2} \]

This differential equation is called “second-order” because it involves the second derivative \(y''\).

Indeed, \((e^t)''=e^t\) and \((e^{-t})''=(-e^{-t})'=e^{-t}\). Furthermore, every solution of Eq. (2) has the form \(y = Ae^{t} + Be^{- t}\) for some constants \(A\) and \(B\) (Exercise 44).

Now let's see what happens when we change Eq. (2) by a minus sign: \[ \boxed{\bbox[#FAF8ED,5pt]{{y'' = -y}}}\tag{3} \]

image
Figure 7.6: Leonhard Euler (1707–1783). Euler (pronounced “oil-er”) ranks among the greatest mathematicians of all time. His work (printed in more than 70 volumes) contains fundamental contributions to almost every aspect of the mathematics and physics of his time. The French mathematician Pierre Simon de Laplace once declared: “Read Euler, he is our master in everything.”

In this case, \(y = \sin t\) and \(y = \cos t\) are solutions because \[ (\sin t)'' = (\cos t)' = -\sin t,\qquad (\cos t)'' = (-\sin t)' = -\cos t \]

And as before, every solution of Eq. (3) has the form \[\boxed{\bbox[#FAF8ED,5pt]{ y = A~\cos t + B \sin t }}\]

This might seem to be the end of the story. However, we can also write down solutions of Eq. (3) using the exponential functions \(y = e^{i t}\) and \(y = e^{-i t}\). Here \[\boxed{\bbox[#FAF8ED,5pt]{ i=\sqrt{-1} }}\]

is the imaginary complex number satisfying \(i^2=-1\). Since \(i\) is not a real number, \(e^{it}\) is not defined without further explanation. But let's assume that \(e^{it}\) can be defined and that the usual rules of calculus apply: \[ \begin{align*} (e^{it})' &= ie^{it}\\ (e^{it})'' &= (ie^{it})' = i^2 e^{it} = -e^{it} \end{align*} \]

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This shows that \(y = e^{it}\) is a solution of \(y'' = -y\), so there must exist constants \(A\) and \(B\) such that \[ e^{it} = A~\cos t + B\sin t\tag{4} \]

The constants are determined by initial conditions. First, set \(t = 0\) in Eq. (4): \[ 1 = e^{i0} = A~\cos 0 + B\sin 0 = A \]

Then take the derivative of Eq. (4) and set \(t = 0\): \[ \begin{eqnarray*} ie^{it} &=& \frac{d}{dt} e^{it} = A~\cos' t + B \sin' t = - A~\sin t + B~\cos t\\ i &=& ie^{i0} = - A~\sin0 + B\cos0 = B \end{eqnarray*} \]

Thus \(A=1\) and \(B=i\), and Eq. (4) yields Euler's Formula: \[ \boxed{\bbox[#FAF8ED,5pt]{e^{it} = \cos t + i\sin t}} \]

Euler proved his formula using power series, which may be used to define \(e^{it}\) in a precise fashion. At \(t = \pi\), Euler's Formula yields \[ \boxed{\bbox[#FAF8ED,5pt]{e^{i\pi} = -1}} \]

Here we have a simple but surprising relation among the four important numbers \(e\), \(i\), \(\pi\), and \(-1\).

Euler's Formula also reveals the source of the analogy between hyperbolic and trigonometric functions. Let us calculate the hyperbolic cosine at \(x = it\): \[ \cosh (it) = \frac{e^{it} + e^{-it}}{2}= \frac{\cos t + i\sin t}{2} + \frac{\cos(-t) + i\sin(-t)}{2} = \cos t \]

A similar calculation shows that \(\sinh(it) = i\sin t\). In other words, the hyperbolic and trigonometric functions are not merely analogous—once we introduce complex numbers, we see that they are very nearly the same functions.

7.4.1 Summary