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APEX Calculus

Section 6.6 Hyperbolic Functions

The hyperbolic functions are a set of functions that have many applications to mathematics, physics, and engineering. Among many other applications, they are used to describe the formation of satellite rings around planets, to describe the shape of a rope hanging from two points, and have application to the theory of special relativity. This section defines the hyperbolic functions and describes many of their properties, especially their usefulness to calculus.
These functions are sometimes referred to as the “hyperbolic trigonometric functions” as there are many, many connections between them and the standard trigonometric functions. Figure 6.6.2 demonstrates one such connection. Just as cosine and sine are used to define points on the circle defined by \(x^2+y^2=1\text{,}\) the functions hyperbolic cosine and hyperbolic sine are used to define points on the hyperbola \(x^2-y^2=1\text{.}\)
Figure 6.6.1. Video introduction to Section 6.6
Graph showing cosine and sine function used to define points on a circle.
The \(y\) and the \(x\) axes are drawn from \(-1\) to \(1\text{.}\) The function \(x^2+y^2=1\) represents a circle of radius \(1\) and center at origin.
A sector in the circle is shaded, it is present in the first quadrant and is drawn with one side on the \(x\) axis. It has an angle \(\theta/2\) drawn from the \(x\) axis and is marked inside the sector. The points between which the sector is drawn on the circumference are \((1,0)\) and \((\cos(\theta), \sin(\theta))\text{.}\)
(a)
Graph showing hyperbolic cosine and hyperbolic sine function used to define points on a hyperbola.
The \(y\) and the \(x\) axes are both drawn from \(-2\) to \(2\text{.}\) The graph of function \(x^2 -y^2=1\) has two \(x\) intercepts at \(x=1\) and \(x=-1\text{.}\) The function represents a hyperbola and has two conic sections facing opposite to each other, opening along the positive and negative \(x\) axis with vertices at the \(x\) intercepts. An angle of \(\theta/2\) is marked at the origin starting from the \(x\) axis, it is drawn from point \((1,0)\) to \((\cos(\theta), \sin(\theta))\) on the conic section to the right of the \(y\) axis.
(b)
Figure 6.6.2. Using trigonometric functions to define points on a circle and hyperbolic functions to define points on a hyperbola. The area of the shaded regions are included in them.

Subsection 6.6.1 The Hyperbolic Functions and their Properties

We begin with their definition.

Definition 6.6.3. Hyperbolic Functions.

  1. \(\ds \cosh(x) = \frac{e^x+e^{-x}}2\)
  2. \(\displaystyle \sinh(x) = \frac{e^x-e^{-x}}2\)
  3. \(\displaystyle \tanh(x) = \frac{\sinh(x) }{\cosh(x) }\)
  4. \(\displaystyle \sech(x) = \frac{1}{\cosh(x) }\)
  5. \(\displaystyle \csch(x) = \frac{1}{\sinh(x) }\)
  6. \(\displaystyle \coth(x) = \frac{\cosh(x) }{\sinh(x) }\)
These hyperbolic functions are graphed in Figure 6.6.4 and Figure 6.6.6.
In the graph of \(\cosh(x)\) in Figure 6.6.4.(a), the graphs of \(e^x/2\) and \(e^{-x}/2\) are included with dashed lines. In the graph of \(\sinh(x)\) in Figure 6.6.4.(b), the graphs of \(e^x/2\) and \(- e^{-x}/2\) are included with dashed lines. As \(x\) gets “large,” \(\cosh(x)\) and \(\sinh(x)\) each act like \(e^x/2\text{;}\) when \(x\) is a large negative number, \(\cosh(x)\) acts like \(e^{-x}/2\) whereas \(\sinh(x)\) acts like \(-e^{-x}/2\text{.}\)
Graph of hyperbolic cosine function.
The \(y\) axis is drawn from \(-10\) to \(10\) and the \(x\) axis is drawn from \(-3\) to \(3\text{.}\) The function \(f(x)=\cosh(x)\) is shown as a U shaped curve that opens upwards along the positive \(y\) axis, it is symmetrical about the \(y\) axis. The graphs of \(e^{x}/2\) and \(e^{-x}/2\) are also included.
For large values of \(x\) the function \(f(x)=\cosh(x)\) is approximately equal to \(e^{x}/2\text{.}\) From left to right, the function \(e^{x}/2\) appears to coincide with the \(x\) axis in the fourth quadrant, it has a positive slope, it is some distance apart from \(f(x)\) then it rises to coincide with \(f(x)\) after \(x=1\text{.}\)
For large negative values of \(x\text{,}\) \(f(x)\) becomes equal to the function \(e^{-x}/2\text{.}\) From right to left, the function \(e^{-x}/2\) appears to start from the first quadrant and enters the second quadrant with a positive slope. It coincides with \(f(x)\) after approximately \(x= -1\text{.}\)
(a)
Graph of hyperbolic sine function.
The \(y\) axis is drawn from \(-10\) to \(10\) and the \(x\) axis is drawn from \(-3\) to \(3\text{.}\) The function \(f(x) = \sinh(x)\text{,}\) \(e^{x}/2\) and \(e^-{x}/2\) is also shown in the graph.
From left to right, the function \(e^{x}/2\) starts in the second quadrant and gets close to the \(x\) axis, it gains the positive slope into the first quadrant. The function \(f(x)\) in the first quadrant, starts at the origin and rises with a positive slope after a bend.It is separated by a small distance from \(e^{x}/2\text{,}\) after approximately \(x=1\) the function \(e^{x}/2\) coincides with \(f(x)\text{.}\)
From right to left, the function \(e^{-x}/2\) starts in the fourth quadrant and coincides with the \(x\) axis, it moves downward after entering the third quadrant. The function \(f(x)\) in the third quadrant, starts at the origin and moves downwards from right to left. It is separated by a small distance from \(e^{-x}/2\) after approximately \(x=-1\) the function \(e^{-x}/2\) coincides with \(f(x)\text{.}\)
(b)
Figure 6.6.4. Graphs of \(\sinh(x)\) and \(\cosh(x)\)
Figure 6.6.5. Video presentation of graphs and basic properties of hyperbolic functions
In Figure Figure 6.6.6, notice the domains of \(\tanh(x)\) and \(\sech(x)\) are \((-\infty,\infty)\text{,}\) whereas both \(\coth(x)\) and \(\csch(x)\) have vertical asymptotes at \(x=0\text{.}\) Also note the ranges of these functions, especially \(\tanh(x)\text{:}\) as \(x\to\infty\text{,}\) both \(\sinh(x)\) and \(\cosh(x)\) approach \(e^{-x}/2\text{,}\) hence \(\tanh(x)\) approaches \(1\text{.}\)
Graph of hyperbolic tangent and hyperbolic cotangent functions.
The \(y\) axis is drawn from \(-2\) to \(2\) and the \(x\) axis is drawn from \(-3\) to \(3\text{.}\) The functions \(\tanh(x)\) and \(\coth(x)\) are shown. There are two lines drawn at \(y=-1\) and \(y=1\text{.}\)
The \(\tanh(x)\) function is drawn in the third and the first quadrant. In the first quadrant the function starts at the origin and gets a positive slope then after \(x=2\) it becomes parallel to the \(x\) axis at \(y=1\text{.}\) In the third quadrant the function starts at the origin and decreases until \(x=-2\) after which it becomes parallel to the \(x\) axis as \(y=-1\text{.}\)
The \(\coth(x)\) function is drawn in the first and the third quadrants. It is hyperbolic in shape with the two parts being symmetrical about the axis \(y=-x\text{.}\) It has a horizontal asymptote at \(x=0\text{.}\) This function coincides with the \(\tanh(x)\) curve after \(x=2\) and extends along the positive \(x\) axis and \(x=-2\) and extends further along the negative \(x\) axis.
(a)
Graph of hyperbolic secant and cosecant functions.
The \(y\) axis is drawn from \(-2\) to \(2\) and the \(x\) axis is drawn from \(-3\) to \(3\text{.}\) The functions \(\sech(x)\) and \(\csch(x)\) are shown.
The \(\sech(x)\) is drawn in the second and the first quadrant. From point \((1,0)\) the function slowly decreases moving left to right, almost touching the \(x\) axis at \(x=3\text{.}\) It is symmetrical about the \(x\) axis and in the third quadrant it decreases from \((1,0)\text{,}\) moving from right to left, and almost touches the \(x\) axis at \(x=-3\text{.}\)
(b)
Figure 6.6.6. Graphs of \(\tanh(x), \coth(x), \csch(x)\) and \(\cosh(x)\)
The following example explores some of the properties of these functions that bear remarkable resemblance to the properties of their trigonometric counterparts.

Example 6.6.7. Exploring properties of hyperbolic functions.

Use Definition 6.6.3 to rewrite the following expressions.
  1. \(\displaystyle \cosh^2(x) -\sinh^2(x)\)
  2. \(\displaystyle \tanh^2(x) +\sech^2(x)\)
  3. \(\displaystyle 2\cosh(x) \sinh(x)\)
  4. \(\displaystyle \frac{d}{dx}\big(\cosh(x) \big)\)
  5. \(\displaystyle \frac{d}{dx}\big(\sinh(x) \big)\)
  6. \(\displaystyle \frac{d}{dx}\big(\tanh(x) \big)\)
Solution 1. Video solution
Solution 2.
  1. \begin{align*} \cosh^2(x) -\sinh^2(x) \amp = \left(\frac{e^x+e^{-x}}2\right)^2 -\left(\frac{e^x-e^{-x}}2\right)^2\\ \amp = \frac{e^{2x}+2e^xe^{-x} + e^{-2x}}4 - \frac{e^{2x}-2e^xe^{-x} + e^{-2x}}4\\ \amp = \frac44=1\text{.} \end{align*}
    So \(\cosh^2(x) -\sinh^2(x) =1\text{.}\)
  2. Again, use Definition 6.6.3
    \begin{align*} \tanh^2(x) +\sech^2(x) \amp =\frac{\sinh^2(x) }{\cosh^2(x) } + \frac{1}{\cosh^2(x) }\\ \amp= \frac{\sinh^2(x) +1}{\cosh^2(x) }\qquad \text{ Now use identity from } \knowl{./knowl/xref/ex_hf1a.html}{\text{Part 1}}\\ \amp = \frac{\cosh^2(x) }{\cosh^2(x) } = 1\text{.} \end{align*}
    So \(\tanh^2(x) +\sech^2(x) =1\text{.}\)
  3. Again, use Definition 6.6.3
    \begin{align*} 2\cosh(x) \sinh(x) \amp = 2\left(\frac{e^x+e^{-x}}2\right)\left(\frac{e^x-e^{-x}}2\right)\\ \amp = 2 \cdot\frac{e^{2x} - e^{-2x}}4\\ \amp = \frac{e^{2x} - e^{-2x}}2 = \sinh(2x)\text{.} \end{align*}
    Thus \(2\cosh(x) \sinh(x) = \sinh(2x)\text{.}\)
  4. Again, use Definition 6.6.3
    \begin{align*} \frac{d}{dx}\big(\cosh(x) \big) \amp = \frac{d}{dx}\left(\frac{e^x+e^{-x}}2\right)\\ \amp = \frac{e^x-e^{-x}}2\\ \amp = \sinh(x) \end{align*}
    So \(\frac{d}{dx}\big(\cosh(x) \big) = \sinh(x)\text{.}\)
  5. Apply derivatives to Definition 6.6.3:
    \begin{align*} \frac{d}{dx}\big(\sinh(x) \big) \amp = \frac{d}{dx}\left(\frac{e^x-e^{-x}}2\right)\\ \amp = \frac{e^x+e^{-x}}2\\ \amp = \cosh(x)\text{.} \end{align*}
    So \(\frac{d}{dx}\big(\sinh(x) \big) = \cosh(x)\text{.}\)
  6. Apply derivatives to Definition 6.6.3:
    \begin{align*} \frac{d}{dx}\big(\tanh(x) \big) \amp = \frac{d}{dx}\left(\frac{\sinh(x) }{\cosh(x) }\right)\\ \amp = \frac{\cosh(x) \cosh(x) - \sinh(x) \sinh(x) }{\cosh^2(x) }\\ \amp = \frac{1}{\cosh^2(x) }\\ \amp = \sech^2(x)\text{.} \end{align*}
    So \(\frac{d}{dx}\big(\tanh(x) \big) = \sech^2(x)\text{.}\)
The following Key Idea summarizes many of the important identities relating to hyperbolic functions. Each can be verified by referring back to Definition 6.6.3.

Key Idea 6.6.8. Useful Hyperbolic Function Properties.

List 6.6.9. Basic Identities
  1. \(\cosh^2(x) -\sinh^2(x) =1\)
  2. \(\displaystyle \tanh^2(x) +\sech^2(x) =1\)
  3. \(\displaystyle \coth^2(x) -\csch^2(x) = 1\)
  4. \(\displaystyle \cosh(2x) =\cosh^2(x) +\sinh^2(x)\)
  5. \(\displaystyle \sinh(2x) = 2\sinh(x) \cosh(x)\)
  6. \(\displaystyle \cosh^2(x) = \frac{\cosh(2x) +1}{2}\)
  7. \(\displaystyle \sinh^2(x) =\frac{\cosh(2x) -1}{2}\)
List 6.6.10. Derivatives
  1. \(\displaystyle \frac{d}{dx}\big(\cosh(x) \big) = \sinh(x)\)
  2. \(\displaystyle \frac{d}{dx}\big(\sinh(x) \big) = \cosh(x)\)
  3. \(\displaystyle \frac{d}{dx}\big(\tanh(x) \big) = \sech^2(x)\)
  4. \(\displaystyle \frac{d}{dx}\big(\sech(x) \big) = -\sech(x) \tanh(x)\)
  5. \(\displaystyle \frac{d}{dx}\big(\csch(x) \big) = -\csch(x) \coth(x)\)
  6. \(\displaystyle \frac{d}{dx}\big(\coth(x) \big) = -\csch^2(x)\)
List 6.6.11. Integrals
  1. \(\displaystyle \int \cosh(x) \, dx = \sinh(x) +C\)
  2. \(\displaystyle \int \sinh(x) \, dx = \cosh(x) +C\)
  3. \(\displaystyle \int \tanh(x) \, dx = \ln(\cosh(x) ) +C\)
  4. \(\displaystyle \int \coth(x) \, dx = \ln\abs{\sinh(x) \,}+C\)
We practice using Key Idea 6.6.8.

Example 6.6.12. Derivatives and integrals of hyperbolic functions.

Evaluate the following derivatives and integrals.
  1. \(\displaystyle \frac{d}{dx}\big(\cosh(2x) \big)\)
  2. \(\displaystyle \int \sech^2(7t-3)\,dt\)
  3. \(\displaystyle \int_0^{\ln(2) } \cosh(x) \, dx\)
Solution 1. Video solution
Solution 2.
  1. Using the Chain Rule directly, we have \(\frac{d}{dx} \big(\cosh(2x) \big) = 2\sinh(2x)\text{.}\) Just to demonstrate that it works, let’s also use the Basic Identity found in Key Idea 6.6.8: \(\cosh(2x) = \cosh^2(x) +\sinh^2(x)\text{.}\)
    \begin{align*} \frac{d}{dx}\big(\cosh(2x) \big) \amp = \frac{d}{dx}\big(\cosh^2(x) +\sinh^2(x) \big)\\ \amp = 2\cosh(x) \sinh(x) + 2\sinh(x) \cosh(x)\\ \amp = 4\cosh(x) \sinh(x) \text{.} \end{align*}
    Using another Basic Identity, we can see that \(4\cosh(x) \sinh(x) = 2\sinh(2x)\text{.}\) We get the same answer either way.
  2. We employ substitution, with \(u = 7t-3\) and \(du = 7dt\text{.}\) Applying Key Ideas 6.1.5 and Key Idea 6.6.8 we have:
    \begin{equation*} \int \sech^2(7t-3)\,dt = \frac17\tanh(7t-3) + C\text{.} \end{equation*}
  3. \begin{align*} \int_0^{\ln(2) } \cosh(x) \, dx \amp = \sinh(x) \Big|_0^{\ln(2) }\\ \amp = \sinh(\ln(2) ) - \sinh(0)\\ \amp = \sinh(\ln(2) )\text{.} \end{align*}
    We can simplify this last expression as \(\sinh(x)\) is based on exponentials:
    \begin{align*} \sinh(\ln(2) ) \amp = \frac{e^{\ln(2) }-e^{-\ln(2) }}2\\ \amp = \frac{2-1/2}{2}\\ \amp = \frac34\text{.} \end{align*}

Subsection 6.6.2 Inverse Hyperbolic Functions

Just as the inverse trigonometric functions are useful in certain applications, the inverse hyperbolic functions are useful with others. Figure 6.6.(a) shows restriction on the domain of \(\cosh(x)\) to make the function one-to-one and the resulting domain and range of its inverse function. Since \(\sinh(x)\) is already one-to-one, no domain restriction is needed as shown in Figure 6.6.(b). Since \(\sech(x)\) is not one to one, it also needs a restricted domain in order to be invertible. Figure 6.6.(d) shows the graph of \(\sech^{-1}(x)\text{.}\) You should carefully compare the graph of this function to the graph given in Figure 6.6.6.(b) to see how this inverse was constructed. The rest of the hyperbolic functions area already one-to-one and need no domain restrictions. Their graphs are also shown in Figure 6.6.15.
Because the hyperbolic functions are defined in terms of exponential functions, their inverses can be expressed in terms of logarithms as shown in Key Idea 6.6.16. It is often more convenient to refer to \(\sinh^{-1}(x)\) than to \(\ln\big(x+\sqrt{x^2+1}\big)\text{,}\) especially when one is working on theory and does not need to compute actual values. On the other hand, when computations are needed, technology is often helpful but many hand-held calculators lack a convenient \(\sinh^{-1}(x)\) button. (Often it can be accessed under a menu system, but not conveniently.) In such a situation, the logarithmic representation is useful. The reader is not encouraged to memorize these, but rather know they exist and know how to use them when needed.
Figure 6.6.13. Finding the inverse of \(f(x)=\sinh(x)\)
Table 6.6.14. Domains and ranges of the hyperbolic and inverse hyperbolic functions
Function Domain Range Function Domain Range
\(\cosh(x)\) \([0,\infty)\) \([1,\infty)\) \(\cosh^{-1}(x)\) \([1,\infty)\) \([0,\infty)\)
\(\sinh(x)\) \((-\infty,\infty)\) \((-\infty,\infty)\) \(\sinh^{-1}(x)\) \((-\infty,\infty)\) \((-\infty,\infty)\)
\(\tanh(x)\) \((-\infty,\infty)\) \((-1,1)\) \(\tanh^{-1}(x)\) \((-1,1)\) \((-\infty,\infty)\)
\(\sech(x)\) \([0,\infty)\) \((0,1]\) \(\sech^{-1}(x)\) \((0,1]\) \([0,\infty)\)
\(\csch(x)\) \((-\infty,0) \cup (0,\infty)\) \((-\infty,0) \cup (0,\infty)\) \(\csch^{-1}(x)\) \((-\infty,0) \cup (0,\infty)\) \((-\infty,0) \cup (0,\infty)\)
\(\coth(x)\) \((-\infty,0) \cup (0,\infty)\) \((-\infty,-1) \cup (1,\infty)\) \(\coth^{-1}(x)\) \((-\infty,-1) \cup (1,\infty)\) \((-\infty,0) \cup (0,\infty)\)
Graph of hyperbolic cosine function and its inverse.
The \(y\) and the \(x\) axes are drawn from \(0\) to \(10\text{.}\) The functions \(y=\cosh(x)\) and \(y=\cosh^{-1}(x)\) are shown. They are symmetrical about the axis \(y=x\text{.}\)
From left to right, the function \(y=\cosh(x)\) starts at point \((0,1)\) then slowly rises from \((1,1)\) then it rises up steeply and it appears to run almost parallel to the \(y\) axis. The function \(\cosh^{-1}(x)\) starts at point \((1,0)\) and curves up steeply until \((2,1)\) then it rises very slowly and appears to almost run parallel to the \(x\) axis.
(a)
Graph of hyperbolic sine function and its inverse.
The \(y\) and the \(x\) axes are drawn from \(-10\) to \(10\text{.}\) The functions \(y=\sinh(x)\) and \(y=\sinh^{-1}(x)\text{.}\) The axis \(y=x\) is shown.
From left to right, the \(\sinh(x)\) function starts in the third quadrant and it rises steeply, very closely to the \(y\) axis. It crosses the origin along the \(y=x\) line, has a dip then increases very steeply and closely to the \(y\) axis in the first quadrant.
From left to right, in the third quadrant, the \(\sinh{-1}(x)\) function runs very closely to the \(x\) axis, it crosses the origin along the \(y=x\) line and bends to move very closely to the \(x\) axis in the first quadrant.
(b)
Graph of inverse of hyperbolic tangent and inverse of hyperbolic cotangent functions.
The \(y\) and the \(x\) axes are drawn from \(-3\) to \(3\text{.}\) There are two functions drawn, \(\coth^{-1}(x)\) and \(\tanh{-1}(x)\) along with two dashed lines \(x=-1\) and \(x=1\text{.}\)
The \(\tanh{-1}(x)\) function is drawn in the third and the first quadrant. From left to right, in the third quadrant the function is aligned with the line \(x=-1\) at around \(y=-2\) it diverges to the right side of the line, it crosses the origin then bends and merges with the line \(x=1\) from its left in the first quadrant.
The \(coth^{-1}(x)\) is also drawn in the third and the first quadrant. From left to right, in the third quadrant, the function appears to be parallel to the \(x\) axis; it diverges and bends down to join the line \(x=-1\text{.}\) In the first quadrant, from left to right the function is along the line \(x=1\text{,}\) it decreases and diverges from the line, there is a bend after \(x=2\) after which it becomes parallel to the \(x\) axis.
(c)
Graph of inverse of hyperbolic cosecant and inverse of hyperbolic secant functions.
The \(sec{-1}(x)\) is drawn only in the first quadrant.From left to right, it starts very close to the \(y\) axis without touching it, at around \(y=3\text{.}\) It moves away from the \(y\) axis while declining then gets a small bend before making an \(x\) intercept at \(x=1\text{.}\)
The \(\csch{-1}(x)\) is drawn in the third and the first quadrant. In the third quadrant from left to right the function appears to be parallel to the \(x\) axis. It bends toward the negative \(y\) axis and comes very close to it at \(y=-3\text{.}\) In the first quadrant from left to right, the function appears to start very close to the \(y\) axis coinciding with the \(sech{-1}(x)\) function. It has a negative slope and it moves down, gets a bend and runs parallel to the \(x\) axis.
(d)
Figure 6.6.15. Graphs of the hyperbolic functions (with restricted domains) and their inverses

Key Idea 6.6.16. Logarithmic definitions of Inverse Hyperbolic Functions.

  1. \(\ds\cosh^{-1}(x) =\ln\big(x+\sqrt{x^2-1}\big);\, x\geq1\)
  2. \(\displaystyle \tanh^{-1}(x) = \frac12\ln\left(\frac{1+x}{1-x}\right);\, \abs{x}\lt 1\)
  3. \(\displaystyle \sech^{-1}(x) = \ln\left(\frac{1+\sqrt{1-x^2}}x\right);\, 0\lt x\leq1\)
  4. \(\displaystyle \sinh^{-1}(x) = \ln\big(x+\sqrt{x^2+1}\big)\)
  5. \(\displaystyle \coth^{-1}(x) = \frac12\ln\left(\frac{x+1}{x-1}\right);\, \abs{x} \gt 1\)
  6. \(\displaystyle \csch^{-1}(x) = \ln\left(\frac1x+\frac{\sqrt{1+x^2}}{\abs{x}}\right);\, x\neq0\)
The following Key Ideas give the derivatives and integrals relating to the inverse hyperbolic functions. In Key Idea 6.6.18, both the inverse hyperbolic and logarithmic function representations of the antiderivative are given, based on Key Idea 6.6.16. Again, these latter functions are often more useful than the former. Note how inverse hyperbolic functions can be used to solve integrals we used Trigonometric Substitution to solve in Section 6.4.

Key Idea 6.6.17. Derivatives Involving Inverse Hyperbolic Functions.

  1. \(\ds\frac{d}{dx}\big(\cosh^{-1}(x) \big) = \frac{1}{\sqrt{x^2-1}};\\ x \gt 1\)
  2. \(\displaystyle \frac{d}{dx}\big(\sinh^{-1}(x) \big) = \frac{1}{\sqrt{x^2+1}}\)
  3. \(\displaystyle \frac{d}{dx}\big(\tanh^{-1}(x) \big) = \frac{1}{1-x^2};\\ \abs{x}\lt 1\)
  4. \(\displaystyle \frac{d}{dx}\big(\sech^{-1}(x) \big) = \frac{-1}{x\sqrt{1-x^2}};\\ 0\lt x\lt 1\)
  5. \(\displaystyle \frac{d}{dx}\big(\csch^{-1}(x) \big) = \frac{-1}{\abs{x}\sqrt{1+x^2}};\\ x\neq0\)
  6. \(\displaystyle \frac{d}{dx}\big(\coth^{-1}(x) \big) = \frac{1}{1-x^2};\\ \abs{x} \gt 1\)

Key Idea 6.6.18. Integrals Involving Inverse Hyperbolic Functions.

Assume \(a\gt 0\text{.}\)
  1. \begin{align*} \int \frac{1}{\sqrt{x^2-a^2}}\, dx \amp = \ln\abs{x+\sqrt{x^2-a^2}}+C\\ (\text{for } 0\lt x\lt a)\, \amp = \cosh^{-1}\left(\frac xa\right)+C \end{align*}
  2. \begin{align*} \int \frac{1}{\sqrt{x^2+a^2}}\, dx \amp =\ln\abs{x+\sqrt{x^2+a^2}}+C\\ \amp = \sinh^{-1}\left(\frac xa\right)+C \end{align*}
  3. \begin{align*} \int \frac{1}{a^2-x^2}\, dx \amp = \frac1{2a}\ln\abs{\frac{a+x}{a-x}}+C\\ \amp = \begin{cases} \frac1a\tanh^{-1}\left(\frac xa\right)+C \amp x^2\lt a^2\\ \frac1a\coth^{-1}\left(\frac xa\right)+C \amp a^2\lt x^2 \end{cases} \end{align*}
  4. \begin{align*} \int \frac{1}{x\sqrt{a^2-x^2}}\, dx \amp =\frac1a \ln\left(\frac{x}{a+\sqrt{a^2-x^2}}\right)+C\\ (\text{for } 0\lt x\lt a)\, \amp = -\frac1a\sech^{-1}\left(\frac xa\right)+C \end{align*}
  5. \begin{align*} \int \frac{1}{x\sqrt{x^2+a^2}}\, dx \amp = \frac1a \ln\abs{\frac{x}{a+\sqrt{a^2+x^2}}}+C\\ \amp = -\frac1a\csch^{-1}\abs{\frac xa} + C \end{align*}
Hyperbolic functions can be used as an alternative to trigonometric substitution, as illustrated in Figure 6.6.19.
Figure 6.6.19. Using a hyperbolic substitution to evaluate an integral
We practice using the derivative and integral formulas in the following example.

Example 6.6.20. Derivatives and integrals involving inverse hyperbolic functions.

Evaluate the following.
  1. \(\displaystyle \frac{d}{dx}\left[\cosh^{-1}\left(\frac{3x-2}{5}\right)\right]\)
  2. \(\displaystyle \int\frac{1}{x^2-1}\, dx\)
  3. \(\displaystyle \int \frac{1}{\sqrt{9x^2+10}}\, dx\)
Solution 1. Video solution
Solution 2.
  1. Applying Key Idea 6.6.17 with the Chain Rule gives:
    \begin{equation*} \frac{d}{dx}\left[\cosh^{-1}\left(\frac{3x-2}5\right)\right] = \frac{1}{\sqrt{\left(\frac{3x-2}5\right)^2-1}}\cdot\frac35\text{.} \end{equation*}
  2. Multiplying the numerator and denominator by \((-1)\) gives: \(\ds \int \frac{1}{x^2-1}\, dx = \int \frac{-1}{1-x^2}\, dx\text{.}\) The second integral can be solved with a direct application of item #3 from Key Idea 6.6.18, with \(a=1\text{.}\) Thus
    \begin{align} \int \frac{1}{x^2-1}\, dx \amp = -\int \frac{1}{1-x^2}\, dx\notag\\ \amp = \left\{\begin{array}{ccc} -\tanh^{-1}\left(x\right)+C \amp \amp x^2\lt 1 \\ \\ -\coth^{-1}\left(x\right)+C \amp \amp 1\lt x^2 \end{array} \right.\notag\\ \amp =-\frac12\ln\abs{\frac{x+1}{x-1}}+C\notag\\ \amp =\frac12\ln\abs{\frac{x-1}{x+1}}+C\text{.}\tag{6.6.1} \end{align}
    We should note that this exact problem was solved at the beginning of Section 6.5. In that example the answer was given as \(\frac12\ln\abs{x-1}-\frac12\ln\abs{x+1}+C\text{.}\) Note that this is equivalent to the answer given in Equation (6.6.1), as \(\ln(a/b) = \ln(a) - \ln(b)\text{.}\)
  3. This requires a substitution, then item #2 of Key Idea 6.6.18 can be applied. Let \(u = 3x\text{,}\) hence \(du = 3dx\text{.}\) We have
    \begin{align*} \int \frac{1}{\sqrt{9x^2+10}}\, dx \amp = \frac13\int\frac{1}{\sqrt{u^2+10}}\,du.\\ \end{align*}

    Note \(a^2=10\text{,}\) hence \(a = \sqrt{10}\text{.}\) Now apply the integral rule.

    \begin{align*} \amp = \frac13 \sinh^{-1}\left(\frac{3x}{\sqrt{10}}\right) + C\\ \amp = \frac13 \ln\abs{3x+\sqrt{9x^2+10}}+C\text{.} \end{align*}
This section covers a lot of ground. New functions were introduced, along with some of their fundamental identities, their derivatives and antiderivatives, their inverses, and the derivatives and antiderivatives of these inverses. Four Key Ideas were presented, each including quite a bit of information.
Do not view this section as containing a source of information to be memorized, but rather as a reference for future problem solving. Key Idea 6.6.18 contains perhaps the most useful information. Know the integration forms it helps evaluate and understand how to use the inverse hyperbolic answer and the logarithmic answer.
The next section takes a brief break from demonstrating new integration techniques. It instead demonstrates a technique of evaluating limits that return indeterminate forms. This technique will be useful in Section 6.8, where limits will arise in the evaluation of certain definite integrals.

Exercises 6.6.3 Exercises

Terms and Concepts

1.
In Key Idea 6.6.8, the equation \(\ds \int \tanh(x) \, dx = \ln(\cosh(x) )+C\) is given. Why is “\(\ln\abs{\cosh(x) }\)” not used — i.e., why are absolute values not necessary?
2.
The hyperbolic functions are used to define points on the right hand portion of the hyperbola \(x^2-y^2=1\text{,}\) as shown in Figure 6.6.2. How can we use the hyperbolic functions to define points on the left hand portion of the hyperbola?

Problems

Exercise Group.
In the following exercises, verify the given identity using Definition 6.6.3, as done in Example 6.6.7.
3.
Verify the identity \(\coth^2(x) -\csch^2(x) =1\) using the definitions of the hyperbolic functions.
4.
Verify the identity \(\cosh(2x) = \cosh^2(x) +\sinh^2(x)\) using the definitions of the hyperbolic functions.
5.
Verify the identity \(\ds\cosh^2(x) = \frac{\cosh(2x) +1}{2}\) using the definitions of the hyperbolic functions.
6.
Verify the identity \(\ds\sinh^2(x) = \frac{\cosh(2x) -1}{2}\) using the definitions of the hyperbolic functions.
7.
Verify the identity \(\ds\frac{d}{dx}\left[\sech(x) \right] = -\sech(x) \tanh(x)\) using the definitions of the hyperbolic functions.
8.
Verify the identity \(\ds\frac{d}{dx}\left[\coth(x) \right] = -\csch^2(x)\) using the definitions of the hyperbolic functions.
9.
Verify the identity \(\ds \int \tanh(x) \, dx = \ln(\cosh(x) ) + C\) using the definitions of the hyperbolic functions.
10.
Verify the identity \(\ds \int \coth(x) \, dx = \ln\abs{\sinh(x) } + C\) using the definitions of the hyperbolic functions.
Exercise Group.
In the following exercises, find the derivative of the given function.
11.
Find the derivative of \(f(x) = \sinh(2x)\text{.}\)
12.
Find the derivative of \(f(x) = \cosh ^2x\text{.}\)
13.
Find the derivative of \(f(x) = \tanh(x^2)\text{.}\)
14.
Find the derivative of \(f(x) = \ln(\sinh(x) )\text{.}\)
15.
Find the derivative of \(f(x) = \sinh(x) \cosh(x)\text{.}\)
16.
Find the derivative of \(f(x) = x\sinh(x) -\cosh(x)\text{.}\)
17.
Find the derivative of \(f(x) = \sech^{-1}(x^2)\text{.}\)
18.
Find the derivative of \(f(x) = \sinh^{-1}(3x)\text{.}\)
19.
Find the derivative of \(f(x) = \cosh^{-1}(2x^2)\text{.}\)
20.
Find the derivative of \(f(x) = \tanh^{-1}(x+5)\text{.}\)
21.
Find the derivative of \(f(x) = \tanh^{-1}(\cos(x) )\text{.}\)
22.
Find the derivative of \(f(x) = \cosh^{-1}(\sec(x) )\text{.}\)
Exercise Group.
In the following exercises, find the equation of the line tangent to the function at the given \(x\)-value.
23.
Find the equation of the tangent line to \(y=f(x)\) at \(x=0\text{,}\) where \(f(x) = \sinh(x)\text{.}\)
\(y =\)
24.
Find the equation of the tangent line to \(y=f(x)\) at \(x=\ln(2)\text{,}\) where \(f(x) = \cosh(x)\text{.}\)
\(y =\)
25.
Find the equation of the tangent line to \(y=f(x)\) at \(x=-\ln(3)\text{,}\) where \(f(x) = \tanh(x)\text{.}\)
\(y =\)
26.
Find the equation of the tangent line to \(y=f(x)\) at \(x=\ln(3)\text{,}\) where \(f(x) = \sech^2(x)\text{.}\)
\(y =\)
27.
Find the equation of the tangent line to \(y=f(x)\) at \(x=0\text{,}\) where \(f(x) = \sinh^{-1}(x)\text{.}\)
\(y =\)
28.
Find the equation of the tangent line to \(y=f(x)\) at \(x=\sqrt 2\text{,}\) where \(f(x) = \cosh^{-1}(x)\text{.}\)
\(y =\)
Exercise Group.
In the following exercises, evaluate the given indefinite integral.
29.
Evaluate the indefinite integral \(\ds \int \tanh(2x)\, dx\text{.}\)
30.
Evaluate the indefinite integral \(\ds \int \cosh(3x-7)\, dx\text{.}\)
31.
Evaluate the indefinite integral \(\ds \int \sinh(x) \cosh(x) \, dx\text{.}\)
32.
Evaluate the indefinite integral \(\ds \int x\cosh(x) \, dx\text{.}\)
33.
Evaluate the indefinite integral \(\ds \int x\sinh(x) \, dx\text{.}\)
34.
Evaluate the indefinite integral \(\ds\int \frac1{\sqrt{x^2+1}}\, dx\text{.}\)
35.
Evaluate the indefinite integral \(\ds\int \frac1{\sqrt{x^2-9}}\, dx\text{.}\)
36.
Evaluate the indefinite integral \(\ds \int \frac{1}{9-x^2}\, dx\text{.}\)
37.
Evaluate the indefinite integral \(\ds \int \frac{2x}{\sqrt{x^4-4}}\, dx\text{.}\)
38.
Evaluate the indefinite integral \(\ds \int \frac{\sqrt{x}}{\sqrt{1+x^3}}\, dx\text{.}\)
39.
Evaluate the indefinite integral \(\ds \int \frac{1}{x^4-16}\, dx\text{.}\)
40.
Evaluate the indefinite integral \(\ds \int \frac{1}{x^2+x}\, dx\text{.}\)
41.
Evaluate the indefinite integral \(\ds \int \frac{e^x}{e^{2x}+1}\, dx\text{.}\)
42.
Evaluate the indefinite integral \(\ds \int \sinh^{-1}(x) \, dx\text{.}\)
43.
Evaluate the indefinite integral \(\ds \int \tanh^{-1}(x) \, dx\text{.}\)
44.
Evaluate the indefinite integral \(\ds \int \sech(x) \, dx\text{.}\)
(Hint: mutiply by \(\frac{\cosh(x) }{\cosh(x) }\text{;}\) set \(u = \sinh(x)\text{.}\))
Exercise Group.
In the following exercises, evaluate the given definite integral.
45.
Evaluate the definite integral \(\ds \int_{-1}^1 \sinh(x) \, dx\text{.}\)
46.
Evaluate the definite integral \(\ds \int_{-\ln(2) }^{\ln(2) } \cosh(x) \, dx\text{.}\)
47.
Evaluate the definite integral \(\ds \int_{0}^{1} \sech^{2}(x) \, dx\text{.}\)
48.
Evaluate the definite integral \(\ds \int_{0}^{2}\frac1{\sqrt{x^2+1}}\, dx\text{.}\)
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