Integration of Hyperbolic Functions

\(\DeclareMathOperator{\sech}{sech}\DeclareMathOperator{\csch}{cosech}\DeclareMathOperator{\coth}{coth}\)Meet Phil the spider. Phil is sitting atop his single strand of web, connected to the walls on either side of the room. He wants to calculate the area beneath his strand, so he knows how long it will take him to fully build his web down to the floor.

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Jetzt kostenlos anmelden\(\DeclareMathOperator{\sech}{sech}\DeclareMathOperator{\csch}{cosech}\DeclareMathOperator{\coth}{coth}\)Meet Phil the spider. Phil is sitting atop his single strand of web, connected to the walls on either side of the room. He wants to calculate the area beneath his strand, so he knows how long it will take him to fully build his web down to the floor.

Luckily Phil is a mathematically inclined spider, so he knows that this area will be the integral of the curve of his web between the start and endpoints. Phil knows that the shape of a web, dangling between two fixed points, is known as a catenoid: the shape created by the hyperbolic cosine function. To answer his burning question, Phil must learn how to integrate hyperbolic functions.

The **hyperbolic trigonometric functions** are similar to the normal trigonometric functions, but instead of mapping out the unit circle, they map out the **unit hyperbola**.

The **standard hyperbolic functions** are,

- Hyperbolic sine: \(\sinh{x}\).
- Hyperbolic cosine: \(\cosh{x}\).
- Hyperbolic tangent: \(\tanh{x}\).

Similarly, the **reciprocal hyperbolic functions** are,

- Hyperbolic secant: \(\sech{x}\).
- Hyperbolic cosecant: \(\csch{x}\).
- Hyperbolic cotangent: \(\coth{x}\).

For more information about hyperbolic functions, including their **exponential forms**, see Hyperbolic Functions.

The integrals of our standard **hyperbolic functions** are,

\[ \begin{align} \int \sinh{x} \,dx & = \cosh{x} + c, \\ \int \cosh{x} \,dx & = \sinh{x} + c, \\ \int \tanh{x} \,dx & = \ln{ |\cosh{x} |} + c. \end{align}\]

Don't forget to add your constant of integration whenever you take an indefinite integral.

The integrals of the **reciprocal hyperbolic functions **are,

\[ \begin{align} \int \sech{x} \,dx & = \tan^{-1}{(\sinh{x})} + c, \\ \int \csch{x} \,dx & = ln{\left| \tanh{ \frac{x}{2} } \right|} + c, \\ \int \coth{x} \,dx & = ln{ \left| \sinh{x} \right| } + c. \end{align}\]

If you have learned about the **derivatives of hyperbolic functions**, you will have seen many standard results created by differentiating hyperbolic tangent and other hyperbolic functions. You can reverse these derivatives to gain formulas for integration,

\[ \begin{align} \int \sech^{2}{x} \,dx & = \tanh{x} + c, \\ \int \csch^{2}{x} \,dx & = - \coth{x} + c, \\ \int \sech{x} \tanh{x} \,dx & = - \sech{x} + c, \\ \int \csch{x} \coth{x} \,dx & = - \csch{x} + c. \end{align} \]

To learn more, see Differentiation of Hyperbolic Functions.

Some questions involving integrating hyperbolic functions will require the use of **Integration by Parts.**

Find \[ \int^{2}_{0} 2 x \cosh{2x} \,dx. \]

**Solution**

To solve this problem, you must first use the **integration by parts formula,**

\[ \int_{a}^{b} u \frac{dv}{dx} \,dx = \left[u v \right]_{a}^{b} - \int_{a}^{b} \frac{du}{dx} v \,dx. \]

In this case, you will want \( u \) to be \( 2x \), and \( \frac{dv}{dx} \) to be \( \cosh{2x} \). This is because you want to differentiate the \(2x\) term so that you are only left with a constant term in its place, so you can then solve the integral normally.

You can differentiate \(u \) in the usual way, and integrate \( \frac{dv}{dx} \) by inspection:

\( u = 2 x \), \(v = \frac{1}{2} \sinh{2x} \),

\( \frac{du}{dx} = 2 \), \(\frac{dv}{dx} = \cosh{2x} \).

Now that you have all of the components for the integration by parts formula, you can plug them in and evaluate the integral,

\[ \begin{align} \int^{2}_{0} 2 x \cosh{2x} \,dx & = \left[ 2 x \cdot \frac{1}{2} \sinh{2x} \right]^{2}_{0} - \int^{2}_{0} 2 \cdot \frac{1}{2} \sinh{2x} \,dx \\ & = \left[x \sinh{2x} \right]^2_0 - \int^2_0 \sinh{2x} \,dx. \end{align} \]

You can expand the first part of the right hand side to get \[ \left[ x \sinh{2x} \right]^2_0 = 2 \sinh{4} .\] Next, you just have to deal with the integral on the right hand side. It's not too difficult to see the solution of this integral, just think about what you would need to differentiate to get this function. This integral will become,

\[ \int_0^2 \sinh {2x} dx =\left[ \frac{1}{2} \cosh{2x} \right]^2_0 = \frac{1}{2} \cosh{4} - \frac{1}{2} \cosh{0} = \frac{1}{2} \cosh{4} - \frac{1}{2}, \]

since \(\cosh{0} =1 \).

Thus, you can plug these back into the initial integral to get,\[ \begin{align} \int^{2}_{0} 2 x \cosh{2x} \,dx & = 2 \sinh{4} - \left(\frac{1}{2} \cosh{4} - \frac{1}{2} \right) \\ & = 2 \sinh{4} - \frac{1}{2} \cosh{4} + \frac{1}{2}. \end{align} \]

This is the final answer. The question may ask you to put this into exponential form, but if it does not, this answer is okay.

The integrals of the inverse hyperbolic functions, (\(\cosh^{-1}{x} \), \(\sinh^{-1}{x}\) and \(\tanh^{-1}{x}\)) are,

\[ \begin{align} \int \sinh^{-1}{x} & = x \sinh^{-1}{x} - \sqrt{x^2 + 1} + c,\\ \int \cosh^{-1}{x} & = x \cosh^{-1}{x} - \sqrt{x^2 - 1} + c,\\ \int \tanh^{-1}{x} & = x \tanh^{-1}{x} + \frac{\ln{\left(1-x^2 \right)}}{2} + c. \end{align}\]

To learn more about the inverse hyperbolic functions, see Inverse Hyperbolic Functions.

Remember that \(\cosh^{-1} \), \(\sinh^{-1}\) and \(\tanh^{-1}\) can also be written as **arcosh**, **arsinh** and **artanh**. Similarly, \(\sech^{-1} \), \(\csch^{-1}\) and \(\coth^{-1}\) can be written as **arsech**, **arcosech** (often **arcsch** for short) and **arcoth**.

A common question about integrating hyperbolic functions is to prove some of the hyperbolic integral formulas above. It is easier to work through questions like this backwards, as working with differentiation is usually easier than working with integration.

Prove

\[ \int \csch^2{x} \,dx = - \coth{x} + c. \]

**Solution**

Since integration is the inverse of differentiating, it is enough to prove that

\[ \frac{d}{dx} (-\coth{x}) = \csch^2{x}. \]

First, write \( \coth{x} \) in terms of hyperbolic sine and cosine,

\[ \frac{d}{dx} (-\coth{x}) = \frac{d}{dx} \left(- \frac{\cosh{x}}{\sinh{x}} \right). \]

Now, use the quotient rule,

\[ \begin{align} \frac{d}{dx} (-\coth{x}) & = - \frac{\frac{d}{dx}(\cosh{x}) \sinh{x} - \cosh{x} \frac{d}{dx} (\sinh{x})}{\sinh^2{x}} \\ & = - \frac{\sinh^2{x} - \cosh^2{x}}{\sinh^2{x}}. \end{align}\]

Finally, use the **hyperbolic identity **\[ \cosh^2{x} - \sinh^2{x} = 1, \] to get,

\[ \begin{align} \frac{d}{dx} (-\coth{x}) & = - \frac{-1}{\sech^2{x}} \\ & = \csch^2{x},\end{align} \]

as required.

Sometimes, a tricky integral can be made much easier by making a substitution with a hyperbolic function.

Find \[ \int \frac{1}{\sqrt{x^2 - 1}} \,dx \] using substitution, for \( x \geq 1.\)

**Solution**

The first thing to decide here is what substitution to use. The most obvious place to use the substitution is on the bottom of the function, within the square root.

Think about the different **hyperbolic function identities:** which would be most useful here? The correct one to use is \[ \cosh^2{u} - \sinh^2{u} = 1 \] as this can be rearranged for \[ \cosh^2{u} - 1 = \sinh^2{u}, \] which looks like the inside of the square root, with \( \cosh{u} \) instead of \(x\). Substituting \( x = \cosh{u} \) in the initial integral, we get

\[\int \frac{1}{\sqrt{x^2 - 1}} \,dx = \int \frac{1}{\sqrt{\cosh^2{u} - 1}} \,dx. \]

Remember that when you make a substitution, you must replace the \(\,dx\) as well. Since \( x = \cosh{u} \), the derivative of \(x\) with respect to \(u\) is \[ \frac{dx}{du} = \sinh{u} \implies dx = \sinh{u} \,du.\] Thus, you can replace \(\,dx\) in the integral with \( \sinh{u} \,du \):

\[\int \frac{1}{\sqrt{x^2 - 1}} \,dx = \int \frac{1}{\sqrt{\cosh^2{u} - 1}} \sinh{u}\,d{u}. \]

Now, notice that on the bottom there is \( \cosh^2{u} - 1 \). Earlier you saw that this is equal to \( \sinh^2{u} \) by the hyperbolic function identities. Thus, you can replace \(\cosh^2{u}-1\) with \( \sinh^2{u} \):

\[ \begin{align} \int \frac{1}{\sqrt{x^2 - 1}} \,dx & = \int \frac{1}{\sqrt{\sinh^2{u}}} \sinh{u}\,d{u} \\ & = \int \frac{1}{\sinh{u}} \sinh{u}\,d{u} \\ & = \int \,du. \end{align} \]

Notice that here, we have written \( \sqrt{\sinh^2{u}} = \sinh{u} \), and not that it is equal to \( \pm \sinh{u}.\) This is because we have \( x \geq 1 \) from the question, and thus \( \frac{1}{\sqrt{x^2 - 1}} \geq 0.\) Hence, the integral of this function must also be positive, so we must take the positive value of this square root.

After doing the steps above, the integrand has been completely eliminated. This integral is direct to solve,

\[ \begin{align} \int \frac{1}{\sqrt{x^2 - 1}} \,dx & = \int \,du = u + c. \end{align} \]

Remember, since the question was in terms of \(x\), the answer must also be expressed in terms of \(x\). Recall that \(x = \cosh{u} \). Taking the inverse hyperbolic cosine of each side will get \( u = \cosh^{-1}{x} \). Finally, you can substitute this for the \(u\) in the formula, to get:

\[ \begin{align} \int \frac{1}{\sqrt{x^2 - 1}} \,dx & = \int \,du = \cosh^{-1}{x} + c, \end{align} \]

and thus, the question is complete.

Often, the substitution won't be so obvious. In this case, you must manipulate the function until it contains one of the **hyperbolic function identities**.

Find \[ \int \frac{1}{\sqrt{x^2 + 8x + 32}} \,dx \] using substitution.

**Solution**

This function doesn't look like any of our hyperbolic identities. The first step to take in simplifying this function is to complete the square of the quadratic inside the square root.

\[ (x + 4)^2 = x^2 + 8x + 16 \] \[ \implies (x+4)^2 + 16 = x^2 + 8x + 32. \]

Now, put the completed square into the original question:

\[ \int \frac{1}{\sqrt{x^2 + 8x + 32}} \,dx = \int \frac{1}{\sqrt{(x+4)^2 + 16}} \,dx.\]

This is a step closer to looking like one of the hyperbolic identities now. You could make the substitution using \(x + 4\), but there is still the 16 in the way making it impossible to simplify. To get rid of the 16, divide the top and bottom of the function by 4:

\[ \begin{align} \int \frac{1}{\sqrt{x^2 + 8x + 32}} \,dx & = \int \frac{\frac{1}{4}}{\frac{1}{4}\sqrt{(x+4)^2 + 16}} \,dx \\ & = \frac{1}{4} \int \frac{1}{\sqrt{\frac{1}{16}\left((x+4)^2 + 16 \right)}} \,dx \\ & = \frac{1}{4} \int \frac{1}{\sqrt{\frac{(x+4)^2}{16} + \frac{16}{16}}} \,dx \\ & = \frac{1}{4} \int \frac{1}{\sqrt{\left(\frac{x+4}{4}\right)^2 + 1}} \,dx. \end{align} \]

Here, the factor of \( \frac{1}{4} \) on the numerator has been taken out of the integral. Next, the \(\frac{1}{4} \) in the denominator has been put inside our square root, so it must be squared to get \(\frac{1}{16} \).

Finally, it has been put inside the square of the first term in the square root to reduce it back to \(\frac{1}{4} \), while the \(\frac{16}{16} \) cancels down to 1.

This looks a lot more like one of the hyperbolic identities if you substitute a hyperbolic function in the place of \(\frac{x+4}{4} \).

Which function should you use though? Recall the hyperbolic identity \[ \cosh^2{u} = \sinh^2{u} + 1, \] whose right-hand side can be identified as the inside of the square root, with the substitution \(\sinh{u}=\frac{x+4}{4} \). To see why this substitution makes sense in more detail, see Calculus of Hyperbolic Functions. After substitution, we get,

\[ \int \frac{1}{\sqrt{x^2 + 8x + 32}} \,dx = \frac{1}{4} \int \frac{1}{\sqrt{ \sinh^2{u} + 1 }} \,dx. \]

As said before, \( \sinh^2{u} + 1 = \cosh^2{u} \), so you can replace this to get,

\[ \begin{align} \int\frac{1}{\sqrt{x^2 + 8x + 32}} \,dx & = \frac{1}{4} \int \frac{1}{\sqrt{ \cosh^2{u}}} \,dx \\ & = \frac{1}{4} \int \frac{1}{\cosh{u}} \,dx. \end{align} \]

Again, we have taken the positive value of the square root and not the negative value. This is because the integrand is always positive, thus the integral itself must also be positive.

You must also remember to replace the \(\,dx\).

The substitution was \(\frac{x+4}{4} = \sinh{u} \), so rearranging this for \(x\) will yield: \( x = 4 \sinh{u} - 4 \). Differentiating this with respect to \(u\) you get, \( \frac{dx}{du} = 4 \cosh{u} \).

Thus, you can replace \( dx \) with \( 4 \cosh{u} \,du \):

\[ \begin{align} \int \frac{1}{\sqrt{x^2 + 8x + 32}} \,dx & = \frac{1}{4} \int \frac{1}{\cosh{u}} 4 \cosh{u} \,du \\ & = \int 1 \,du. \end{align} \]

From here, again the integral is direct. On other similar questions, you may be left with a scalar value different than 1, but here it just so happens to have cancelled out.

\[ \begin{align} \int \frac{1}{\sqrt{x^2 + 8x + 32}} \,dx & = \int \,du \\ & = u+c. \end{align} \]

Recall that the substitution was \( \sinh{u} = \frac{x+4}{4} \implies u = \sinh^{-1}{\frac{x+4}{4}} \). Thus, the answer is

\[ \int \frac{1}{\sqrt{x^2 + 8x + 32}} \,dx = \sinh^{-1}{\left( \frac{x+4}{4} \right)} +c.\]

- The integrals of the standard
**hyperbolic functions**are:\[ \begin{align} \int \sinh{x} \,dx & = \cosh{x} + c,\\ \int \cosh{x} \,dx & = \sinh{x} + c, \\ \int \tanh{x} \,dx & = \ln{ \left( \cosh{x} \right)} + c. \end{align}\]

The integrals of the

**reciprocal hyperbolic functions**are:\[ \begin{align} \int \sech{x} \,dx & = \tan^{-1}{(\sinh{x})} + c, \\ \int \csch{x} \,dx & = ln{\left| \tanh{ \frac{x}{2} } \right|} + c, \\ \int \coth{x} \,dx & = ln{ \left| \sinh{x} \right| } + c. \end{align}\]

The integrals of the

**inverse hyperbolic functions**are:\[ \begin{align} \int \sinh^{-1}{x} & = x \sinh^{-1}{x} - \sqrt{x^2 + 1} + c. \\ \int \cosh^{-1}{x} & = x \cosh^{-1}{x} - \sqrt{x^2 - 1} + c, \\ \int \tanh^{-1}{x} & = x \tanh^{-1}{x} + \frac{\ln{\left(1-x^2 \right)}}{2} + c. \end{align} \]

You can also use your knowledge of the

**derivatives of hyperbolic functions**to solve integrals as well, since integration is the opposite of differentiation. This gives the following formulas: \[ \begin{align} \int \sech^{2}{x} \,dx & = \tanh{x} + c, \\ \int \csch^{2}{x} \,dx & = - \coth{x} + c, \\ \int \sech{x} \tanh{x} \,dx & = - \sech{x} + c, \\ \int \csch{x} \coth{x} \,dx & = - \csch{x} + c. \end{align} \]- You can solve certain complicated integrals using
**integration by substitution**, taking a hyperbolic function as the substitution. You may have to manipulate this function first until it looks like one of the**hyperbolic identities**. These integrals will predominantly consist of quadratics and square roots.

The integration of sinh x is cosh x +C. Similarly, the integration of cosh x is sinh x+C.

The integral of tanh(x) is ln|cosh(x)| +C.

The integral of sinx is -cosx+C.

The integral of sin(3x) is -1/3 cos(3x)+C.

The integration of cos theta is sin theta+C.

More about Integration of Hyperbolic Functions

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