List of Integrals of Hyperbolic Functions: Full Tables

What Is The List Of Integrals Of Hyperbolic Functions?

The list of integrals of hyperbolic functions is the set of antiderivatives for $\sinh x$, $\cosh x$, $\tanh x$, $\coth x$, $\operatorname{sech} x$ and $\operatorname{csch} x$, together with their powers, products, inverse functions and the algebraic integrals whose answers are inverse hyperbolic functions. Every entry below was checked by differentiating the answer back to the integrand.

The hyperbolic functions are built from the exponential function:

$$\sinh x = \frac{e^{x} - e^{-x}}{2}, \qquad \cosh x = \frac{e^{x} + e^{-x}}{2},$$

with $\tanh x = \frac{\sinh x}{\cosh x}$, $\coth x = \frac{\cosh x}{\sinh x}$, $\operatorname{sech} x = \frac{1}{\cosh x}$ and $\operatorname{csch} x = \frac{1}{\sinh x}$. The full background on their graphs and identities sits on the hyperbolic functions page, so this list keeps to integration.

Three conventions run through every table:

• $C$ is the arbitrary constant of integration, and it appears on every indefinite integral.

• $a$ is a positive constant unless a row says otherwise.

• "Valid for" gives the interval of $x$ on which the formula is a true antiderivative. Outside it, the integrand or the answer is undefined.

What Are The Integrals Of The Six Hyperbolic Functions?

The six basic integrals are below. The first two reverse the derivative rules directly; the other four each need one substitution, explained under the table.

With a constant inside, divide by that constant: $\int \sinh(ax),dx = \frac{1}{a}\cosh(ax) + C$ and $\int \cosh(ax),dx = \frac{1}{a}\sinh(ax) + C$.

These match NIST's reference list in the Digital Library of Mathematical Functions, section 4.40, which writes the coth and csch rows for $x > 0$ without the absolute value.

How Do You Integrate Tanh And Coth?

Write each as a ratio and substitute the denominator. This answers a question that comes up on Quora as "How to evaluate an integral with tangent hyperbolic or cotangent hyperbolic".

For $\tanh x = \frac{\sinh x}{\cosh x}$, let $u = \cosh x$, so $du = \sinh x,dx$:

$$\int \tanh x,dx = \int \frac{\sinh x}{\cosh x},dx$$

$$= \int \frac{du}{u}$$

$$= \ln\lvert u\rvert + C$$

$$= \ln(\cosh x) + C$$

The absolute value can go because $\cosh x \geq 1$ for every $x$. For $\coth x$ the same move with $u = \sinh x$ gives $\ln\lvert\sinh x\rvert + C$, and here the absolute value must stay, since $\sinh x < 0$ whenever $x < 0$. This is the integration by substitution pattern $\int \frac{f'(x)}{f(x)},dx = \ln\lvert f(x)\rvert + C$.

Why Does Sech Have Several Different Antiderivatives?

A Pearson Q&A puts it plainly: "There are several ways to express the indefinite integral of sech x". All of them are correct, because they differ only by a constant.

Multiply top and bottom by $\cosh x$ and use $\cosh^2 x = 1 + \sinh^2 x$:

$$\int \operatorname{sech} x,dx = \int \frac{\cosh x}{\cosh^2 x},dx$$

$$= \int \frac{\cosh x}{1 + \sinh^2 x},dx$$

$$= \arctan(\sinh x) + C$$

with $u = \sinh x$ in the last step. Three other forms appear in textbooks:

• $2\arctan\left(e^{x}\right) + C$, from the exponential form $\operatorname{sech} x = \frac{2e^{x}}{e^{2x} + 1}$.

• $2\arctan\left(\tanh\frac{x}{2}\right) + C$, from a half-argument substitution.

• $\operatorname{gd} x + C$, the Gudermannian function, which is how NIST writes it.

Checking these against each other is worth doing once. $\arctan(\sinh x)$ and $2\arctan\left(e^{x}\right)$ differ by exactly $\frac{\pi}{2}$ at every $x$, so they are the same antiderivative with different constants. The rule this teaches: two answers that look nothing alike can both be right, and the test is to differentiate them, not to compare them by eye.

How Do You Integrate Csch?

Rewrite $\operatorname{csch} x$ in exponentials and substitute $u = e^{x}$, so $dx = \frac{du}{u}$:

$$\int \operatorname{csch} x,dx = \int \frac{2e^{x}}{e^{2x} - 1},dx$$

$$= \int \frac{2}{u^{2} - 1},du$$

$$= \ln\left\lvert\frac{u - 1}{u + 1}\right\rvert + C$$

$$= \ln\left\lvert\frac{e^{x} - 1}{e^{x} + 1}\right\rvert + C$$

The step from the second line to the third is integration by partial fractions, since $\frac{2}{u^{2} - 1} = \frac{1}{u - 1} - \frac{1}{u + 1}$. The fraction $\frac{e^{x} - 1}{e^{x} + 1}$ equals $\tanh\frac{x}{2}$, which gives the table entry $\ln\left\lvert\tanh\frac{x}{2}\right\rvert + C$.

Which Integrals Come From Reading The Derivative Table Backwards?

Every derivative rule is also an integral rule read from right to left. Four hyperbolic derivatives give four integrals that are worth knowing on sight.

The minus signs in this table copy the minus signs in the derivative table. Only the reciprocal functions ($\coth$, $\operatorname{sech}$, $\operatorname{csch}$) have negative derivatives, so only their reversed rows carry a minus.

What Are The Integrals Of Powers And Products Of Hyperbolic Functions?

Powers and products reduce to the basic six through an identity, a substitution or integration by parts. The derivation column says which.

Two notes on this table. The $\sinh x\cosh x$ row has other correct forms, such as $\frac{\cosh^2 x}{2} + C$ and $\frac{\cosh 2x}{4} + C$, and each differs from the others by a constant. The square rows come from the double-argument identity $\cosh 2x = 1 + 2\sinh^2 x = 2\cosh^2 x - 1$, the hyperbolic twin of the one used for $\int \sin^2 x,dx$ in integration of trigonometric functions.

What Are The Integrals Of Inverse Hyperbolic Functions?

Each inverse hyperbolic function integrates by parts with $u$ equal to the inverse function and $dv = dx$. For $\sinh^{-1} x$ (also written $\operatorname{arsinh} x$):

$$\int \sinh^{-1} x,dx = x\sinh^{-1} x - \int \frac{x}{\sqrt{x^{2} + 1}},dx$$

$$= x\sinh^{-1} x - \sqrt{x^{2} + 1} + C$$

The remaining integral uses $w = x^{2} + 1$. The same two steps produce the whole table.

The csch row is the one most lists get half right. NIST gives $x\operatorname{csch}^{-1} x + \sinh^{-1} x$ for $x > 0$ only; for negative $x$ the sign of the second term flips, and writing $\sinh^{-1}\lvert x\rvert$ covers both sides at once.

Which Integrals Give Inverse Hyperbolic Functions As The Answer?

This is the part of the list you will use most. Square roots of $x^{2} + a^{2}$ and $x^{2} - a^{2}$, and the fraction $\frac{1}{a^{2} - x^{2}}$, all integrate to inverse hyperbolic functions, with $a > 0$:

The two forms in each row agree up to a constant. The logarithm form is what most exam answer keys expect, and the inverse hyperbolic form is shorter to write; the general list of integrals uses the logarithm versions.

Why do these integrals land on hyperbolic functions at all? Substitute $x = a\sinh t$ in the first row. Then $dx = a\cosh t,dt$ and $\sqrt{x^{2} + a^{2}} = a\cosh t$, so the integrand collapses to $1$ and the integral is $t = \sinh^{-1}\frac{x}{a}$. This is the hyperbolic version of trigonometric substitution, where $x = a\tan\theta$ does the same job with a messier answer.

Note the two $\frac{1}{a^{2} - x^{2}}$ rows. The integrand is the same, but $\tanh^{-1}$ only accepts inputs between $-1$ and $1$, so the answer switches to $\coth^{-1}$ outside that interval.

How Do Hyperbolic Integrals Compare With Trigonometric Integrals?

The hyperbolic list mirrors the trig list, with sign changes in predictable places. Side by side:

The sign differences trace back to one fact: $\frac{d}{dx}\cos x = -\sin x$ but $\frac{d}{dx}\cosh x = \sinh x$. Every trig minus sign that comes from differentiating cosine disappears in the hyperbolic column.

Examples Of Integrals Of Hyperbolic Functions

Example 1

Evaluate $\int \cosh(3x),dx$.

Use the $\cosh(ax)$ row with $a = 3$:

$$\int \cosh(3x),dx = \frac{1}{3}\sinh(3x) + C$$

Check: $\frac{d}{dx}\left[\frac{1}{3}\sinh(3x)\right] = \frac{1}{3}\cdot 3\cosh(3x) = \cosh(3x)$.

Final answer: $\frac{1}{3}\sinh(3x) + C$

Example 2

Evaluate $\int_{0}^{\ln 2} \sinh x,dx$.

The trig habit says the integral of a sine is minus a cosine, so a first attempt reads:

$$\left[-\cosh x\right]_{0}^{\ln 2} = -\cosh(\ln 2) + \cosh 0$$

With $\cosh(\ln 2) = \frac{2 + \frac{1}{2}}{2} = \frac{5}{4}$, this gives $-\frac{5}{4} + 1 = -\frac{1}{4}$. But $\sinh x > 0$ for every $x > 0$, so the area under it from $0$ to $\ln 2$ must be positive. A negative answer cannot be right.

The minus sign belongs to $\int \sin x,dx$, not to $\int \sinh x,dx$. Since $\frac{d}{dx}\cosh x = \sinh x$, the antiderivative is $\cosh x$:

$$\int_{0}^{\ln 2} \sinh x,dx = \left[\cosh x\right]_{0}^{\ln 2}$$

$$= \cosh(\ln 2) - \cosh 0$$

$$= \frac{5}{4} - 1$$

$$= \frac{1}{4}$$

Final answer: $\frac{1}{4}$

Example 3

Evaluate $\int \operatorname{sech} x\tanh x\operatorname{sech} x,dx$.

This is a Vedantu Q&A question, "How do you integrate sech x tanh x sech x dx". Collect the factors first: the integrand is $\operatorname{sech}^2 x\tanh x$. Let $u = \tanh x$, so $du = \operatorname{sech}^2 x,dx$:

$$\int \operatorname{sech}^2 x\tanh x,dx = \int u,du$$

$$= \frac{u^{2}}{2} + C$$

$$= \frac{\tanh^2 x}{2} + C$$

Choosing $u = \operatorname{sech} x$ instead gives $-\frac{\operatorname{sech}^2 x}{2} + C$. Both are correct, because $\frac{\tanh^2 x}{2} = \frac{1}{2} - \frac{\operatorname{sech}^2 x}{2}$, so they differ by the constant $\frac{1}{2}$.

Final answer: $\frac{\tanh^2 x}{2} + C$ (equivalently $-\frac{\operatorname{sech}^2 x}{2} + C$)

Example 4

Evaluate $\int x\cosh(2x),dx$.

Use integration by parts with $u = x$ and $dv = \cosh(2x),dx$, so $du = dx$ and $v = \frac{1}{2}\sinh(2x)$:

$$\int x\cosh(2x),dx = \frac{x}{2}\sinh(2x) - \int \frac{1}{2}\sinh(2x),dx$$

$$= \frac{x}{2}\sinh(2x) - \frac{1}{4}\cosh(2x) + C$$

The step that gets skipped here is the second factor of $\frac{1}{2}$: integrating $\frac{1}{2}\sinh(2x)$ needs another division by $2$, which is where the $\frac{1}{4}$ comes from.

Final answer: $\frac{x}{2}\sinh(2x) - \frac{1}{4}\cosh(2x) + C$

Example 5

Evaluate $\int_{0}^{1} \frac{dx}{4 - x^{2}}$.

Here $a = 2$ and every $x$ in $[0, 1]$ satisfies $\lvert x\rvert < 2$, so the $\tanh^{-1}$ row applies:

$$\int_{0}^{1} \frac{dx}{4 - x^{2}} = \left[\frac{1}{2}\tanh^{-1}\frac{x}{2}\right]_{0}^{1}$$

$$= \frac{1}{2}\tanh^{-1}\frac{1}{2} - 0$$

$$= \frac{1}{2}\cdot\frac{1}{2}\ln\frac{1 + \frac{1}{2}}{1 - \frac{1}{2}}$$

$$= \frac{1}{4}\ln 3$$

$$\approx 0.2747$$

Final answer: $\frac{\ln 3}{4} \approx 0.2747$

Example 6

Evaluate $\int_{3}^{5} \frac{dx}{\sqrt{x^{2} - 4}}$.

Here $a = 2$ and the interval $[3, 5]$ lies inside $x > 2$, so the $\cosh^{-1}$ row applies:

$$\int_{3}^{5} \frac{dx}{\sqrt{x^{2} - 4}} = \left[\cosh^{-1}\frac{x}{2}\right]_{3}^{5}$$

$$= \cosh^{-1}\frac{5}{2} - \cosh^{-1}\frac{3}{2}$$

$$= \ln\frac{5 + \sqrt{21}}{2} - \ln\frac{3 + \sqrt{5}}{2}$$

$$\approx 1.5668 - 0.9624$$

$$\approx 0.6044$$

Final answer: $\cosh^{-1}\frac{5}{2} - \cosh^{-1}\frac{3}{2} \approx 0.6044$

Why Do We Need Hyperbolic Integrals At All?

"Some square-root integrals have no tidy answer without them."

Without the hyperbolic list, $\int \frac{dx}{\sqrt{x^{2} + 1}}$ is still solvable, but only by a trig substitution with $\tan\theta$ and $\sec\theta$ followed by a long logarithm simplification. With it, the answer is one symbol: $\sinh^{-1} x + C$.

The hyperbolic list closes three gaps in the ordinary table:

• Square roots with a plus sign. $\sqrt{x^{2} + a^{2}}$ and $\sqrt{x^{2} - a^{2}}$ have no inverse-trig antiderivatives; $\sinh^{-1}$ and $\cosh^{-1}$ fill exactly that space.

• Integrals that would otherwise need partial fractions every time. $\int \frac{dx}{a^{2} - x^{2}}$ becomes a single $\tanh^{-1}$, with the logarithm form available when an answer key wants it.

• Clean definite values. For example, $\int_{0}^{1} \frac{dx}{\sqrt{x^{2} + 1}} = \sinh^{-1} 1 = \ln\left(1 + \sqrt{2}\right) \approx 0.8814$, and the improper integral $\int_{0}^{\infty} \operatorname{sech} x,dx$ equals exactly $\frac{\pi}{2}$, since $\arctan(\sinh x)$ climbs from $0$ to $\frac{\pi}{2}$.

• Harder improper integrals with named answers. Two Quora threads ask how to evaluate $\int_{0}^{\infty} \left(\ln(\cosh x) - \ln(\sinh x)\right)dx$ and $\int_{0}^{\infty} \frac{\sinh x}{x\cosh^2 x},dx$. For the first, write $\ln\frac{\cosh x}{\sinh x} = \ln\frac{1 + e^{-2x}}{1 - e^{-2x}} = 2\sum_{n \text{ odd}} \frac{e^{-2nx}}{n}$ and integrate term by term: each term gives $\frac{1}{n^{2}}$, and the sum over odd $n$ is $\frac{\pi^{2}}{8} \approx 1.2337$. The second equals $\frac{4G}{\pi} \approx 1.1662$, where $G \approx 0.9160$ is Catalan's constant; neither has an elementary antiderivative to plug limits into, which is where the basic list stops.

Where Are Integrals Of Hyperbolic Functions Used In Real Life?

The two most common applications are the length of a hanging cable and the distance an object falls against air resistance.

• Length of a hanging cable. A cable of uniform weight hangs in the shape $y = a\cosh\frac{x}{a}$. Its arc length uses $1 + \sinh^2 u = \cosh^2 u$, so the square root in the arc-length formula disappears and the integral is $\int \cosh\frac{x}{a},dx$. For $y = 10\cosh\frac{x}{10}$ from $x = -15$ to $x = 15$ (in feet), the length is $20\sinh(1.5) \approx 42.5856$ feet, the worked problem in OpenStax Calculus Volume 1, section 6.9.

• Distance fallen with drag. A body falling from rest with drag proportional to $v^{2}$ has speed $v(t) = \sqrt{\frac{g}{k}}\tanh\left(\sqrt{gk},t\right)$, where $g$ is gravitational acceleration and $k$ is the drag constant per unit mass. Integrating with the $\tanh$ row gives the distance fallen: $s(t) = \frac{1}{k}\ln\left(\cosh\left(\sqrt{gk},t\right)\right)$.

• Exact values for physics tables. Integrals such as $\int_{0}^{\infty} \frac{\tanh(ax) - \tanh(bx)}{x},dx = \ln\frac{a}{b}$ for $a, b > 0$ appear in NIST's reference list, where hyperbolic forms turn otherwise awkward integrals into closed values.

Where Solutions Go Off The Rails With Hyperbolic Integrals

Most errors on this topic come from carrying a trig habit across, or from using a formula outside its domain.

Mistake 1: Copying The Minus Sign From The Sine Integral

Where it slips in: Any $\int \sinh x,dx$ or $\int \sinh(ax),dx$, usually right after a trig integration unit.

Don't do this: Write $\int \sinh x,dx = -\cosh x + C$ because $\int \sin x,dx = -\cos x + C$.

The correct way: Write $\int \sinh x,dx = \cosh x + C$, and check by differentiating: $\frac{d}{dx}\cosh x = \sinh x$, with no sign.

The first instinct on $\int \sinh x,dx$ is almost always $-\cosh x$, carried over from sine. The memoriser who learned "integral of sine is minus cosine" as a sound rather than as a reversed derivative is the one who makes it, and the one-line derivative check is the habit that catches it.

Mistake 2: Using Tanh Inverse Outside Its Domain

Where it slips in: Integrals like $\int \frac{dx}{1 - x^{2}}$ evaluated on an interval such as $[2, 3]$.

Don't do this: Write $\left[\tanh^{-1} x\right]_{2}^{3}$. $\tanh^{-1} 2$ and $\tanh^{-1} 3$ do not exist as real numbers.

The correct way: For $\lvert x\rvert > 1$ use $\coth^{-1} x$, or the logarithm form $\frac{1}{2}\ln\frac{x + 1}{x - 1}$. On $[2, 3]$ the answer is $\frac{1}{2}\ln 2 - \frac{1}{2}\ln 3 = \frac{1}{2}\ln\frac{2}{3} \approx -0.2027$, which is negative because $\frac{1}{1 - x^{2}} < 0$ there.

Mistake 3: Dropping The Absolute Value In Coth And Csch

Where it slips in: $\int \coth x,dx$ or $\int \operatorname{csch} x,dx$ on an interval of negative $x$.

Don't do this: Write $\ln(\sinh x)$ for $x < 0$, where $\sinh x$ is negative and the logarithm is undefined.

The correct way: Keep $\ln\lvert\sinh x\rvert + C$ and $\ln\left\lvert\tanh\frac{x}{2}\right\rvert + C$. The $\tanh$ integral is the only one of the three where dropping the bars is safe, since $\cosh x \geq 1$.

The rusher who drops absolute values to save a stroke gets away with it on positive inputs, so the habit survives until the first problem set on negative intervals.

Mistake 4: Forgetting To Divide By The Inner Constant

Where it slips in: $\int \cosh(ax),dx$, $\int \sinh(ax),dx$ and the $x\sinh(ax)$ rows.

Don't do this: Write $\int \cosh(5x),dx = \sinh(5x) + C$.

The correct way: Divide by $a$: $\int \cosh(5x),dx = \frac{1}{5}\sinh(5x) + C$, since differentiating $\sinh(5x)$ produces a factor of $5$.

Swapping a hyperbolic shape for a more familiar one has a real cost. For centuries the hanging chain was assumed to be a parabola, a claim Galileo made and that was refuted in 1669, as the MacTutor history of the catenary records. For the OpenStax cable above, a parabola through the same lowest point and the same two end points has length about $42.1756$ feet against the catenary's $42.5856$, so treating the cable as a parabola under-orders it by about $0.41$ feet.

Practice Problems On The List Of Integrals Of Hyperbolic Functions

Evaluate each integral.

1. $\int \sinh(4x),dx$

2. $\int_{0}^{\ln 2} \cosh x,dx$

3. $\int x\sinh x,dx$

4. $\int \frac{dx}{\sqrt{x^{2} + 4}}$

5. $\int \cosh^2 x\sinh x,dx$

6. $\int \operatorname{sech} x\tanh x,dx$

Answer to Question 1: $\frac{1}{4}\cosh(4x) + C$

Answer to Question 2: $\sinh(\ln 2) = \frac{3}{4}$

Answer to Question 3: $x\cosh x - \sinh x + C$

Answer to Question 4: $\sinh^{-1}\frac{x}{2} + C$, or $\ln\left(x + \sqrt{x^{2} + 4}\right) + C$

Answer to Question 5: $\frac{\cosh^3 x}{3} + C$

Answer to Question 6: $-\operatorname{sech} x + C$

What To Remember About Integrals Of Hyperbolic Functions

• The list of integrals of hyperbolic functions starts with $\int \sinh x,dx = \cosh x + C$ and $\int \cosh x,dx = \sinh x + C$, with no minus sign.

• Tanh and coth integrate to logarithms of cosh and sinh, and only the tanh answer can drop its absolute value.

• Sech has several correct antiderivatives, and any two of them differ by a constant.

• Square roots of $x^{2} + a^{2}$ and $x^{2} - a^{2}$ integrate to inverse hyperbolic functions, each valid only on its stated domain.

• Differentiating the answer back is the fastest check for every row in the list.

Your Next Move

Work through practice problems 1 to 6 before checking the answers; if problem 4 trips you up, return to the section on integrals that give inverse hyperbolic functions and redo the $x = a\sinh t$ substitution with $a = 2$. Book a free Bhanzu demo class:

Read More

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