Notes · Further MathematicsUK · A-Levels

Hyperbolic functions

The hyperbolic functions $\cosh$ , $\sinh$ and $\tanh$ are defined from the exponential function and behave in many ways like the trigonometric functions, with identities related by Osborn's rule. This topic covers their definitions and graphs, the hyperbolic identities, the inverse hyperbolic functions and their logarithmic forms, and the differentiation, integration and use of hyperbolic substitutions.

4 sections~10 min reading time3 competenciesLevel Standard 1 · Advanced 3

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Exam profile

AO1 · Differentiate, integrate and manipulate hyperbolic functions accuratelyAO2 · Derive hyperbolic identities and the logarithmic forms of the inverse functionsAO3 · Choose and carry out hyperbolic substitutions to evaluate integrals

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basic level

AS Further Mathematics introduces the definitions, graphs and basic identities of the hyperbolic functions.

higher level

The full A-Level adds the inverse hyperbolic functions with their logarithmic forms, their calculus, and hyperbolic substitutions in integration.

Depth

Text

Contents · 4 sections

Hyperbolic functions

Definitions and graphs#

StandardAQA 7367 §3.8DfE Further Mathematics

Graphs of cosh, sinh and tanh

Fig. 1

\cosh x \geq 1

with minimum at

(0,1)

;

\sinh x

is odd through the origin;

\tanh x

is bounded by the asymptotes

y = \pm 1

The hyperbolic functions are defined directly from the exponential function:

\cosh x = \dfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}

\sinh x = \dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}

, and

\tanh x = \dfrac{\sinh x}{\cosh x} = \dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{\mathrm{e}^x + \mathrm{e}^{-x}}

. In effect

\cosh

is the even part of

\mathrm{e}^x

and

\sinh

its odd part, so that

\cosh x + \sinh x = \mathrm{e}^x

— a decomposition worth remembering, as it makes many results immediate.

Their graphs reflect these definitions. Because

\cosh(-x) = \cosh x

, the cosh curve is even, symmetric about the

y

-axis, with a minimum value of

1

x = 0

and rising like

\tfrac{1}{2}\mathrm{e}^{|x|}

for large

|x|

; it is the shape of a hanging chain (the catenary). Since

\sinh(-x) = -\sinh x

, the sinh curve is odd, passing through the origin and increasing everywhere. The tanh curve is also odd, passing through the origin, and has horizontal asymptotes

y = 1

and

y = -1

, which it approaches but never reaches as

x \to \pm\infty

The reciprocal hyperbolic functions complete the family:

\operatorname{sech} x = \dfrac{1}{\cosh x}

\operatorname{cosech} x = \dfrac{1}{\sinh x}

and

\coth x = \dfrac{1}{\tanh x}

. Their domains exclude the points where the denominator is zero (so

\operatorname{cosech}

and

\coth

are undefined at

x = 0

), and their graphs follow from those of the primary functions by taking reciprocals.

The naming — hyperbolic 'cosine' and 'sine' — reflects a deep analogy with the circular functions: just as

(\cos t, \sin t)

parametrises the unit circle

x^2 + y^2 = 1

, the pair

(\cosh t, \sinh t)

parametrises the right branch of the unit hyperbola

x^2 - y^2 = 1

. This is the source of the identity

\cosh^2 x - \sinh^2 x = 1

and of the whole system of hyperbolic identities that mirrors the trigonometric ones.

\cosh x = \frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}, \quad \sinh x = \frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}, \quad \tanh x = \frac{\mathrm{e}^x - \mathrm{e}^{-x}}{\mathrm{e}^x + \mathrm{e}^{-x}}

Definitions

cosh and sinh are the even and odd parts of $\mathrm{e}^x$ .

cosh and sinh as parts of the exponential

Fig. 2The average of

\mathrm{e}^x

and

\mathrm{e}^{-x}

\cosh x

; half their difference is

\sinh x

, and

\cosh x + \sinh x = \mathrm{e}^x

Worked example

cosh and sinh from the exponential

Prove that $\cosh x + \sinh x = \mathrm{e}^x$ and hence show $2\sinh x\cosh x = \sinh 2x$ .

Add the definitions
$\cosh x + \sinh x = \dfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2} + \dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2} = \dfrac{2\mathrm{e}^x}{2} = \mathrm{e}^x$ .
Form the product
$2\sinh x\cosh x = 2\cdot\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\cdot\dfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2} = \dfrac{(\mathrm{e}^x)^2 - (\mathrm{e}^{-x})^2}{2}$ .
$2\sinh x\cosh x = \frac{\mathrm{e}^{2x} - \mathrm{e}^{-2x}}{2}$
Recognise sinh 2x
$\dfrac{\mathrm{e}^{2x} - \mathrm{e}^{-2x}}{2} = \sinh 2x$ , exactly the definition with $x$ replaced by $2x$ .

Result: $\cosh x + \sinh x = \mathrm{e}^x$ and $2\sinh x\cosh x = \sinh 2x$ .

Typical mistakes

Sign slips in the definitions — confusing the $+$ in $\cosh$ with the $-$ in $\sinh$ .
Believing that $\cosh$ can be negative; its minimum value is $1$ , so $\cosh x \geq 1$ for all real $x$ .

Active revision

Prove from the definitions that $\cosh x + \sinh x = \mathrm{e}^x$ and that $\cosh x - \sinh x = \mathrm{e}^{-x}$ , and hence find $\cosh x\,\sinh x$ in terms of $\sinh 2x$ .

Active recall

Recall the key points — then reveal.

Sources: Further mathematics: AS and A level content (GCE subject content) (Department for Education)

Hyperbolic identities and Osborn's rule#

AdvancedAQA 7367 §3.8DfE Further Mathematics

The fundamental hyperbolic identity is

\cosh^2 x - \sinh^2 x = 1

, the analogue of

\cos^2 x + \sin^2 x = 1

but with a crucial minus sign. It is proved directly from the definitions, and dividing it through by

\cosh^2 x

gives

1 - \tanh^2 x = \operatorname{sech}^2 x

, the analogue of

1 + \tan^2 x = \sec^2 x

— again with a changed sign. These two identities are the ones used most often.

The double-angle and addition identities also mirror the trigonometric ones:

\sinh 2x = 2\sinh x\cosh x

\cosh 2x = \cosh^2 x + \sinh^2 x = 2\cosh^2 x - 1 = 1 + 2\sinh^2 x

, and

\sinh(x \pm y) = \sinh x\cosh y \pm \cosh x\sinh y

. Each is proved from the exponential definitions, and the different forms of

\cosh 2x

are useful for linearising

\cosh^2

\sinh^2

before integration.

Osborn's rule is the quick way to convert a trigonometric identity into the corresponding hyperbolic one: replace each

\cos

\cosh

and each

\sin

\sinh

, but change the sign of any term that contains a product of two sines (including such a product hidden inside another function, for example the

\sin^2

implicit in

\tan^2

). Thus

\cos^2 + \sin^2 = 1

becomes

\cosh^2 - \sinh^2 = 1

, the sign of the sine-squared term flipping.

Osborn's rule is a memory aid, not a proof: any identity it suggests should, if a proof is asked for, be established from the exponential definitions. The rule's value is in generating the correct hyperbolic form instantly and in checking one's algebra. Fluency with these identities is essential for the integration techniques of the next sections, where

\cosh 2x

and

\operatorname{sech}^2 x

appear constantly.

\cosh^2 x - \sinh^2 x = 1, \qquad 1 - \tanh^2 x = \operatorname{sech}^2 x

Fundamental identities

The minus signs distinguish these from the trigonometric analogues.

\cosh 2x = 2\cosh^2 x - 1 = 1 + 2\sinh^2 x, \qquad \sinh 2x = 2\sinh x\cosh x

Double-angle identities

Used to linearise $\cosh^2 x$ and $\sinh^2 x$ before integrating.

Worked example

Proving a double-angle identity

Prove from the definitions that $\cosh 2x = 2\cosh^2 x - 1$ .

Write cosh 2x
$\cosh 2x = \dfrac{\mathrm{e}^{2x} + \mathrm{e}^{-2x}}{2}$ .
Write 2cosh²x
$2\cosh^2 x = 2\left(\dfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^2 = \dfrac{\mathrm{e}^{2x} + 2 + \mathrm{e}^{-2x}}{2}$ .
$2\cosh^2 x = \frac{\mathrm{e}^{2x} + 2 + \mathrm{e}^{-2x}}{2}$
Subtract 1
$2\cosh^2 x - 1 = \dfrac{\mathrm{e}^{2x} + 2 + \mathrm{e}^{-2x}}{2} - 1 = \dfrac{\mathrm{e}^{2x} + \mathrm{e}^{-2x}}{2} = \cosh 2x$ .

Result: $\cosh 2x = 2\cosh^2 x - 1$ , proved from the definitions.

Typical mistakes

Carrying the trigonometric sign across unchanged — writing $\cosh^2 x + \sinh^2 x = 1$ instead of the correct minus.
Forgetting that Osborn's rule also affects sine-squared terms hidden inside $\tan^2$ or a product.

Active revision

Prove from the definitions that $\cosh 2x = 2\cosh^2 x - 1$ , and use Osborn's rule to write down the hyperbolic form of $\sin(A+B) = \sin A\cos B + \cos A\sin B$ .

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Further Mathematics 7367 specification (AQA)

Inverse hyperbolic functions and logarithmic forms#

AdvancedAQA 7367 §3.8DfE Further Mathematics

Each hyperbolic function has an inverse, written

\operatorname{arsinh}

\operatorname{arcosh}

and

\operatorname{artanh}

(or

\sinh^{-1}

etc.). Because

\sinh

and

\tanh

are one-to-one on the whole real line, their inverses are defined for all appropriate inputs; but

\cosh

is even and so is only one-to-one for

x \geq 0

, so

\operatorname{arcosh}

is defined for arguments

\geq 1

and takes non-negative values.

Because the hyperbolic functions are built from exponentials, their inverses can be written in closed form using logarithms. Solving

x = \sinh y

for

y

leads to a quadratic in

\mathrm{e}^y

, and choosing the positive root gives

\operatorname{arsinh} x = \ln\!\left(x + \sqrt{x^2 + 1}\right)

, valid for all

x

. The same method yields

\operatorname{arcosh} x = \ln\!\left(x + \sqrt{x^2 - 1}\right)

for

x \geq 1

and

\operatorname{artanh} x = \tfrac{1}{2}\ln\!\left(\dfrac{1+x}{1-x}\right)

for

|x| < 1

The derivation is a template worth mastering. To find

\operatorname{arsinh} x

, set

y = \operatorname{arsinh} x

so that

x = \sinh y = \tfrac{1}{2}(\mathrm{e}^y - \mathrm{e}^{-y})

; writing

t = \mathrm{e}^y

gives

t^2 - 2xt - 1 = 0

, whose positive root is

t = x + \sqrt{x^2 + 1}

(the other root is negative and rejected since

\mathrm{e}^y > 0

); then

y = \ln t

. The whole method is: substitute the definition, form a quadratic in

\mathrm{e}^y

, take the valid root, and take logarithms.

These logarithmic forms are more than curiosities: they let you evaluate inverse hyperbolic functions exactly and, crucially, they turn integrals whose answers are inverse hyperbolic functions into logarithms, and vice versa. The domain restrictions must be respected —

\operatorname{arcosh}

needs

x \geq 1

and

\operatorname{artanh}

needs

|x| < 1

— since outside these ranges the square root or logarithm is not real.

\operatorname{arsinh} x = \ln\!\big(x + \sqrt{x^2+1}\big) \ (\text{all } x), \qquad \operatorname{arcosh} x = \ln\!\big(x + \sqrt{x^2-1}\big) \ (x \geq 1)

Logarithmic forms

Obtained by solving the defining exponential equation for the inverse.

\operatorname{artanh} x = \frac{1}{2}\ln\!\left(\frac{1+x}{1-x}\right) \quad (|x| < 1)

Inverse tanh

Defined only on $(-1,1)$ , matching the range of $\tanh$ .

Worked example

The logarithmic form of arsinh

Show that $\operatorname{arsinh} x = \ln\!\left(x + \sqrt{x^2 + 1}\right)$ for all real $x$ .

Set up the equation
Let $y = \operatorname{arsinh} x$ , so $x = \sinh y = \dfrac{\mathrm{e}^y - \mathrm{e}^{-y}}{2}$ .
Form a quadratic in eʸ
Put $t = \mathrm{e}^y$ : $2x = t - \dfrac{1}{t}$ , so $t^2 - 2xt - 1 = 0$ .
$t^2 - 2xt - 1 = 0$
Solve and choose the root
$t = \dfrac{2x \pm \sqrt{4x^2 + 4}}{2} = x \pm \sqrt{x^2 + 1}$ . Since $t = \mathrm{e}^y > 0$ and $\sqrt{x^2+1} > |x|$ , take the $+$ root: $t = x + \sqrt{x^2+1}$ .
Take logarithms
$y = \ln t = \ln\!\left(x + \sqrt{x^2 + 1}\right)$ , valid for all $x$ because $x + \sqrt{x^2+1} > 0$ always.

Result: $\operatorname{arsinh} x = \ln\!\left(x + \sqrt{x^2 + 1}\right)$ .

Typical mistakes

Taking the wrong (negative) root of the quadratic in $\mathrm{e}^y$ , which cannot equal a positive exponential.
Using $\operatorname{arcosh}$ or $\operatorname{artanh}$ outside its domain, where the logarithmic form is not real.

Active revision

Show that $\operatorname{artanh} x = \tfrac{1}{2}\ln\!\left(\dfrac{1+x}{1-x}\right)$ for $|x| < 1$ , and hence evaluate $\operatorname{artanh}\tfrac{1}{2}$ exactly.

Active recall

Recall the key points — then reveal.

Sources: Further mathematics: AS and A level content (GCE subject content) (Department for Education)

Calculus of hyperbolic functions#

AdvancedAQA 7367 §3.8DfE Further Mathematics

Differentiating the definitions gives the derivative rules, which resemble the trigonometric ones but without the awkward sign:

\dfrac{\mathrm{d}}{\mathrm{d}x}\sinh x = \cosh x

and

\dfrac{\mathrm{d}}{\mathrm{d}x}\cosh x = \sinh x

— note there is no minus sign, unlike

\dfrac{\mathrm{d}}{\mathrm{d}x}\cos x = -\sin x

. Also

\dfrac{\mathrm{d}}{\mathrm{d}x}\tanh x = \operatorname{sech}^2 x

. Reversing these gives the basic integrals

\int\cosh x\,\mathrm{d}x = \sinh x + c

and

\int\sinh x\,\mathrm{d}x = \cosh x + c

The inverse hyperbolic functions have derivatives that are algebraic — no exponentials or logarithms appear — and these are exactly the antiderivatives needed for a common class of surd integrals:

\dfrac{\mathrm{d}}{\mathrm{d}x}\operatorname{arsinh} x = \dfrac{1}{\sqrt{x^2 + 1}}

\dfrac{\mathrm{d}}{\mathrm{d}x}\operatorname{arcosh} x = \dfrac{1}{\sqrt{x^2 - 1}}

and

\dfrac{\mathrm{d}}{\mathrm{d}x}\operatorname{artanh} x = \dfrac{1}{1 - x^2}

. Reading them backwards,

\int\dfrac{1}{\sqrt{x^2 + a^2}}\,\mathrm{d}x = \operatorname{arsinh}\dfrac{x}{a} + c

and

\int\dfrac{1}{\sqrt{x^2 - a^2}}\,\mathrm{d}x = \operatorname{arcosh}\dfrac{x}{a} + c

Hyperbolic substitutions are the technique that ties everything together. An integrand containing

\sqrt{x^2 + a^2}

is simplified by the substitution

x = a\sinh u

, because then

x^2 + a^2 = a^2\sinh^2 u + a^2 = a^2\cosh^2 u

, so the square root becomes

a\cosh u

and the identity

\cosh^2 - \sinh^2 = 1

has done the work. Similarly

\sqrt{x^2 - a^2}

suggests

x = a\cosh u

. This is the hyperbolic counterpart of the trigonometric substitution

x = a\sin\theta

for

\sqrt{a^2 - x^2}

In practice, choosing the right substitution comes from matching the surd to an identity that turns it into a perfect square. After substituting and simplifying, the integral in

u

is usually elementary; the final step is to convert back to

x

, either through the inverse hyperbolic function or, using the logarithmic forms, into a logarithm. Being able to move between the inverse-hyperbolic answer and the logarithmic answer is frequently required, since exam answers may be quoted in either form.

\frac{\mathrm{d}}{\mathrm{d}x}\sinh x = \cosh x, \quad \frac{\mathrm{d}}{\mathrm{d}x}\cosh x = \sinh x, \quad \frac{\mathrm{d}}{\mathrm{d}x}\tanh x = \operatorname{sech}^2 x

Derivatives (no sign change)

Unlike $\cos$ , the derivative of $\cosh$ carries no minus sign.

\int \frac{\mathrm{d}x}{\sqrt{x^2 + a^2}} = \operatorname{arsinh}\frac{x}{a} + c, \qquad \int \frac{\mathrm{d}x}{\sqrt{x^2 - a^2}} = \operatorname{arcosh}\frac{x}{a} + c

Standard surd integrals

Solved by the substitutions $x = a\sinh u$ and $x = a\cosh u$ .

Worked example

An integral by hyperbolic substitution

Find $\displaystyle\int \frac{1}{\sqrt{x^2 + 9}}\,\mathrm{d}x$ .

Substitute
Let $x = 3\sinh u$ , so $\mathrm{d}x = 3\cosh u\,\mathrm{d}u$ and $x^2 + 9 = 9\sinh^2 u + 9 = 9\cosh^2 u$ .
Simplify the surd
$\sqrt{x^2 + 9} = 3\cosh u$ , so the integral becomes $\displaystyle\int \frac{3\cosh u}{3\cosh u}\,\mathrm{d}u = \int 1\,\mathrm{d}u = u + c$ .
$\int \frac{1}{\sqrt{x^2+9}}\,\mathrm{d}x = u + c$
Convert back
Since $x = 3\sinh u$ , $u = \operatorname{arsinh}\dfrac{x}{3}$ , so the integral is $\operatorname{arsinh}\dfrac{x}{3} + c$ .
Logarithmic form
Using the log form, $= \ln\!\left(\dfrac{x}{3} + \sqrt{\dfrac{x^2}{9} + 1}\right) + c = \ln\!\left(x + \sqrt{x^2 + 9}\right) + c'$ (absorbing the constant $\ln 3$ ).

Result: $\displaystyle\int \frac{1}{\sqrt{x^2 + 9}}\,\mathrm{d}x = \operatorname{arsinh}\frac{x}{3} + c = \ln\!\left(x + \sqrt{x^2 + 9}\right) + c'$ .

Typical mistakes

Inserting a spurious minus sign into $\tfrac{\mathrm{d}}{\mathrm{d}x}\cosh x$ by false analogy with $\cos$ .
Choosing the wrong substitution ( $\sinh$ for a $\sqrt{x^2 - a^2}$ integrand, say), so the surd fails to simplify.

Active revision

Find $\displaystyle\int \frac{1}{\sqrt{x^2 + 9}}\,\mathrm{d}x$ using a hyperbolic substitution, giving the answer in terms of $\operatorname{arsinh}$ and also as a logarithm.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Further Mathematics 7367 specification (AQA)

References & sources

Sources

Department for Education

Further mathematics: AS and A level content (GCE subject content)

AQA

AQA A-level Further Mathematics 7367 specification

Hyperbolic functions

Key points

cosh and sinh from the exponential

Key points

Proving a double-angle identity

Key points

The logarithmic form of arsinh

Key points

An integral by hyperbolic substitution

Sources