Inverse Hyperbolic Functions

When you solve an equation such as $\sinh y=x$sinhy=x for $y$y, the answer is an inverse hyperbolic function, written $y=\operatorname{arsinh} x$y=arsinhx (read "ar-sinh", from area — see §13). The defining feature that makes these functions so valuable — and so different from inverse trig functions — is that each one can be written explicitly as a natural logarithm. That logarithmic form is the key that unlocks a whole family of standard integrals later in the course. This lesson derives the three logarithmic forms from the exponential definitions, examines the domains and graphs, and connects them to the integrals they generate.

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1 Spec mapping

Inverse hyperbolic functions are compulsory pure content for Papers 1 and 2 of AQA Further Mathematics (7367). The derivations of the logarithmic forms are textbook AO2 "show that" material — short, self-contained proofs that examiners return to year after year — while evaluating them and using the resulting standard integrals is AO1. The topic sits between the hyperbolic-identities lesson (which supplies the algebra) and the integration lesson (which is its main application), and it leans on the logarithm laws and the quadratic formula from A-Level Maths. Mastering the three derivations is the single highest-value thing you can do here: the same algebra reappears verbatim whenever an integral produces an inverse hyperbolic function.

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2 Core theory — the three logarithmic forms

The three inverse hyperbolic functions and their logarithmic forms are:

$\boxed{\ \operatorname{arsinh} x=\ln\!\big(x+\sqrt{x^2+1}\big)\ } \qquad (x\in\mathbb R)$ arsinhx=ln(x+x2+1​) ​(x∈R)

$\boxed{\ \operatorname{arcosh} x=\ln\!\big(x+\sqrt{x^2-1}\big)\ } \qquad (x\ge1)$ arcoshx=ln(x+x2−1​) ​(x≥1)

$\boxed{\ \operatorname{artanh} x=\tfrac12\ln\!\left(\frac{1+x}{1-x}\right)\ } \qquad (-1<x<1)$ artanhx=21​ln(1−x1+x​) ​(−1<x<1)

Function	Logarithmic form	Domain	Range
$\operatorname{arsinh} x$arsinhx	$\ln(x+\sqrt{x^2+1})$ln(x+x2+1​)	$\mathbb R$R	$\mathbb R$R
$\operatorname{arcosh} x$arcoshx	$\ln(x+\sqrt{x^2-1})$ln(x+x2−1​)	$[1,\infty)$[1,∞)	$[0,\infty)$[0,∞)
$\operatorname{artanh} x$artanhx	$\tfrac12\ln\!\frac{1+x}{1-x}$21​ln1−x1+x​	$(-1,1)$(−1,1)	$\mathbb R$R

Two of these domains and ranges need care, and both reasons trace straight back to the parent functions. Because $\sinh$sinh is a bijection from $\mathbb R$R onto $\mathbb R$R (strictly increasing, no repeats), its inverse $\operatorname{arsinh}$arsinh has no restriction. But $\cosh$cosh is even and only $\ge1$≥1, so it is not one-to-one; to invert it we restrict to the increasing branch $x\ge0$x≥0, which is why $\operatorname{arcosh}$arcosh is defined only for inputs $\ge1$≥1 and returns only non-negative outputs. And $\tanh$tanh maps $\mathbb R$R onto the open interval $(-1,1)$(−1,1), so $\operatorname{artanh}$artanh accepts only inputs strictly between $-1$−1 and $1$1 — quoting $\operatorname{artanh}(1)$artanh(1) is meaningless, as the logarithm $\tfrac12\ln\tfrac{2}{0}$21​ln02​ diverges.

A general principle is doing the work here: the domain of an inverse is the range of the original, and the range of the inverse is the domain (or chosen branch) of the original. Reading off the table from the previous lesson — $\sinh$sinh has range $\mathbb R$R, $\cosh$cosh has range $[1,\infty)$[1,∞), $\tanh$tanh has range $(-1,1)$(−1,1) — immediately gives the input domains of the three inverses, with no extra work. This is worth internalising because exam questions frequently assume a value is defined and award a mark for spotting when it is not: an integral whose limits stray below $x=a$x=a in an $\operatorname{arcosh}(x/a)$arcosh(x/a) problem, for instance, has no real value, and saying so is the point of the question.

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3 Deriving $\operatorname{arsinh} x$arsinhx

This is the model derivation; learn it cold, because the other two are variations on it. Let $y=\operatorname{arsinh} x$y=arsinhx, so by definition $\sinh y=x$sinhy=x. Write $\sinh y$sinhy in exponential form and clear the fraction:

$\frac{e^y-e^{-y}}{2}=x\ \implies\ e^y-e^{-y}=2x.$2ey−e−y​=x ⟹ ey−e−y=2x.

Multiply through by $e^y$ey (always positive, so no solutions are lost) to clear the negative power and reveal a quadratic in $e^y$ey:

$e^{2y}-2x\,e^y-1=0.$e2y−2xey−1=0.

By the quadratic formula in $e^y$ey,

$e^y=\frac{2x\pm\sqrt{4x^2+4}}{2}=x\pm\sqrt{x^2+1}.$ey=22x±4x2+4​​=x±x2+1​.

Now the crucial sign decision: $e^y>0$ey>0 always, but $x-\sqrt{x^2+1}<0$x−x2+1​<0 for every $x$x (since $\sqrt{x^2+1}>\sqrt{x^2}=|x|\ge x$x2+1​>x2​=∣x∣≥x), so we must take the $+$+ root. Hence $e^y=x+\sqrt{x^2+1}$ey=x+x2+1​ and taking logarithms gives the boxed result $\operatorname{arsinh} x=\ln(x+\sqrt{x^2+1})$arsinhx=ln(x+x2+1​).

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4 Deriving $\operatorname{arcosh} x$arcoshx

Let $y=\operatorname{arcosh} x$y=arcoshx with $y\ge0$y≥0, so $\cosh y=x$coshy=x (and necessarily $x\ge1$x≥1). The same steps give

$\frac{e^y+e^{-y}}{2}=x\ \implies\ e^{2y}-2x\,e^y+1=0\ \implies\ e^y=x\pm\sqrt{x^2-1}.$2ey+e−y​=x ⟹ e2y−2xey+1=0 ⟹ ey=x±x2−1​.

This time both roots are positive (their product is $+1$+1 by the constant term), so the sign choice is governed instead by the branch $y\ge0$y≥0, i.e. $e^y\ge1$ey≥1. The larger root $x+\sqrt{x^2-1}\ge1$x+x2−1​≥1 gives $y\ge0$y≥0; the smaller root $x-\sqrt{x^2-1}$x−x2−1​ lies in $(0,1]$(0,1] and would give $y\le0$y≤0, the rejected branch. So $\operatorname{arcosh} x=\ln(x+\sqrt{x^2-1})$arcoshx=ln(x+x2−1​). (The two roots are reciprocals — $(x+\sqrt{x^2-1})(x-\sqrt{x^2-1})=x^2-(x^2-1)=1$(x+x2−1​)(x−x2−1​)=x2−(x2−1)=1 — which is exactly why $\operatorname{arcosh}$arcosh is "double-valued" before the branch restriction: $\pm\operatorname{arcosh} x$±arcoshx both satisfy $\cosh y=x$coshy=x.)

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5 Deriving $\operatorname{artanh} x$artanhx

Let $y=\operatorname{artanh} x$y=artanhx, so $\tanh y=x$tanhy=x. Using the exponential form $\tanh y=\dfrac{e^{2y}-1}{e^{2y}+1}$tanhy=e2y+1e2y−1​ (divide numerator and denominator of the definition by $e^{-y}$e−y):

$\frac{e^{2y}-1}{e^{2y}+1}=x\ \implies\ e^{2y}-1=x(e^{2y}+1)\ \implies\ e^{2y}(1-x)=1+x.$e2y+1e2y−1​=x ⟹ e2y−1=x(e2y+1) ⟹ e2y(1−x)=1+x.

$e^{2y}=\frac{1+x}{1-x}\ \implies\ 2y=\ln\!\left(\frac{1+x}{1-x}\right)\ \implies\ \operatorname{artanh} x=\tfrac12\ln\!\left(\frac{1+x}{1-x}\right).$e2y=1−x1+x​ ⟹ 2y=ln(1−x1+x​) ⟹ artanhx=21​ln(1−x1+x​).

The condition $-1<x<1$−1<x<1 is forced here: the argument $\tfrac{1+x}{1-x}$1−x1+x​ must be positive for the logarithm to exist, and it is positive precisely when $x\in(-1,1)$x∈(−1,1). It is worth pausing on the step $e^{2y}(1-x)=1+x$e2y(1−x)=1+x: we divided by $1-x$1−x to isolate $e^{2y}$e2y, which is legitimate exactly because $x<1$x<1 keeps $1-x>0$1−x>0. The factor of $\tfrac12$21​ out front is the single feature that distinguishes this derivation from the other two — it appears because the exponential form of $\tanh$tanh naturally produces $e^{2y}$e2y rather than $e^y$ey, so taking logarithms gives $2y$2y, which must then be halved. Dropping that $\tfrac12$21​ is the most common slip in the whole derivation; it also propagates into the corresponding integral, where $\int\tfrac{1}{a^2-x^2}\,\mathrm dx=\tfrac1a\operatorname{artanh}\tfrac xa+c$∫a2−x21​dx=a1​artanhax​+c carries its own $\tfrac1a$a1​ for a related reason.

A small but useful observation: all three derivations share the same skeleton — write the hyperbolic function in exponential form, multiply through to clear negative powers, recognise a quadratic (in $e^y$ey for $\operatorname{arsinh}$arsinh and $\operatorname{arcosh}$arcosh) or a linear equation (in $e^{2y}$e2y for $\operatorname{artanh}$artanh), solve, justify the root or sign, and take logarithms. If you remember the skeleton you never need to memorise the three answers separately; you can reconstruct any of them under exam pressure in under a minute.

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6 Worked examples with mark scheme

Worked Example 1 — evaluate $\operatorname{arsinh}$arsinh

Find the exact value of $\operatorname{arsinh}\tfrac34$arsinh43​.

$\operatorname{arsinh}\tfrac34=\ln\!\left(\tfrac34+\sqrt{\tfrac{9}{16}+1}\right)=\ln\!\left(\tfrac34+\sqrt{\tfrac{25}{16}}\right)=\ln\!\left(\tfrac34+\tfrac54\right)=\ln 2.$arsinh43​=ln(43​+169​+1​)=ln(43​+1625​​)=ln(43​+45​)=ln2.

Mark scheme: M1 substitute into $\ln(x+\sqrt{x^2+1})$ln(x+x2+1​); A1 simplify $\sqrt{\tfrac{25}{16}}=\tfrac54$1625​​=45​; A1 $\ln 2$ln2 (exact, not $0.693$0.693).

Worked Example 2 — evaluate $\operatorname{arcosh}$arcosh

Find the exact value of $\operatorname{arcosh}\tfrac53$arcosh35​.

$\operatorname{arcosh}\tfrac53=\ln\!\left(\tfrac53+\sqrt{\tfrac{25}{9}-1}\right)=\ln\!\left(\tfrac53+\sqrt{\tfrac{16}{9}}\right)=\ln\!\left(\tfrac53+\tfrac43\right)=\ln 3.$arcosh35​=ln(35​+925​−1​)=ln(35​+916​​)=ln(35​+34​)=ln3.

Mark scheme: M1 use $\ln(x+\sqrt{x^2-1})$ln(x+x2−1​); A1 $\sqrt{\tfrac{16}{9}}=\tfrac43$916​​=34​; A1 $\ln 3$ln3.

Worked Example 3 — evaluate $\operatorname{artanh}$artanh

Find the exact value of $\operatorname{artanh}\tfrac13$artanh31​.

$\operatorname{artanh}\tfrac13=\tfrac12\ln\!\left(\frac{1+\tfrac13}{1-\tfrac13}\right)=\tfrac12\ln\!\left(\frac{4/3}{2/3}\right)=\tfrac12\ln 2.$artanh31​=21​ln(1−31​1+31​​)=21​ln(2/34/3​)=21​ln2.

Mark scheme: M1 substitute into $\tfrac12\ln\tfrac{1+x}{1-x}$21​ln1−x1+x​; A1 simplify the inner fraction to $2$2; A1 $\tfrac12\ln 2$21​ln2.

Worked Example 4 — solve an equation via $\operatorname{arcosh}$arcosh

Solve $\cosh x=2$coshx=2 for $x\ge0$x≥0, giving the answer exactly.

$x=\operatorname{arcosh} 2=\ln\!\big(2+\sqrt{4-1}\big)=\ln\!\big(2+\sqrt3\big).$x=arcosh2=ln(2+4−1​)=ln(2+3​).

(The full solution set is $x=\pm\ln(2+\sqrt3)$x=±ln(2+3​) since $\cosh$cosh is even; the $x\ge0$x≥0 restriction selects the principal value.)

Mark scheme: M1 recognise $x=\operatorname{arcosh} 2$x=arcosh2; A1 $\ln(2+\sqrt3)$ln(2+3​). B1 (if not restricted) state both $\pm$± values.

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7 Specimen-style exam question

(Specimen-style.) (a) Show that $\operatorname{arsinh} x=\ln\!\big(x+\sqrt{x^2+1}\big)$arsinhx=ln(x+x2+1​). (4) (b) Hence find the exact value of $\operatorname{arsinh} 1$arsinh1. (2)

Model solution.

(a) Let $y=\operatorname{arsinh} x$y=arsinhx, so $\sinh y=x$sinhy=x. Then $\dfrac{e^y-e^{-y}}{2}=x$2ey−e−y​=x, and multiplying by $2e^y$2ey gives $e^{2y}-2x\,e^y-1=0$e2y−2xey−1=0. Solving the quadratic in $e^y$ey: $e^y=x\pm\sqrt{x^2+1}$ey=x±x2+1​. Since $e^y>0$ey>0 and $x-\sqrt{x^2+1}<0$x−x2+1​<0, take the $+$+ root: $e^y=x+\sqrt{x^2+1}$ey=x+x2+1​, so $y=\ln(x+\sqrt{x^2+1})$y=ln(x+x2+1​), as required.

(b) $\operatorname{arsinh} 1=\ln(1+\sqrt{1+1})=\ln(1+\sqrt2)$arsinh1=ln(1+1+1​)=ln(1+2​).

The "Hence" in (b) is an instruction to use the result of (a) — substitute into the logarithmic form rather than starting a fresh exponential calculation. Ignoring "Hence" and re-deriving wastes time and risks the method mark.

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8 The graphs

Each inverse graph is the reflection of the corresponding hyperbolic graph in the line $y=x$y=x (restricted to the invertible branch).

• $y=\operatorname{arsinh} x$y=arsinhx: domain $\mathbb R$R, range $\mathbb R$R; an odd function through the origin, strictly increasing, gradient $1$1 at the origin, flattening to a slow logarithmic climb (for large $x$x, $\operatorname{arsinh} x\approx\ln 2x$arsinhx≈ln2x).
• $y=\operatorname{arcosh} x$y=arcoshx: domain $[1,\infty)$[1,∞), range $[0,\infty)$[0,∞); starts at $(1,0)$(1,0) with a vertical tangent there, then increases. It is not odd or even — it only exists for $x\ge1$x≥1.
• $y=\operatorname{artanh} x$y=artanhx: domain $(-1,1)$(−1,1), range $\mathbb R$R; an odd function through the origin with vertical asymptotes at $x=\pm1$x=±1, shooting to $\pm\infty$±∞ as $x\to\pm1$x→±1.

The reflection idea is worth dwelling on because it explains every feature at a glance. Reflecting a curve in $y=x$y=x swaps the roles of horizontal and vertical, so a horizontal asymptote of the original becomes a vertical asymptote of the inverse: $\tanh x$tanhx has horizontal asymptotes $y=\pm1$y=±1, so $\operatorname{artanh} x$artanhx has vertical asymptotes $x=\pm1$x=±1 — no separate calculation needed. Likewise, $\cosh$cosh has a minimum turning point at $(0,1)$(0,1) where its tangent is horizontal; reflecting, $\operatorname{arcosh}$arcosh starts at $(1,0)$(1,0) with a vertical tangent, which is why $\operatorname{arcosh}$arcosh is so steep near $x=1$x=1 and why the integral $\int\tfrac{1}{\sqrt{x^2-a^2}}$∫x2−a2​1​ it generates is improper at the lower limit $x=a$x=a. And because reflecting an odd curve in $y=x$y=x leaves it odd, $\operatorname{arsinh}$arsinh and $\operatorname{artanh}$artanh inherit the oddness of $\sinh$sinh and $\tanh$tanh, whereas $\operatorname{arcosh}$arcosh — coming from a curve that was never odd — has no symmetry at all. Reading the inverse's behaviour off the parent's graph is far quicker, and far more reliable, than trying to plot the logarithmic form directly.