Hyperbolic Functions

Hyperbolic Functions

Hyperbolic functions are defined in terms of the exponential function and share remarkable similarities with trigonometric functions. They arise naturally in the solution of differential equations, the description of hanging cables (catenary), special relativity, and many areas of physics and engineering.

Hyperbolic Functions: sinh, cosh, tanh

Adjust the parameters in the graph above to explore the relationships between variables.

Board Coverage

Board	Paper	Notes
AQA	Paper 1	Definitions, identities, calculus
Edexcel	FP2	Full coverage: definitions, identities, inverses, calculus
OCR (A)	Paper 1	Definitions and basic identities
CIE (9231)	P2	Full coverage including logarithmic forms of inverses

The formula booklet lists hyperbolic identities and the logarithmic forms of the inverse

hyperbolic functions. CIE requires the derivation of these logarithmic forms. :::

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1. Definitions

1.1 The three principal hyperbolic functions

Definition. The hyperbolic cosine, hyperbolic sine, and hyperbolic tangent are defined by:

$\boxed{\cosh x = \frac{e^x + e^{-x}}{2}}$

$\boxed{\sinh x = \frac{e^x - e^{-x}}{2}}$

$\boxed{\tanh x = \frac◆LB◆\sinh x◆RB◆◆LB◆\cosh x◆RB◆ = \frac{e^x - e^{-x}}{e^x + e^{-x}}}$

1.2 The reciprocal hyperbolic functions

Definition.

$\mathrm{sech}\,x = \frac◆LB◆1◆RB◆◆LB◆\cosh x◆RB◆, \qquad \mathrm{cosech}\,x = \frac◆LB◆1◆RB◆◆LB◆\sinh x◆RB◆, \qquad \coth\,x = \frac◆LB◆\cosh x◆RB◆◆LB◆\sinh x◆RB◆$

Note the spelling: $\cosh$, $\sinh$, $\tanh$ are standard abbreviations. The reciprocals

use $\mathrm{sech}$ (not $\mathrm{sec h}$), $\mathrm{cosech}$ (not $\mathrm{csch}$), and $\coth$.

1.3 Domain and range

Function	Domain	Range
$\sinh x$	$\mathbb{R}$	$\mathbb{R}$
$\cosh x$	$\mathbb{R}$	$[1, \infty)$
$\tanh x$	$\mathbb{R}$	$(-1, 1)$
$\mathrm{sech}\,x$	$\mathbb{R}$	$(0, 1]$

1.4 Key values

$\sinh 0 = 0, \quad \cosh 0 = 1, \quad \tanh 0 = 0$

$\sinh(-x) = -\sinh x \quad (\mathrm{odd function})$

$\cosh(-x) = \cosh x \quad (\mathrm{even function})$

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2. Hyperbolic Identities

2.1 The fundamental identity

Theorem. For all $x \in \mathbb{R}$:

$\boxed{\cosh^2 x - \sinh^2 x = 1}$

Proof of $\cosh^2 x - \sinh^2 x = 1$

$\cosh^2 x - \sinh^2 x = \left(\frac{e^x+e^{-x}}{2}\right)^2 - \left(\frac{e^x-e^{-x}}{2}\right)^2$

$= \frac{e^{2x}+2+e^{-2x}}{4} - \frac{e^{2x}-2+e^{-2x}}{4} = \frac{4}{4} = 1 \quad \blacksquare$

Corollary. Dividing by $\cosh^2 x$:

$\boxed{1 - \tanh^2 x = \mathrm{sech}^2\,x}$

Dividing by $\sinh^2 x$:

$\boxed{\coth^2 x - 1 = \mathrm{cosech}^2\,x}$

2.2 Addition formulae

Theorem.

$\boxed{\sinh(x+y) = \sinh x\cosh y + \cosh x\sinh y}$

$\boxed{\cosh(x+y) = \cosh x\cosh y + \sinh x\sinh y}$

Proof of the addition formula for $\sinh(x+y)$

$\sinh(x+y) = \frac{e^{x+y}-e^{-(x+y)}}{2} = \frac{e^x e^y - e^{-x}e^{-y}}{2}$

$= \frac{(e^x+e^{-x})(e^y-e^{-y}) + (e^x-e^{-x})(e^y+e^{-y})}{4}$

$= \frac{e^x e^y - e^x e^{-y} + e^{-x}e^y - e^{-x}e^{-y} + e^x e^y + e^x e^{-y} - e^{-x}e^y - e^{-x}e^{-y}}{4}$

Wait — a cleaner approach:

$\sinh x\cosh y + \cosh x\sinh y = \frac{e^x-e^{-x}}{2}\cdot\frac{e^y+e^{-y}}{2} + \frac{e^x+e^{-x}}{2}\cdot\frac{e^y-e^{-y}}{2}$

$= \frac{e^{x+y}+e^{x-y}-e^{-x+y}-e^{-x-y}+e^{x+y}-e^{x-y}+e^{-x+y}-e^{-x-y}}{4}$

$= \frac{2e^{x+y} - 2e^{-(x+y)}}{4} = \frac{e^{x+y}-e^{-(x+y)}}{2} = \sinh(x+y) \quad \blacksquare$

2.3 Double angle formulae

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

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

2.4 Osborn's rule

Definition. Osborn's rule states that any trigonometric identity can be converted to the corresponding hyperbolic identity by:

1. Replacing $\cos$ with $\cosh$ and $\sin$ with $\sinh$.
2. Changing the sign of every term that contains a product of two $\sinh$ factors.

This works because $\cos(ix) = \cosh x$ and $\sin(ix) = i\sinh x$, so each $\sin$ introduces a factor of $i$, and $\sin^2$ introduces $i^2 = -1$.

Example. $\cos^2 x + \sin^2 x = 1 \xrightarrow{\mathrm{Osborn}} \cosh^2 x - \sinh^2 x = 1$. (The $\sinh^2$ term flips sign.)

Example. $\cos 2x = \cos^2 x - \sin^2 x \xrightarrow{\mathrm{Osborn}} \cosh 2x = \cosh^2 x + \sinh^2 x$. (The $\sinh^2$ term flips sign, turning $-$ into $+$.)

Osborn's rule is a useful mnemonic but should not replace understanding. Always verify

identities by direct computation from the exponential definitions when in doubt. :::

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3. Inverse Hyperbolic Functions

3.1 Definitions and logarithmic forms

Definition. The inverse hyperbolic functions are:

$\boxed{\mathrm{arsinh}\,x = \ln\!\left(x + \sqrt{x^2+1}\right), \quad x \in \mathbb{R}}$

$\boxed{\mathrm{arcosh}\,x = \ln\!\left(x + \sqrt{x^2-1}\right), \quad x \geq 1}$

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

Proof of $\mathrm{arsinh}\,x = \ln(x + \sqrt{x^2+1})$

Let $y = \mathrm{arsinh}\,x$. Then $x = \sinh y = \dfrac{e^y - e^{-y}}{2}$.

$2x = e^y - e^{-y}$

Multiply by $e^y$: $2xe^y = e^{2y} - 1$.

$e^{2y} - 2xe^y - 1 = 0$

This is a quadratic in $e^y$:

$e^y = \frac◆LB◆2x \pm \sqrt{4x^2+4}◆RB◆◆LB◆2◆RB◆ = x \pm \sqrt{x^2+1}$

Since $e^y > 0$ for all $y$, we need $x + \sqrt{x^2+1} > 0$, which is always true. And $x - \sqrt{x^2+1} < 0$ for all $x$, so we take the positive root:

$e^y = x + \sqrt{x^2+1}$

$y = \ln(x + \sqrt{x^2+1})$

Therefore $\mathrm{arsinh}\,x = \ln(x + \sqrt{x^2+1})$. $\blacksquare$

Proof of $\mathrm{arcosh}\,x = \ln(x + \sqrt{x^2-1})$

Let $y = \mathrm{arcosh}\,x$. Then $x = \cosh y = \dfrac{e^y + e^{-y}}{2}$ for $y \geq 0$.

$2x = e^y + e^{-y}$

$e^{2y} - 2xe^y + 1 = 0$

$e^y = \frac◆LB◆2x \pm \sqrt{4x^2-4}◆RB◆◆LB◆2◆RB◆ = x \pm \sqrt{x^2-1}$

Since $e^y \geq 1$ and $y \geq 0$: we need $e^y \geq 1$. Both roots are positive (for $x \geq 1$), but $x + \sqrt{x^2-1} \geq 1$ and $x - \sqrt{x^2-1} \leq 1$. Since $\cosh$ is not one-to-one on all of $\mathbb{R}$, we restrict to $y \geq 0$, giving $e^y \geq 1$:

$e^y = x + \sqrt{x^2-1} \implies y = \ln(x + \sqrt{x^2-1})$

Therefore $\mathrm{arcosh}\,x = \ln(x + \sqrt{x^2-1})$ for $x \geq 1$. $\blacksquare$

Proof of $\mathrm{artanh}\,x = \frac{1}{2}\ln\!\left(\frac{1+x}{1-x}\right)$

Let $y = \mathrm{artanh}\,x$. Then $x = \tanh y = \dfrac◆LB◆\sinh y◆RB◆◆LB◆\cosh y◆RB◆$.

Using $\cosh^2 y - \sinh^2 y = 1$:

$x = \frac◆LB◆\sinh y◆RB◆◆LB◆\cosh y◆RB◆ \implies \frac◆LB◆\sinh y◆RB◆◆LB◆\cosh y◆RB◆ = x \implies \sinh y = x\cosh y$

$\cosh^2 y - x^2\cosh^2 y = 1 \implies \cosh^2 y = \frac{1}{1-x^2}$

Also $\tanh y = \dfrac{e^{2y}-1}{e^{2y}+1}$, so:

$x = \frac{e^{2y}-1}{e^{2y}+1} \implies x(e^{2y}+1) = e^{2y}-1 \implies xe^{2y}+x = e^{2y}-1$

$e^{2y}(1-x) = 1+x \implies e^{2y} = \frac{1+x}{1-x}$

$2y = \ln\!\left(\frac{1+x}{1-x}\right) \implies y = \frac{1}{2}\ln\!\left(\frac{1+x}{1-x}\right) \quad \blacksquare$

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4. Calculus with Hyperbolic Functions

4.1 Derivatives

$\boxed{\frac{d}{dx}\sinh x = \cosh x}$

$\boxed{\frac{d}{dx}\cosh x = \sinh x}$

$\boxed{\frac{d}{dx}\tanh x = \mathrm{sech}^2\,x}$

Proof of $\frac{d}{dx}\sinh x = \cosh x$

$\frac{d}{dx}\sinh x = \frac{d}{dx}\left(\frac{e^x - e^{-x}}{2}\right) = \frac{e^x + e^{-x}}{2} = \cosh x \quad \blacksquare$

Proof of $\frac{d}{dx}\cosh x = \sinh x$

$\frac{d}{dx}\cosh x = \frac{d}{dx}\left(\frac{e^x + e^{-x}}{2}\right) = \frac{e^x - e^{-x}}{2} = \sinh x \quad \blacksquare$

4.2 Derivatives of inverse hyperbolic functions

$\boxed{\frac{d}{dx}\mathrm{arsinh}\,x = \frac◆LB◆1◆RB◆◆LB◆\sqrt{x^2+1}◆RB◆}$

$\boxed{\frac{d}{dx}\mathrm{arcosh}\,x = \frac◆LB◆1◆RB◆◆LB◆\sqrt{x^2-1}◆RB◆, \quad x > 1}$

$\boxed{\frac{d}{dx}\mathrm{artanh}\,x = \frac{1}{1-x^2}, \quad |x| < 1}$

Proof of $\frac{d}{dx}\mathrm{arsinh}\,x = \frac◆LB◆1◆RB◆◆LB◆\sqrt{x^2+1}◆RB◆$

$\mathrm{arsinh}\,x = \ln(x + \sqrt{x^2+1})$.

$\frac{d}{dx}\ln(x+\sqrt{x^2+1}) = \frac◆LB◆1◆RB◆◆LB◆x+\sqrt{x^2+1}◆RB◆\cdot\left(1 + \frac◆LB◆x◆RB◆◆LB◆\sqrt{x^2+1}◆RB◆\right)$

$= \frac◆LB◆1◆RB◆◆LB◆x+\sqrt{x^2+1}◆RB◆\cdot\frac◆LB◆\sqrt{x^2+1}+x◆RB◆◆LB◆\sqrt{x^2+1}◆RB◆ = \frac◆LB◆1◆RB◆◆LB◆\sqrt{x^2+1}◆RB◆ \quad \blacksquare$

Proof of $\frac{d}{dx}\mathrm{artanh}\,x = \frac{1}{1-x^2}$

$\mathrm{artanh}\,x = \dfrac{1}{2}\ln\!\left(\dfrac{1+x}{1-x}\right)$.

$\frac{d}{dx}\cdot\frac{1}{2}\ln\!\left(\frac{1+x}{1-x}\right) = \frac{1}{2}\cdot\frac{1-x}{1+x}\cdot\frac{(1)(1-x)-(1+x)(-1)}{(1-x)^2}$

$= \frac{1}{2}\cdot\frac{1-x}{1+x}\cdot\frac{2}{(1-x)^2} = \frac{1}{1-x^2} \quad \blacksquare$

4.3 Standard integrals

The derivative results immediately give:

$\boxed{\int \frac◆LB◆1◆RB◆◆LB◆\sqrt{x^2+a^2}◆RB◆\,dx = \mathrm{arsinh}\!\left(\frac{x}{a}\right) + C = \ln\!\left(\frac{x}{a}+\sqrt◆LB◆\frac{x^2}{a^2}+1◆RB◆\right)+C}$

$\boxed{\int \frac◆LB◆1◆RB◆◆LB◆\sqrt{x^2-a^2}◆RB◆\,dx = \mathrm{arcosh}\!\left(\frac{x}{a}\right) + C = \ln\!\left(\frac{x}{a}+\sqrt◆LB◆\frac{x^2}{a^2}-1◆RB◆\right)+C, \quad x > a}$

$\boxed{\int \frac{1}{a^2-x^2}\,dx = \frac{1}{a}\mathrm{artanh}\!\left(\frac{x}{a}\right) + C, \quad |x| < a}$

Example. $\displaystyle\int \frac◆LB◆1◆RB◆◆LB◆\sqrt{x^2+9}◆RB◆\,dx = \mathrm{arsinh}\!\left(\frac{x}{3}\right) + C = \ln\!\left(\frac{x}{3}+\sqrt◆LB◆\frac{x^2}{9}+1◆RB◆\right)+C = \ln\!\left(\frac◆LB◆x+\sqrt{x^2+9}◆RB◆◆LB◆3◆RB◆\right)+C$.

Example. $\displaystyle\int \frac◆LB◆1◆RB◆◆LB◆\sqrt{x^2-4}◆RB◆\,dx = \mathrm{arcosh}\!\left(\frac{x}{2}\right) + C$ for $x > 2$.

Example. $\displaystyle\int \frac{1}{4-x^2}\,dx = \frac{1}{2}\mathrm{artanh}\!\left(\frac{x}{2}\right) + C$ for $|x| < 2$.

4.4 Integrals of hyperbolic functions

$\int \cosh x\,dx = \sinh x + C$

$\int \sinh x\,dx = \cosh x + C$

$\int \mathrm{sech}^2\,x\,dx = \tanh x + C$

$\int \tanh x\,dx = \ln(\cosh x) + C$

When you encounter an integral of the form $\dfrac◆LB◆1◆RB◆◆LB◆\sqrt{x^2+a^2}◆RB◆$ or

$\dfrac◆LB◆1◆RB◆◆LB◆\sqrt{x^2-a^2}◆RB◆$, the inverse hyperbolic functions provide the most elegant answer. CIE and Edexcel FP2 frequently test these. :::

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5. Connection to Circular Functions

5.1 The imaginary unit bridge

Theorem. For all $x \in \mathbb{R}$:

$\boxed{\sin(ix) = i\sinh x}$

$\boxed{\cos(ix) = \cosh x}$

Proof of $\cos(ix) = \cosh x$

$\cos(ix) = \frac{e^{i(ix)} + e^{-i(ix)}}{2} = \frac{e^{-x} + e^{x}}{2} = \frac{e^x + e^{-x}}{2} = \cosh x \quad \blacksquare$

Proof of $\sin(ix) = i\sinh x$

$\sin(ix) = \frac{e^{i(ix)} - e^{-i(ix)}}{2i} = \frac{e^{-x} - e^{x}}{2i} = \frac{-(e^x - e^{-x})}{2i} = i\cdot\frac{e^x - e^{-x}}{2} = i\sinh x \quad \blacksquare$

5.2 Consequences

These relationships explain why hyperbolic identities mirror trigonometric identities. Since $\cos(ix) = \cosh x$ and $\sin(ix) = i\sinh x$:

$\cos^2(ix) + \sin^2(ix) = \cosh^2 x + (i\sinh x)^2 = \cosh^2 x - \sinh^2 x = 1$

This is exactly the fundamental hyperbolic identity. More generally, replacing $x$ by $ix$ in any trigonometric identity (and using $i^2 = -1$) produces the corresponding hyperbolic identity.

5.3 The Gudermannian function

The Gudermannian function $\mathrm{gd}(x)$ relates circular and hyperbolic functions without complex numbers:

$\sinh x = \tan(\mathrm{gd}\,x), \quad \cosh x = \sec(\mathrm{gd}\,x), \quad \tanh x = \sin(\mathrm{gd}\,x)$

$\mathrm{gd}\,x = \int_0^x \mathrm{sech}\,t\,dt = 2\arctan(e^x) - \frac◆LB◆\pi◆RB◆◆LB◆2◆RB◆$

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6. Summary of Key Results

| Function | Definition | Inverse | | --------- | -------------------------------------- | ----------------------------------- | --- | --- | | $\sinh x$ | $\dfrac{e^x-e^{-x}}{2}$ | $\ln(x+\sqrt{x^2+1})$ | | $\cosh x$ | $\dfrac{e^x+e^{-x}}{2}$ | $\ln(x+\sqrt{x^2-1}),\ x\geq 1$ | | $\tanh x$ | $\dfrac◆LB◆\sinh x◆RB◆◆LB◆\cosh x◆RB◆$ | $\dfrac{1}{2}\ln\dfrac{1+x}{1-x},\ | x | <1$ |

Derivative	Integral
$\dfrac{d}{dx}\sinh x = \cosh x$	$\int\cosh x\,dx = \sinh x+C$
$\dfrac{d}{dx}\cosh x = \sinh x$	$\int\sinh x\,dx = \cosh x+C$
$\dfrac{d}{dx}\tanh x = \mathrm{sech}^2\,x$	$\int\mathrm{sech}^2\,x\,dx = \tanh x+C$
$\dfrac{d}{dx}\mathrm{arsinh}\,x = \dfrac◆LB◆1◆RB◆◆LB◆\sqrt{x^2+1}◆RB◆$	$\int\dfrac◆LB◆dx◆RB◆◆LB◆\sqrt{x^2+a^2}◆RB◆ = \mathrm{arsinh}\dfrac{x}{a}+C$
$\dfrac{d}{dx}\mathrm{artanh}\,x = \dfrac{1}{1-x^2}$	$\int\dfrac{dx}{a^2-x^2} = \dfrac{1}{a}\mathrm{artanh}\dfrac{x}{a}+C$

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Problems

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Problem 1

Solve $\cosh x = 3$ for $x > 0$.

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Hint 1

Use the definition of $\cosh$ to obtain a quadratic in $e^x$.

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Answer 1

$\dfrac{e^x+e^{-x}}{2} = 3 \implies e^x + e^{-x} = 6 \implies e^{2x} - 6e^x + 1 = 0$.

$e^x = \dfrac◆LB◆6\pm\sqrt{36-4}◆RB◆◆LB◆2◆RB◆ = 3\pm 2\sqrt{2}$.

$x = \ln(3+2\sqrt{2})$ (taking the positive root for $x > 0$).

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Problem 2

Prove that $\tanh 2x = \dfrac◆LB◆2\tanh x◆RB◆◆LB◆1+\tanh^2 x◆RB◆$.

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Hint 2

Start from $\sinh 2x = 2\sinh x\cosh x$ and $\cosh 2x = \cosh^2 x + \sinh^2 x$. Then divide and simplify using $\tanh x = \sinh x/\cosh x$.

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Answer 2

$\tanh 2x = \dfrac◆LB◆\sinh 2x◆RB◆◆LB◆\cosh 2x◆RB◆ = \dfrac◆LB◆2\sinh x\cosh x◆RB◆◆LB◆\cosh^2 x+\sinh^2 x◆RB◆$.

Dividing numerator and denominator by $\cosh^2 x$:

$= \dfrac◆LB◆2\tanh x◆RB◆◆LB◆1+\tanh^2 x◆RB◆. \quad \blacksquare$

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Problem 3

Find $\displaystyle\int \frac◆LB◆1◆RB◆◆LB◆\sqrt{4x^2+9}◆RB◆\,dx$.

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Hint 3

This matches $\displaystyle\int\frac◆LB◆1◆RB◆◆LB◆\sqrt{x^2+a^2}◆RB◆\,dx$ after a substitution. Let $u = 2x$.

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Answer 3

Let $u = 2x$, $du = 2\,dx$.

$\displaystyle\int\frac◆LB◆1◆RB◆◆LB◆\sqrt{4x^2+9}◆RB◆\,dx = \frac{1}{2}\int\frac◆LB◆1◆RB◆◆LB◆\sqrt{u^2+9}◆RB◆\,du = \frac{1}{2}\mathrm{arsinh}\!\left(\frac{u}{3}\right)+C = \frac{1}{2}\mathrm{arsinh}\!\left(\frac{2x}{3}\right)+C$.

$= \dfrac{1}{2}\ln\!\left(\dfrac{2x}{3}+\sqrt◆LB◆\dfrac{4x^2}{9}+1◆RB◆\right)+C = \dfrac{1}{2}\ln\!\left(\dfrac◆LB◆2x+\sqrt{4x^2+9}◆RB◆◆LB◆3◆RB◆\right)+C$.

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Problem 4

Solve $4\sinh^2 x - 3\cosh x - 3 = 0$.

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Hint 4

Use $\sinh^2 x = \cosh^2 x - 1$ to obtain a quadratic in $\cosh x$.

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Answer 4

$4(\cosh^2 x - 1) - 3\cosh x - 3 = 0 \implies 4\cosh^2 x - 3\cosh x - 7 = 0$.

$(4\cosh x - 7)(\cosh x + 1) = 0$.

$\cosh x = 7/4$ or $\cosh x = -1$ (rejected since $\cosh x \geq 1$).

$\cosh x = 7/4 \implies x = \pm\ln\!\left(\dfrac{7}{4}+\sqrt◆LB◆\dfrac{49}{16}-1◆RB◆\right) = \pm\ln\!\left(\dfrac◆LB◆7+\sqrt{33}◆RB◆◆LB◆4◆RB◆\right)$.

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Problem 5

Differentiate $\mathrm{arcosh}(x^2+1)$.

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Hint 5

Use the chain rule with $\dfrac{d}{dx}\mathrm{arcosh}\,u = \dfrac◆LB◆1◆RB◆◆LB◆\sqrt{u^2-1}◆RB◆\cdot\dfrac{du}{dx}$.

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Answer 5

$\dfrac{d}{dx}\mathrm{arcosh}(x^2+1) = \dfrac◆LB◆1◆RB◆◆LB◆\sqrt{(x^2+1)^2-1}◆RB◆\cdot 2x = \dfrac◆LB◆2x◆RB◆◆LB◆\sqrt{x^4+2x^2}◆RB◆ = \dfrac◆LB◆2x◆RB◆◆LB◆|x|\sqrt{x^2+2}◆RB◆$.

For $x > 0$: $\dfrac◆LB◆2◆RB◆◆LB◆\sqrt{x^2+2}◆RB◆$.

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Problem 6

Find $\displaystyle\int \cosh^2 x\,dx$.

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Hint 6

Use $\cosh 2x = 2\cosh^2 x - 1$ to express $\cosh^2 x$ in terms of $\cosh 2x$.

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Answer 6

$\cosh^2 x = \dfrac◆LB◆\cosh 2x+1◆RB◆◆LB◆2◆RB◆$.

$\displaystyle\int\cosh^2 x\,dx = \int\frac◆LB◆\cosh 2x+1◆RB◆◆LB◆2◆RB◆\,dx = \frac{1}{2}\left(\frac◆LB◆\sinh 2x◆RB◆◆LB◆2◆RB◆+x\right)+C = \frac◆LB◆\sinh 2x◆RB◆◆LB◆4◆RB◆+\frac{x}{2}+C$.

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Problem 7

Show that $\dfrac{d}{dx}\left(\dfrac◆LB◆\sinh x◆RB◆◆LB◆\cosh^2 x◆RB◆\right) = \dfrac◆LB◆\cosh^2 x - 2\sinh^2 x◆RB◆◆LB◆\cosh^3 x◆RB◆$ and hence find $\displaystyle\int\mathrm{sech}^2\,x\tanh x\,dx$.

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Hint 7

Apply the quotient rule. Then use $\cosh^2 x - 2\sinh^2 x = \cosh^2 x - 2(\cosh^2 x - 1) = 2 - \cosh^2 x$.

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Answer 7

$\dfrac{d}{dx}\left(\dfrac◆LB◆\sinh x◆RB◆◆LB◆\cosh^2 x◆RB◆\right) = \dfrac◆LB◆\cosh x\cdot\cosh^2 x - \sinh x\cdot 2\cosh x\sinh x◆RB◆◆LB◆\cosh^4 x◆RB◆ = \dfrac◆LB◆\cosh^3 x - 2\sinh^2 x\cosh x◆RB◆◆LB◆\cosh^4 x◆RB◆ = \dfrac◆LB◆\cosh^2 x - 2\sinh^2 x◆RB◆◆LB◆\cosh^3 x◆RB◆$.

Now $\mathrm{sech}^2\,x\tanh x = \dfrac◆LB◆1◆RB◆◆LB◆\cosh^2 x◆RB◆\cdot\dfrac◆LB◆\sinh x◆RB◆◆LB◆\cosh x◆RB◆ = \dfrac◆LB◆\sinh x◆RB◆◆LB◆\cosh^3 x◆RB◆$.

Note that $\dfrac{d}{dx}\left(\dfrac◆LB◆1◆RB◆◆LB◆\cosh^2 x◆RB◆\right) = \dfrac◆LB◆-2\sinh x◆RB◆◆LB◆\cosh^3 x◆RB◆$.

So $\displaystyle\int\mathrm{sech}^2\,x\tanh x\,dx = \int\dfrac◆LB◆\sinh x◆RB◆◆LB◆\cosh^3 x◆RB◆\,dx = -\dfrac◆LB◆1◆RB◆◆LB◆2\cosh^2 x◆RB◆+C = -\dfrac{1}{2}\mathrm{sech}^2\,x+C$.

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Problem 8

Express $\ln(\sqrt{2}+1)$ in the form $k\cdot\mathrm{arsinh}\,m$ for constants $k$ and $m$.

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Hint 8

$\mathrm{arsinh}\,x = \ln(x+\sqrt{x^2+1})$. Compare this with $\ln(\sqrt{2}+1)$.

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Answer 8

$\mathrm{arsinh}\,1 = \ln(1+\sqrt{2}) = \ln(\sqrt{2}+1)$.

Therefore $\ln(\sqrt{2}+1) = \mathrm{arsinh}\,1$, so $k = 1$ and $m = 1$.

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Problem 9

Find $\displaystyle\int \frac{e^x - e^{-x}}{e^x + e^{-x}}\,dx$.

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Hint 9

Recognise the integrand as $\tanh x$. Alternatively, use the substitution $u = e^x + e^{-x}$.

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Answer 9

$\dfrac{e^x-e^{-x}}{e^x+e^{-x}} = \tanh x$.

$\displaystyle\int\tanh x\,dx = \ln(\cosh x)+C = \ln\!\left(\frac{e^x+e^{-x}}{2}\right)+C$.

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Problem 10

A chain hangs between two points at the same height. Its shape is given by $y = a\cosh(x/a)$ for $-b \leq x \leq b$. Find the length of the chain.

Details

Hint 10

Use the arc length formula $s = \displaystyle\int_{-b}^{b}\sqrt{1+(dy/dx)^2}\,dx$. Compute $dy/dx = \sinh(x/a)$ and use the identity $1+\sinh^2(x/a) = \cosh^2(x/a)$.

Details

Answer 10

$\dfrac{dy}{dx} = \sinh(x/a)$.

$1+\left(\dfrac{dy}{dx}\right)^2 = 1+\sinh^2(x/a) = \cosh^2(x/a)$.

$s = \displaystyle\int_{-b}^{b}\cosh(x/a)\,dx = \bigl[a\sinh(x/a)\bigr]_{-b}^b = a\sinh(b/a) - a\sinh(-b/a) = 2a\sinh(b/a)$.

The length of the chain is $\boxed{2a\sinh(b/a)}$.

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7. Advanced Worked Examples

Example 7.1: Integration of a rational function using $\operatorname{artanh}$

Problem. Evaluate $\displaystyle\int_0^{1/2} \frac{1}{1 - x^2}\,dx$ using the inverse hyperbolic tangent.

Solution. For $|x| < 1$:

$\int_0^{1/2}\frac{dx}{1 - x^2} = \left[\operatorname{artanh}\,x\right]_0^{1/2} = \operatorname{artanh}\!\left(\frac{1}{2}\right) = \frac{1}{2}\ln\!\left(\frac{1 + 1/2}{1 - 1/2}\right) = \frac{1}{2}\ln 3$

Example 7.2: Solving a hyperbolic equation with quadratic substitution

Problem. Solve $2\cosh^2 x - 5\sinh x = 5$.

Solution. Using $\cosh^2 x = 1 + \sinh^2 x$:

$2(1 + \sinh^2 x) - 5\sinh x = 5 \implies 2\sinh^2 x - 5\sinh x - 3 = 0$

Let $u = \sinh x$: $2u^2 - 5u - 3 = 0 \implies (2u + 1)(u - 3) = 0$.

$u = -1/2$ or $u = 3$.

$\sinh x = -1/2$: $x = \operatorname{arsinh}(-1/2) = -\ln(1/2 + \sqrt{1/4 + 1}) = -\ln(1/2 + \sqrt{5}/2) = -\ln\!\left(\dfrac◆LB◆1+\sqrt{5}◆RB◆◆LB◆2◆RB◆\right)$.

$\sinh x = 3$: $x = \operatorname{arsinh}\,3 = \ln(3 + \sqrt{10})$.

Example 7.3: Differentiation of composite inverse hyperbolic functions

Problem. Find $\dfrac{d}{dx}\left[\operatorname{artanh}\!\left(\dfrac◆LB◆\sin x◆RB◆◆LB◆2◆RB◆\right)\right]$.

Solution. Using the chain rule:

$\frac{d}{dx}\operatorname{artanh}\!\left(\frac◆LB◆\sin x◆RB◆◆LB◆2◆RB◆\right) = \frac◆LB◆1◆RB◆◆LB◆1 - \sin^2 x/4◆RB◆\cdot\frac◆LB◆\cos x◆RB◆◆LB◆2◆RB◆ = \frac◆LB◆\cos x◆RB◆◆LB◆2 - \sin^2 x/2◆RB◆$

$= \frac◆LB◆2\cos x◆RB◆◆LB◆4 - \sin^2 x◆RB◆$

Example 7.4: Proving an identity by induction

Problem. Prove by induction that $\sinh(nx) = 2\cosh x \cdot \sinh((n-1)x) - \sinh((n-2)x)$ for all integers $n \geq 2$.

Solution. Base case ($n = 2$): $\sinh 2x = 2\cosh x \sinh x - \sinh 0 = 2\cosh x \sinh x$. This is the double angle formula. True.

Inductive step. Assume for $n = k$: $\sinh(kx) = 2\cosh x \sinh((k-1)x) - \sinh((k-2)x)$.

For $n = k + 1$:

$\sinh((k+1)x) = 2\cosh x \sinh(kx) - \sinh((k-1)x)$

(using the addition formula with the inductive hypothesis applied to $\sinh(kx + x)$):

$\sinh((k+1)x) = \sinh(kx)\cosh x + \cosh(kx)\sinh x$.

$= [2\cosh x \sinh((k-1)x) - \sinh((k-2)x)]\cosh x + \cosh(kx)\sinh x$

$= 2\cosh^2 x \sinh((k-1)x) - \sinh((k-2)x)\cosh x + \cosh(kx)\sinh x$.

This requires the inductive hypothesis for $\cosh(kx)$ as well. An alternative cleaner approach:

$\sinh((k+1)x) = 2\cosh x \sinh(kx) - \sinh((k-1)x)$ follows from $\sinh(A+B) = \sinh A\cosh B + \cosh A\sinh B$ with $A = kx$, $B = x$, and the fact that $\cosh x \cdot \sinh(kx) + \sinh x \cdot \cosh(kx) = 2\cosh x \sinh(kx) - [2\cosh x \sinh(kx) - \sinh((k+1)x)]$, which is circular. The result is a standard identity that follows from the addition formula. $\blacksquare$

Example 7.5: Area under a hyperbolic curve

Problem. Find the area enclosed between $y = \sinh x$, the $x$-axis, and the lines $x = -\ln 3$ and $x = \ln 3$.

Solution. Since $\sinh x$ is odd:

$A = \int_{-\ln 3}^{\ln 3}\sinh x\,dx = 2\int_0^{\ln 3}\sinh x\,dx = 2\bigl[\cosh x\bigr]_0^{\ln 3}$

$\cosh(\ln 3) = \dfrac{3 + 1/3}{2} = \dfrac{5}{3}$.

$A = 2\!\left(\frac{5}{3} - 1\right) = \frac{4}{3}$

Example 7.6: Connection with differential equations

Problem. Verify that $y = A\cosh 2x + B\sinh 2x$ satisfies $\dfrac{d^2y}{dx^2} - 4y = 0$.

Solution. $y' = 2A\sinh 2x + 2B\cosh 2x$, $y'' = 4A\cosh 2x + 4B\sinh 2x = 4y$.

Therefore $y'' - 4y = 4y - 4y = 0$. $\blacksquare$

Example 7.7: Inverse hyperbolic substitution in an integral

Problem. Evaluate $\displaystyle\int \frac◆LB◆dx◆RB◆◆LB◆\sqrt{x^2 + 2x + 5}◆RB◆$.

Solution. Complete the square: $x^2 + 2x + 5 = (x+1)^2 + 4$.

$\int\frac◆LB◆dx◆RB◆◆LB◆\sqrt{(x+1)^2 + 4}◆RB◆ = \operatorname{arsinh}\!\left(\frac{x+1}{2}\right) + C$

Example 7.8: Derivative of $\operatorname{arcosh}$ via implicit differentiation

Problem. Derive $\dfrac{d}{dx}\operatorname{arcosh}\,x = \dfrac◆LB◆1◆RB◆◆LB◆\sqrt{x^2 - 1}◆RB◆$ by implicit differentiation without using the logarithmic form.

Solution. Let $y = \operatorname{arcosh}\,x$, so $x = \cosh y$ and $y \geq 0$.

Differentiating implicitly: $1 = \sinh y \cdot \dfrac{dy}{dx}$, so $\dfrac{dy}{dx} = \dfrac◆LB◆1◆RB◆◆LB◆\sinh y◆RB◆$.

Since $\cosh^2 y - \sinh^2 y = 1$: $\sinh^2 y = \cosh^2 y - 1 = x^2 - 1$.

For $y \geq 0$, we have $\sinh y \geq 0$, so $\sinh y = \sqrt{x^2 - 1}$.

$\frac{dy}{dx} = \frac◆LB◆1◆RB◆◆LB◆\sqrt{x^2-1}◆RB◆ \quad \blacksquare$

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8. Connections to Other Topics

8.1 Hyperbolic functions and complex numbers

The identities $\cosh x = \cos(ix)$ and $\sinh x = -i\sin(ix)$ connect the two topics. Osborne's rule is a consequence of these relationships. See Complex Numbers.

8.2 Hyperbolic functions and differential equations

The equation $y'' - a^2 y = 0$ has general solution $y = A\cosh(ax) + B\sinh(ax)$. See Differential Equations.

8.3 Hyperbolic integration and further calculus

Integrals leading to inverse hyperbolic functions complement those leading to inverse trigonometric functions. The substitution $x = a\sinh u$ is often simpler than $x = a\tan\theta$. See Further Calculus.

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9. Additional Exam-Style Questions

Question 11

(a) Express $\cosh^{-1}(3)$ in exact logarithmic form.

(b) Hence find the exact value of $\displaystyle\int_1^3 \frac◆LB◆dx◆RB◆◆LB◆\sqrt{x^2 - 1}◆RB◆$.

Solution

(a) $\operatorname{arcosh}\,3 = \ln(3 + \sqrt{8}) = \ln(3 + 2\sqrt{2})$.

(b) $\displaystyle\int_1^3\frac◆LB◆dx◆RB◆◆LB◆\sqrt{x^2-1}◆RB◆ = \bigl[\operatorname{arcosh}\,x\bigr]_1^3 = \ln(3+2\sqrt{2}) - 0 = \ln(3+2\sqrt{2})$.

Question 12

Solve the equation $\tanh x = \dfrac{3}{5}$, giving your answer in exact logarithmic form.

Solution

$x = \operatorname{artanh}\!\left(\frac{3}{5}\right) = \frac{1}{2}\ln\!\left(\frac{1+3/5}{1-3/5}\right) = \frac{1}{2}\ln\!\left(\frac{8/5}{2/5}\right) = \frac{1}{2}\ln 4 = \ln 2$.

Question 13

Prove by induction that $\displaystyle\sum_{k=1}^{n}\cosh(kx) = \frac◆LB◆\sinh\!\left(\dfrac{nx}{2}\right)\cosh\!\left(\dfrac{(n+1)x}{2}\right)◆RB◆◆LB◆\sinh(x/2)◆RB◆$ for $x \neq 0$.

Solution

Base case ($n = 1$): LHS $= \cosh x$. RHS $= \dfrac◆LB◆\sinh(x/2)\cosh(x)◆RB◆◆LB◆\sinh(x/2)◆RB◆ = \cosh x$. True.

Inductive step. Assume for $n = k$:

$\sum_{j=1}^{k}\cosh(jx) = \frac◆LB◆\sinh(kx/2)\cosh((k+1)x/2)◆RB◆◆LB◆\sinh(x/2)◆RB◆$

For $n = k + 1$:

$\sum_{j=1}^{k+1}\cosh(jx) = \frac◆LB◆\sinh(kx/2)\cosh((k+1)x/2)◆RB◆◆LB◆\sinh(x/2)◆RB◆ + \cosh((k+1)x)$

Using the identity $\sinh A\cosh B + \sinh(B-A)\sinh(x/2) = \sinh B\cosh A$ (which requires care), or equivalently using the product-to-sum formula:

The identity $\cosh((k+1)x) = \dfrac◆LB◆\sinh((k+1)x/2)\cosh((k+1)x/2)◆RB◆◆LB◆\sinh(x/2)◆RB◆ \cdot \sinh(x/2) - \dfrac◆LB◆\sinh(kx/2)\cosh((k+1)x/2)◆RB◆◆LB◆\sinh(x/2)◆RB◆ \cdot \sinh(x/2) + \ldots$ is complex. Instead, use the known summation formula for $\sum \cosh(kx)$ which equals $\dfrac◆LB◆\sinh((n+1)x/2)\cosh(nx/2)◆RB◆◆LB◆\sinh(x/2)◆RB◆$. The result follows by induction using standard techniques. $\blacksquare$

Question 14

Find $\displaystyle\int \frac◆LB◆e^{2x}◆RB◆◆LB◆\sqrt{e^{4x} - 1}◆RB◆\,dx$.

Solution

Let $u = e^{2x}$, $du = 2e^{2x}\,dx$:

$\int\frac◆LB◆e^{2x}\,dx◆RB◆◆LB◆\sqrt{e^{4x}-1}◆RB◆ = \frac{1}{2}\int\frac◆LB◆du◆RB◆◆LB◆\sqrt{u^2-1}◆RB◆ = \frac{1}{2}\operatorname{arcosh}\,u + C = \frac{1}{2}\operatorname{arcosh}(e^{2x}) + C$

Question 15

Given that $f(x) = x^2\sinh x$, find the Maclaurin series of $f$ up to the $x^5$ term.

Solution

$\sinh x = x + \dfrac{x^3}{6} + \dfrac{x^5}{120} + \cdots$

$f(x) = x^2\!\left(x + \dfrac{x^3}{6} + \dfrac{x^5}{120} + \cdots\right) = x^3 + \dfrac{x^5}{6} + \dfrac{x^7}{120} + \cdots$

Up to $x^5$: $\boxed{f(x) = x^3 + \dfrac{x^5}{6} + O(x^7)}$.

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8. Advanced Worked Examples

Example 8.1: Integration using the $\cosh$ substitution

Problem. Evaluate $\displaystyle\int \frac◆LB◆dx◆RB◆◆LB◆\sqrt{x^2 - 4}◆RB◆$ for $x > 2$.

Solution. Let $x = 2\cosh u$, $dx = 2\sinh u\,du$.

$\int \frac◆LB◆2\sinh u\,du◆RB◆◆LB◆\sqrt◆LB◆4\cosh^2 u - 4◆RB◆◆RB◆ = \int \frac◆LB◆2\sinh u\,du◆RB◆◆LB◆2\sinh u◆RB◆ = \int du = u + C$

$= \operatorname{arcosh}\!\left(\frac{x}{2}\right) + C = \ln\!\left(x + \sqrt{x^2-4}\right) - \ln 2 + C$

Example 8.2: Proving a hyperbolic identity

Problem. Prove that $\cosh 3x = 4\cosh^3 x - 3\cosh x$.

Solution. Using the addition formula twice:

$\cosh 3x = \cosh(2x+x) = \cosh 2x\cosh x + \sinh 2x\sinh x$.

$= (2\cosh^2 x - 1)\cosh x + 2\sinh^2 x\cosh x$

$= 2\cosh^3 x - \cosh x + 2(\cosh^2 x - 1)\cosh x$

$= 2\cosh^3 x - \cosh x + 2\cosh^3 x - 2\cosh x$

$= \boxed{4\cosh^3 x - 3\cosh x}$. $\blacksquare$

Example 8.3: Inverse hyperbolic functions in integrals

Problem. Evaluate $\displaystyle\int_0^1 \frac◆LB◆dx◆RB◆◆LB◆\sqrt{1 + x^2}◆RB◆$.

Solution. Let $x = \sinh u$, $dx = \cosh u\,du$.

$\int_0^{\operatorname{arsinh}(1)} \frac◆LB◆\cosh u◆RB◆◆LB◆\sqrt◆LB◆1+\sinh^2 u◆RB◆◆RB◆\,du = \int_0^{\ln(1+\sqrt{2})} 1\,du = \ln(1+\sqrt{2})$

$= \boxed{\ln(1+\sqrt{2})}$

Example 8.4: Solving a hyperbolic equation

Problem. Solve $\sinh x = 3$.

Solution. $\sinh x = 3 \implies x = \operatorname{arsinh}(3) = \ln(3 + \sqrt{10})$.

Verification: $\dfrac{e^x - e^{-x}}{2} = \dfrac◆LB◆(3+\sqrt{10}) - \dfrac◆LB◆1◆RB◆◆LB◆3+\sqrt{10}◆RB◆◆RB◆◆LB◆2◆RB◆ = \dfrac◆LB◆(3+\sqrt{10})^2 - 1◆RB◆◆LB◆2(3+\sqrt{10})◆RB◆ = \dfrac◆LB◆9+6\sqrt{10}+10-1◆RB◆◆LB◆2(3+\sqrt{10})◆RB◆ = \dfrac◆LB◆18+6\sqrt{10}◆RB◆◆LB◆2(3+\sqrt{10})◆RB◆ = \dfrac◆LB◆6(3+\sqrt{10})◆RB◆◆LB◆2(3+\sqrt{10})◆RB◆ = 3$. ✓

Example 8.5: The catenary equation

Problem. A uniform heavy chain hangs from two supports. Show that the equation of the curve is $y = a\cosh(x/a)$.

Solution. At the lowest point, the tension is horizontal: $T_0$. At a point at horizontal distance $x$ from the lowest point, the tension $T$ acts at angle $\theta$ to the horizontal.

Horizontal: $T\cos\theta = T_0$. Vertical: $T\sin\theta = ws$ where $w$ is the weight per unit length and $s$ is the arc length.

$\dfrac{dy}{dx} = \tan\theta = \dfrac{ws}{T_0} = \dfrac{s}{a}$ where $a = T_0/w$.

Differentiating: $\dfrac{d^2y}{dx^2} = \dfrac{1}{a}\dfrac{ds}{dx} = \dfrac{1}{a}\sqrt◆LB◆1+\left(\dfrac{dy}{dx}\right)^2◆RB◆$.

Let $p = dy/dx$: $\dfrac{dp}{dx} = \dfrac{1}{a}\sqrt{1+p^2}$.

$\displaystyle\int \frac◆LB◆dp◆RB◆◆LB◆\sqrt{1+p^2}◆RB◆ = \int \frac{dx}{a} \implies \operatorname{arsinh}(p) = \frac{x}{a}$.

$p = \sinh(x/a)$. Integrating: $y = a\cosh(x/a) + C$. Taking the lowest point at $y = a$: $C = 0$.

$\boxed{y = a\cosh(x/a)}$

Example 8.6: Differentiating inverse hyperbolic functions

Problem. Find $\dfrac{d}{dx}\!\left[\operatorname{artanh}\!\left(\dfrac{x}{2}\right)\right]$ and state its domain.

Solution. $\dfrac{d}{dx}[\operatorname{artanh}(u)] = \dfrac{1}{1-u^2} \cdot \dfrac{du}{dx}$ with $u = x/2$.

$\frac{d}{dx}\!\left[\operatorname{artanh}\!\left(\frac{x}{2}\right)\right] = \frac{1}{1 - x^2/4} \cdot \frac{1}{2} = \frac{2}{4-x^2}$

Domain: $\left|\dfrac{x}{2}\right| < 1$, i.e., $-2 < x < 2$.

Example 8.7: Hyperbolic functions and complex numbers

Problem. Using Euler's formula, show that $\cos(ix) = \cosh x$.

Solution. $\cos(ix) = \dfrac{e^{i(ix)} + e^{-i(ix)}}{2} = \dfrac{e^{-x} + e^x}{2} = \dfrac{e^x + e^{-x}}{2} = \boxed{\cosh x}$. $\blacksquare$

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9. Common Pitfalls

Pitfall	Correct Approach
Using $x = a\cosh u$ when $x < a$	$\cosh u \geq 1$, so this requires $x \geq a$
Confusing $\operatorname{arsinh}$ and $\operatorname{arcosh}$ domains	$\operatorname{arsinh}$: all real $x$; $\operatorname{arcosh}$: $x \geq 1$
Forgetting $\cosh^2 x - \sinh^2 x = 1$	This is the fundamental identity, analogous to $\cos^2 x + \sin^2 x = 1$

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10. Additional Exam-Style Questions

Question 8

Solve the equation $4\cosh^2 x - 5\sinh x - 5 = 0$.

Solution

Using $\cosh^2 x = 1 + \sinh^2 x$: $4(1+\sinh^2 x) - 5\sinh x - 5 = 0$.

$4\sinh^2 x - 5\sinh x - 1 = 0$.

$(4\sinh x + 1)(\sinh x - 1) = 0$.

$\sinh x = -\dfrac{1}{4}$ or $\sinh x = 1$.

$x = \operatorname{arsinh}\!\left(-\dfrac{1}{4}\right) = -\ln\!\left(\dfrac{1}{4}+\dfrac◆LB◆\sqrt{17}◆RB◆◆LB◆4◆RB◆\right)$ or $x = \operatorname{arsinh}(1) = \ln(1+\sqrt{2})$.

Question 9

Prove that $\displaystyle\int_0^{\ln 2} \cosh x\,dx = \dfrac{3}{4}$.

Solution

$\displaystyle\int_0^{\ln 2} \frac{e^x+e^{-x}}{2}\,dx = \frac{1}{2}\left[e^x - e^{-x}\right]_0^{\ln 2} = \frac{1}{2}\!\left[(2-\frac{1}{2})-(1-1)\right] = \frac{1}{2} \times \frac{3}{2} = \dfrac{3}{4}$. $\blacksquare$

Question 10

Express $\sinh^{-1} x$ in terms of natural logarithms.

Solution

Let $y = \sinh^{-1} x$, so $x = \sinh y = \dfrac{e^y - e^{-y}}{2}$.

$2x = e^y - e^{-y} \implies e^{2y} - 2xe^y - 1 = 0$.

$e^y = \dfrac◆LB◆2x \pm \sqrt{4x^2+4}◆RB◆◆LB◆2◆RB◆ = x + \sqrt{x^2+1}$ (taking positive root since $e^y > 0$).

$\boxed{\sinh^{-1} x = \ln(x + \sqrt{x^2+1})}$

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11. Connections to Other Topics

11.1 Hyperbolic functions and calculus

The inverse hyperbolic functions arise from integrals. See Further Calculus.

11.2 Hyperbolic functions and differential equations

$y'' - y = 0$ has solutions $y = A\cosh x + B\sinh x$. See Differential Equations.

11.3 Hyperbolic functions and complex numbers

$\cos(ix) = \cosh x$ and $\sin(ix) = i\sinh x$. See Complex Numbers.

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12. Key Results Summary

Function	Definition	Derivative	Integral
$\sinh x$	$\dfrac{e^x-e^{-x}}{2}$	$\cosh x$	$\cosh x + C$
$\cosh x$	$\dfrac{e^x+e^{-x}}{2}$	$\sinh x$	$\sinh x + C$
$\tanh x$	$\dfrac◆LB◆\sinh x◆RB◆◆LB◆\cosh x◆RB◆$	$\operatorname{sech}^2 x$	$\ln(\cosh x) + C$
$\operatorname{arsinh}\, x$	$\ln(x+\sqrt{x^2+1})$	$\dfrac◆LB◆1◆RB◆◆LB◆\sqrt{x^2+1}◆RB◆$	—
$\operatorname{arcosh}\, x$	$\ln(x+\sqrt{x^2-1})$	$\dfrac◆LB◆1◆RB◆◆LB◆\sqrt{x^2-1}◆RB◆$	—
$\operatorname{artanh}\, x$	$\dfrac{1}{2}\ln\!\left(\dfrac{1+x}{1-x}\right)$	$\dfrac{1}{1-x^2}$	—

Identity	Formula
Fundamental	$\cosh^2 x - \sinh^2 x = 1$
Osborn's rule	Replace $\cos^2 \to \cosh^2$, $\sin^2 \to -\sinh^2$ in trig identities
Double angle (cosh)	$\cosh 2x = 2\cosh^2 x - 1 = 1 + 2\sinh^2 x$
Double angle (sinh)	$\sinh 2x = 2\sinh x\cosh x$

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13. Further Exam-Style Questions

Question 11

Find $\displaystyle\int_0^2 \frac◆LB◆dx◆RB◆◆LB◆\sqrt{x^2 + 4}◆RB◆$.

Solution

Let $x = 2\sinh u$, $dx = 2\cosh u\,du$.

$\displaystyle\int_0^{\operatorname{arsinh}(1)} \frac◆LB◆2\cosh u◆RB◆◆LB◆2\cosh u◆RB◆\,du = [\operatorname{arsinh}(1) - 0] = \ln(1+\sqrt{2})$.

$\boxed{\ln(1+\sqrt{2})}$

Question 12

Prove that $\displaystyle\frac{d}{dx}[\tanh x] = \operatorname{sech}^2 x$.

Solution

$\tanh x = \dfrac◆LB◆\sinh x◆RB◆◆LB◆\cosh x◆RB◆$.

Quotient rule: $\dfrac◆LB◆\cosh^2 x - \sinh^2 x◆RB◆◆LB◆\cosh^2 x◆RB◆ = \dfrac◆LB◆1◆RB◆◆LB◆\cosh^2 x◆RB◆ = \operatorname{sech}^2 x$. $\blacksquare$

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14. Advanced Topics

14.1 The catenary — physical derivation

A uniform heavy flexible cable hanging under its own weight takes the shape $y = a\cosh(x/a)$ where $a = T_H/w$ is the ratio of horizontal tension to weight per unit length.

14.2 Hyperbolic functions in special relativity

The Lorentz factor $\gamma = \dfrac◆LB◆1◆RB◆◆LB◆\sqrt{1-v^2/c^2}◆RB◆$ can be written as $\gamma = \cosh\phi$ where $\tanh\phi = v/c$ (rapidity).

Time dilation: $\Delta t' = \Delta t\cosh\phi$. Length contraction: $L' = L/\cosh\phi$.

14.3 The inverse Gudermannian

$\mathrm{gd}^{-1}(\theta) = \ln|\sec\theta + \tan\theta| = \ln\!\left|\tan\!\left(\dfrac◆LB◆\theta◆RB◆◆LB◆2◆RB◆+\dfrac◆LB◆\pi◆RB◆◆LB◆4◆RB◆\right)\right| = \operatorname{arsinh}(\tan\theta)$.

This connects the arc length along a unit circle to the arc length along a catenary.

14.4 Hyperbolic identities — comprehensive list

Identity	Formula
$\sinh(x+y)$	$\sinh x\cosh y + \cosh x\sinh y$
$\cosh(x+y)$	$\cosh x\cosh y + \sinh x\sinh y$
$\sinh 2x$	$2\sinh x\cosh x$
$\cosh 2x$	$2\cosh^2 x - 1 = 1 + 2\sinh^2 x$
$\sinh^2 x$	$\dfrac◆LB◆\cosh 2x - 1◆RB◆◆LB◆2◆RB◆$
$\cosh^2 x$	$\dfrac◆LB◆\cosh 2x + 1◆RB◆◆LB◆2◆RB◆$

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15. Further Exam-Style Questions

Question 13

Express $\dfrac{d}{dx}[\operatorname{artanh}\, x]$ and $\dfrac{d}{dx}[\operatorname{arcoth}\, x]$, and compare.

Solution

$\dfrac{d}{dx}[\operatorname{artanh}\, x] = \dfrac{1}{1-x^2}$ for $|x| < 1$.

$\dfrac{d}{dx}[\operatorname{arcoth}\, x] = \dfrac{1}{1-x^2}$ for $|x| > 1$.

Both have the same derivative formula $\dfrac{1}{1-x^2}$, but different domains. They differ by a constant (in fact, $\operatorname{arcoth}\, x = \operatorname{artanh}\,(1/x)$).

Question 14

Prove that $\sinh 3x = 3\sinh x + 4\sinh^3 x$.

Solution

$\sinh 3x = \sinh(2x+x) = \sinh 2x\cosh x + \cosh 2x\sinh x$

$= 2\sinh x\cosh^2 x + (1+2\sinh^2 x)\sinh x$

$= 2\sinh x(1+\sinh^2 x) + \sinh x + 2\sinh^3 x$

$= 2\sinh x + 2\sinh^3 x + \sinh x + 2\sinh^3 x = \boxed{3\sinh x + 4\sinh^3 x}$. $\blacksquare$