Hyperbolic Functions (Extended)

Hyperbolic Functions (Extended Treatment)

This document provides a rigorous treatment of hyperbolic functions, their identities, inverses, and calculus.

Hyperbolic functions are analogues of trigonometric functions defined using exponentials

rather than circles. They arise naturally in many areas including differential equations, special relativity, and catenary curves. :::

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1. Definitions

1.1 The three fundamental hyperbolic functions

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

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

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

1.2 Reciprocal functions

$\mathrm{cosech}\,x = \frac◆LB◆1◆RB◆◆LB◆\sinh x◆RB◆ = \frac{2}{e^x - e^{-x}}, \quad x \neq 0$

$\mathrm{sech}\,x = \frac◆LB◆1◆RB◆◆LB◆\cosh x◆RB◆ = \frac{2}{e^x + e^{-x}}$

$\coth x = \frac◆LB◆1◆RB◆◆LB◆\tanh x◆RB◆ = \frac{e^x + e^{-x}}{e^x - e^{-x}}, \quad x \neq 0$

1.3 Basic properties

• $\sinh 0 = 0$, $\cosh 0 = 1$, $\tanh 0 = 0$.
• $\sinh x$ is odd: $\sinh(-x) = -\sinh x$.
• $\cosh x$ is even: $\cosh(-x) = \cosh x$.
• $\tanh x$ is odd.
• As $x \to +\infty$: $\sinh x \to +\infty$, $\cosh x \to +\infty$, $\tanh x \to 1$.
• As $x \to -\infty$: $\sinh x \to -\infty$, $\cosh x \to +\infty$, $\tanh x \to -1$.

1.4 Connection with Euler's formula

$\cosh x = \cos(ix), \qquad \sinh x = -i\sin(ix)$

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

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

1.5 Graphs

• $\sinh x$: passes through the origin, increasing, resembles $y = x/2$ near the origin and $y = e^x/2$ for large positive $x$.
• $\cosh x$: minimum at $(0, 1)$, symmetric about the $y$-axis, resembles $y = 1 + x^2/2$ near the origin and $y = e^x/2$ for large positive $x$.
• $\tanh x$: S-shaped curve with horizontal asymptotes at $y = \pm 1$, passing through the origin with gradient 1.

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2. Hyperbolic Identities

2.1 Fundamental identity

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

Proof.

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

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

2.2 Identity for $\tanh$

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

Proof. Divide the fundamental identity by $\cosh^2 x$:

$1 - \tanh^2 x = \frac◆LB◆1◆RB◆◆LB◆\cosh^2 x◆RB◆ = \mathrm{sech}^2\,x \quad \blacksquare$

2.3 Osborne's rule

Every trigonometric identity has a corresponding hyperbolic identity, obtained by:

• Replacing $\cos$ with $\cosh$.
• Replacing $\sin$ with $\sinh$.
• Negating every product (or power of 2 or more) of $\sinh$.

Example: $\cos^2 x + \sin^2 x = 1$ becomes $\cosh^2 x - \sinh^2 x = 1$ (the $\sinh^2$ term is negated).

Example: $\sin 2x = 2\sin x\cos x$ becomes $\sinh 2x = 2\sinh x\cosh x$ (the product $\sinh x \cosh x$ is negated since it contains $\sinh$, but $\sinh 2x$ on the left is not a product, so the overall sign changes on both sides cancel).

2.4 Double angle formulas

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

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

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

$\cosh 2x = \frac{e^{2x} + e^{-2x}}{2} = \frac{(e^x + e^{-x})^2 - 2}{2} = \frac◆LB◆4\cosh^2 x - 2◆RB◆◆LB◆2◆RB◆ = 2\cosh^2 x - 1 \quad \blacksquare$

2.5 Addition formulas

$\sinh(A \pm B) = \sinh A\cosh B \pm \cosh A\sinh B$

$\cosh(A \pm B) = \cosh A\cosh B \pm \sinh A\sinh B$

Proof of the addition formula for $\sinh$:

$\sinh(A + B) = \frac{e^{A+B} - e^{-(A+B)}}{2} = \frac{e^A e^B - e^{-A}e^{-B}}{2}$

$= \frac{(e^A - e^{-A})(e^B + e^{-B}) + (e^A + e^{-A})(e^B - e^{-B})}{4}$

$= \sinh A\cosh B + \cosh A\sinh B \quad \blacksquare$

2.6 Worked example

Problem. Given $\sinh x = 3$, find $\cosh x$ and $\tanh x$ without finding $x$.

From $\cosh^2 x - \sinh^2 x = 1$:

$\cosh^2 x = 1 + 9 = 10 \implies \cosh x = \sqrt{10}$

(We take the positive root since $\cosh x \geq 1$ for all $x$.)

$\tanh x = \frac◆LB◆\sinh x◆RB◆◆LB◆\cosh x◆RB◆ = \frac◆LB◆3◆RB◆◆LB◆\sqrt{10}◆RB◆ = \frac◆LB◆3\sqrt{10}◆RB◆◆LB◆10◆RB◆$

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3. Inverse Hyperbolic Functions

3.1 Definitions in logarithmic form

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

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

$\operatorname{artanh}\,x = \frac{1}{2}\ln\!\left(\frac{1 + x}{1 - x}\right), \quad -1 \lt x \lt 1$

3.2 Derivation of $\operatorname{arsinh}\,x = \ln(x + \sqrt{x^2 + 1})$

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

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

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 $e^y \gt 0$ and $\sqrt{x^2 + 1} \gt |x|$, we take the positive root:

$e^y = x + \sqrt{x^2 + 1} \implies y = \ln\!\left(x + \sqrt{x^2 + 1}\right) \quad \blacksquare$

3.3 Derivation of $\operatorname{arcosh}\,x = \ln(x + \sqrt{x^2 - 1})$

Let $y = \operatorname{arcosh}\,x$, so $x = \cosh y = \dfrac{e^y + e^{-y}}{2}$.

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

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

Since $e^y \geq 1$ and $x \geq 1$, we need $e^y \geq 1$. Both roots are positive when $x \geq 1$. The convention is to take $e^y = x + \sqrt{x^2 - 1}$ (which gives $y \geq 0$):

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

3.4 Derivation of $\operatorname{artanh}\,x$

Let $y = \operatorname{artanh}\,x$, so $x = \tanh y$.

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

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

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

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

3.5 Worked example

Problem. Evaluate $\operatorname{arsinh}\,2$ and $\operatorname{artanh}\,\dfrac{1}{3}$ in exact logarithmic form.

$\operatorname{arsinh}\,2 = \ln(2 + \sqrt{5})$

$\operatorname{artanh}\,\frac{1}{3} = \frac{1}{2}\ln\!\left(\frac{4/3}{2/3}\right) = \frac{1}{2}\ln 2$

3.6 Domains and ranges

Function	Domain	Range
$\operatorname{arsinh}\,x$	$\mathbb{R}$	$\mathbb{R}$
$\operatorname{arcosh}\,x$	$x \geq 1$	$y \geq 0$
$\operatorname{artanh}\,x$	$-1 \lt x \lt 1$	$\mathbb{R}$

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4. Calculus of Hyperbolic Functions

4.1 Differentiation

$\frac{d}{dx}(\sinh x) = \cosh x$

$\frac{d}{dx}(\cosh x) = \sinh x$

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

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

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

4.2 Differentiation of inverse hyperbolic functions

$\frac{d}{dx}(\operatorname{arsinh}\,x) = \frac◆LB◆1◆RB◆◆LB◆\sqrt{x^2 + 1}◆RB◆$

$\frac{d}{dx}(\operatorname{arcosh}\,x) = \frac◆LB◆1◆RB◆◆LB◆\sqrt{x^2 - 1}◆RB◆, \quad x \gt 1$

$\frac{d}{dx}(\operatorname{artanh}\,x) = \frac{1}{1 - x^2}, \quad |x| \lt 1$

Proof for $\operatorname{arsinh}\,x$. Let $y = \operatorname{arsinh}\,x$, so $x = \sinh y$.

$\frac{dy}{dx} = \frac◆LB◆1◆RB◆◆LB◆\dfrac{dx}{dy}◆RB◆ = \frac◆LB◆1◆RB◆◆LB◆\cosh y◆RB◆ = \frac◆LB◆1◆RB◆◆LB◆\sqrt◆LB◆1 + \sinh^2 y◆RB◆◆RB◆ = \frac◆LB◆1◆RB◆◆LB◆\sqrt{1 + x^2}◆RB◆ \quad \blacksquare$

Proof for $\operatorname{artanh}\,x$. Let $y = \operatorname{artanh}\,x$, so $x = \tanh y$.

$\frac{dy}{dx} = \frac◆LB◆1◆RB◆◆LB◆\mathrm{sech}^2\,y◆RB◆ = \frac◆LB◆1◆RB◆◆LB◆1 - \tanh^2 y◆RB◆ = \frac{1}{1 - x^2} \quad \blacksquare$

4.3 Integration

The differentiation results give standard integrals:

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

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

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

4.4 Integrals leading to inverse hyperbolic functions

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

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

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

Proof of the first formula. Let $u = x/a$:

$\int \frac◆LB◆dx◆RB◆◆LB◆\sqrt{x^2 + a^2}◆RB◆ = \int \frac◆LB◆a\,du◆RB◆◆LB◆a\sqrt{u^2 + 1}◆RB◆ = \operatorname{arsinh}\,u + C = \operatorname{arsinh}\!\left(\frac{x}{a}\right) + C \quad \blacksquare$

4.5 Worked example: differentiation

Problem. Differentiate $f(x) = \sinh(3x^2)$.

$f'(x) = 6x\cosh(3x^2)$

4.6 Worked example: integration

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

$\int \frac◆LB◆dx◆RB◆◆LB◆\sqrt{4x^2 + 9}◆RB◆ = \frac{1}{2}\int \frac◆LB◆d(2x)◆RB◆◆LB◆\sqrt{(2x)^2 + 9}◆RB◆ = \frac{1}{2}\operatorname{arsinh}\!\left(\frac{2x}{3}\right) + C$

4.7 Worked example: definite integral

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

$= \left[\operatorname{arsinh}\,x\right]_0^1 = \operatorname{arsinh}\,1 - \operatorname{arsinh}\,0 = \ln(1 + \sqrt{2}) - 0 = \ln(1 + \sqrt{2})$

4.8 Worked example: integration by substitution with hyperbolic functions

Problem. Evaluate $\displaystyle\int \sqrt{x^2 + 4}\,dx$.

Use the substitution $x = 2\sinh u$, $dx = 2\cosh u\,du$:

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

Using $\cosh^2 u = \dfrac◆LB◆1 + \cosh 2u◆RB◆◆LB◆2◆RB◆$:

$= 4\int \frac◆LB◆1 + \cosh 2u◆RB◆◆LB◆2◆RB◆\,du = 2u + \sinh 2u + C$

$= 2u + 2\sinh u\cosh u + C$

Since $x = 2\sinh u$: $\sinh u = \dfrac{x}{2}$, $\cosh u = \sqrt◆LB◆1 + \dfrac{x^2}{4}◆RB◆ = \dfrac◆LB◆\sqrt{x^2 + 4}◆RB◆◆LB◆2◆RB◆$, $u = \operatorname{arsinh}\!\left(\dfrac{x}{2}\right)$.

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

Common Pitfall The substitution $x = a\sinh u$ is a powerful technique for integrals

involving $\sqrt{x^2 + a^2}$. Similarly, $x = a\cosh u$ handles $\sqrt{x^2 - a^2}$ and $x = a\tanh u$ handles expressions with $a^2 - x^2$. The choice of substitution mirrors the trigonometric substitutions but is often simpler algebraically. :::

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5. Practice Problems

Problem 1

Solve the equation $4\sinh x - 3\cosh x = 0$.

Solution

$4\sinh x = 3\cosh x \implies \tanh x = \dfrac{3}{4}$.

$x = \operatorname{artanh}\!\left(\dfrac{3}{4}\right) = \dfrac{1}{2}\ln\!\left(\dfrac{1 + 3/4}{1 - 3/4}\right) = \dfrac{1}{2}\ln 7$.

Problem 2

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

Solution

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

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

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

$= 2\sinh x + 2\sinh^3 x + \sinh x + 2\sinh^3 x = 3\sinh x + 4\sinh^3 x$.

Problem 3

Evaluate $\displaystyle\int_0^2 \frac◆LB◆dx◆RB◆◆LB◆\sqrt{x^2 + 16}◆RB◆$.

Solution

$= \left[\operatorname{arsinh}\!\left(\dfrac{x}{4}\right)\right]_0^2 = \operatorname{arsinh}\!\left(\dfrac{1}{2}\right) = \ln\!\left(\dfrac{1}{2} + \sqrt◆LB◆\dfrac{5}{4}◆RB◆\right) = \ln\!\left(\dfrac◆LB◆1 + \sqrt{5}◆RB◆◆LB◆2◆RB◆\right)$.

Problem 4

Differentiate $f(x) = x\,\operatorname{arcosh}\,x$ for $x \gt 1$.

Solution

Using the product rule:

$f'(x) = \operatorname{arcosh}\,x + x \cdot \dfrac◆LB◆1◆RB◆◆LB◆\sqrt{x^2 - 1}◆RB◆ = \operatorname{arcosh}\,x + \dfrac◆LB◆x◆RB◆◆LB◆\sqrt{x^2 - 1}◆RB◆$.

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6. Further Proofs and Key Results

6.1 Proof: $\int \frac◆LB◆1◆RB◆◆LB◆\sqrt{x^2 - a^2}◆RB◆\,dx = \operatorname{arcosh}\!\left(\frac{x}{a}\right) + C$

Proof. Let $u = x/a$, so $dx = a\,du$:

$\int \frac◆LB◆dx◆RB◆◆LB◆\sqrt{x^2 - a^2}◆RB◆ = \int \frac◆LB◆a\,du◆RB◆◆LB◆a\sqrt{u^2 - 1}◆RB◆ = \int \frac◆LB◆du◆RB◆◆LB◆\sqrt{u^2 - 1}◆RB◆$

Now let $u = \cosh t$, so $du = \sinh t\,dt$:

$= \int \frac◆LB◆\sinh t\,dt◆RB◆◆LB◆\sqrt◆LB◆\cosh^2 t - 1◆RB◆◆RB◆ = \int \frac◆LB◆\sinh t\,dt◆RB◆◆LB◆\sinh t◆RB◆ = \int 1\,dt = t + C = \operatorname{arcosh}\,u + C$

$= \operatorname{arcosh}\!\left(\frac{x}{a}\right) + C \quad \blacksquare$

6.2 Proof: $\int \frac{1}{a^2 - x^2}\,dx = \frac{1}{a}\operatorname{artanh}\!\left(\frac{x}{a}\right) + C$

Proof. Let $u = x/a$, so $dx = a\,du$:

$\int \frac{dx}{a^2 - x^2} = \frac{1}{a}\int \frac{du}{1 - u^2} = \frac{1}{a}\operatorname{artanh}\,u + C = \frac{1}{a}\operatorname{artanh}\!\left(\frac{x}{a}\right) + C \quad \blacksquare$

6.3 Proof: the catenary equation

Theorem. A uniform chain hanging under gravity takes the shape $y = a\cosh\!\left(\dfrac{x}{a}\right) + c$.

Proof (sketch). Consider a small element of the chain between horizontal positions $x$ and $x + \delta x$. Let the tension at position $x$ be $T$, making angle $\theta$ with the horizontal.

Horizontal equilibrium: $T\cos\theta = T_0$ (constant).

Vertical equilibrium: $\dfrac{d}{dx}(T\sin\theta) = w$ where $w$ is the weight per unit length.

Since $T = T_0\sec\theta$ and $T\sin\theta = T_0\tan\theta$:

$\frac{d}{dx}(T_0\tan\theta) = w \implies T_0\sec^2\theta\,\frac◆LB◆d\theta◆RB◆◆LB◆dx◆RB◆ = w$

Let $y' = \tan\theta$, so $\dfrac{dy'}{dx} = \sec^2\theta\,\dfrac◆LB◆d\theta◆RB◆◆LB◆dx◆RB◆ = \dfrac{w}{T_0}$.

Integrating: $y' = \dfrac{w}{T_0}\,x + C_1$. Taking $C_1 = 0$ by symmetry:

$y' = \frac{x}{a} \quad \text{where } a = \frac{T_0}{w}$

Integrating again: $y = a\cosh\!\left(\dfrac{x}{a}\right) + C$. $\blacksquare$

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7. Common Pitfalls

Common Pitfall

1. Sign in the fundamental identity: Unlike $\cos^2 x + \sin^2 x = 1$, the hyperbolic identity is $\cosh^2 x - \sinh^2 x = 1$. The minus sign is crucial and is the source of many errors.
2. Domain of $\operatorname{arcosh}$: The domain is $x \geq 1$ (not $x > 0$). Attempting to evaluate $\operatorname{arcosh}(0.5)$ is undefined.
3. $\cosh x \geq 1$ always: When solving $\cosh^2 x = k$ and taking the square root, always take the positive root $\cosh x = +\sqrt{k}$ since $\cosh x \geq 1 > 0$ for all real $x$.
4. Integration: artanh vs ln: When $|x| > a$ in $\displaystyle\int \frac{dx}{a^2 - x^2}$, the result involves $\operatorname{arcoth}$ (or an alternative logarithmic form), not $\operatorname{artanh}$. Check the domain of the integrand carefully. :::

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8. Additional Exam-Style Questions

Question 5

(a) Solve the equation $\cosh x = 2.5$, giving your answer in exact logarithmic form.

(b) Hence solve $\cosh 2x = 2.5$.

Solution

(a) $\cosh x = \dfrac{e^x + e^{-x}}{2} = 2.5$

$e^x + e^{-x} = 5 \implies e^{2x} - 5e^x + 1 = 0$

$e^x = \dfrac◆LB◆5 \pm \sqrt{25 - 4}◆RB◆◆LB◆2◆RB◆ = \dfrac◆LB◆5 \pm \sqrt{21}◆RB◆◆LB◆2◆RB◆$

$x = \ln\!\left(\dfrac◆LB◆5 \pm \sqrt{21}◆RB◆◆LB◆2◆RB◆\right)$

Since $\cosh$ is even, both $\pm$ give valid solutions (one positive, one negative).

(b) $\cosh 2x = 2.5 \implies 2x = \ln\!\left(\dfrac◆LB◆5 \pm \sqrt{21}◆RB◆◆LB◆2◆RB◆\right)$

$x = \dfrac{1}{2}\ln\!\left(\dfrac◆LB◆5 \pm \sqrt{21}◆RB◆◆LB◆2◆RB◆\right)$

Alternatively, using $\cosh 2x = 2\cosh^2 x - 1 = 2.5 \implies \cosh^2 x = 1.75$, which gives the same result.

Question 6

(a) Prove that $\dfrac{d}{dx}(\operatorname{arcosh}\,x) = \dfrac◆LB◆1◆RB◆◆LB◆\sqrt{x^2 - 1}◆RB◆$ for $x > 1$.

(b) Evaluate $\displaystyle\int_2^3 \frac◆LB◆dx◆RB◆◆LB◆\sqrt{x^2 - 1}◆RB◆$ in exact form.

Solution

(a) Let $y = \operatorname{arcosh}\,x$, so $x = \cosh y$.

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

Since $\cosh^2 y - \sinh^2 y = 1$: $\sinh y = \sqrt◆LB◆\cosh^2 y - 1◆RB◆ = \sqrt{x^2 - 1}$.

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

(b) $\displaystyle\int_2^3 \frac◆LB◆dx◆RB◆◆LB◆\sqrt{x^2 - 1}◆RB◆ = \bigl[\operatorname{arcosh}\,x\bigr]_2^3$

$= \ln(3 + \sqrt{8}) - \ln(2 + \sqrt{3}) = \ln(3 + 2\sqrt{2}) - \ln(2 + \sqrt{3})$

$= \ln\!\left(\dfrac◆LB◆3 + 2\sqrt{2}◆RB◆◆LB◆2 + \sqrt{3}◆RB◆\right)$.

Question 7

A curve $C$ has equation $y = \sinh^{-1}(2x - 1)$.

(a) Find $\dfrac{dy}{dx}$.

(b) Find the equation of the tangent to $C$ at the point where $x = 1$.

Solution

(a) $y = \operatorname{arsinh}(2x - 1)$.

$\dfrac{dy}{dx} = \dfrac◆LB◆2◆RB◆◆LB◆\sqrt{(2x - 1)^2 + 1}◆RB◆ = \dfrac◆LB◆2◆RB◆◆LB◆\sqrt{4x^2 - 4x + 2}◆RB◆$.

(b) At $x = 1$: $y = \operatorname{arsinh}(1) = \ln(1 + \sqrt{2})$.

$\dfrac{dy}{dx}\bigg|_{x=1} = \dfrac◆LB◆2◆RB◆◆LB◆\sqrt{4 - 4 + 2}◆RB◆ = \dfrac◆LB◆2◆RB◆◆LB◆\sqrt{2}◆RB◆ = \sqrt{2}$.

Equation of tangent: $y - \ln(1 + \sqrt{2}) = \sqrt{2}(x - 1)$, i.e.

$y = \sqrt{2}\,x - \sqrt{2} + \ln(1 + \sqrt{2})$.

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9. Advanced Worked Examples

Example 9.1: Solving hyperbolic equations

Problem. Solve $3\sinh x + 4\cosh x = 5$, giving your answer in exact logarithmic form.

Solution. Using the exponential definitions:

$3\cdot\frac{e^x - e^{-x}}{2} + 4\cdot\frac{e^x + e^{-x}}{2} = 5$

$3e^x - 3e^{-x} + 4e^x + 4e^{-x} = 10$

$7e^x + e^{-x} = 10$

Multiplying by $e^x$: $7e^{2x} + 1 = 10e^x$

$7e^{2x} - 10e^x + 1 = 0$

This is a quadratic in $e^x$:

$e^x = \frac◆LB◆10 \pm \sqrt{100 - 28}◆RB◆◆LB◆14◆RB◆ = \frac◆LB◆10 \pm \sqrt{72}◆RB◆◆LB◆14◆RB◆ = \frac◆LB◆10 \pm 6\sqrt{2}◆RB◆◆LB◆14◆RB◆ = \frac◆LB◆5 \pm 3\sqrt{2}◆RB◆◆LB◆7◆RB◆$

$x = \ln\!\left(\frac◆LB◆5 + 3\sqrt{2}◆RB◆◆LB◆7◆RB◆\right) \quad \text{or} \quad x = \ln\!\left(\frac◆LB◆5 - 3\sqrt{2}◆RB◆◆LB◆7◆RB◆\right)$

Since $\dfrac◆LB◆5 - 3\sqrt{2}◆RB◆◆LB◆7◆RB◆ \approx 0.109 > 0$, both solutions are valid.

Example 9.2: Integration using hyperbolic substitution

Problem. Evaluate $\displaystyle\int \sqrt{x^2 - 9}\,dx$ for $x \geq 3$.

Solution. Use the substitution $x = 3\cosh u$, $dx = 3\sinh u\,du$:

$\int \sqrt◆LB◆9\cosh^2 u - 9◆RB◆\cdot 3\sinh u\,du = \int 3\sinh u \cdot 3\sinh u\,du = 9\int \sinh^2 u\,du$

Using $\sinh^2 u = \dfrac◆LB◆\cosh 2u - 1◆RB◆◆LB◆2◆RB◆$:

$= 9\int \frac◆LB◆\cosh 2u - 1◆RB◆◆LB◆2◆RB◆\,du = \frac{9}{2}\left(\frac◆LB◆\sinh 2u◆RB◆◆LB◆2◆RB◆ - u\right) + C = \frac{9}{4}\sinh 2u - \frac{9}{2}u + C$

Since $x = 3\cosh u$: $\cosh u = \dfrac{x}{3}$, $\sinh u = \sqrt◆LB◆\dfrac{x^2}{9} - 1◆RB◆ = \dfrac◆LB◆\sqrt{x^2 - 9}◆RB◆◆LB◆3◆RB◆$.

$\sinh 2u = 2\sinh u\cosh u = \dfrac◆LB◆2x\sqrt{x^2 - 9}◆RB◆◆LB◆9◆RB◆$.

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

$= \frac◆LB◆x\sqrt{x^2 - 9}◆RB◆◆LB◆2◆RB◆ - \frac{9}{2}\operatorname{arcosh}\!\left(\frac{x}{3}\right) + C$

Example 9.3: Proving an identity using Osborne's rule

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

Solution. From the trigonometric identity $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$, applying Osborne's rule: since $\cos^3\theta$ contains no products of $\sin$, it remains unchanged. Therefore:

$\cosh 3x = 4\cosh^3 x - 3\cosh x$

Direct proof. Starting from $\cosh 3x = \cosh(2x + x)$:

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

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

$= 4\cosh^3 x - 3\cosh x \quad \blacksquare$

Example 9.4: Differentiation involving multiple hyperbolic functions

Problem. Find $\dfrac{dy}{dx}$ where $y = \dfrac◆LB◆\sinh x◆RB◆◆LB◆1 + \cosh x◆RB◆$ and simplify your answer.

Solution. Using the quotient rule:

$\frac{dy}{dx} = \frac◆LB◆\cosh x(1 + \cosh x) - \sinh x \cdot \sinh x◆RB◆◆LB◆(1 + \cosh x)^2◆RB◆ = \frac◆LB◆\cosh x + \cosh^2 x - \sinh^2 x◆RB◆◆LB◆(1 + \cosh x)^2◆RB◆$

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

$= \frac◆LB◆\cosh x + 1◆RB◆◆LB◆(1 + \cosh x)^2◆RB◆ = \frac◆LB◆1◆RB◆◆LB◆1 + \cosh x◆RB◆ = \mathrm{sech}^2\!\left(\frac{x}{2}\right)$

The final simplification uses the identity $1 + \cosh x = 2\cosh^2(x/2)$.

Example 9.5: Definite integral with inverse hyperbolic functions

Problem. Evaluate $\displaystyle\int_0^1 \frac◆LB◆dx◆RB◆◆LB◆\sqrt{1 + 4x^2}◆RB◆$.

Solution. Write $\sqrt{1 + 4x^2} = \sqrt{4(x^2 + 1/4)} = 2\sqrt{x^2 + (1/2)^2}$.

$\int_0^1 \frac◆LB◆dx◆RB◆◆LB◆\sqrt{1 + 4x^2}◆RB◆ = \frac{1}{2}\int_0^1 \frac◆LB◆dx◆RB◆◆LB◆\sqrt{x^2 + 1/4}◆RB◆ = \frac{1}{2}\left[\operatorname{arsinh}\!\left(\frac{x}{1/2}\right)\right]_0^1$

$= \frac{1}{2}\bigl[\operatorname{arsinh}\,2 - \operatorname{arsinh}\,0\bigr] = \frac{1}{2}\ln(2 + \sqrt{5})$

Example 9.6: Parametric differentiation with hyperbolic functions

Problem. A curve is given parametrically by $x = 2\cosh t$, $y = 3\sinh t$. Find the value of $\dfrac{dy}{dx}$ at $t = \ln 2$.

Solution. $\dfrac{dx}{dt} = 2\sinh t$, $\dfrac{dy}{dt} = 3\cosh t$.

$\frac{dy}{dx} = \frac◆LB◆3\cosh t◆RB◆◆LB◆2\sinh t◆RB◆ = \frac{3}{2}\coth t$

At $t = \ln 2$: $\cosh(\ln 2) = \dfrac{2 + 1/2}{2} = \dfrac{5}{4}$, $\sinh(\ln 2) = \dfrac{2 - 1/2}{2} = \dfrac{3}{4}$.

$\frac{dy}{dx}\bigg|_{t = \ln 2} = \frac◆LB◆3 \cdot 5/4◆RB◆◆LB◆2 \cdot 3/4◆RB◆ = \frac{15/4}{3/2} = \frac{5}{2}$

Example 9.7: Verifying a reduction formula

Problem. Let $I_n = \displaystyle\int_0^{\pi/2} \cosh^n(\sin x)\,dx$ is not well-formed. Instead, verify that $\displaystyle\int \tanh^3 x\,dx = \ln(\cosh x) - \dfrac{1}{2}\mathrm{sech}^2\,x\tanh x + C$ is incorrect, and find the correct integral.

Solution. Let us compute $\displaystyle\int \tanh^3 x\,dx$ correctly.

Write $\tanh^3 x = \tanh x \cdot \tanh^2 x = \tanh x(1 - \mathrm{sech}^2\,x) = \tanh x - \tanh x\,\mathrm{sech}^2\,x$.

$\int \tanh^3 x\,dx = \int \tanh x\,dx - \int \tanh x\,\mathrm{sech}^2\,x\,dx$

The first integral: $\displaystyle\int\tanh x\,dx = \ln(\cosh x) + C$.

For the second integral, let $u = \tanh x$, $du = \mathrm{sech}^2\,x\,dx$:

$\int u\,du = \frac{u^2}{2} + C = \frac◆LB◆\tanh^2 x◆RB◆◆LB◆2◆RB◆ + C$

Therefore:

$\boxed{\int \tanh^3 x\,dx = \ln(\cosh x) - \frac◆LB◆\tanh^2 x◆RB◆◆LB◆2◆RB◆ + C}$

Example 9.8: Arc length of a hyperbolic cosine curve

Problem. Find the arc length of $y = a\cosh(x/a)$ from $x = 0$ to $x = b$.

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

$s = \int_0^b \sqrt◆LB◆1 + \sinh^2(x/a)◆RB◆\,dx = \int_0^b \cosh(x/a)\,dx = \bigl[a\sinh(x/a)\bigr]_0^b = a\sinh(b/a)$

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10. Connections to Other Topics

10.1 Hyperbolic functions and differential equations

The differential equation $\dfrac{d^2y}{dx^2} - y = 0$ has general solution $y = A\cosh x + B\sinh x$, which can also be written $y = Ce^x + De^{-x}$. See Differential Equations.

10.2 Hyperbolic functions and complex numbers

The identities $\cosh x = \cos(ix)$ and $\sinh x = -i\sin(ix)$ connect the two topics. See Complex Numbers.

10.3 Hyperbolic functions and integration techniques

Hyperbolic substitutions ($x = a\sinh u$, $x = a\cosh u$) are powerful alternatives to trigonometric substitutions. See Further Calculus.

10.4 The catenary and mechanics

The shape $y = a\cosh(x/a)$ describes a hanging chain. The arc length is $s = a\sinh(x/a)$. See Further Mechanics.

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11. Additional Exam-Style Questions

Question 8

(a) Given $\sinh x = \dfrac{12}{5}$, find the exact value of $\cosh x$ and $\tanh x$.

(b) Hence evaluate $\operatorname{arcosh}\!\left(\dfrac{13}{5}\right)$ in exact logarithmic form.

Solution

(a) $\cosh^2 x = 1 + \sinh^2 x = 1 + \dfrac{144}{25} = \dfrac{169}{25}$, so $\cosh x = \dfrac{13}{5}$ (positive root).

$\tanh x = \dfrac◆LB◆\sinh x◆RB◆◆LB◆\cosh x◆RB◆ = \dfrac{12/5}{13/5} = \dfrac{12}{13}$.

(b) Since $\cosh x = \dfrac{13}{5}$:

$\operatorname{arcosh}\!\left(\frac{13}{5}\right) = x = \ln\!\left(\frac{13}{5} + \sqrt◆LB◆\frac{169}{25} - 1◆RB◆\right) = \ln\!\left(\frac{13}{5} + \frac{12}{5}\right) = \ln 5$

Question 9

Find the exact value of $\displaystyle\int_0^{\ln 2} \cosh 2x\,dx$.

Solution

$\int_0^{\ln 2}\cosh 2x\,dx = \left[\frac◆LB◆\sinh 2x◆RB◆◆LB◆2◆RB◆\right]_0^{\ln 2} = \frac◆LB◆\sinh(2\ln 2)◆RB◆◆LB◆2◆RB◆$

$\sinh(2\ln 2) = \dfrac◆LB◆e^{2\ln 2} - e^{-2\ln 2}◆RB◆◆LB◆2◆RB◆ = \dfrac{4 - 1/4}{2} = \dfrac{15}{8}$.

$\int_0^{\ln 2}\cosh 2x\,dx = \frac{15}{16}$

Question 10

Prove by induction that $\dfrac{d^n}{dx^n}(\sinh x) = \sinh x$ when $n$ is even and $\dfrac{d^n}{dx^n}(\sinh x) = \cosh x$ when $n$ is odd.

Solution

Base case ($n = 0$): $\dfrac{d^0}{dx^0}(\sinh x) = \sinh x$. Correct for $n = 0$ (even).

Base case ($n = 1$): $\dfrac{d}{dx}(\sinh x) = \cosh x$. Correct for $n = 1$ (odd).

Inductive step. Assume the result holds for $n = k$.

If $k$ is even: $\dfrac{d^k}{dx^k}(\sinh x) = \sinh x$. Then $\dfrac{d^{k+1}}{dx^{k+1}}(\sinh x) = \dfrac{d}{dx}(\sinh x) = \cosh x$. Since $k + 1$ is odd, the result holds.

If $k$ is odd: $\dfrac{d^k}{dx^k}(\sinh x) = \cosh x$. Then $\dfrac{d^{k+1}}{dx^{k+1}}(\sinh x) = \dfrac{d}{dx}(\cosh x) = \sinh x$. Since $k + 1$ is even, the result holds. $\blacksquare$

Question 11

Evaluate $\displaystyle\int_{3/2}^2 \frac◆LB◆3◆RB◆◆LB◆\sqrt{4x^2 - 9}◆RB◆\,dx$ in exact form.

Solution

$\int_{3/2}^2 \frac◆LB◆3◆RB◆◆LB◆\sqrt{4x^2 - 9}◆RB◆\,dx = \frac{3}{2}\int_{3/2}^2 \frac◆LB◆dx◆RB◆◆LB◆\sqrt{x^2 - 9/4}◆RB◆ = \frac{3}{2}\left[\operatorname{arcosh}\!\left(\frac{2x}{3}\right)\right]_{3/2}^2$

$= \frac{3}{2}\left[\operatorname{arcosh}\!\left(\frac{4}{3}\right) - \operatorname{arcosh}(1)\right] = \frac{3}{2}\operatorname{arcosh}\!\left(\frac{4}{3}\right)$

$= \frac{3}{2}\ln\!\left(\frac{4}{3} + \sqrt◆LB◆\frac{16}{9} - 1◆RB◆\right) = \frac{3}{2}\ln\!\left(\frac◆LB◆4 + \sqrt{7}◆RB◆◆LB◆3◆RB◆\right)$

Question 12

Given that $y = \ln(\sinh x)$, show that $\dfrac{d^2y}{dx^2} = -\mathrm{cosech}^2\,x$.

Solution

$\frac{dy}{dx} = \frac◆LB◆\cosh x◆RB◆◆LB◆\sinh x◆RB◆ = \coth x$

$\frac{d^2y}{dx^2} = \frac{d}{dx}(\coth x) = -\mathrm{cosech}^2\,x = \frac◆LB◆-1◆RB◆◆LB◆\sinh^2 x◆RB◆ \quad \blacksquare$

Question 13

(a) Express $\sinh(2\ln 3)$ in the form $\dfrac{a}{b}$ where $a, b$ are integers.

(b) Hence find $\ln 3 - \ln 2$ in terms of inverse hyperbolic functions.

Solution

(a) $\sinh(2\ln 3) = 2\sinh(\ln 3)\cosh(\ln 3)$.

$\sinh(\ln 3) = \dfrac{3 - 1/3}{2} = \dfrac{4}{3}$, $\cosh(\ln 3) = \dfrac{3 + 1/3}{2} = \dfrac{5}{3}$.

$\sinh(2\ln 3) = 2 \cdot \dfrac{4}{3} \cdot \dfrac{5}{3} = \dfrac{40}{9}$.

(b) $\ln 3 - \ln 2 = \ln(3/2)$. Since $\operatorname{arsinh}\,x = \ln(x + \sqrt{x^2+1})$:

$\operatorname{arsinh}\!\left(\dfrac{3}{4}\right) = \ln\!\left(\dfrac{3}{4} + \sqrt◆LB◆\dfrac{9}{16}+1◆RB◆\right) = \ln\!\left(\dfrac{3}{4} + \dfrac{5}{4}\right) = \ln 2$.

Therefore $\ln 2 = \operatorname{arsinh}(3/4)$ and $\ln 3 = \ln 2 + \ln(3/2) = \operatorname{arsinh}(3/4) + \operatorname{artanh}(1/5)$ (using $\operatorname{artanh}(1/5) = \frac{1}{2}\ln(6/4) = \frac{1}{2}\ln(3/2)$).

So $\ln 3 - \ln 2 = \operatorname{artanh}(1/5)$.

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12. Summary of Extended Results

Identity	Formula
Triple angle	$\sinh 3x = 3\sinh x + 4\sinh^3 x$
Triple angle	$\cosh 3x = 4\cosh^3 x - 3\cosh x$
Half angle	$\cosh^2\!\left(\dfrac{x}{2}\right) = \dfrac◆LB◆1 + \cosh x◆RB◆◆LB◆2◆RB◆$
Half angle	$\sinh^2\!\left(\dfrac{x}{2}\right) = \dfrac◆LB◆\cosh x - 1◆RB◆◆LB◆2◆RB◆$
Integral of $\tanh$	$\displaystyle\int\tanh x\,dx = \ln(\cosh x) + C$
Integral of $\coth$	$\displaystyle\int\coth x\,dx = \ln(\sinh x) + C$
Integral of $\tanh^3$	$\displaystyle\int\tanh^3 x\,dx = \ln(\cosh x) - \dfrac◆LB◆\tanh^2 x◆RB◆◆LB◆2◆RB◆ + C$
Arc length of catenary	$s = a\sinh(b/a)$ for $y = a\cosh(x/a)$

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13. Further Common Pitfalls

Common Pitfall

1. Substitution domain errors: When using $x = a\cosh u$, the substitution requires $x \geq a$ (since $\cosh u \geq 1$). Attempting to use $x = a\cosh u$ for $x < a$ leads to an error. Use $x = a\sinh u$ for $\sqrt{x^2 + a^2}$ and $x = a\cosh u$ for $\sqrt{x^2 - a^2}$.
2. Confusing $\operatorname{artanh}$ and $\ln$ forms: The formula $\displaystyle\int\frac{dx}{a^2 - x^2} = \frac{1}{2a}\ln\!\left|\frac{a+x}{a-x}\right|$ is valid for all $|x| \neq a$, but $\dfrac{1}{a}\operatorname{artanh}(x/a)$ is only valid for $|x| < a$. For $|x| > a$, use the logarithmic form or $\operatorname{arcoth}$.
3. No absolute value needed for $\cosh$: Unlike $|\cos x|$, $\sqrt◆LB◆\cosh^2 x◆RB◆ = \cosh x$ (no absolute value needed) since $\cosh x \geq 1 > 0$ for all real $x$.
4. Differential equation solutions: The equation $y'' - y = 0$ has solutions in both exponential and hyperbolic forms. When boundary conditions involve $y(0)$ and $y'(0)$, the hyperbolic form $y = A\cosh x + B\sinh x$ is often more convenient since $\cosh 0 = 1$ and $\sinh 0 = 0$. :::

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14. Advanced Worked Examples

Example 14.1: Integration using $\sinh$ substitution

Problem. Evaluate $\displaystyle\int \sqrt{x^2 + 9}\,dx$.

Solution. Let $x = 3\sinh u$, $dx = 3\cosh u\,du$.

$\int 3\cosh u \cdot 3\cosh u\,du = 9\int \cosh^2 u\,du = 9\int \frac◆LB◆1+\cosh 2u◆RB◆◆LB◆2◆RB◆\,du = \frac{9}{2}\!\left(u + \frac◆LB◆\sinh 2u◆RB◆◆LB◆2◆RB◆\right)$

$= \frac{9u}{2} + \frac◆LB◆9\sinh u\cosh u◆RB◆◆LB◆2◆RB◆ = \frac{9}{2}\operatorname{arsinh}\!\left(\frac{x}{3}\right) + \frac◆LB◆x\sqrt{x^2+9}◆RB◆◆LB◆2◆RB◆ + C$

Example 14.2: Solving $y'' - 4y = 0$ with hyperbolic functions

Problem. Solve $y'' - 4y = 0$ with $y(0) = 3$ and $y'(0) = 8$.

Solution. Auxiliary: $m^2 - 4 = 0 \implies m = \pm 2$.

$y = A\cosh 2x + B\sinh 2x$.

$y(0) = A = 3$. $y'(0) = 2B = 8 \implies B = 4$.

$\boxed{y = 3\cosh 2x + 4\sinh 2x}$

In exponential form: $y = 3 \cdot \dfrac{e^{2x}+e^{-2x}}{2} + 4 \cdot \dfrac{e^{2x}-e^{-2x}}{2} = \dfrac{7}{2}e^{2x} - \dfrac{1}{2}e^{-2x}$.

Example 14.3: Arc length of a catenary

Problem. Find the arc length of $y = \cosh x$ from $x = 0$ to $x = 1$.

Solution. $y' = \sinh x$. $ds = \sqrt◆LB◆1 + \sinh^2 x◆RB◆\,dx = \cosh x\,dx$.

$s = \int_0^1 \cosh x\,dx = [\sinh x]_0^1 = \sinh 1 = \frac{e - e^{-1}}{2} \approx \boxed{1.175}$

Example 14.4: Osborn's rule applied to $\tan 2x$

Problem. Using Osborn's rule, find $\tanh 2x$ in terms of $\tanh x$.

Solution. $\tan 2x = \dfrac◆LB◆2\tan x◆RB◆◆LB◆1-\tan^2 x◆RB◆$. Apply Osborn's rule (change $\tan^2 x$ to $-\tanh^2 x$):

$\boxed{\tanh 2x = \frac◆LB◆2\tanh x◆RB◆◆LB◆1+\tanh^2 x◆RB◆}$

Example 14.5: Deriving the Gudermannian function relationship

Problem. The Gudermannian function relates circular and hyperbolic functions: $\sec\theta = \cosh u$ where $u = \operatorname{gd}^{-1}(\theta)$. Show that $\dfrac◆LB◆d\theta◆RB◆◆LB◆du◆RB◆ = \operatorname{sech}\, u$.

Solution. $\sec\theta = \cosh u \implies \cos\theta = \operatorname{sech}\, u$.

Differentiating implicitly with respect to $u$: $-\sin\theta \cdot \dfrac◆LB◆d\theta◆RB◆◆LB◆du◆RB◆ = -\operatorname{sech}\,u \tanh u$.

$\dfrac◆LB◆d\theta◆RB◆◆LB◆du◆RB◆ = \dfrac◆LB◆\operatorname{sech}\,u \tanh u◆RB◆◆LB◆\sin\theta◆RB◆$.

Since $\cos\theta = \operatorname{sech}\,u$: $\sin\theta = \sqrt◆LB◆1-\operatorname{sech}^2 u◆RB◆ = \sqrt◆LB◆1-\dfrac◆LB◆1◆RB◆◆LB◆\cosh^2 u◆RB◆◆RB◆ = \dfrac◆LB◆\sqrt◆LB◆\cosh^2 u - 1◆RB◆◆RB◆◆LB◆\cosh u◆RB◆ = \dfrac◆LB◆\sinh u◆RB◆◆LB◆\cosh u◆RB◆ = \tanh u$.

$\dfrac◆LB◆d\theta◆RB◆◆LB◆du◆RB◆ = \dfrac◆LB◆\operatorname{sech}\,u \tanh u◆RB◆◆LB◆\tanh u◆RB◆ = \boxed{\operatorname{sech}\,u}$. $\blacksquare$

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15. Additional Exam-Style Questions

Question 16

Evaluate $\displaystyle\int_1^2 \frac◆LB◆dx◆RB◆◆LB◆\sqrt{x^2 - 1}◆RB◆$.

Solution

$= [\operatorname{arcosh}\, x]_1^2 = \ln(2+\sqrt{3}) - \ln 1 = \boxed{\ln(2+\sqrt{3})}$.

Question 17

Prove that $\sinh(x+y) = \sinh x\cosh y + \cosh x\sinh y$.

Solution

$\sinh x\cosh y + \cosh x\sinh y = \dfrac{(e^x-e^{-x})(e^y+e^{-y}) + (e^x+e^{-x})(e^y-e^{-y})}{4}$

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

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

Question 18

Find $\dfrac{d}{dx}[\operatorname{arcosh}\, x]$ and state the domain.

Solution

Let $y = \operatorname{arcosh}\, x$, so $x = \cosh y$ and $x \geq 1$.

$1 = \sinh y \cdot \dfrac{dy}{dx}$.

$\dfrac{dy}{dx} = \dfrac◆LB◆1◆RB◆◆LB◆\sinh y◆RB◆ = \dfrac◆LB◆1◆RB◆◆LB◆\sqrt◆LB◆\cosh^2 y - 1◆RB◆◆RB◆ = \dfrac◆LB◆1◆RB◆◆LB◆\sqrt{x^2-1}◆RB◆$.

$\boxed{\dfrac{d}{dx}[\operatorname{arcosh}\, x] = \dfrac◆LB◆1◆RB◆◆LB◆\sqrt{x^2-1}◆RB◆}$ for $x > 1$.

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16. Advanced Topics

16.1 The Gudermannian function

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

$\sin(\mathrm{gd}\,x) = \tanh x$, $\cos(\mathrm{gd}\,x) = \mathrm{sech}\,x$, $\tan(\mathrm{gd}\,x) = \sinh x$.

$\dfrac{d}{dx}\mathrm{gd}(x) = \mathrm{sech}\,x$.

16.2 Hyperbolic functions and the Lorentz transformation

In special relativity, the Lorentz transformation uses hyperbolic functions. If $\beta = v/c$ and $\gamma = (1-\beta^2)^{-1/2} = \cosh\phi$ where $\tanh\phi = \beta$, then:

$t' = t\cosh\phi - x\sinh\phi/c$, $x' = x\cosh\phi - ct\sinh\phi$.

16.3 Inverse hyperbolic functions in logarithmic form

| Function | Logarithmic Form | Domain | | -------------------------- | -------------------------------------------------------------- | -------------- | --- | ---- | | $\operatorname{arsinh}\,x$ | $\ln(x+\sqrt{x^2+1})$ | all real $x$ | | $\operatorname{arcosh}\,x$ | $\ln(x+\sqrt{x^2-1})$ | $x \geq 1$ | | $\operatorname{artanh}\,x$ | $\dfrac{1}{2}\ln\!\left(\dfrac{1+x}{1-x}\right)$ | $| x | < 1$ | | $\operatorname{arcoth}\,x$ | $\dfrac{1}{2}\ln\!\left(\dfrac{x+1}{x-1}\right)$ | $| x | > 1$ | | $\operatorname{arsech}\,x$ | $\ln\!\left(\dfrac◆LB◆1+\sqrt{1-x^2}◆RB◆◆LB◆x◆RB◆\right)$ | $0 < x \leq 1$ | | $\operatorname{arcsch}\,x$ | $\ln\!\left(\dfrac{1}{x}+\sqrt◆LB◆\dfrac{1}{x^2}+1◆RB◆\right)$ | $x \neq 0$ |

16.4 Hyperbolic functions and catenary applications

The catenary $y = a\cosh(x/a)$ appears in:

• Suspension bridges (cables hang in a catenary)
• Arch shapes (the inverted catenary is the strongest arch)
• Power lines between poles

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17. Further Exam-Style Questions

Question 19

Solve the equation $2\cosh x - \sinh x = 3$.

Solution

$2\cdot\dfrac{e^x+e^{-x}}{2} - \dfrac{e^x-e^{-x}}{2} = 3$.

$e^x+e^{-x} - \dfrac{e^x}{2} + \dfrac{e^{-x}}{2} = 3$.

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

$e^{2x} - 6e^x + 3 = 0$.

$e^x = \dfrac◆LB◆6 \pm \sqrt{36-12}◆RB◆◆LB◆2◆RB◆ = 3 \pm \sqrt{6}$.

$x = \ln(3+\sqrt{6})$ or $x = \ln(3-\sqrt{6})$.

Since $3-\sqrt{6} > 0$, both are valid.

Question 20

Prove that $\operatorname{arcosh}\,x = \ln(x+\sqrt{x^2-1})$ for $x \geq 1$.

Solution

Let $y = \operatorname{arcosh}\,x$, so $x = \cosh y = \dfrac{e^y+e^{-y}}{2}$.

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

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

Since $e^y \geq 1$ and $x \geq 1$: $e^y = x + \sqrt{x^2-1}$ (the positive root, since $x-\sqrt{x^2-1} \leq 1$).

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

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17. Further Advanced Topics

17.1 Hyperbolic substitutions in integration

Standard substitutions:

• $\sqrt{x^2+a^2}$: use $x = a\cosh t$
• $\sqrt{x^2-a^2}$: use $x = a\cosh t$ (for $x \geq a$)
• $\sqrt{a^2-x^2}$: use $x = a\cos t$ (circular, not hyperbolic)

Example: $\displaystyle\int \frac◆LB◆dx◆RB◆◆LB◆\sqrt{x^2-4}◆RB◆$ with $x = 2\cosh t$:

$dx = 2\sinh t\,dt$, $\sqrt{x^2-4} = 2\sinh t$.

$\displaystyle\int \frac◆LB◆2\sinh t◆RB◆◆LB◆2\sinh t◆RB◆\,dt = t = \operatorname{arcosh}\!\left(\frac{x}{2}\right) + C$.

17.2 Gudermannian function

The Gudermannian function relates circular and hyperbolic functions without complex numbers:

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

Key identities:

• $\sin(\operatorname{gd}\,x) = \tanh x$
• $\cos(\operatorname{gd}\,x) = \operatorname{sech} x$
• $\tan(\operatorname{gd}\,x) = \sinh x$

17.3 Hyperbolic functions and catenary

The catenary curve $y = a\cosh\!\left(\dfrac{x}{a}\right)$ describes a hanging chain.

Arc length from the vertex: $s = a\sinh\!\left(\dfrac{x}{a}\right)$.

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18. Further Exam-Style Questions

Question 16

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

Solution

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

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

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

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

Question 17

Use the substitution $x = 3\sinh t$ to evaluate $\displaystyle\int_0^{\operatorname{arsinh}(4/3)} \sqrt{x^2+9}\,dx$.

Solution

$dx = 3\cosh t\,dt$, $\sqrt{x^2+9} = 3\cosh t$.

$= \displaystyle\int_0^{\operatorname{arsinh}(4/3)} 9\cosh^2 t\,dt = \frac{9}{2}\int_0^{\operatorname{arsinh}(4/3)} (1+\cosh 2t)\,dt$

$= \frac{9}{2}\!\left[t + \frac◆LB◆\sinh 2t◆RB◆◆LB◆2◆RB◆\right]_0^{\operatorname{arsinh}(4/3)}$.

At $t = \operatorname{arsinh}(4/3)$: $\sinh t = 4/3$, $\cosh t = 5/3$, $\sinh 2t = 2\cdot\dfrac{4}{3}\cdot\dfrac{5}{3} = \dfrac{40}{9}$.

$= \frac{9}{2}\!\left(\operatorname{arsinh}\!\frac{4}{3} + \frac{20}{9}\right) = \boxed{\frac{9}{2}\operatorname{arsinh}\!\frac{4}{3} + 10}$.