Hyperbolic Functions (Extended)

This topic explores fundamental concepts that shape our understanding of the world.

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

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

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

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

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

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

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

• \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.

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

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

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

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

Proof.

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

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

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

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

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

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).

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

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

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

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

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

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

Proof of the addition formula for $\sinh$:

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

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

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

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 $x$.)

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

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

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

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

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 $e^y$:

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

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

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

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

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

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

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 $y \geq 0$):

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The differentiation results give standard integrals:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\int \sqrt{4\sinh^2 u + 4}\cdot 2\cosh u\,du = \int 2\cosh u \cdot 2\cosh u\,du = 4\int \cosh^2 u\,du$

Using $\cosh^2 u = \dfrac{1 + \cosh 2u}{2}$:

$= 4\int \frac{1 + \cosh 2u}{2}\,du = 2u + \sinh 2u + C$

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

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

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

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

Solution

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

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

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

Solution

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

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

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

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

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

Solution

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

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

Solution

Using the product rule:

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

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

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

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

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

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

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

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

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 $T\sin\theta = T_0\tan\theta$:

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

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

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

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

Solution

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

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

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

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

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

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

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

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

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

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

Solution

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

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

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

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

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

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

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

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

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

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

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

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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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 $e^x$:

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

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

Since $\dfrac{5 - 3\sqrt{2}}{7} \approx 0.109 > 0$Both solutions are valid.

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

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

$\int \sqrt{9\cosh^2 u - 9}\cdot 3\sinh u\,du = \int 3\sinh u \cdot 3\sinh u\,du = 9\int \sinh^2 u\,du$

Using $\sinh^2 u = \dfrac{\cosh 2u - 1}{2}$:

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

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

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

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

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

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

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

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

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

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

Solution. Using the quotient rule:

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

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

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

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

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

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

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

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

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

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

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

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

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 $u = \tanh x$, $du = \mathrm{sech}^2\,x\,dx$:

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

Therefore:

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

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

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

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

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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.

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

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

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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(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).

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

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

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

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

Solution

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

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

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

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. $\blacksquare$

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

Solution

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

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

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

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

Solution

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

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

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

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

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

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

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

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

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

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

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

Solution. $y' = \sinh x$. $ds = \sqrt{1 + \sinh^2 x}\,dx = \cosh x\,dx$.

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

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

Solution. $\tan 2x = \dfrac{2\tan x}{1-\tan^2 x}$. Apply Osborn’s rule (change $\tan^2 x$ to $-\tanh^2 x$):

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

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

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

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

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

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

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

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

Solution

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

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

Solution

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

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

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

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

Solution

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

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

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

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

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

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

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

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:

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

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

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

Solution

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

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

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

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

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

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

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

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

Solution

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

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

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

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

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

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

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

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

Key identities:

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

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

Solution

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

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

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

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

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

Solution

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

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

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

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

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

• Further Calculus: Hyperbolic functions use calculus
• Pure Mathematics: Hyperbolic functions are a pure maths topic
• Complex Numbers: Hyperbolic functions relate to complex exponentials