Hyperbolic Functions (AQA A-Level Further Maths): Revision Notes

Worked Example: Differentiating sech x

Question: Differentiate $\operatorname{sech} x$ with respect to $x$.

Solution:

We start by writing sech x in terms of cosh x:

$\operatorname{sech} x = (\cosh x)^{-1}$

Using the chain rule, we differentiate:

$\frac{\text{d}(\operatorname{sech} x)}{\text{d}x} = -(\cosh x)^{-2} \cdot \sinh x$

Simplifying:

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

We can rewrite this using our reciprocal definitions:

$= -\frac{\sinh x}{\cosh x} \cdot \frac{1}{\cosh x}$

$= -\tanh x \cdot \operatorname{sech} x$

Therefore: $\frac{\text{d}(\operatorname{sech} x)}{\text{d}x} = -\operatorname{sech} x \tanh x$

Exam tip: Notice the difference in sign compared with the derivative of sec x, which is $\tan x \sec x$ (positive). With hyperbolic functions, the derivative of sech x is negative.

Worked Example: Definite integral of sinh²x

Question: Work out the exact value of $\int_0^{\ln 3} \sinh^2 x \, \text{d}x$

Solution:

First, we need to use the hyperbolic identity: $\cosh(2x) = 1 + 2\sinh^2 x$

Rearranging for $\sinh^2 x$:

$\sinh^2 x = \frac{1}{2}(\cosh(2x) - 1)$

Substituting this into our integral:

$\int_0^{\ln 3} \sinh^2 x \, \text{d}x = \int_0^{\ln 3} \frac{1}{2}(\cosh(2x) - 1) \, \text{d}x$

Integrating and substituting the limits:

$= \left[\frac{1}{4}\sinh(2x) - \frac{x}{2}\right]_0^{\ln 3}$

$= \frac{1}{4}\left(\frac{e^{2\ln 3} - e^{-2\ln 3}}{2}\right) - \frac{1}{2}\ln 3 - 0$

Using logarithm rules to simplify: $e^{2\ln 3} = e^{\ln 3^2} = 9$ and $e^{-2\ln 3} = \frac{1}{9}$

$= \frac{1}{4}\left(\frac{9 - \frac{1}{9}}{2}\right) - \frac{1}{2}\ln 3$

$= \frac{1}{8}\left(9 - \frac{1}{9}\right) - \frac{1}{2}\ln 3$

$= \frac{1}{8} \cdot \frac{80}{9} - \frac{1}{2}\ln 3$

$= \frac{10}{9} - \frac{1}{2}\ln 3$

Exam tip: You can check this on your calculator, but it will not give an exact answer. Always show full algebraic working for exact value questions.

Worked Example: Differentiating arcosh x

Question: Find $\frac{\text{d}}{\text{d}x}(\text{arcosh } x)$

Solution:

Let $y = \text{arcosh } x$

Then $x = \cosh y$

Differentiating both sides with respect to x using the chain rule:

$1 = \sinh y \cdot \frac{\text{d}y}{\text{d}x}$

Therefore:

$\frac{\text{d}y}{\text{d}x} = \frac{1}{\sinh y}$

We now need to express sinh y in terms of x. Using the hyperbolic identity $\cosh^2 y - \sinh^2 y = 1$:

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

Since $x = \cosh y$:

$\sinh^2 y = x^2 - 1$

$\sinh y = \sqrt{x^2 - 1}$

(We take the positive square root because sinh y is positive for y > 0, which is the range of arcosh x)

Therefore:

$\frac{\text{d}(\text{arcosh } x)}{\text{d}x} = \frac{1}{\sqrt{x^2 - 1}}$

Worked Example: Integration with √(x² + a²)

Question: Work out $\int \frac{6}{\sqrt{36 + 4x^2}} \, \text{d}x$

Solution:

Step 1: Simplify by dividing numerator and denominator by 2:

$\int \frac{6}{\sqrt{36 + 4x^2}} \, \text{d}x = \int \frac{3}{\sqrt{9 + x^2}} \, \text{d}x$

Step 2: Identify the form. We have $\sqrt{9 + x^2} = \sqrt{a^2 + x^2}$ where $a^2 = 9$, so $a = 3$.

Use the substitution x = 3sinh u

Then $\frac{\text{d}x}{\text{d}u} = 3\cosh u$, so $\text{d}x = 3\cosh u \, \text{d}u$

Step 3: Substitute into the integral:

$\int \frac{3}{\sqrt{9 + (3\sinh u)^2}} \cdot 3\cosh u \, \text{d}u$

$= \int \frac{9\cosh u}{\sqrt{9 + 9\sinh^2 u}} \, \text{d}u$

$= \int \frac{9\cosh u}{\sqrt{9(1 + \sinh^2 u)}} \, \text{d}u$

Using the identity 1 + sinh²u = cosh²u:

$= \int \frac{9\cosh u}{\sqrt{9\cosh^2 u}} \, \text{d}u$

$= \int \frac{9\cosh u}{3\cosh u} \, \text{d}u$

$= \int 3 \, \text{d}u$

$= 3u + c$

Step 4: Rearrange $x = 3\sinh u$ to write the answer in terms of x:

$\sinh u = \frac{x}{3}$

$u = \text{arsinh}\left(\frac{x}{3}\right)$

Therefore:

$\int \frac{6}{\sqrt{36 + 4x^2}} \, \text{d}x = 3\text{arsinh}\left(\frac{x}{3}\right) + c$

Worked Example: Integration with √(x² - a²) after completing the square

Question: Work out $\int \frac{1}{\sqrt{x^2 - 8x - 20}} \, \text{d}x$

Solution:

Step 1: Complete the square on the denominator:

$x^2 - 8x - 20 = (x - 4)^2 - 16 - 20 = (x - 4)^2 - 36$

So the integral becomes:

$\int \frac{1}{\sqrt{(x - 4)^2 - 36}} \, \text{d}x$

Step 2: Choose the substitution. We have the form $\sqrt{(x-4)^2 - 36}$ which matches $\sqrt{u^2 - a^2}$ where $a^2 = 36$, so $a = 6$.

Let x - 4 = 6cosh u

Then $\frac{\text{d}x}{\text{d}u} = 6\sinh u$, so $\text{d}x = 6\sinh u \, \text{d}u$

Step 3: Substitute:

$\int \frac{6\sinh u}{\sqrt{(6\cosh u)^2 - 36}} \, \text{d}u$

$= \int \frac{6\sinh u}{\sqrt{36\cosh^2 u - 36}} \, \text{d}u$

$= \int \frac{6\sinh u}{\sqrt{36(\cosh^2 u - 1)}} \, \text{d}u$

Using the identity $\cosh^2 u - \sinh^2 u = 1$, we have cosh²u - 1 = sinh²u:

$= \int \frac{6\sinh u}{\sqrt{36\sinh^2 u}} \, \text{d}u$

$= \int \frac{6\sinh u}{6\sinh u} \, \text{d}u$

$= \int 1 \, \text{d}u$

$= u + c$

Step 4: Rearrange $x - 4 = 6\cosh u$ to write the answer in terms of x:

$\cosh u = \frac{x - 4}{6}$

$u = \text{arcosh}\left(\frac{x - 4}{6}\right)$

Therefore:

$\int \frac{1}{\sqrt{x^2 - 8x - 20}} \, \text{d}x = \text{arcosh}\left(\frac{x - 4}{6}\right) + c$

Exam tip: When completing the square appears necessary, don't forget this crucial first step. Many students try to apply substitution directly without simplifying first, which leads to much more complicated algebra.