Why is "Writing $\int\sinh x\,dx = -\cosh x$ (importing a trig minus sign)" a common mistake in Integrate Hyperbolic Functions and Recognise Integrals of Functions?

Why it happens: Muscle memory from x\,dx = - x and x\,dx = x. How to avoid it: Hyperbolic integrals carry NO minus: = +, = +. Check against the derivatives d/dx x = x (also no minus). Source: 9231 Paper 2 Examiner Reports

Why is "Omitting the $\frac{1}{a}$ in $\int\frac{dx}{a^2+x^2} = \frac{1}{a}\tan^{-1}\frac{x}{a}$" a common mistake in Integrate Hyperbolic Functions and Recognise Integrals of Functions?

Why it happens: The other three standard forms have no leading factor, so candidates apply the same blank pattern to the ^-1 case. How to avoid it: Memorise that the ^-1 form is the ONLY one with a 1/a in front. Differentiate your answer mentally to check the chain-rule factor. Source: 9231 Paper 2 mark schemes

Why is "Using $a^2$ where $a$ is required (e.g. answer $\sin^{-1}\frac{x}{9}$ for $\sqrt{9 - x^2}$)" a common mistake in Integrate Hyperbolic Functions and Recognise Integrals of Functions?

Why it happens: Reading the constant directly out of the denominator without taking the square root. How to avoid it: The denominator gives a^2. Always take the root: a^2 = 9 a = 3, so the answer is ^-1x/3. Source: 9231 Paper 2 Examiner Reports

Why is "Sign error when completing the square on a negative-$x^2$ quadratic" a common mistake in Integrate Hyperbolic Functions and Recognise Integrals of Functions?

Why it happens: Forgetting to distribute the -1 through the completed bracket, e.g. writing 5 - 2x - x^2 = (x+1)^2 + 4 instead of 6 - (x+1)^2. How to avoid it: Factor -1 from the x-terms FIRST, complete the square inside, then carefully distribute the minus: 5 - (x^2 + 2x) = 5 - [(x+1)^2 - 1] = 6 - (x+1)^2. Source: 9231 Paper 2 Examiner Reports

Why is "Folding the $\frac{1}{\sqrt c}$ or $\frac{1}{c}$ scaling factor incorrectly into the standard form" a common mistake in Integrate Hyperbolic Functions and Recognise Integrals of Functions?

Why it happens: Trying to absorb the coefficient of x^2 into a instead of pulling it cleanly outside the integral. How to avoid it: Pull the leading constant (√ c from a root, c otherwise) ALL the way outside the integral, then apply the standard form to the clean x^2 + (·)^2 that remains. Source: 9231 Paper 2 mark schemes

Why is "Omitting $+C$ on an indefinite integral" a common mistake in Integrate Hyperbolic Functions and Recognise Integrals of Functions?

Why it happens: Treating the standard-form result as a finished answer rather than a family of antiderivatives. How to avoid it: Every indefinite integral ends in +C. (For a definite integral, drop +C and substitute the limits instead.) Source: 9231 Paper 2 Examiner Reports

Cambridge International A Levels Further Mathematics (9231)

Integrate Hyperbolic Functions and Recognise Integrals of Functions

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Detailed Study Notes

Detailed notes on Integration for Cambridge International A Levels Further Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Integrate Hyperbolic Functions and Recognise Standard Inverse-Form Integrals — Cambridge International AS & A Level Further Mathematics 9231 Paper 2 (2026-2027 syllabus)

Learn the integrals of $\sinh x$ , $\cosh x$ and $\text{sech}^2 x$ , then master the four standard inverse-form results in MF19 — $\int\frac{dx}{\sqrt{a^2-x^2}}$ , $\int\frac{dx}{\sqrt{x^2+a^2}}$ , $\int\frac{dx}{\sqrt{x^2-a^2}}$ and $\int\frac{dx}{a^2+x^2}$ . The exam skill is recognising which form applies and completing the square to reach it.

At a glance

Hyperbolic integrals mirror the trig ones: $\int\sinh x\,dx = \cosh x + C$ , $\int\cosh x\,dx = \sinh x + C$ , $\int\text{sech}^2 x\,dx = \tanh x + C$ .
Four standard inverse-form integrals ARE in MF19 — quote them, do not re-derive them.
The shape of the denominator tells you which form: $\sqrt{a^2 - x^2}\to\sin^{-1}$ , $\sqrt{x^2 + a^2}\to\sinh^{-1}$ , $\sqrt{x^2 - a^2}\to\cosh^{-1}$ , $a^2 + x^2\to\frac{1}{a}\tan^{-1}$ .
Completing the square turns a messy quadratic (e.g. $x^2 + 4x + 13$ ) into a recognisable $u^2 \pm a^2$ form.
Read off $a$ , not $a^2$ : in $\sqrt{9 - x^2}$ the constant is $a^2 = 9$ , so $a = 3$ .

What you’ll learn

Mapped to the Cambridge International A Level 9231 syllabus (2026-2027).

2.4a — Integrate hyperbolic functions ( $\sinh x$ , $\cosh x$ , $\text{sech}^2 x$ and related forms).
2.4b — Recognise integrals of functions of the form $\frac{1}{\sqrt{a^2-x^2}}$ , $\frac{1}{\sqrt{x^2+a^2}}$ , $\frac{1}{\sqrt{x^2-a^2}}$ and $\frac{1}{a^2+x^2}$ , and integrate associated functions using completing the square and linear substitutions.
2.4c — Decide which standard form (and value of $a$ ) applies, including after completing the square.

Integrating the hyperbolic functions

sinh and cosh swap on integration just like sin and cos — but with no sign change.

The basic results. Because $\frac{d}{dx}\sinh x = \cosh x$ and $\frac{d}{dx}\cosh x = \sinh x$ (note: no minus sign, unlike trig), reversing these gives:

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

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

These three are the workhorses. They are not separately listed in MF19, so you must know them — but they follow instantly from the derivatives, which are in MF19.

Linear inner functions. For $\sinh(ax+b)$ or $\cosh(ax+b)$ , divide by the inner coefficient $a$ :

$\int \cosh(3x)\,dx = \frac{1}{3}\sinh(3x) + C$

Using hyperbolic identities first. Integrals like $\int \cosh^2 x\,dx$ have no direct result, so rewrite using $\cosh^2 x = \frac{1}{2}(\cosh 2x + 1)$ :

$\int \cosh^2 x\,dx = \int \tfrac{1}{2}(\cosh 2x + 1)\,dx = \tfrac{1}{4}\sinh 2x + \tfrac{1}{2}x + C$

cosh x is even with minimum 1; sinh x is odd through the origin. Integrating one gives the other (no sign flip).

Cambridge tip. The single most common slip is importing a minus sign from trigonometry. Repeat to yourself: $\int\sinh = +\cosh$ and $\int\cosh = +\sinh$ — both positive.

$\int\sinh x\,dx = \cosh x + C$ and $\int\cosh x\,dx = \sinh x + C$ (no minus).
$\int\text{sech}^2 x\,dx = \tanh x + C$ .
For $\cosh^2 x$ / $\sinh^2 x$ , use $\cosh^2 x = \frac12(\cosh 2x + 1)$ , $\sinh^2 x = \frac12(\cosh 2x - 1)$ first.

See the full worked example for integrate hyperbolic functions and recognise integrals of functions →

The four standard inverse-form integrals (MF19)

Match the denominator shape to the result. All four are printed in MF19 — quote them.

Four integrals appear so often that Cambridge prints them in the MF19 list of standard integrals. You may quote them directly — but you must recognise which one applies and read off $a$ correctly.

$\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac{x}{a} + C$

$\int \frac{dx}{\sqrt{x^2 + a^2}} = \sinh^{-1}\frac{x}{a} + C$

$\int \frac{dx}{\sqrt{x^2 - a^2}} = \cosh^{-1}\frac{x}{a} + C \quad (x > a)$

$\int \frac{dx}{a^2 + x^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C$

How to choose — read the denominator:

$\sqrt{a^2 - x^2}$ (constant minus square root term): $\sin^{-1}$ .
$\sqrt{x^2 + a^2}$ (square plus constant, under a root): $\sinh^{-1}$ .
$\sqrt{x^2 - a^2}$ (square minus constant, under a root): $\cosh^{-1}$ .
$a^2 + x^2$ (square plus constant, no root): $\frac{1}{a}\tan^{-1}$ , and don't forget the $\frac{1}{a}$ in front.

Find $a$ , not $a^2$ . In $\int\frac{dx}{\sqrt{16 - x^2}}$ the constant is $a^2 = 16$ , so $a = 4$ and the answer is $\sin^{-1}\frac{x}{4} + C$ .

Match the denominator to its result. Only the last (no square root) carries the extra 1/a factor.

Cambridge tip. The $\frac{1}{a}\tan^{-1}$ result is the one candidates botch — the $\frac{1}{a}$ is easy to forget because the other three have no leading factor.

$\sqrt{a^2-x^2}\to\sin^{-1}\frac{x}{a}$ ; $\sqrt{x^2+a^2}\to\sinh^{-1}\frac{x}{a}$ .
$\sqrt{x^2-a^2}\to\cosh^{-1}\frac{x}{a}$ ; $a^2+x^2\to\frac{1}{a}\tan^{-1}\frac{x}{a}$ .
Only the no-root $\tan^{-1}$ form has a $\frac{1}{a}$ in front.

See the full worked example for integrate hyperbolic functions and recognise integrals of functions →

Completing the square to reach a standard form

A general quadratic in the denominator becomes a perfect-square ± constant — then it matches.

Most exam integrals don't arrive pre-packaged as $a^2 \pm x^2$ . Instead you meet a general quadratic such as $x^2 + 4x + 13$ or $5 - 2x - x^2$ . Complete the square to convert it, then substitute $u = x + (\text{shift})$ .

The procedure:

Complete the square on the quadratic: write it as $(x + p)^2 \pm a^2$ (or $a^2 - (x+p)^2$ ).
Substitute $u = x + p$ , so $du = dx$ — the integral becomes a standard form in $u$ .
Quote the MF19 result, then substitute back $u = x + p$ .

Worked shape. For $\int \frac{dx}{x^2 + 4x + 13}$ :

$x^2 + 4x + 13 = (x+2)^2 + 9 = (x+2)^2 + 3^2$

So with $u = x + 2$ this is $\int\frac{du}{u^2 + 3^2}$ — the $\frac{1}{a}\tan^{-1}$ form with $a = 3$ :

$= \frac{1}{3}\tan^{-1}\frac{u}{3} + C = \frac{1}{3}\tan^{-1}\frac{x+2}{3} + C$

Watch the sign of the $x^2$ term. If the quadratic is $5 - 2x - x^2$ , factor out $-1$ from the $x$ -terms first:

$5 - 2x - x^2 = 5 - (x^2 + 2x) = 5 - [(x+1)^2 - 1] = 6 - (x+1)^2$

This is the $a^2 - u^2$ shape ( $a = \sqrt{6}$ ), giving a $\sin^{-1}$ when under a root.

Cambridge tip. Always state your substitution explicitly ("let $u = x + 2$ ") and remember that $du = dx$ for these linear shifts — the working is short but every line earns a method mark.

Complete the square to turn a general quadratic into $(x+p)^2 \pm a^2$ .
Substitute $u = x + p$ ( $du = dx$ ), apply the matching MF19 form, then substitute back.
If the $x^2$ term is negative, factor out $-1$ from the $x$ -terms before completing the square.

See the full worked example for integrate hyperbolic functions and recognise integrals of functions →

Linear substitutions and scaling the variable

A coefficient on x² is handled by factoring it out before matching a standard form.

When the $x^2$ term has a coefficient other than $1$ — say $4x^2 + 9$ or $\sqrt{9x^2 + 4}$ — factor that coefficient out so the bracket starts with a clean $x^2$ , then match a standard form.

Coefficient on $x^2$ (the $\tan^{-1}$ case). For $\int\frac{dx}{4x^2 + 9}$ :

$4x^2 + 9 = 4\left(x^2 + \tfrac{9}{4}\right) = 4\left(x^2 + \left(\tfrac{3}{2}\right)^2\right)$

$\int \frac{dx}{4\left(x^2 + (3/2)^2\right)} = \frac{1}{4}\cdot\frac{1}{3/2}\tan^{-1}\frac{x}{3/2} + C = \frac{1}{6}\tan^{-1}\frac{2x}{3} + C$

Coefficient under a root (the $\sinh^{-1}$ case). For $\int\frac{dx}{\sqrt{9x^2 + 4}}$ :

$\sqrt{9x^2 + 4} = \sqrt{9}\,\sqrt{x^2 + \tfrac{4}{9}} = 3\sqrt{x^2 + \left(\tfrac{2}{3}\right)^2}$

$\int \frac{dx}{3\sqrt{x^2 + (2/3)^2}} = \frac{1}{3}\sinh^{-1}\frac{x}{2/3} + C = \frac{1}{3}\sinh^{-1}\frac{3x}{2} + C$

Definite integrals. For a definite integral, substitute the limits at the end into the evaluated antiderivative; no $+C$ is needed. Keep non-exact answers to 3 significant figures unless an exact form (e.g. $\sinh^{-1} 2$ or $\ln(2+\sqrt 5)$ ) is expected.

Cambridge tip. Pull the leading constant all the way outside the integral before quoting the standard form — mixing the $\frac{1}{\sqrt c}$ or $\frac{1}{c}$ scaling factor into the formula is where marks vanish.

Factor a coefficient out of $x^2$ so the bracket reads $x^2 + (\text{const})^2$ .
A constant under the root contributes a $\frac{1}{\sqrt c}$ scaling; outside, a $\frac{1}{c}$ scaling.
For definite integrals, no $+C$ ; give non-exact results to 3 s.f.

See the full worked example for integrate hyperbolic functions and recognise integrals of functions →

Quick recap

$\int\sinh x\,dx = \cosh x + C$ , $\int\cosh x\,dx = \sinh x + C$ , $\int\text{sech}^2 x\,dx = \tanh x + C$ — no minus signs.
$\sqrt{a^2-x^2}\to\sin^{-1}\frac{x}{a}$ ; $\sqrt{x^2+a^2}\to\sinh^{-1}\frac{x}{a}$ ; $\sqrt{x^2-a^2}\to\cosh^{-1}\frac{x}{a}$ .
$\int\frac{dx}{a^2+x^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C$ — remember the $\frac{1}{a}$ .
All four inverse-form integrals are printed in MF19; quote them after identifying $a$ .
Complete the square to convert a general quadratic; factor out the $x^2$ coefficient when it isn't 1.
Read off $a$ from $a^2$ (e.g. $a^2 = 9 \Rightarrow a = 3$ ), and add $+C$ on every indefinite integral.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

$\int\sinh x\,dx = \cosh x + C$ , $\int\cosh x\,dx = \sinh x + C$ , $\int\text{sech}^2 x\,dx = \tanh x + C$ .
$\int\frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\frac{x}{a} + C$ (MF19).
$\int\frac{dx}{\sqrt{x^2+a^2}} = \sinh^{-1}\frac{x}{a} + C$ and $\int\frac{dx}{\sqrt{x^2-a^2}} = \cosh^{-1}\frac{x}{a} + C$ (MF19).
$\int\frac{dx}{a^2+x^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C$ (MF19) — note the $\frac{1}{a}$ .
$\cosh^2 x = \frac12(\cosh 2x + 1)$ , $\sinh^2 x = \frac12(\cosh 2x - 1)$ for squared integrands.

How it’s examined

Topic 2.4 is a staple of Paper 2 (Further Pure Mathematics 2), worth 4-8 marks either as a standalone integration or embedded in an area / arc-length question. Examiners expect you to identify the standard form and read off $a$ correctly; a common 4-mark question requires completing the square first. Marks are lost through (i) importing a spurious minus sign into the hyperbolic integrals, (ii) forgetting the $\frac{1}{a}$ in the $\tan^{-1}$ result, and (iii) using $a^2$ where $a$ is required. Because all four inverse-form results are in MF19, the marks reward recognition and clean substitution, not memorisation.

Sources: Cambridge International AS & A Level Further Mathematics 9231 syllabus (2026-2027); Cambridge List of Formulae MF19 — Standard integrals; 9231/21 & 9231/23 Further Pure Mathematics 2 — past papers and mark schemes. Last reviewed 2026-06-23.

Take this whole topic with you

Step-by-step worked examples — Integrate Hyperbolic Functions and Recognise Integrals of Functions

Step-by-step solutions to past-paper-style questions on integrate hyperbolic functions and recognise integrals of functions, written exactly the way a tutor would explain them at the board.

Question type:

1Integrating sinh and cosh directly
Getting startedDirect calculation• hyperbolic, basic, easy
Question
Find $\int \left(\cosh x + 3\sinh x\right)\,dx$ .
Step-by-step solution
1. Step 1
  Recall the basic results. $\int\cosh x\,dx = \sinh x$ and $\int\sinh x\,dx = \cosh x$ — both with a plus sign (no minus, unlike trig).
2. Step 2
  Integrate term by term.
  $\int \cosh x\,dx + 3\int \sinh x\,dx = \sinh x + 3\cosh x + C$
Answer
$\sinh x + 3\cosh x + C$ .
Examiner tip
Both terms keep a positive sign. The classic error is writing $-3\cosh x$ by analogy with $\int\sin x\,dx = -\cos x$ .
2Recognising a standard inverse-form integral
Getting startedIdentify & classify• standard-form, mf19, easy
Question
Find $\int \frac{dx}{\sqrt{25 - x^2}}$ and $\int \frac{dx}{16 + x^2}$ .
Step-by-step solution
1. Step 1
  First integral. Denominator $\sqrt{25 - x^2}$ is the $\sqrt{a^2 - x^2}$ shape, so the result is $\sin^{-1}\frac{x}{a}$ . Here $a^2 = 25$ , so $a = 5$ .
  $\int \frac{dx}{\sqrt{25 - x^2}} = \sin^{-1}\frac{x}{5} + C$
2. Step 2
  Second integral. Denominator $16 + x^2$ (no root) is the $a^2 + x^2$ shape, giving $\frac{1}{a}\tan^{-1}\frac{x}{a}$ with $a = 4$ .
  $\int \frac{dx}{16 + x^2} = \frac{1}{4}\tan^{-1}\frac{x}{4} + C$
Answer
$\sin^{-1}\frac{x}{5} + C$ and $\frac{1}{4}\tan^{-1}\frac{x}{4} + C$ .
Examiner tip
Read $a$ from $a^2$ : $25\to 5$ , $16\to 4$ . Don't forget the $\frac{1}{4}$ on the $\tan^{-1}$ result — only that form carries a leading factor.
3Completing the square to reach a tan⁻¹ form
Building confidenceMulti-step problem• complete-the-square, tan-inverse, medium
Question
Find $\int \frac{dx}{x^2 + 6x + 13}$ .
Step-by-step solution
1. Step 1
  Complete the square on the denominator.
  $x^2 + 6x + 13 = (x+3)^2 + 4 = (x+3)^2 + 2^2$
2. Step 2
  Substitute $u = x + 3$ , so $du = dx$ . The integral becomes the $\frac{1}{a}\tan^{-1}$ form with $a = 2$ .
  $\int \frac{du}{u^2 + 2^2} = \frac{1}{2}\tan^{-1}\frac{u}{2} + C$
3. Step 3
  Substitute back $u = x + 3$ .
  $= \frac{1}{2}\tan^{-1}\frac{x+3}{2} + C$
Answer
$\frac{1}{2}\tan^{-1}\frac{x+3}{2} + C$ .
Examiner tip
State 'let $u = x + 3$ ' explicitly. Since $du = dx$ the substitution is clean, but the line earns a method mark.
4A coefficient on x² under a square root
Building confidenceMulti-step problem• scaling, sinh-inverse, medium
Question
Find $\int \frac{dx}{\sqrt{9x^2 + 4}}$ .
Step-by-step solution
1. Step 1
  Factor the coefficient of $x^2$ out of the root so the bracket starts with a clean $x^2$ .
  $\sqrt{9x^2 + 4} = 3\sqrt{x^2 + \tfrac{4}{9}} = 3\sqrt{x^2 + \left(\tfrac{2}{3}\right)^2}$
2. Step 2
  Pull the constant outside and match the $\sqrt{x^2 + a^2}\to\sinh^{-1}$ form with $a = \frac{2}{3}$ .
  $\frac{1}{3}\int \frac{dx}{\sqrt{x^2 + (2/3)^2}} = \frac{1}{3}\sinh^{-1}\frac{x}{2/3} + C$
3. Step 3
  Simplify $\frac{x}{2/3} = \frac{3x}{2}$ .
  $= \frac{1}{3}\sinh^{-1}\frac{3x}{2} + C$
Answer
$\frac{1}{3}\sinh^{-1}\frac{3x}{2} + C$ .
Examiner tip
The $\frac{1}{3}$ comes from $\sqrt{9} = 3$ pulled out of the root. Keep it outside the formula — folding it in is a frequent slip.
5A definite integral giving cosh⁻¹
StretchMulti-step problem• definite, cosh-inverse, hard
Question
Evaluate $\int_{5}^{10} \frac{dx}{\sqrt{x^2 - 16}}$ , giving your answer in an exact logarithmic form.
Step-by-step solution
1. Step 1
  Identify the form. $\sqrt{x^2 - 16}$ is the $\sqrt{x^2 - a^2}$ shape with $a = 4$ , giving $\cosh^{-1}\frac{x}{4}$ .
  $\int \frac{dx}{\sqrt{x^2 - 16}} = \cosh^{-1}\frac{x}{4} + C$
2. Step 2
  Apply the limits.
  $\left[\cosh^{-1}\frac{x}{4}\right]_{5}^{10} = \cosh^{-1}\frac{10}{4} - \cosh^{-1}\frac{5}{4}$
3. Step 3
  Convert to logarithms using $\cosh^{-1} t = \ln\!\left(t + \sqrt{t^2 - 1}\right)$ . For $t = \frac{5}{2}$ : $\sqrt{\frac{25}{4} - 1} = \sqrt{\frac{21}{4}} = \frac{\sqrt{21}}{2}$ . For $t = \frac{5}{4}$ : $\sqrt{\frac{25}{16} - 1} = \frac{3}{4}$ .
  $= \ln\!\left(\frac{5 + \sqrt{21}}{2}\right) - \ln\!\left(\frac{5/4 + 3/4}{1}\right) = \ln\!\left(\frac{5 + \sqrt{21}}{2}\right) - \ln 2$
4. Step 4
  Combine the logarithms.
  $= \ln\!\left(\frac{5 + \sqrt{21}}{4}\right)$
Answer
$\ln\!\left(\dfrac{5 + \sqrt{21}}{4}\right) \approx 0.864$ (3 s.f.).
Examiner tip
The logarithmic form $\cosh^{-1} t = \ln(t + \sqrt{t^2-1})$ is in MF19. Note $\cosh^{-1}\frac{5}{4} = \ln 2$ exactly — spotting that simplification earns the final accuracy mark.
6Negative x² coefficient — complete the square then sin⁻¹
StretchMulti-step problem• complete-the-square, sin-inverse, hard
Question
Find $\int \frac{dx}{\sqrt{5 + 4x - x^2}}$ .
Step-by-step solution
1. Step 1
  Factor $-1$ from the $x$ -terms first, then complete the square inside.
  $5 + 4x - x^2 = 5 - (x^2 - 4x) = 5 - \left[(x-2)^2 - 4\right] = 9 - (x-2)^2$
2. Step 2
  Substitute $u = x - 2$ ( $du = dx$ ). This is the $\sqrt{a^2 - u^2}\to\sin^{-1}$ form with $a = 3$ .
  $\int \frac{du}{\sqrt{9 - u^2}} = \sin^{-1}\frac{u}{3} + C$
3. Step 3
  Substitute back $u = x - 2$ .
  $= \sin^{-1}\frac{x - 2}{3} + C$
Answer
$\sin^{-1}\frac{x-2}{3} + C$ .
Examiner tip
Pulling $-1$ from $(x^2 - 4x)$ flips the sign of the constant correction: $-[(x-2)^2 - 4] = -(x-2)^2 + 4$ . Mishandling this sign is the most common error on the negative- $x^2$ case.

Model Answers — Integrate Hyperbolic Functions and Recognise Integrals of Functions

High-scoring sample answers for integrate hyperbolic functions and recognise integrals of functions on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
3 marks
Find $\int \left(2\sinh x - \text{sech}^2 x\right)\,dx$ .
Model answer
Integrate term by term, using $\int\sinh x\,dx = \cosh x$ and $\int\text{sech}^2 x\,dx = \tanh x$ :

$\int \left(2\sinh x - \text{sech}^2 x\right)\,dx = 2\cosh x - \tanh x + C$
Why this scores
M1 for $\int\sinh x = \cosh x$ , M1 for $\int\text{sech}^2 x = \tanh x$ , A1 for the complete answer with $+C$ . Each integral keeps its natural sign — no spurious minus.
Question 2
2 marks
Find $\int \frac{dx}{\sqrt{x^2 + 49}}$ .
Model answer
The denominator $\sqrt{x^2 + 49}$ is the standard form $\sqrt{x^2 + a^2}$ with $a^2 = 49$ , so $a = 7$ . From MF19:

$\int \frac{dx}{\sqrt{x^2 + 49}} = \sinh^{-1}\frac{x}{7} + C$
Why this scores
B1 for identifying the $\sinh^{-1}$ form, B1 for $a = 7$ and the correct answer. Reading $a^2 = 49 \Rightarrow a = 7$ (not $a = 49$ ) is essential.
Question 3
5 marks
Find $\int \frac{dx}{2x^2 + 4x + 10}$ .
Model answer
Factor out the leading coefficient $2$ :

$2x^2 + 4x + 10 = 2\left(x^2 + 2x + 5\right)$

Complete the square inside the bracket:

$x^2 + 2x + 5 = (x+1)^2 + 4 = (x+1)^2 + 2^2$

So the integral is

$\frac{1}{2}\int \frac{dx}{(x+1)^2 + 2^2}$

Substitute $u = x + 1$ ( $du = dx$ ), giving the $\frac{1}{a}\tan^{-1}$ form with $a = 2$ :

$\frac{1}{2}\cdot\frac{1}{2}\tan^{-1}\frac{u}{2} = \frac{1}{4}\tan^{-1}\frac{x+1}{2} + C$
Why this scores
M1 for factoring out $2$ , M1 for completing the square to $(x+1)^2 + 2^2$ , M1 for the $\frac{1}{a}\tan^{-1}$ form, A1 for combining the $\frac{1}{2}$ and $\frac{1}{2}$ to $\frac{1}{4}$ , A1 for the final answer in $x$ . The two separate $\frac{1}{2}$ factors are the common loss point.
Question 4
5 marks
Evaluate $\int_{0}^{3/2} \frac{dx}{\sqrt{9 - 4x^2}}$ , giving your answer exactly.
Model answer
Factor the $4$ out of the root so the bracket starts with $x^2$ :

$\sqrt{9 - 4x^2} = \sqrt{4}\,\sqrt{\tfrac{9}{4} - x^2} = 2\sqrt{\left(\tfrac{3}{2}\right)^2 - x^2}$

So

$\int \frac{dx}{\sqrt{9 - 4x^2}} = \frac{1}{2}\int \frac{dx}{\sqrt{(3/2)^2 - x^2}} = \frac{1}{2}\sin^{-1}\frac{x}{3/2} = \frac{1}{2}\sin^{-1}\frac{2x}{3}$

Apply the limits $x = 0$ to $x = \frac{3}{2}$ :

$\frac{1}{2}\left[\sin^{-1}\frac{2x}{3}\right]_{0}^{3/2} = \frac{1}{2}\left(\sin^{-1} 1 - \sin^{-1} 0\right) = \frac{1}{2}\cdot\frac{\pi}{2} = \frac{\pi}{4}$
Why this scores
M1 for factoring $\sqrt 4 = 2$ out, M1 for the $\sin^{-1}$ form with $a = \frac{3}{2}$ , A1 for the antiderivative $\frac{1}{2}\sin^{-1}\frac{2x}{3}$ , M1 for substituting the limits, A1 for $\frac{\pi}{4}$ . Note $\sin^{-1} 1 = \frac{\pi}{2}$ .
Question 5
4 marks
Find $\int \frac{dx}{\sqrt{x^2 + 2x + 5}}$ .
Model answer
Complete the square under the root:

$x^2 + 2x + 5 = (x+1)^2 + 4 = (x+1)^2 + 2^2$

Substitute $u = x + 1$ ( $du = dx$ ). This is the $\sqrt{u^2 + a^2}\to\sinh^{-1}$ form with $a = 2$ :

$\int \frac{du}{\sqrt{u^2 + 2^2}} = \sinh^{-1}\frac{u}{2} + C$

Substitute back:

$= \sinh^{-1}\frac{x+1}{2} + C$

(Equivalently, in logarithmic form, $\ln\!\left(x + 1 + \sqrt{x^2 + 2x + 5}\right) + C'$ .)
Why this scores
M1 for completing the square to $(x+1)^2 + 2^2$ , M1 for recognising the $\sinh^{-1}$ form, A1 for $\sinh^{-1}\frac{u}{2}$ , A1 for the answer in $x$ . The logarithmic alternative is fully acceptable.
Question 6
4 marks
By using a suitable identity, find $\int \sinh^2 x\,dx$ .
Model answer
Use the identity $\sinh^2 x = \frac{1}{2}\left(\cosh 2x - 1\right)$ (from $\cosh 2x = 1 + 2\sinh^2 x$ ):

$\int \sinh^2 x\,dx = \int \frac{1}{2}\left(\cosh 2x - 1\right)\,dx$

Integrate, remembering $\int\cosh 2x\,dx = \frac{1}{2}\sinh 2x$ :

$= \frac{1}{2}\left(\frac{1}{2}\sinh 2x - x\right) + C = \frac{1}{4}\sinh 2x - \frac{1}{2}x + C$
Why this scores
M1 for the correct double-angle identity, M1 for $\int\cosh 2x = \frac{1}{2}\sinh 2x$ , A1 A1 for the two terms with $+C$ . Using $\sinh^2 x = \frac12(1 - \cosh 2x)$ (wrong sign) is the typical error — $\cosh 2x \geq 1$ but $\sinh^2 x \geq 0$ , so the minus must sit on the $1$ .

Key Formulae — Integrate Hyperbolic Functions and Recognise Integrals of Functions

The formulae you need to memorise for integrate hyperbolic functions and recognise integrals of functions on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Integrals of sinh, cosh, sech²
$\int\sinh x\,dx = \cosh x + C,\quad \int\cosh x\,dx = \sinh x + C,\quad \int\text{sech}^2 x\,dx = \tanh x + C$
$C$
arbitrary constant of integration
When to use
Direct integration of the basic hyperbolic functions. No minus signs — unlike the trigonometric analogues.
Inverse-sine standard integral (MF19)
$\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac{x}{a} + C$
$a$
positive constant; read off from $a^2$ in the denominator
When to use
Denominator is $\sqrt{a^2 - x^2}$ (constant minus square, under a root). Match after completing the square if needed.
Inverse-sinh standard integral (MF19)
$\int \frac{dx}{\sqrt{x^2 + a^2}} = \sinh^{-1}\frac{x}{a} + C$
$a$
positive constant; $a^2$ is the added constant under the root
When to use
Denominator is $\sqrt{x^2 + a^2}$ . Equivalent logarithmic form $\ln(x + \sqrt{x^2 + a^2})$ is also acceptable.
Inverse-cosh standard integral (MF19)
$\int \frac{dx}{\sqrt{x^2 - a^2}} = \cosh^{-1}\frac{x}{a} + C$
$a$
positive constant; valid for $x > a$
When to use
Denominator is $\sqrt{x^2 - a^2}$ . Use $\cosh^{-1} t = \ln(t + \sqrt{t^2 - 1})$ to give an exact log form.
Inverse-tan standard integral (MF19)
$\int \frac{dx}{a^2 + x^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C$
$a$
positive constant; read off from $a^2$
When to use
Denominator is $a^2 + x^2$ with NO square root. This is the only standard form carrying a leading $\frac{1}{a}$ .
Hyperbolic double-angle identities
$\cosh^2 x = \tfrac{1}{2}(\cosh 2x + 1),\qquad \sinh^2 x = \tfrac{1}{2}(\cosh 2x - 1)$
When to use
Before integrating $\cosh^2 x$ or $\sinh^2 x$ , which have no direct integral.

Key Definitions and Keywords — Integrate Hyperbolic Functions and Recognise Integrals of Functions

Definitions to memorise and the exact keywords mark schemes credit for integrate hyperbolic functions and recognise integrals of functions answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Standard integral (MF19)
Examiner keyword
An integral whose result is printed in the Cambridge List of Formulae MF19. It may be quoted directly in an exam without derivation, provided you identify the correct value of $a$ .
Completing the square
Examiner keyword
Rewriting a quadratic $x^2 + bx + c$ as $(x + \frac{b}{2})^2 + (c - \frac{b^2}{4})$ , used to convert a general quadratic denominator into a recognisable $(x+p)^2 \pm a^2$ standard form.
Inverse hyperbolic function
Examiner keyword
The inverse of a hyperbolic function, e.g. $\sinh^{-1} x$ and $\cosh^{-1} x$ . Each has a logarithmic form: $\sinh^{-1} x = \ln(x + \sqrt{x^2 + 1})$ , $\cosh^{-1} x = \ln(x + \sqrt{x^2 - 1})$ .
Linear substitution
A substitution of the form $u = x + p$ (where $du = dx$ ), used after completing the square to shift a quadratic into a pure standard form before integrating.
$\text{sech}\,x$
The hyperbolic secant, $\text{sech}\,x = \frac{1}{\cosh x}$ . Its square integrates to $\tanh x$ , mirroring $\int\sec^2 x\,dx = \tan x$ .
Scaling factor
The constant pulled outside an integral when the $x^2$ coefficient is not $1$ : $\frac{1}{\sqrt c}$ when factored from under a root, $\frac{1}{c}$ when factored from a non-root denominator.

Common Mistakes and Misconceptions — Integrate Hyperbolic Functions and Recognise Integrals of Functions

The traps other students keep falling into on integrate hyperbolic functions and recognise integrals of functions questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Writing $\int\sinh x\,dx = -\cosh x$ (importing a trig minus sign)
9231 Paper 2 Examiner Reports
Why it happens
Muscle memory from $\int\sin x\,dx = -\cos x$ and $\int\cos x\,dx = \sin x$ .
How to avoid it
Hyperbolic integrals carry NO minus: $\int\sinh = +\cosh$ , $\int\cosh = +\sinh$ . Check against the derivatives $\frac{d}{dx}\sinh x = \cosh x$ (also no minus).
✕Omitting the $\frac{1}{a}$ in $\int\frac{dx}{a^2+x^2} = \frac{1}{a}\tan^{-1}\frac{x}{a}$
9231 Paper 2 mark schemes
Why it happens
The other three standard forms have no leading factor, so candidates apply the same blank pattern to the $\tan^{-1}$ case.
How to avoid it
Memorise that the $\tan^{-1}$ form is the ONLY one with a $\frac{1}{a}$ in front. Differentiate your answer mentally to check the chain-rule factor.
✕Using $a^2$ where $a$ is required (e.g. answer $\sin^{-1}\frac{x}{9}$ for $\sqrt{9 - x^2}$ )
9231 Paper 2 Examiner Reports
Why it happens
Reading the constant directly out of the denominator without taking the square root.
How to avoid it
The denominator gives $a^2$ . Always take the root: $a^2 = 9 \Rightarrow a = 3$ , so the answer is $\sin^{-1}\frac{x}{3}$ .
✕Sign error when completing the square on a negative- $x^2$ quadratic
9231 Paper 2 Examiner Reports
Why it happens
Forgetting to distribute the $-1$ through the completed bracket, e.g. writing $5 - 2x - x^2 = (x+1)^2 + 4$ instead of $6 - (x+1)^2$ .
How to avoid it
Factor $-1$ from the $x$ -terms FIRST, complete the square inside, then carefully distribute the minus: $5 - (x^2 + 2x) = 5 - [(x+1)^2 - 1] = 6 - (x+1)^2$ .
✕Folding the $\frac{1}{\sqrt c}$ or $\frac{1}{c}$ scaling factor incorrectly into the standard form
9231 Paper 2 mark schemes
Why it happens
Trying to absorb the coefficient of $x^2$ into $a$ instead of pulling it cleanly outside the integral.
How to avoid it
Pull the leading constant ( $\sqrt c$ from a root, $c$ otherwise) ALL the way outside the integral, then apply the standard form to the clean $x^2 + (\cdot)^2$ that remains.
✕Omitting $+C$ on an indefinite integral
9231 Paper 2 Examiner Reports
Why it happens
Treating the standard-form result as a finished answer rather than a family of antiderivatives.
How to avoid it
Every indefinite integral ends in $+C$ . (For a definite integral, drop $+C$ and substitute the limits instead.)

Integrate Hyperbolic Functions and Recognise Integrals of Functions

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Integrating sinh and cosh directly

2Recognising a standard inverse-form integral

3Completing the square to reach a tan⁻¹ form

4A coefficient on x² under a square root

5A definite integral giving cosh⁻¹

6Negative x² coefficient — complete the square then sin⁻¹

Integrals of sinh, cosh, sech²

Inverse-sine standard integral (MF19)

Inverse-sinh standard integral (MF19)

Inverse-cosh standard integral (MF19)

Inverse-tan standard integral (MF19)

Hyperbolic double-angle identities

Standard integral (MF19)

Completing the square

Inverse hyperbolic function

Linear substitution

sech x\text{sech}\,xsechx

Scaling factor

✕Writing ∫sinh⁡x dx=−cosh⁡x\int\sinh x\,dx = -\cosh x∫sinhxdx=−coshx (importing a trig minus sign)

✕Omitting the 1a\frac{1}{a}a1​ in ∫dxa2+x2=1atan⁡−1xa\int\frac{dx}{a^2+x^2} = \frac{1}{a}\tan^{-1}\frac{x}{a}∫a2+x2dx​=a1​tan−1ax​

✕Using a2a^2a2 where aaa is required (e.g. answer sin⁡−1x9\sin^{-1}\frac{x}{9}sin−19x​ for 9−x2\sqrt{9 - x^2}9−x2​)

✕Sign error when completing the square on a negative-x2x^2x2 quadratic

✕Folding the 1c\frac{1}{\sqrt c}c​1​ or 1c\frac{1}{c}c1​ scaling factor incorrectly into the standard form

✕Omitting +C+C+C on an indefinite integral

$\text{sech}\,x$

✕Writing $\int\sinh x\,dx = -\cosh x$ (importing a trig minus sign)

✕Omitting the $\frac{1}{a}$ in $\int\frac{dx}{a^2+x^2} = \frac{1}{a}\tan^{-1}\frac{x}{a}$

✕Using $a^2$ where $a$ is required (e.g. answer $\sin^{-1}\frac{x}{9}$ for $\sqrt{9 - x^2}$ )

✕Sign error when completing the square on a negative- $x^2$ quadratic

✕Folding the $\frac{1}{\sqrt c}$ or $\frac{1}{c}$ scaling factor incorrectly into the standard form

✕Omitting $+C$ on an indefinite integral