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Learning objectives
Core idea
Key formulae
Worked examples
Explore the graph
Common mistakes
Exam tip
Guided practice
Practice questions
Challenge
Summary
Quick quiz
Same topic, different board
Past paper questions

Learn7367Paper 2 · Compulsory contentFurther calculusInverse trig and hyperbolic functions in calculus

Further calculus · Subtopic 2 of 3

Calculus with Inverse Trig and Hyperbolic Functions

Differentiating $y=\arcsin x$ implicitly gives $\frac1{\sqrt{1-x^2}}$ ; similar results hold for $\arctan$ , $\operatorname{arsinh}$ , $\operatorname{arcosh}$ . Read backwards, these give the integrals of $\frac1{\sqrt{a^2-x^2}}$ , $\frac1{a^2+x^2}$ and $\frac1{\sqrt{x^2\pm a^2}}$ , which algebra alone can't do.

Cambridge 9231 Paper 2Edexcel YFM01 WFM03Edexcel 9FM0 Paper 1Edexcel 9FM0 Paper 2AQA 7367 Paper 1AQA 7367 Paper 2≈ 45 minRevision 1 · updated 2026-10-27

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Section 1 of 14

Learning objectives

Derive and use the derivatives of inverse trig and inverse hyperbolic functions.
Integrate $\frac1{a^2+x^2}$ , $\frac1{\sqrt{a^2-x^2}}$ and $\frac1{\sqrt{x^2\pm a^2}}$ .
Complete the square to reach a standard form.

Pause and think

What function has derivative $\frac1{1+x^2}$ ?

Section 2 of 14

Core idea

Start hereIntegrals with no algebraic answer

$\int\frac{dx}{1+x^2}$ looks harmless, yet no polynomial, power or logarithm differentiates to it. The answer is an angle: $\arctan x$ . Inverse trig and hyperbolic functions are exactly the missing antiderivatives for a whole family of integrals.

The big idea

Differentiate inverse functions implicitly. Reversed: $\int\frac{dx}{a^2+x^2}=\frac1a\arctan\frac xa$ , $\int\frac{dx}{\sqrt{a^2-x^2}}=\arcsin\frac xa$ , $\int\frac{dx}{\sqrt{x^2+a^2}}=\operatorname{arsinh}\frac xa$ , $\int\frac{dx}{\sqrt{x^2-a^2}}=\operatorname{arcosh}\frac xa$ .

1. Derivatives

\frac{d}{dx}\arcsin x

$y=\arcsin x\Rightarrow\sin y=x$
Rewrite.
$\cos y\frac{dy}{dx}=1$
Differentiate implicitly.
$\frac{dy}{dx}=\frac1{\cos y}=\frac1{\sqrt{1-x^2}}$
$\cos y=\sqrt{1-\sin^2y}$ (positive on $[-\frac\pi2,\frac\pi2]$ ).

\frac{d}{dx}\operatorname{arsinh}x

$\sinh y=x\Rightarrow\cosh y\frac{dy}{dx}=1$
Same method.
$\frac{dy}{dx}=\frac1{\sqrt{1+x^2}}$
$\cosh y=\sqrt{1+\sinh^2y}$ .

Likewise

\frac{d}{dx}\arctan x=\frac1{1+x^2}

and

\frac{d}{dx}\operatorname{arcosh}x=\frac1{\sqrt{x^2-1}}

Worked example $\arctan(2x)$

Differentiate.

Goal: $y'$ .
Strategy: Chain rule.

$\frac1{1+(2x)^2}\cdot2$
Outer derivative × inner derivative.
$=\frac2{1+4x^2}$
Simplify.

Answer

\frac2{1+4x^2}

✓ CheckAt

x=0

gradient

2

, matching

\arctan2x\approx2x

near

0

✓.

2. Integrals

Where the

\frac1a

comes from

$x=au,\ dx=a\,du$
Scale the variable.
$\int\frac{a\,du}{a^2+a^2u^2}=\frac1a\int\frac{du}{1+u^2}$
Factor out $a^2$ .
$=\frac1a\arctan\frac xa+c$
Standard result.

For the root forms the

a

cancels:

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

Worked example $\int_0^4\frac{dx}{\sqrt{x^2+9}}$

Evaluate exactly.

Goal: A logarithm.
Strategy: arsinh, then log form.

$\left[\operatorname{arsinh}\frac x3\right]_0^4=\operatorname{arsinh}\frac43$
$a=3$ .
$=\ln\left(\tfrac43+\sqrt{\tfrac{16}9+1}\right)$
Log form.
$=\ln\left(\tfrac43+\tfrac53\right)=\ln3$
Simplify.

Answer

\ln3

✓ Check

\sinh(\ln3)=\frac{3-\frac13}2=\frac43

✓.

Worked exampleCompleting the square

$\int\frac{dx}{x^2+4x+13}$ .

Goal: Antiderivative.
Strategy: $(x+2)^2+9$ .

$\int\frac{dx}{(x+2)^2+3^2}$
Complete the square.
$=\frac13\arctan\frac{x+2}3+c$
Standard form with $u=x+2$ .

Answer

\frac13\arctan\frac{x+2}3+c

✓ CheckDifferentiate:

\frac13\cdot\frac{1/3}{1+\frac{(x+2)^2}9}=\frac1{9+(x+2)^2}

✓.

Common misconceptionThe $\frac1a$ Only arctan carries

\frac1a

; arcsin, arsinh and arcosh don't.

Summary

Implicit differentiation for derivatives.
Match the integrand's form.
Complete the square when needed.

Principal value: The single output chosen for an inverse function.

HistoryLeibniz series. Integrating

\frac1{1+x^2}=1-x^2+x^4-\cdots

gives

\arctan1=\frac\pi4=1-\frac13+\frac15-\cdots

, found by Madhava in India around 1400 and later Leibniz.

Fun factπ from a fraction.

\int_0^1\frac{4}{1+x^2}dx=\pi

: one of the simplest integrals in mathematics produces

\pi

Section 3 of 14

Key formulae

Inverse trig

\frac{d}{dx}\arcsin x=\frac1{\sqrt{1-x^2}},\quad\frac{d}{dx}\arctan x=\frac1{1+x^2}

Inverse hyperbolic

\frac{d}{dx}\operatorname{arsinh}x=\frac1{\sqrt{x^2+1}},\quad\frac{d}{dx}\operatorname{arcosh}x=\frac1{\sqrt{x^2-1}}

Integrals

\int\frac{dx}{a^2+x^2}=\frac1a\arctan\frac xa,\ \ \int\frac{dx}{\sqrt{a^2-x^2}}=\arcsin\frac xa,\ \ \int\frac{dx}{\sqrt{x^2+a^2}}=\operatorname{arsinh}\frac xa

TrickPlus or minus decides. Under a root:

a^2-x^2

→ arcsin;

x^2+a^2

→ arsinh;

x^2-a^2

→ arcosh. No root,

a^2+x^2

→ arctan (with the

\frac1a

Section 4 of 14

Worked examples

Example 1 · A derivative0 / 2 steps

Differentiate

\arctan(2x)

Example 2 · Completing the square0 / 2 steps

Find

\int\frac{dx}{x^2+4x+13}

Section 5 of 14

Explore the graph

Area under 1/(1 + x²)

Loading the interactive graph…

Try this

The area under $\frac1{1+x^2}$ from $0$ to $b$ equals the height of $\arctan$ at $b$ .
As $b\to\infty$ , the area tends to $\frac\pi2$ .

$\int_0^b\frac{dx}{1+x^2}=\arctan b$ .

In the real worldElectric fields. The field of a long charged rod comes from

\int\frac{dx}{(x^2+a^2)^{3/2}}

-type integrals; related inverse-function integrals appear throughout physics.

Section 6 of 14

Common mistakes

Incorrect

\int\frac{dx}{4+x^2}=\arctan\frac x2

Correct

\frac12\arctan\frac x2

The factor

\frac1a

is needed for arctan.

Incorrect

\int\frac{dx}{\sqrt{4-x^2}}=\frac12\arcsin\frac x2

Correct

\arcsin\frac x2

\frac1a

factor for arcsin (the root absorbs it).

IncorrectForgetting the chain rule factor

Correct

\frac{d}{dx}\arctan(kx)=\frac k{1+k^2x^2}

Differentiate the inside.

∫ 1/(1+x²) dx with only polynomials and logs

arctan: I've been waiting for you.

Section 7 of 14

Exam tip

Exam tipUse the formula book. The integrals are in the formula booklet, but you must be able to adapt them: substitute

x=a\sinh u

if asked to "show".

Section 8 of 14

Guided practice

SA-FP-IC-G011 markL1 FoundationNon-calculatorDifferentiating

Find the gradient of

y=\arcsin x

x=0

Salam Academy Original Practice · Proposed mark scheme

Section 9 of 14

Practice questions

SA-FP-IC-0012 marksL2 StandardNon-calculatorDifferentiating

Find

\frac{d}{dx}\arctan(2x)

x=0

Salam Academy Original Practice · Proposed mark scheme

SA-FP-IC-0022 marksL2 StandardCalculatorDifferentiating

Find the gradient of

y=\operatorname{arcosh}x

x=2

(to 3 s.f.).

Salam Academy Original Practice · Proposed mark scheme

SA-FP-IC-0032 marksL2 StandardCalculatorIntegrating

Evaluate

\displaystyle\int_0^1\frac{dx}{1+x^2}

(to 4 s.f.).

Salam Academy Original Practice · Proposed mark scheme

SA-FP-IC-0043 marksL3 ExamCalculatorIntegrating

Evaluate

\displaystyle\int_0^{3/2}\frac{dx}{\sqrt{9-x^2}}

exactly (enter to 4 s.f.).

Salam Academy Original Practice · Proposed mark scheme

More questions on every skill in this subtopic are in the question bank under Practise.

Section 10 of 14

Challenge

SA-FP-IC-0054 marksL4 ChallengingCalculatorIntegrating

Show that

\displaystyle\int_0^4\frac{dx}{\sqrt{x^2+9}}=\ln k

and find

k

Salam Academy Original Practice · Proposed mark scheme

Section 11 of 14

Summary

Derive by implicit differentiation.
Root forms: arcsin, arsinh, arcosh.
$\frac1{a^2+x^2}$ : $\frac1a\arctan$ .
Complete the square first.

Section 12 of 14

Quick quiz

Q1 · SA-FP-IC-Q011 markL1 FoundationNon-calculatorIntegrating

\displaystyle\int\frac{dx}{\sqrt{x^2+a^2}}=

Salam Academy Original Practice · Proposed mark scheme

Section 13 of 14

Same topic, different board

Specification: 7367 A-level Further Mathematics
Where: Paper 1 and 2 · Further calculus
Expected: Integrals giving inverse trig and hyperbolic functions, including completing the square.
Calculator: Permitted on all papers.
Typical style: AQA mixes short multiple-choice questions with longer structured ones.
Open in: 7367 · Paper 1 7367 · Paper 2

Section 14 of 14

Past paper questions

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Salam Academy original content · Content status: reviewed and published. Syllabus mapping: AQA 7367, First teaching 2017 (official specification).