What is the fundamental concept of integration in Pure Mathematics 3 for Cambridge International A Levels?

Integration is a mathematical process used to find the area under a curve or the accumulation of quantities. In Pure Mathematics 3, it is often introduced as the reverse process of differentiation. The fundamental theorem of calculus links these two concepts, stating that integration can be used to find the original function given its derivative.

How do I apply integration techniques to solve problems in the Cambridge International A Levels curriculum?

In the Cambridge International A Levels, you will learn various integration techniques such as substitution, integration by parts, and partial fractions. These methods help solve complex integrals by simplifying them into more manageable forms. Practice is key to mastering these techniques, as they are essential for solving problems involving areas, volumes, and differential equations.

What are definite and indefinite integrals, and how are they different?

An indefinite integral represents a family of functions and includes a constant of integration, denoted as 'C'. It is the antiderivative of a function. A definite integral, on the other hand, calculates the exact area under a curve between two specific points, providing a numerical value. In Pure Mathematics 3, understanding both types is crucial for solving various mathematical problems.

How is integration used to find the area between curves in Pure Mathematics 3?

To find the area between two curves, you first determine the points of intersection. Then, you integrate the difference between the two functions over the interval defined by these points. This process involves setting up the integral of the upper function minus the lower function, which gives the area between the curves over the specified range.

What role does integration play in solving differential equations in the Cambridge International A Levels Mathematics syllabus?

Integration is a crucial tool for solving differential equations, which are equations involving derivatives. In Pure Mathematics 3, you will learn to use integration to find general solutions to first-order differential equations. This involves integrating the derivative to find the original function, often requiring techniques like separation of variables or integrating factors.

Why is "Wrong choice of $u$ in by-parts" a common mistake in Integration?

Why it happens: Not using LIATE. How to avoid it: Apply LIATE — u should be the one earlier in the LIATE order. Source: 9709 Examiner Reports 2022-2024

Why is "Forgetting to change limits in definite integral substitution" a common mistake in Integration?

Why it happens: Habit from indefinite. How to avoid it: For definite \int_a^b, change limits from x values to u values via u = g(x). Source: 9709 Examiner Reports 2022-2024

Cambridge International A Levels Mathematics (9709)

Integration

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Short Notes - Integration

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Detailed Study Notes

Detailed notes on Pure Mathematics 3 - Paper 3 for Cambridge International A Levels Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Integration P3 Study Notes — Cambridge International A Level Mathematics 9709 (2024-2026 syllabus)

Integration by parts. Substitution. Partial fractions. The advanced toolkit for the P3 integration questions.

At a glance

By parts: $\int u \, dv = uv - \int v \, du$ .
LIATE: priority for choosing $u$ .
Substitution: $u = g(x)$ , $du = g'(x) dx$ .
Partial fractions before integrating.
Change limits for definite substitution.

What you’ll learn

Mapped to the Cambridge International A Level 9709 syllabus (2024-2026).

P3.5.1 — Apply integration by parts.
P3.5.2 — Apply integration by substitution.
P3.5.3 — Integrate via partial fractions.

Integration by parts

$\int u \, dv = uv - \int v \, du$ .

Formula. $\int u \, dv = uv - \int v \, du$ . Reverse of product rule.

Method.

Choose $u$ and $dv$ from the integrand.
Compute $du$ by differentiating $u$ .
Compute $v$ by integrating $dv$ .
Apply formula.

Choosing $u$ . Use LIATE priority:

Logarithmic
Inverse trig
Algebraic (powers of $x$ )
Trigonometric
Exponential

The one EARLIER in LIATE becomes $u$ .

Example. $\int x \cos x \, dx$ .

LIATE: $x$ is Algebraic, $\cos x$ is Trig. $u = x$ , $dv = \cos x \, dx$ .
$du = dx$ , $v = \sin x$ .
$\int x \cos x \, dx = x \sin x - \int \sin x \, dx = x \sin x + \cos x + c$ .

Cambridge tip. Mark scheme often awards 1 mark for stating $u, dv, du, v$ .

For ∫ u dv, pick the function that appears higher in the LIATE list as u; the other becomes dv.

$\int u \, dv = uv - \int v \, du$ .
LIATE for choosing $u$ .
State $u, dv, du, v$ .

See the full worked example for integration →

Integration by substitution

$u = g(x)$ , rewrite integral.

Idea. Substitute $u = g(x)$ to simplify integrand. Reverse of chain rule.

Method.

Set $u = g(x)$ .
Compute $du = g'(x) \, dx$ .
Rewrite integrand entirely in $u$ (including $dx \to du$ ).
Integrate.
Back-substitute to $x$ (for indefinite).

For definite integrals. Change limits to $u$ values — don't back-substitute.

Recognition. Look for $g(x)$ together with its derivative $g'(x)$ .

Example. $\int 2x(x^2 + 1)^4 \, dx$ .

Notice: $2x$ is derivative of $x^2 + 1$ . Substitute $u = x^2 + 1$ .
$du = 2x \, dx$ .
$\int u^4 \, du = \frac{u^5}{5} + c = \frac{(x^2 + 1)^5}{5} + c$ .

Cambridge tip. Show substitution explicitly: 'Let $u = \ldots$ , then $du = \ldots$ '.

Spot $g'(x)$ alongside $g(x)$ .
Substitute, simplify, integrate.
Change limits for definite.

See the full worked example for integration →

Integration via partial fractions

Decompose first.

Method.

Decompose into partial fractions (P3 algebra).
Integrate each simpler fraction.

Standard integrals.

$\int \frac{1}{x - a} \, dx = \ln|x - a| + c$ .
$\int \frac{1}{(x - a)^2} \, dx = -\frac{1}{x - a} + c$ .
$\int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a}\tan^{-1}\frac{x}{a} + c$ (P3 if covered).

Example. $\int \frac{3x + 5}{(x+1)(x-2)} \, dx$ .

Partial fractions: $\frac{3x+5}{(x+1)(x-2)} = -\frac{2}{3(x+1)} + \frac{11}{3(x-2)}$ .
Integrate: $-\frac{2}{3}\ln|x+1| + \frac{11}{3}\ln|x-2| + c$ .

Cambridge tip. State partial fractions BEFORE integrating — mark scheme awards them separately.

Decompose first.
Integrate each fraction.
Use log integral repeatedly.

See the full worked example for integration →

Quick recap

By parts: LIATE for $u$ choice.
Substitution: $u = g(x)$ , simplify.
Partial fractions before integrating.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

By parts formula.
LIATE priority.
Substitution procedure.
Change limits for definite.

How it’s examined

P3 integration is heavily tested — usually 2 questions per paper. Most-tested: by parts (8 marks), substitution (8 marks), partial fractions (8 marks).

Sources: Cambridge International A Level Mathematics 9709 syllabus (2024-2026); 9709 Examiner Reports 2022-2024; 9709/32 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-11.

Take this whole topic with you

Step-by-step worked examples — Integration

Step-by-step solutions to past-paper-style questions on integration, written exactly the way a tutor would explain them at the board.

1Integration by parts (8 marks)
Extended• Adapted from 9709/32 May/Jun 2024• by parts
Question
Find $\int x \cos x \, dx$ . (8 marks)
Step-by-step solution
1. Step 1
  Choose $u, dv$ . Use 'LIATE' priority: $u$ logarithmic/Inverse/Algebraic preferred. $u = x$ , $dv = \cos x \, dx$ .
2. Step 2
  Compute $du, v$ . $du = dx$ , $v = \sin x$ .
3. Step 3
  Apply formula. $\int u \, dv = uv - \int v \, du$ .
  $\int x \cos x \, dx = x \sin x - \int \sin x \, dx$
4. Step 4
  Integrate.
  $= x \sin x + \cos x + c$
Answer
$x \sin x + \cos x + c$ .
2Integration by substitution (8 marks)
Extended• substitution
Question
Find $\int 2x(x^2 + 1)^4 \, dx$ using the substitution $u = x^2 + 1$ . (8 marks)
Step-by-step solution
1. Step 1
  Substitute. $u = x^2 + 1$ , $\frac{du}{dx} = 2x$ , so $du = 2x \, dx$ .
2. Step 2
  Rewrite integral entirely in $u$ .
  $\int 2x(x^2+1)^4 \, dx = \int u^4 \, du$
3. Step 3
  Integrate.
  $= \dfrac{u^5}{5} + c$
4. Step 4
  Back-substitute.
  $= \dfrac{(x^2+1)^5}{5} + c$
Answer
$\dfrac{(x^2+1)^5}{5} + c$ .
3Integration via partial fractions (8 marks)
Extended• partial fractions
Question
Find $\int \dfrac{3x + 5}{(x+1)(x-2)} \, dx$ . (8 marks)
Step-by-step solution
1. Step 1
  Partial fractions. $\frac{3x+5}{(x+1)(x-2)} = -\frac{2}{3(x+1)} + \frac{11}{3(x-2)}$ .
2. Step 2
  Integrate term-by-term.
  $\int = -\dfrac{2}{3}\ln|x+1| + \dfrac{11}{3}\ln|x-2| + c$
Answer
$-\dfrac{2}{3}\ln|x+1| + \dfrac{11}{3}\ln|x-2| + c$ .

Key Formulae — Integration

The formulae you need to memorise for integration on the Cambridge International A Level 9709 paper, with every variable defined in plain English and a note on when to use it.

Integration by parts
$\displaystyle\int u \, dv = uv - \int v \, du$
$u, v$
two parts of the integrand
When to use
Integrating a product. Choose $u$ via LIATE: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential — priority for $u$ .
Integration by substitution
$\displaystyle\int f(g(x)) g'(x) \, dx = \int f(u) \, du \text{ where } u = g(x)$
When to use
Recognise $g'(x)$ multiplied with something containing $g(x)$ . Set $u = g(x)$ .

Key Definitions and Keywords — Integration

Definitions to memorise and the exact keywords mark schemes credit for integration answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9709 sitting.

Integration by parts
Examiner keyword
Technique using $\int u \, dv = uv - \int v \, du$ . Reverse of product rule.
Integration by substitution
Examiner keyword
Technique using $u = g(x)$ to convert integral to easier form. Reverse of chain rule.
LIATE rule
Heuristic for choosing $u$ in by-parts: Logarithmic, Inverse trig, Algebraic, Trig, Exponential.

Common Mistakes and Misconceptions — Integration

The traps other students keep falling into on integration questions — taken from recent Cambridge International A Level 9709 examiner reports and mark schemes — and how to avoid them.

✕Wrong choice of $u$ in by-parts
9709 Examiner Reports 2022-2024
Why it happens
Not using LIATE.
How to avoid it
Apply LIATE — $u$ should be the one earlier in the LIATE order.
✕Forgetting to change limits in definite integral substitution
9709 Examiner Reports 2022-2024
Why it happens
Habit from indefinite.
How to avoid it
For definite $\int_a^b$ , change limits from $x$ values to $u$ values via $u = g(x)$ .

Integration — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Integration

Short Notes - Integration

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Integration by parts (8 marks)

2Integration by substitution (8 marks)

3Integration via partial fractions (8 marks)

Integration by parts

Integration by substitution

Integration by parts

Integration by substitution

LIATE rule

✕Wrong choice of uuu in by-parts

✕Forgetting to change limits in definite integral substitution

Integration — frequently asked questions

✕Wrong choice of $u$ in by-parts