# Integration

> CIE A-Level Mathematics · Pure Mathematics 2 & 3 (§2.5 / §3.5)
> Source: https://www.owlsprep.com/study/cie-9709-u2-integration/

This module covers the integration techniques for CIE A-Level Pure 2 & 3: the standard integrals shared by both papers, integration using trigonometric identities, and the Pure 3-only methods — the arctan result, recognising $\frac{kf'(x)}{f(x)}$, integration by substitution and by parts, and partial-fraction integration — with worked examples and common pitfalls.

**Prerequisites:** [Core differentiation rules from Pure 1](https://www.owlsprep.com/study/cie-9709-u1-differentiation/); [Partial fraction decomposition of rational functions](https://www.owlsprep.com/study/cie-9709-u2-partial-fractions/)

## Learning objectives

- Recall and apply the standard integrals shared by Pure 2 and 3, including $e^{ax+b}$, $\frac{1}{ax+b}$ and $\sin/\cos/\sec^2(ax+b)$
- Integrate $\sin^2 x$, $\cos^2 2x$ and $\tan^2 x$ by first applying a trigonometric identity
- Use the Pure 3 arctan result and recognise integrals of the form $\frac{kf'(x)}{f(x)} = k\ln|f(x)|$
- Use integration by a given substitution for definite and indefinite integrals (Pure 3)
- Apply integration by parts and integrate rational functions via partial fractions (Pure 3)
- Avoid common exam traps in integration questions

## How often is this tested?

Based on OwlsPrep's analysis of official CIE 9709 P2 past papers (2016–2025): **Integration** appears **232 times** in the last 10 years — **26.2%** of all P2 questions (232 of 455).

Most-tested forms: Integrals of 1/x, e^x, sin/cos/sec^2 etc (122), Integration by parts (40), Integration by substitution (definite + indefinite) (31), Numerical integration (trapezium rule) (24), Integration using partial fractions (24).

## Standard Integrals (Pure 2 & 3)

Across Pure 2 and Pure 3 you extend the standard integrals from Pure 1 to exponential, reciprocal and trigonometric forms of a linear argument $ax+b$. All of these are shared by Papers 2 and 3; Pure 3 then adds one further standard result, the inverse-tangent integral $\int\frac{1}{a^2+x^2}\,dx$.

**Indefinite Integral (Antiderivative)** — A function $F(x)$ whose derivative equals the original integrand $f(x)$, with $C$ the arbitrary constant of integration.

*Notation:* \int f(x) dx = F(x) + C

*Example:* The antiderivative of $2x$ is $x^2 + C$

$$\begin{aligned}
\int e^{ax+b} dx &= \frac{1}{a}e^{ax+b} + C \\
\int \frac{1}{ax+b} dx &= \frac{1}{a}\ln|ax+b| + C \\
\int \sin(ax+b) dx &= -\frac{1}{a}\cos(ax+b) + C \\
\int \cos(ax+b) dx &= \frac{1}{a}\sin(ax+b) + C \\
\int \sec^2 x dx &= \tan x + C \\
\int \frac{1}{a^2+x^2} dx &= \frac{1}{a}\arctan\left(\frac{x}{a}\right) + C
\end{aligned}$$

**Worked example:** Find the indefinite integral $\int \frac{3}{4 + x^2} dx$

1. Match the integrand to the standard form for $\int \frac{1}{a^2+x^2} dx$, with $a^2 = 4$ so $a=2$.
2. $$3 \int \frac{1}{2^2 + x^2} dx$$
3. Apply the standard integral formula, adding the constant of integration:
4. $$3 \times \frac{1}{2} \arctan\left(\frac{x}{2}\right) + C = \frac{3}{2}\arctan\left(\frac{x}{2}\right) + C$$

> **Exam tip:** Always write the constant of integration $C$ for indefinite integrals, you will lose a mark for omitting it.

## Integration Using Trigonometric Identities

Powers of sine, cosine and tangent have no direct standard integral. Before integrating, rewrite them with a double-angle or Pythagorean identity so that only $\sin(kx)$, $\cos(kx)$ or $\sec^2 x$ terms remain.

$$\begin{aligned}
\int \sin^2 x\,dx &= \int \frac{1-\cos 2x}{2}\,dx = \frac{x}{2} - \frac{1}{4}\sin 2x + C \\
\int \cos^2 2x\,dx &= \int \frac{1+\cos 4x}{2}\,dx = \frac{x}{2} + \frac{1}{8}\sin 4x + C \\
\int \tan^2 x\,dx &= \int (\sec^2 x - 1)\,dx = \tan x - x + C
\end{aligned}$$

> **tip**
>
> Learn the three rewrites: $\sin^2 x=\frac{1-\cos 2x}{2}$, $\cos^2 x=\frac{1+\cos 2x}{2}$ and $\tan^2 x=\sec^2 x-1$. For $\cos^2 2x$ double the angle: $\cos^2 2x=\frac{1+\cos 4x}{2}$.

**Worked example:** Find $\int \cos^2 2x\,dx$.

1. There is no direct integral for $\cos^2 2x$, so apply the double-angle identity $\cos^2\theta = \frac{1+\cos 2\theta}{2}$ with $\theta = 2x$:
2. $$\cos^2 2x = \frac{1 + \cos 4x}{2}$$
3. Integrate term by term, remembering to divide by the coefficient of $x$ inside the cosine:
4. $$\int \frac{1+\cos 4x}{2}\,dx = \frac{1}{2}\left(x + \frac{1}{4}\sin 4x\right) + C = \frac{x}{2} + \frac{1}{8}\sin 4x + C$$

> **Exam tip:** You cannot integrate $\sin^2 x$, $\cos^2 2x$ or $\tan^2 x$ directly — always convert with a trigonometric identity first. These integrals are examinable in both Paper 2 and Paper 3.

## Recognising kf'(x)/f(x) → ln|f(x)|

When the top of a fraction is (a constant multiple of) the derivative of the bottom, the integral is a natural logarithm. This is really integration by substitution with $u=f(x)$ done by inspection, and it saves time in the exam.

$$\begin{aligned}
\int \frac{f'(x)}{f(x)}\,dx &= \ln|f(x)| + C \\
\int \frac{kf'(x)}{f(x)}\,dx &= k\ln|f(x)| + C \\
\int \frac{2x}{x^2+3}\,dx &= \ln(x^2+3) + C \\
\int \tan x\,dx &= \int \frac{\sin x}{\cos x}\,dx = -\ln|\cos x| + C
\end{aligned}$$

> **tip**
>
> Spot the pattern first: is the numerator the derivative of the denominator, up to a constant? If the constant is off, fix it with a scalar. For $\int\tan x\,dx$ the numerator $\sin x$ is $-1$ times the derivative of $\cos x$, which produces the minus sign.

**Worked example:** Find $\int \tan x\,dx$.

1. Write $\tan x$ as a quotient so the pattern is visible:
2. $$\int \tan x\,dx = \int \frac{\sin x}{\cos x}\,dx$$
3. The derivative of the denominator $\cos x$ is $-\sin x$, so the numerator is $-1$ times $f'(x)$ with $f(x)=\cos x$:
4. $$= -\int \frac{-\sin x}{\cos x}\,dx = -\ln|\cos x| + C$$

> **Exam tip:** This is a Pure 3 technique. Before setting up a substitution, check whether the numerator is a constant multiple of the derivative of the denominator — if so, the integral is a logarithm.

## Integration by Substitution

Integration by substitution simplifies integrals of composite functions by changing the variable of integration. It is the reverse of the chain rule for differentiation.

$$\int f(x) dx = \int f(u(x)) \times \frac{dx}{du} du$$

> **tip**
>
> In 9709 the substitution is always **given** in the question — you are never asked to choose it yourself. To use the given $u$: differentiate it to replace $dx$, rewrite every part of the integrand (and, for a definite integral, the limits) in terms of $u$, then integrate.

**Worked example:** Use the substitution $u = x^2 + 1$ to find $\int \frac{x}{\sqrt{x^2 + 1}} dx$

1. Differentiate $u$ with respect to $x$ to find $dx$ in terms of $du$:
2. $$u = x^2 + 1 \implies \frac{du}{dx} = 2x \implies dx = \frac{du}{2x}$$
3. Substitute into the original integral, cancel common terms:
4. $$\int \frac{x}{\sqrt{u}} \times \frac{du}{2x} = \int \frac{1}{2\sqrt{u}} du$$
5. Integrate the simplified expression with respect to $u$:
6. $$\frac{1}{2} \int u^{-1/2} du = \frac{1}{2} \times 2 u^{1/2} + C = \sqrt{u} + C$$
7. Substitute back to the original variable $x$:
8. $$= \sqrt{x^2 + 1} + C$$

> **Exam tip:** For definite integrals, change the limits to the new variable immediately to avoid substitution errors later.

## Integration by Parts

Integration by parts is the reverse of the product rule for differentiation, used to integrate products of two different types of functions (e.g. $x \sin x$, $x e^x$).

$$\int u \frac{dv}{dx} dx = uv - \int v \frac{du}{dx} dx$$

> **LIATE Rule for choosing $u$**
>
> Choose $u$ as the function that comes first in this order: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. This always gives the correct choice.

**Worked example:** Find $\int x e^{2x} dx$

1. Apply the LIATE rule: $x$ is algebraic, $e^{2x}$ is exponential, so set $u = x$ and $\frac{dv}{dx} = e^{2x}$.
2. Calculate $\frac{du}{dx}$ and $v$:
3. $$\frac{du}{dx} = 1, \quad v = \int e^{2x} dx = \frac{1}{2}e^{2x}$$
4. Substitute into the integration by parts formula:
5. $$uv - \int v \frac{du}{dx} dx = \frac{1}{2}x e^{2x} - \int \frac{1}{2}e^{2x} dx$$
6. Integrate the remaining term and add the constant:
7. $$= \frac{1}{2}x e^{2x} - \frac{1}{4}e^{2x} + C = \frac{1}{4}e^{2x}(2x - 1) + C$$

> **Exam tip:** For repeated integration by parts, keep track of signs to avoid arithmetic errors.

## Integration via Partial Fractions

To integrate a rational function (a fraction of two polynomials), you first decompose it into partial fractions, then integrate each term separately using the standard result for $\frac{1}{ax+b}$.

> **warning**
>
> If the degree of the numerator is greater than or equal to the degree of the denominator (improper rational function), you must do polynomial division first before decomposing into partial fractions.

**Worked example:** Find $\int \frac{3x + 1}{(x-1)(x+2)} dx$

1. The degree of the numerator (1) is less than the degree of the denominator (2), so it is proper. Decompose into partial fractions:
2. $$\frac{3x + 1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$$
3. Solve for constants $A$ and $B$ by substituting roots of the denominator:
4. $$3x+1 = A(x+2) + B(x-1) \implies A = \frac{4}{3}, B = \frac{5}{3}$$
5. Rewrite the integral and integrate term by term:
6. $$\int \left(\frac{4}{3(x-1)} + \frac{5}{3(x+2)}\right) dx$$
7. $$= \frac{4}{3} \ln|x-1| + \frac{5}{3} \ln|x+2| + C$$

> **Exam tip:** Always check if the rational function is improper before decomposing into partial fractions.

## Common pitfalls

- **Wrong:** Omitting the constant of integration $C$ for indefinite integrals
  - Why it fails: Examiners always penalize missing constants, even if the rest of the answer is correct
  - Correct: Always write $+C$ at the end of every indefinite integral
- **Wrong:** Forgetting to change limits of integration for substitution with definite integrals
  - Why it fails: This leads to common arithmetic errors when substituting back to the original variable
  - Correct: Calculate new limits for the substituted variable immediately after choosing the substitution
- **Wrong:** Choosing the wrong $u$ in integration by parts, leading to a more complex integral
  - Why it fails: Picking $u$ as the exponential/trigonometric function instead of algebraic leads to unnecessary complexity
  - Correct: Always use the LIATE mnemonic to select the correct $u$
- **Wrong:** Forgetting to divide by the coefficient of $x$ for composite functions, e.g. $\int e^{2x} dx = e^{2x} + C$
  - Why it fails: You forget to apply the reverse chain rule for linear inner functions
  - Correct: Always divide by the derivative of the linear term, so $\int e^{2x} dx = \frac{1}{2}e^{2x} + C$
- **Wrong:** Skipping polynomial division for improper rational functions
  - Why it fails: You cannot decompose an improper rational function into partial fractions correctly, leading to an incorrect result
  - Correct: Always check the degrees of numerator and denominator, divide first if the fraction is improper

## Cheatsheet

| Integral Type | Formula / Rule | Key Notes |
| --- | --- | --- |
| Composite $f(ax+b)$ | $\frac{1}{a}F(ax+b) + C$ | Divide by $a$, add $C$ |
| Integration by substitution | $\int f(x) dx = \int f(u(x)) \frac{dx}{du} du$ | Change limits for definite integrals |
| Integration by parts | $\int u \frac{dv}{dx} dx = uv - \int v \frac{du}{dx} dx$ | Choose $u$ by LIATE rule |
| $\int \frac{1}{ax+b} dx$ | $\frac{1}{a} \ln\|ax+b\| + C$ | Keep absolute value |
| $\int \frac{1}{a^2+x^2} dx$ | $\frac{1}{a} \arctan(\frac{x}{a}) + C$ | Recall for P3 |
| Improper rational function | Divide first, then decompose to partial fractions | Don't skip the division step |
| $\int \sin^2 x\,dx$ | $\frac{x}{2} - \frac{1}{4}\sin 2x + C$ | Use $\sin^2 x=\frac{1-\cos 2x}{2}$ |
| $\int \cos^2 2x\,dx$ | $\frac{x}{2} + \frac{1}{8}\sin 4x + C$ | Use $\cos^2 2x=\frac{1+\cos 4x}{2}$ |
| $\int \tan^2 x\,dx$ | $\tan x - x + C$ | Use $\tan^2 x=\sec^2 x-1$ |
| $\int \frac{kf'(x)}{f(x)}\,dx$ | $k\ln\|f(x)\| + C$ | Top = derivative of bottom (P3) |

## What's next

Mastering these integration techniques underpins almost all remaining calculus in CIE Pure 2 & 3, including solving differential equations and further applications. You will also combine them with the volume-of-revolution method you already met in Pure 1 (§1.8) — for example integrating a partial-fraction or trigonometric-identity expression to find a volume. Integration is weighted heavily in Papers 2 and 3, and questions frequently combine several techniques (e.g. substitution after partial fractions, or repeated integration by parts) in one problem. Building fluency now will help you tackle harder, higher-mark questions efficiently.

- [Numerical methods](https://www.owlsprep.com/study/cie-9709-u2-numerical-methods/)
- [3D Vector Geometry (Lines and Planes)](https://www.owlsprep.com/study/cie-9709-u2-vectors/)
- [Differential equations](https://www.owlsprep.com/study/cie-9709-u2-differential-equations/)

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