Integration by Substitution

Builds on: aqa-alevel-maths-pure-2 / integration-techniques — this lesson is the A2-level deep dive.

This lesson covers integration by substitution (also called u-substitution), a powerful technique for evaluating integrals that cannot be found by simple inspection. It is the reverse of the chain rule for differentiation and is a key topic in A-Level Mathematics.

Spec Mapping — AQA 7357 Section H Integration. This lesson covers the integration-by-substitution content of Section H, at the depth required for A2-level synoptic questions and modelling-style problems. Refer to the official AQA specification document for exact wording.

The Idea

If we can write an integral in the form:

$\int f(g(x)) \times g'(x)\,dx$

then by substituting u = g(x), du = g'(x) dx, the integral simplifies to:

int f(u) du

which is often much easier to integrate.

Method for Indefinite Integrals

Choose a substitution u = g(x).
Find du/dx and rearrange to express dx in terms of du.
Replace all x terms in the integral with u terms.
Integrate with respect to u.
Substitute back to express the answer in terms of x.

Example 1: Basic Substitution

Find ∫ 2x(x² + 3)⁴ dx.

Let u = x² + 3. Then du/dx = 2x, so du = 2x dx.

\int 2x(x^{2} + 3)^{4}\, dx = \int u^{4}\, du = \frac{u^{5}}{5} + C = \frac{(x^{2} + 3)^{5}}{5} + C

Example 2: Adjusting the Constant

Find ∫ x(x² + 1)³ dx.

Let u = x² + 1. Then du = 2x dx, so x dx = du/2.

\int x(x^{2} + 1)^{3}\, dx = \int u^{3} \times \frac{du}{2} = \frac{1}{2} \int u^{3}\, du = \frac{1}{2} \times \frac{u^{4}}{4} + C = \frac{(x^{2} + 1)^{4}}{8} + C

Example 3: Linear Substitution

Find ∫ (3x + 2)⁵ dx.

Let u = 3x + 2. Then du = 3 dx, so dx = du/3.

\int (3x + 2)^{5}\, dx = \int u^{5} \times \frac{du}{3} = \frac{1}{3} \times \frac{u^{6}}{6} + C = \frac{(3x + 2)^{6}}{18} + C

Method for Definite Integrals

When using substitution with definite integrals, you must change the limits to match the new variable u:

When x = a, u = g(a).
When x = b, u = g(b).

This means you do not need to substitute back to x at the end.

Example 4

Evaluate $\int_0^1 x\sqrt{1 - x^2}\, dx$ .

Let $u = 1 - x^2$ . Then $du = -2x\, dx$ , so $x\, dx = -\dfrac{du}{2}$ .

Change limits: When x = 0, u = 1. When x = 1, u = 0.

\begin{aligned} \int_{0}^{1} x \sqrt{1 - x^{2}}\,dx &= \int_{1}^{0} \sqrt{u} \times \Bigl(-\tfrac{du}{2}\Bigr) \\ &= \tfrac{1}{2} \int_{0}^{1} \sqrt{u}\,du \quad \text{(reversing limits removes the negative sign)} \\ &= \tfrac{1}{2} \Bigl[\tfrac{u^{3/2}}{3/2}\Bigr]_{0}^{1} \\ &= \tfrac{1}{2} \times \tfrac{2}{3} \bigl[u^{3/2}\bigr]_{0}^{1} \\ &= \tfrac{1}{3} [1 - 0] \\ &= \tfrac{1}{3} \end{aligned}

Recognising f'(x)/f(x) Patterns

A very important special case: if the integrand has the form f'(x)/f(x), then:

\int \frac{f'(x)}{f(x)}\, dx = \ln|f(x)| + C

This is because substituting u = f(x) gives du = f'(x) dx, and ∫ du/u = ln|u| + C.

Example 5

Find ∫ 2x/(x² + 5) dx.

The numerator 2x is the derivative of the denominator x² + 5. Therefore:

\int \frac{2x}{x^{2} + 5}\, dx = \ln|x^{2} + 5| + C = \ln(x^{2} + 5) + C

(We can drop the absolute value since x² + 5 > 0 for all x.)

Example 6

Find ∫ cos x/sin x dx = ∫ cot x dx.

Here f(x) = sin x, f'(x) = cos x:

\int \frac{\cos x}{\sin x}\, dx = \ln|\sin x| + C

Recognising f'(x) × [f(x)]ⁿ Patterns

If the integrand has the form f'(x) × [f(x)]ⁿ, then:

\int f'(x) \times [f(x)]^{n}\, dx = \frac{[f(x)]^{n+1}}{n + 1} + C \qquad (n \neq -1)

Example 7

Find ∫ cos x × sin³x dx.

Here f(x) = sin x, f'(x) = cos x, n = 3:

\int \cos x \times \sin^{3} x\, dx = \frac{\sin^{4} x}{4} + C

Alternatively, let u = sin x, du = cos x dx:

\int u^{3}\, du = \frac{u^{4}}{4} + C = \frac{\sin^{4} x}{4} + C \quad \checkmark

Example 8: More Complex Substitution

Find $\int \dfrac{x}{\sqrt{3x + 1}}\, dx$ .

Let $u = 3x + 1$ . Then $x = \dfrac{u - 1}{3}$ and $dx = \dfrac{du}{3}$ .

\begin{aligned} \int \frac{x}{\sqrt{3x + 1}}\, dx &= \int \frac{u - 1}{3} \times \frac{1}{\sqrt{u}} \times \frac{du}{3} \\ &= \frac{1}{9} \int \frac{u - 1}{\sqrt{u}}\, du \\ &= \frac{1}{9} \int \left( u^{1/2} - u^{-1/2} \right) du \\ &= \frac{1}{9} \left[ \frac{2u^{3/2}}{3} - 2u^{1/2} \right] + C \\ &= \frac{2}{27}(3x + 1)^{3/2} - \frac{2}{9}(3x + 1)^{1/2} + C \end{aligned}

We can factorise:

\begin{aligned} &= \frac{2}{27}\sqrt{3x + 1}\, [3x + 1 - 3] + C \\ &= \frac{2}{27}\sqrt{3x + 1}\,(3x - 2) + C \end{aligned}

Standard Substitutions Worth Knowing

Integrand contains	Try the substitution
$(ax + b)^n$	$u = ax + b$
$f(x^2) \times x$	$u = x^2$
$f(\sin x) \times \cos x$	$u = \sin x$
$f(\cos x) \times \sin x$	$u = \cos x$
$f(e^x) \times e^x$	$u = e^x$
$f(\ln x) \times \dfrac{1}{x}$	$u = \ln x$
$\sqrt{a^2 - x^2}$	$x = a\sin\theta$ (trigonometric substitution)

Example 9: Definite Integral with eˣ

Evaluate ∫₀¹ eˣ/(eˣ + 1) dx.

Let u = eˣ + 1. Then du = eˣ dx.

Limits: When x = 0, u = 2. When x = 1, u = e + 1.

\begin{aligned} \int_{0}^{1} \frac{e^{x}}{e^{x} + 1}\, dx &= \int_{2}^{e+1} \frac{1}{u}\, du \\ &= \left[ \ln u \right]_{2}^{e+1} \\ &= \ln(e + 1) - \ln 2 \\ &= \ln\left[ \frac{e + 1}{2} \right] \end{aligned}

Common Mistakes

Forgetting to change dx: When substituting u = g(x), you must replace dx with du/g'(x).
Not changing limits in definite integrals.
Failing to substitute back in indefinite integrals — the final answer must be in terms of x.
Choosing a poor substitution — if the substitution does not simplify the integral, try a different one.

Synoptic Links

This topic connects to:

Integration techniques (aqa-alevel-maths-pure-2 / integration-techniques) — substitution is one of the two standard techniques alongside parts.
Chain rule (aqa-alevel-maths-pure-1 / differentiation) — substitution is the inverse operation, so confident chain-rule recognition unlocks the right $u$.
Areas under curves (aqa-alevel-maths-calculus-applications / areas-under-curves) — many definite-area integrals require substitution before they can be evaluated.

Summary

Integration by substitution reverses the chain rule.
Choose u to simplify the integrand. Common choices: the "inside function", the denominator, or an expression under a root.
For definite integrals, change the limits when you change variable.
Recognise the patterns f'(x)/f(x) → ln|f(x)| and f'(x)[f(x)]ⁿ → [f(x)]^(n+1)/(n+1).
Always check your answer by differentiating.

Exam Tip: In AQA papers, substitution questions often tell you what substitution to use (e.g. "Use the substitution u = x² + 1"). When they do, follow their instruction. When they do not, look for a "function and its derivative" pattern. Always show the substitution for du/dx, the replacement of dx, the change of limits (for definite integrals), and the final integration. Each of these stages earns method marks.

A-Level Deep Dive: Integration by Substitution

Spec mapping

AQA 7357 specification, Paper 2 — Pure Mathematics, Section H (Integration), Year 2 content covers using a given substitution (refer to the official specification document for exact wording). Substitution is the structural reverse of the chain rule and the most-used integration technique beyond direct anti-differentiation. The AQA formula booklet lists standard integrals ( $\int \frac{1}{x}\,dx = \ln|x| + C$ , $\int \sec^2 x\,dx = \tan x + C$ , etc.) but does not list substitution templates — students must recognise the pattern $\int f(g(x))\,g'(x)\,dx$ themselves. Substitution interacts with Section G (Differentiation, chain rule), Section H (integration by parts, partial fractions) and feeds Section P (Mechanics, work-energy via $\int F\,dx$ ). On Paper 2 it is examined both as a directed technique ("using the substitution $u = \ldots$ ") and as an open-ended choice on harder integrals.

Specimen Question (Modelled on the AQA Paper Format)

Note: this question is constructed to model AQA Paper 1/2/3 style; it is not a reproduction of any published past paper.

Worked example with full mark scheme

Question (8 marks):

(a) Use the substitution $u = x^2 + 1$ to find $\int x(x^2 + 1)^4\,dx$ . (5)

(b) Hence, or otherwise, evaluate $\int_0^{\pi/2} \sin^3 x \cos x\,dx$ using the substitution $u = \sin x$ . (3)

Solution with mark scheme:

(a) Step 1 — set up the substitution.

Let $u = x^2 + 1$ . Differentiating: $\dfrac{du}{dx} = 2x$ , so $du = 2x\,dx$ , equivalently $x\,dx = \dfrac{1}{2}\,du$ .

M1 — correct expression for $du$ in terms of $dx$ (or vice versa). The frequent error is writing $du = 2x$ without the $dx$ , which conflates the derivative with the differential and breaks the substitution algebra.

Step 2 — rewrite the integral entirely in $u$ .

$\int x(x^2 + 1)^4\,dx = \int (x^2 + 1)^4 \cdot x\,dx = \int u^4 \cdot \dfrac{1}{2}\,du = \dfrac{1}{2}\int u^4\,du$

M1 — every $x$ replaced. A common loss of marks: leaving a stray $x$ in the integrand (e.g. writing $\int xu^4\,du$ ). The integral must be entirely in the new variable before integrating.

A1 — correct integrand $\dfrac{1}{2}u^4$ .

Step 3 — integrate with respect to $u$ .

$\dfrac{1}{2}\int u^4\,du = \dfrac{1}{2} \cdot \dfrac{u^5}{5} + C = \dfrac{u^5}{10} + C$

M1 — correct application of the power rule $\int u^n\,du = \dfrac{u^{n+1}}{n+1} + C$ .

Step 4 — back-substitute.

$\dfrac{(x^2 + 1)^5}{10} + C$

A1 — final answer in terms of $x$ , with constant of integration. Forgetting to back-substitute (leaving the answer in $u$ ) loses the final A1 — the question asked for the integral with respect to $x$ , so the answer must reference $x$ .

(b) Step 1 — set up substitution and change limits.

Let $u = \sin x$ , so $\dfrac{du}{dx} = \cos x$ and $du = \cos x\,dx$ .

When $x = 0$ , $u = \sin 0 = 0$ . When $x = \dfrac{\pi}{2}$ , $u = \sin\left(\dfrac{\pi}{2}\right) = 1$ .

M1 — correct $du$ and correctly transformed limits. For a definite integral, changing the limits is mandatory if you do not intend to back-substitute; failing to change limits and then evaluating from $0$ to $\pi/2$ in the $u$ integral is dimensionally wrong and scores zero on this M1.

Step 2 — rewrite and evaluate.

$\int_0^{\pi/2} \sin^3 x \cos x\,dx = \int_0^1 u^3\,du = \left[\dfrac{u^4}{4}\right]_0^1 = \dfrac{1}{4} - 0 = \dfrac{1}{4}$

M1 — correct anti-derivative. A1 — final exact value $\dfrac{1}{4}$ .

Total: 8 marks (M5 A3, split as shown).

Specimen question modelled on the AQA 7357 Paper 2 format

Question (6 marks): Use the substitution $u = 1 + \sqrt{x}$ to find $\displaystyle\int \dfrac{1}{(1 + \sqrt{x})\sqrt{x}}\,dx$ , simplifying your answer.

Mark scheme decomposition by AO:

M1 (AO1.1a) — differentiating: $\dfrac{du}{dx} = \dfrac{1}{2\sqrt{x}}$ , hence $du = \dfrac{1}{2\sqrt{x}}\,dx$ , equivalently $\dfrac{1}{\sqrt{x}}\,dx = 2\,du$ .
M1 (AO1.1b) — recognising the integrand as $\dfrac{1}{u} \cdot \dfrac{1}{\sqrt{x}}\,dx$ so that the $\dfrac{1}{\sqrt{x}}\,dx$ cluster collapses cleanly to $2\,du$ .
A1 (AO1.1b) — transformed integral $\displaystyle\int \dfrac{2}{u}\,du$ .
M1 (AO1.1b) — integrating: $2\ln|u| + C$ .
A1 (AO1.1b) — back-substituting to give $2\ln|1 + \sqrt{x}| + C$ .
A1 (AO2.5) — observing that $1 + \sqrt{x} > 0$ for $x \geq 0$ so the modulus is redundant: $2\ln(1 + \sqrt{x}) + C$ .

Total: 6 marks split AO1 = 5, AO2 = 1. The AO2 mark rewards the domain observation that justifies dropping the modulus — a small but characteristic A* presentation move.

Synoptic links

Connects to:

Section G — Chain rule (reverse direction): the substitution rule is the chain rule played backwards. If $F'(u) = f(u)$ and $u = g(x)$ , then $\dfrac{d}{dx}F(g(x)) = f(g(x))\,g'(x)$ by the chain rule, so $\int f(g(x))\,g'(x)\,dx = F(g(x)) + C$ . Recognising the chain-rule "fingerprint" $f(g(x)) \cdot g'(x)$ in an integrand is the trigger for substitution.
Section H — Integration by parts: when substitution fails (no clean $g'(x)$ factor present), parts $\int u\,dv = uv - \int v\,du$ becomes the next tool. Some integrals — $\int x e^{x^2}\,dx$ , for instance — yield instantly to substitution; others — $\int x e^x\,dx$ — demand parts. Diagnosing which is an A* skill.
Trigonometric substitution $x = \sin\theta$ : integrals containing $\sqrt{1 - x^2}$ are tackled by $x = \sin\theta$ , exploiting $1 - \sin^2\theta = \cos^2\theta$ . The AQA 7357 specification includes this technique within Section H. The substitution is "inverse" — $x$ is written as a function of $\theta$ rather than $\theta$ as a function of $x$ .
Partial fractions (Section H): $\displaystyle\int \dfrac{1}{(x-1)(x+2)}\,dx$ resists direct substitution but yields after partial-fraction decomposition. Partial fractions and substitution are complementary: PF prepares the integrand, substitution then handles each term.

Integration by Substitution

Integration by Substitution

The Idea

Method for Indefinite Integrals

Example 1: Basic Substitution

Example 2: Adjusting the Constant

Example 3: Linear Substitution

Method for Definite Integrals

Example 4

Recognising f'(x)/f(x) Patterns

Example 5

Example 6

Recognising f'(x) × [f(x)]ⁿ Patterns

Example 7

Example 8: More Complex Substitution

Standard Substitutions Worth Knowing

Example 9: Definite Integral with eˣ

Common Mistakes

Synoptic Links

Summary

A-Level Deep Dive: Integration by Substitution

Spec mapping

Specimen Question (Modelled on the AQA Paper Format)

Worked example with full mark scheme

Specimen question modelled on the AQA 7357 Paper 2 format

Synoptic links

More in Mathematics