Integration by Substitution

This lesson covers integration by substitution — a powerful technique for integrating composite functions — as required by the Edexcel A-Level Mathematics specification (9MA0). You need to be able to choose an appropriate substitution, change variables (and limits for definite integrals), and recognise patterns that suggest substitution.

---

Why Substitution?

Many integrals cannot be evaluated directly using the basic rules. For example:

∫ 2x(x² + 1)⁵ dx

You cannot expand (x² + 1)⁵ easily. Instead, we use substitution to transform the integral into something simpler.

The method is the reverse of the chain rule for differentiation.

---

The Method

Step-by-Step Process

1. Choose a substitution u = g(x) (usually the "inner function" of a composite).
2. Find du/dx and rearrange to express dx in terms of du.
3. Replace all x-terms in the integral with expressions involving u.
4. Integrate with respect to u.
5. Substitute back to express the answer in terms of x (for indefinite integrals).

---

Worked Examples — Indefinite Integrals

Example 1: ∫ 2x(x² + 1)⁵ dx

Step 1: Let u = x² + 1

Step 2: du/dx = 2x, so du = 2x dx

Step 3: The integral becomes ∫ u⁵ du

Step 4: Integrate: u⁶/6 + c

Step 5: Substitute back: (x² + 1)⁶ / 6 + c

Example 2: ∫ 3x²√(x³ + 4) dx

Step 1: Let u = x³ + 4

Step 2: du/dx = 3x², so du = 3x² dx

Step 3: The integral becomes ∫ √u du = ∫ u^(1/2) du

Step 4: Integrate: (2/3)u^(3/2) + c

Step 5: Substitute back: (2/3)(x³ + 4)^(3/2) + c

Example 3: ∫ cos(3x + 1) dx

Step 1: Let u = 3x + 1

Step 2: du/dx = 3, so dx = du/3

Step 3: The integral becomes ∫ cos(u) × (1/3) du = (1/3) ∫ cos u du

Step 4: Integrate: (1/3) sin u + c

Step 5: Substitute back: (1/3) sin(3x + 1) + c

Example 4: ∫ xe^(x²) dx

Step 1: Let u = x²

Step 2: du/dx = 2x, so x dx = du/2

Step 3: The integral becomes ∫ e^u × (1/2) du = (1/2) ∫ e^u du

Step 4: Integrate: (1/2)e^u + c

Step 5: Substitute back: (1/2)e^(x²) + c

---

Definite Integrals — Changing the Limits

For definite integrals, you have two choices:

Method A (recommended): Change the limits when you substitute, then evaluate directly in terms of u — no need to substitute back.

Method B: Integrate in terms of u, substitute back to x, then use the original limits.

Example: Evaluate ∫ from 0 to 2 of x(x² + 1)³ dx

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

Step 2: Change limits:

• When x = 0: u = 0² + 1 = 1
• When x = 2: u = 2² + 1 = 5

Step 3: The integral becomes: ∫ from 1 to 5 of u³ × (1/2) du = (1/2)[u⁴/4] from 1 to 5

Step 4: Evaluate: (1/2)(625/4 - 1/4) = (1/2)(624/4) = (1/2)(156) = 78

Exam Tip: If you change the limits, you must NOT substitute back to x. The new limits are u-values, not x-values.

---

Choosing the Right Substitution

General Guidelines

Pattern in Integrand	Suggested Substitution
f(ax + b)	u = ax + b
f(g(x)) × g'(x)	u = g(x)
xⁿ × f(xⁿ⁺¹)	u = xⁿ⁺¹
√(ax + b)	u = ax + b
Expression inside brackets raised to a power	u = expression inside brackets

The Key Insight

Look for a function and its derivative appearing together. If you see something that looks like f(g(x)) × g'(x), the substitution u = g(x) will work because du = g'(x) dx.

---

More Complex Examples

Example: ∫ x/√(2x + 3) dx with u = 2x + 3

Step 1: u = 2x + 3, so x = (u - 3)/2 and du = 2 dx, giving dx = du/2.

Step 2: Substitute: ∫ [(u - 3)/2] × u^(-1/2) × (1/2) du = (1/4) ∫ (u - 3)u^(-1/2) du

Step 3: Expand: (1/4) ∫ (u^(1/2) - 3u^(-1/2)) du

Step 4: Integrate: (1/4)[(2/3)u^(3/2) - 3(2)u^(1/2)] + c = (1/4)[(2/3)u^(3/2) - 6u^(1/2)] + c

Step 5: Substitute back: (1/6)(2x + 3)^(3/2) - (3/2)(2x + 3)^(1/2) + c

Example: ∫ tan x dx

Rewrite: ∫ (sin x / cos x) dx

Let u = cos x, then du = -sin x dx, so sin x dx = -du.

∫ (-1/u) du = -ln|u| + c = -ln|cos x| + c = ln|sec x| + c

---

Recognising "Reverse Chain Rule" Patterns

Many substitution integrals can be done by inspection (without writing out the full substitution) once you recognise the pattern:

∫ f'(g(x)) × g'(x) dx = f(g(x)) + c

Example: ∫ 2x × cos(x²) dx

Recognise: the derivative of x² is 2x, and cos is being applied to x². By the reverse chain rule:

= sin(x²) + c

Example: ∫ 3x² / (x³ + 5) dx

Recognise: the derivative of x³ + 5 is 3x², and we have 1/(x³ + 5). This is the pattern for ln:

= ln|x³ + 5| + c

Exam Tip: The "reverse chain rule" approach saves time in exams. The pattern ∫ f'(x)/f(x) dx = ln|f(x)| + c is especially common.

---

Common Mistakes

1. Forgetting to change limits — If you change variable in a definite integral, change the limits too.
2. Not replacing ALL x-terms — Every occurrence of x must be replaced with an expression in u.
3. Leaving the answer in terms of u — For indefinite integrals, always substitute back to x.
4. Wrong sign on du — Be careful when du/dx is negative (e.g., u = cos x gives du = -sin x dx).

---

Summary

• Integration by substitution is the reverse of the chain rule.
• Choose u to be the "inner function" or the most complicated part.
• Replace everything (including dx) with expressions in u.
• For definite integrals, change the limits to u-values.
• Recognise the reverse chain rule pattern: ∫ f'(g(x)) × g'(x) dx = f(g(x)) + c.
• Common patterns: ∫ f'(x)/f(x) dx = ln|f(x)| + c.

A-Level Deep Dive: Integration by Substitution

Spec mapping

Edexcel 9MA0-02 specification section 10 — Integration, sub-strand 10.5 covers using the substitution $u = g(x)$u=g(x), including changing limits for definite integrals (refer to the official specification document for exact wording). Substitution is the Year 2 Pure technique that reverses the chain rule of differentiation, and it sits at the centre of Paper 2 — almost every Year 2 integration question that is not a standard form requires either substitution, integration by parts, or partial fractions. Substitution also feeds into section 8 (Differential equations) where separable ODEs use the same algebraic move, and into 9MA0-03 Mechanics, where work-done integrals $W = \int F(x)\,dx$W=∫F(x)dx frequently require substitution when $F$F is composite.

---

Worked example with full mark scheme

Question (8 marks): Find $\displaystyle\int_0^{\pi/2} \sin^3 x \cos x\,dx$∫0π/2​sin3xcosxdx, showing all steps, including the change of limits.

Solution with mark scheme:

Step 1 — identify the substitution. The integrand contains $\sin^3 x$sin3x and a factor of $\cos x$cosx, which is the derivative of $\sin x$sinx. This is the chain-rule signature: choose $u = \sin x$u=sinx.

M1 — correct identification of $u = g(x)$u=g(x) such that $g'(x)$g′(x) appears (up to a constant) as a factor.

Step 2 — compute $du$du in terms of $dx$dx.

$u = \sin x \implies \dfrac{du}{dx} = \cos x \implies du = \cos x\,dx$u=sinx⟹dxdu​=cosx⟹du=cosxdx

M1 — correct differentiation and conversion of the differential.

Step 3 — change the limits.

When $x = 0$x=0: $u = \sin 0 = 0$u=sin0=0. When $x = \pi/2$x=π/2: $u = \sin(\pi/2) = 1$u=sin(π/2)=1.

B1 — both new limits stated correctly. Examiners explicitly check that the limits are converted; leaving the original $x$x-limits with the new $u$u-integrand is an automatic A-mark loss.

Step 4 — rewrite the integral entirely in $u$u.

$\int_0^{\pi/2} \sin^3 x \cos x\,dx = \int_0^{1} u^3\,du$∫0π/2​sin3xcosxdx=∫01​u3du

A1 — integral correctly expressed in $u$u, with the $\cos x\,dx$cosxdx collapsed to $du$du and $\sin^3 x$sin3x rewritten as $u^3$u3.

Step 5 — integrate.

$\int_0^{1} u^3\,du = \left[\dfrac{u^4}{4}\right]_0^{1}$∫01​u3du=[4u4​]01​

M1 — correct application of the power rule in $u$u.

Step 6 — evaluate.

$\left[\dfrac{u^4}{4}\right]_0^{1} = \dfrac{1}{4} - 0 = \dfrac{1}{4}$[4u4​]01​=41​−0=41​

A1 — correct numerical evaluation.

Step 7 — present in exact form. The answer $\tfrac{1}{4}$41​ is already exact and rational; no further simplification is needed.

A1 — final exact answer $\dfrac{1}{4}$41​.

Total: 8 marks (M3 A3 B1, plus a final A1 for exact presentation). The most common loss-pattern at this level is mixing $x$x-limits with a $u$u-integrand — a single line of "muddled limits" can cost up to three A-marks.

---

Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks): Use the substitution $u = x^2 + 1$u=x2+1 to find $\displaystyle\int x(x^2 + 1)^4\,dx$∫x(x2+1)4dx, giving your answer in terms of $x$x.

Mark scheme decomposition by AO:

• M1 (AO1.1a) — differentiating: $\dfrac{du}{dx} = 2x$dxdu​=2x, so $x\,dx = \tfrac{1}{2}\,du$xdx=21​du.
• A1 (AO1.1b) — rewriting the integral as $\displaystyle\int u^4 \cdot \tfrac{1}{2}\,du = \tfrac{1}{2}\int u^4\,du$∫u4⋅21​du=21​∫u4du.
• M1 (AO1.1b) — applying the power rule: $\tfrac{1}{2} \cdot \tfrac{u^5}{5} = \tfrac{u^5}{10}$21​⋅5u5​=10u5​.
• A1 (AO1.1b) — back-substitution: $\dfrac{(x^2 + 1)^5}{10}$10(x2+1)5​.
• B1 (AO2.5) — adding the constant of integration $+ C$+C.
• A1 (AO2.5) — fully simplified final answer $\dfrac{(x^2 + 1)^5}{10} + C$10(x2+1)5​+C, expressed in $x$x as the question demanded.

Total: 6 marks split AO1 = 4, AO2 = 2. The AO2 marks are reserved for the back-substitution (returning to the original variable) and the constant of integration — both presentation-discipline marks rather than computational ones.

---

Synoptic links

Connects to:

1. Section 9 — Differentiation (chain rule reverse): substitution is the chain rule run backwards. If $F'(u) = f(u)$F′(u)=f(u) and $u = g(x)$u=g(x), then $\frac{d}{dx}F(g(x)) = f(g(x))g'(x)$dxd​F(g(x))=f(g(x))g′(x), so $\int f(g(x))g'(x)\,dx = F(g(x)) + C$∫f(g(x))g′(x)dx=F(g(x))+C. Recognising this pattern is the synoptic skill that unlocks the technique.

2. Section 10 — Integration by parts: parts ($\int u\,dv = uv - \int v\,du$∫udv=uv−∫vdu) and substitution are the two big Year 2 integration techniques. Choosing between them is itself an AO3 problem-solving skill — products of an algebraic and a transcendental factor (e.g. $\int x e^x\,dx$∫xexdx) call for parts; products containing a function and its derivative call for substitution.

3. Section 10 — Trig substitution preview: for integrals like $\int \sqrt{1 - x^2}\,dx$∫1−x2​dx, the substitution $x = \sin\theta$x=sinθ converts the radical $\sqrt{1 - \sin^2\theta} = \cos\theta$1−sin2θ​=cosθ via the Pythagorean identity. This is substitution in reverse — choosing $x = g(\theta)$x=g(θ) rather than $u = g(x)$u=g(x) — and previews undergraduate technique.

4. Section 4 — Partial fractions: integrating $\dfrac{1}{x^2 - 1}$x2−11​ requires splitting via partial fractions before integrating term-by-term; substitution is often the next step on each fragment, e.g. $\int \dfrac{1}{x - 1}\,dx$∫x−11​dx via $u = x - 1$u=x−1 gives $\ln|x - 1| + C$ln∣x−1∣+C.

5. 9MA0-03 Mechanics — work-energy: the work done by a variable force $F(x)$F(x) over displacement is $W = \int_a^b F(x)\,dx$W=∫ab​F(x)dx. When $F$F is composite (e.g. $F(x) = x e^{-x^2}$F(x)=xe−x2), substitution $u = -x^2$u=−x2 or $u = x^2$u=x2 is the standard route, linking Pure technique to Mechanics applications.

---

Mark-scheme literacy

Substitution questions on 9MA0-02 split AO marks across all three objectives:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	55–65%	Differentiating the substitution, rewriting $dx$dx in terms of $du$du, applying the power/exp/log rules in $u$u, back-substituting at the end
AO2 (reasoning / interpretation)	25–35%	Justifying the choice of substitution, presenting the change of limits explicitly, including $+ C$+C for indefinite integrals, expressing the final answer in the variable the question requests
AO3 (problem-solving)	5–15%	Selecting substitution from a menu of techniques (vs parts, partial fractions, standard forms) when the question does not prescribe one