Learn9MA0Paper 2 · Pure Mathematics 2IntegrationIntegration by substitution

Integration · Subtopic 8 of 14

Integration by Substitution

The reverse of the chain rule: replace part of the integrand by $u$ , change $dx$ into $du$ , change the limits, and turn a hard integral into an easy one.

Cambridge 9709 Paper 3Edexcel YMA01 WMA14Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 50 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Use a given substitution $u=g(x)$ , replacing $dx$ by $\dfrac{du}{g'(x)}$ .
Express the whole integrand in terms of $u$ .
Change the limits for a definite integral, so there is no need to return to $x$ .
Recognise integrals of the form $\int f'(x)\big(f(x)\big)^n\,dx$ and $\int f'(x)e^{f(x)}\,dx$ .

Pause and think

Differentiate $(x^2+1)^5$ . Now look at $\displaystyle\int2x(x^2+1)^4\,dx$ . What do you notice?

Section 2 of 14

Core idea

Start hereChange the variable, simplify the problem

$\int x(x^2+1)^3dx$ looks awkward. But notice: $x$ is (half) the derivative of $x^2+1$ . If we call $x^2+1$ a new letter $u$ , the integral collapses to $\frac12\int u^3du$ . Substitution is the chain rule run backwards: rename the inner part and the mess disappears.

The big idea

Choose $u=g(x)$ ; replace $dx$ using $\frac{du}{dx}=g'(x)$ ; replace every $x$ ; change the limits to $u$ -values; integrate in $u$ .

1. The method

Same area, new variable

The two regions look different but have the same area,

\tfrac{15}{4}

: substitution reshapes the problem without changing its value.

Why substitution works

$\frac{d}{dx}F(g(x))=F'(g(x))g'(x)$
Chain rule.
$\int f(g(x))g'(x)\,dx=F(g(x))+c$
Integrate both sides, with $F'=f$ .
$=\int f(u)\,du\ \text{ with }u=g(x),\ du=g'(x)\,dx$
The same thing written in $u$ .

Worked example $\int_0^2x(x^2+1)^3dx$ , $u=x^2+1$

Evaluate.

Goal: Number.
Strategy: Convert $x\,dx$ and limits.

$du=2x\,dx\Rightarrow x\,dx=\tfrac12du$
Differentiate $u$ .
$x=0\to u=1;\quad x=2\to u=5$
New limits.
$\tfrac12\int_1^5u^3du=\tfrac12\left[\tfrac{u^4}4\right]_1^5=\tfrac18(625-1)=78$
Integrate in $u$ .

Answer

78

✓ CheckExpand:

x(x^6+3x^4+3x^2+1)

integrates to

\frac{x^8}8+\frac{x^6}2+\frac{3x^4}4+\frac{x^2}2

: at

2

32+32+12+2=78

✓.

Worked example $\int\frac{x}{\sqrt{x+3}}dx$ , $u=x+3$

Integrate.

Goal: In terms of $x$ .
Strategy: Express $x$ in $u$ too.

$x=u-3,\quad dx=du$
From the substitution.
$\int\frac{u-3}{\sqrt u}du=\int\left(u^{1/2}-3u^{-1/2}\right)du$
Split.
$=\tfrac23u^{3/2}-6u^{1/2}+c$
Integrate.
$=\tfrac23(x+3)^{3/2}-6\sqrt{x+3}+c$
Back to $x$ .

Answer

\frac23(x+3)^{3/2}-6\sqrt{x+3}+c

✓ CheckDifferentiate:

\sqrt{x+3}-\frac3{\sqrt{x+3}}=\frac{x}{\sqrt{x+3}}

✓.

Worked exampleTrig substitution: $\int_0^1\frac{dx}{1+x^2}$ , $x=\tan\theta$

Evaluate.

Goal: Exact value.
Strategy: $1+\tan^2=\sec^2$ .

$dx=\sec^2\theta\,d\theta;\ \ 0\to0,\ 1\to\tfrac\pi4$
Convert.
$\int_0^{\pi/4}\frac{\sec^2\theta}{\sec^2\theta}d\theta=\int_0^{\pi/4}1\,d\theta=\tfrac\pi4$
Simplify.

Answer

\frac\pi4

✓ Check

\approx0.785

; integrand from

1

\frac12

over width

1

✓.

Common misconceptionForgetting to change limits Either change the limits to

u

-values, or substitute back to

x

before using the original limits. Never mix.

Pause and think

$\int2xe^{x^2}dx$ ?

Summary

Reverse chain rule.
Convert $x$ , $dx$ and limits.

Substitution: Changing the integration variable.

HistoryNotation that guides the hand. Leibniz designed

\dfrac{du}{dx}

to behave like a fraction, so that "

du=g'(x)\,dx

" looks natural. This was deliberate: he wrote that good notation "wonderfully reduces the work of the mind". Substitution is where that design pays off most.

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