Year 12 Maths Extension 1 (2027) Calculus

Integration by substitution

Q: How does integration by substitution work?

Let u equal g of x, so du equals g prime of x dx; rewrite the integral entirely in u, integrate, then back-substitute or change limits.

Q: What do you do with the limits in a definite integral?

Change them: x equals a gives u equals g of a, x equals b gives u equals g of b, then evaluate in u without back-substituting.

Q: How do you choose u?

Usually the inner function whose derivative also appears, up to a constant, in the integrand.

Q: What if the derivative isn't exactly there?

Adjust by a constant; for example if you need 2x dx but have x dx, write x dx equals one half du.

20 practice questions 2 video lessons Theory + worked examples

Master integration by substitution for NSW Year 12 Mathematics Extension 1. A well-chosen change of variable turns a difficult integral into a straightforward one.

You will learn to apply a given substitution, adjust the limits of integration for definite integrals, and reverse the change of variable to complete the answer — a powerful and widely used integration technique in the Extension 1 course.

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Practice questions

Every question with a fully worked solution.

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Integration by Substitution Introduction Watch
Integration by substitution worked examples part 1 Watch

Theory

Integration by substitution reverses the chain rule: put $u = g (x)$ , replace every $x$ including $d x$ , and integrate in $u$ . This NSW Year 12 Mathematics Extension 1 topic (NESA outcome ME1-12-04) covers both indefinite integrals and definite integrals with changed limits.

Integration by substitution reverses the chain rule: replace an inner expression with a new variable $u$ to turn a hard integral into a standard one.

If $u = g (x)$ then $d u = g^{'} (x) d x$ . Replace every $x$ — including $d x$ — so the integral is entirely in $u$ : $\int f (g (x)) g^{'} (x) d x = \int f (u) d u$ .

For an indefinite integral, back-substitute $u = g (x)$ at the end. For a definite integral, change the limits to $u$ -values and do not back-substitute.

NESA link. Part of the Year 12 Calculus topic, outcome ME1-12-04 ("selects and applies differentiation and integration techniques to solve problems") with MAO-WM-01. The syllabus asks students to use substitution to evaluate definite and indefinite integrals.

Substitute, integrate in

u

, then back-substitute (indefinite) or change limits (definite).

For a definite integral, convert the

x

-limits to

u

-limits with

u = g (x)

\int f (g (x)) g^{'} (x) d x = \int f (u) d u, u = g (x), d u = g^{'} (x) d x .

integral f(g(x)) g'(x) dx = integral f(u) du

For a definite integral, change the limits: $x = a \Rightarrow u = g (a)$ and $x = b \Rightarrow u = g (b)$ .

\int_{a}^{b} f (g (x)) g^{'} (x) d x = \int_{g (a)}^{g (b)} f (u) d u .

integral a to b f(g(x)) g'(x) dx = integral g(a) to g(b) f(u) du

Every $x$ must go. If any $x$ remains after substituting, the substitution is incomplete — adjust $d u$ by a constant if $g^{'} (x)$ is not exactly present.

How to integrate by substitution

Choose $u = g (x)$ (usually the inner function) and write $d u = g^{'} (x) d x$ , solving for $d x$ if needed.
Substitute so the integrand is entirely in $u$ .
Integrate in $u$ .
Finish: indefinite — back-substitute $u = g (x)$ ; definite — change the limits and evaluate.

Example 1 — Spot the inner function

Find

\int 3 x^{2} (x^{3} + 1)^{4} d x

Solution

$u = x^{3} + 1$ , $d u = 3 x^{2} d x$ .

$\int 3 x^{2} (x^{3} + 1)^{4} d x$	$=$	$\int u^{4} d u = \frac{u^{5}}{5} + C$
	$=$	$\frac{(x^{3} + 1)^{5}}{5} + C$

integral 3x^2 (x^3+1)^4 dx = (x^3+1)^5 / 5 + C

So the integral is $\frac{(x^{3} + 1)^{5}}{5} + C$ .

Example 2 — Root in the denominator

Find

\int \frac{x}{\sqrt{x^{2} + 4}} d x

Solution

$u = x^{2} + 4$ , $d u = 2 x d x$ , so $x d x = \frac{1}{2} d u$ .

$\int \frac{x}{\sqrt{x^{2} + 4}} d x$	$=$	$\frac{1}{2} \int u^{- 1 / 2} d u$
	$=$	$\sqrt{x^{2} + 4} + C$

integral x / sqrt(x^2 + 4) dx = sqrt(x^2 + 4) + C

So the integral is $\sqrt{x^{2} + 4} + C$ .

Example 3 — Definite: change the limits

Evaluate

\int_{1}^{2} 2 x (x^{2} - 1)^{3} d x

Solution

$u = x^{2} - 1$ , $d u = 2 x d x$ ; $x : 1 \to 2$ gives $u : 0 \to 3$ .

$\int_{1}^{2} 2 x (x^{2} - 1)^{3} d x$	$=$	$\int_{0}^{3} u^{3} d u$
	$=$	$[\frac{u^{4}}{4}]_{0}^{3} = \frac{81}{4}$

integral 1 to 2 of 2x (x^2 - 1)^3 dx = 81/4

Value: $\frac{81}{4}$ .

Example 4 — A logarithm inside

Find

\int \frac{(\ln x)^{2}}{x} d x

Solution

$u = \ln x$ , $d u = \frac{1}{x} d x$ .

$\int \frac{(\ln x)^{2}}{x} d x$	$=$	$\int u^{2} d u = \frac{u^{3}}{3} + C$
	$=$	$\frac{(\ln x)^{3}}{3} + C$

integral (ln x)^2 / x dx = (ln x)^3 / 3 + C

So the integral is $\frac{(\ln x)^{3}}{3} + C$ .

Common pitfalls

Leftover $x$ . Every

x

must disappear — including

d x

. If any

x

remains, the substitution is incomplete.

Forgetting to change limits. For a definite integral, convert the limits to

u

-values; then do not back-substitute.

Mismatched $d u$ . Adjust by a constant if

g^{'} (x)

is not exactly present in the integrand.

Back-substituting a definite integral. Once the limits are in

u

, just evaluate — don't return to

x

Frequently asked questions

How does integration by substitution work?

Let $u = g (x)$ , so $d u = g^{'} (x) d x$ ; rewrite the integral entirely in $u$ , integrate, then back-substitute or change limits.

What do you do with the limits in a definite integral?

Change them: $x = a \Rightarrow u = g (a)$ , $x = b \Rightarrow u = g (b)$ , then evaluate in $u$ without back-substituting.

How do you choose u?

Usually the inner function whose derivative also appears (up to a constant) in the integrand.

What if the derivative isn't exactly there?

Adjust by a constant — e.g. if you need $2 x d x$ but have $x d x$ , write $x d x = \frac{1}{2} d u$ .

Do you always back-substitute?

Only for indefinite integrals. For definite integrals with changed limits, you evaluate directly in $u$ .

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Integration by substitution

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How to integrate by substitution

Common pitfalls

Frequently asked questions

Theory