Integration by substitution (SACE Stage 2 Specialist Mathematics)

Syllabus dot point

“Evaluate integrals using the method of substitution, including changing the limits for definite integrals.”

8 min readUpdated 2 Jun 2026

Quick answer

Substitution reverses the chain rule. Choose $u = g(x)$ so that its derivative also appears in the integrand, then replace $dx$ using $du = g'(x)\,dx$ . The integral becomes one in $u$ that you can do directly. For a definite integral, either convert the limits to $u$ -values, or convert back to $x$ before substituting the original limits.

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What this dot point is asking
The method
A worked indefinite integral
Why substitution works
Handling a constant factor
Definite integrals: change the limits
Trigonometric substitutions
Exam-style questions

What this dot point is asking

You need to evaluate integrals by substitution, including correctly handling the limits of definite integrals.

The method

If $u = g(x)$ then $\dfrac{du}{dx} = g'(x)$ , so $du = g'(x)\,dx$ . The substitution rule is

\int f(g(x))\,g'(x)\,dx = \int f(u)\,du.

The aim is to pick

u

so the integrand turns into

f(u)\,du

with no

x

left over. Look for a function whose derivative (up to a constant factor) is also in the integrand.

Choosing the substitution

A good $u$ is usually the "inside" of a composite function, or whatever sits under a root, inside a power, or in an exponent - provided its derivative appears (possibly times a constant) elsewhere in the integrand. After substituting, the integral must contain only $u$ and $du$ .

A worked indefinite integral

Substitution for an indefinite integral

Find $\displaystyle\int 2x\sqrt{x^{2} + 1}\,dx$

Step 1: Choose $u$ , compute $du$ , and note it matches the integrand

The expression inside the square root, $x^2 + 1$ , is the natural candidate for $u$ because its derivative $2x$ appears as the other factor in the integrand. Set $u = x^2 + 1$ , so $\dfrac{du}{dx} = 2x$ and $du = 2x\,dx$ . The $2x\,dx$ in the integrand is exactly $du$ , so the match is perfect with no extra constant needed.

Step 2: Substitute and integrate in terms of $u$

Replacing $\sqrt{x^2 + 1}$ with $\sqrt{u}$ and $2x\,dx$ with $du$ converts the integral into a standard power rule.

\int 2x\sqrt{x^2 + 1}\,dx = \int \sqrt{u}\,du = \int u^{1/2}\,du = \frac{2}{3}u^{3/2} + C.

Step 3: Substitute back to express the answer in terms of $x$

Replace $u$ with $x^2 + 1$ to return to the original variable.

= \frac{2}{3}(x^2 + 1)^{3/2} + C.

Why substitution works

Substitution is the chain rule run in reverse. When you differentiate $F(g(x))$ the chain rule produces $F'(g(x))g'(x)$ , so any integrand of that shape antidifferentiates back to $F(g(x))$ . Setting $u = g(x)$ simply renames the inside function so the integral looks like the standard $\int F'(u)\,du$ . This is why the search for a good substitution is really a search for an inside function whose derivative is already present, up to a constant, in the integrand. If you can spot that pairing, the substitution is guaranteed to collapse the integral.

Handling a constant factor

Often $du$ differs from the integrand by a constant. Solve for $dx$ and carry the constant through.

For $\displaystyle\int x e^{x^2}\,dx$ , let $u = x^2$ , so $du = 2x\,dx$ , hence $x\,dx = \tfrac{1}{2}du$ :

\int x e^{x^2}\,dx = \int e^{u}\cdot\tfrac{1}{2}\,du = \tfrac{1}{2}e^{u} + C = \tfrac{1}{2}e^{x^2} + C.

Definite integrals: change the limits

For a definite integral, the cleanest approach changes the limits to $u$ -values so you never convert back.

Definite integral by substitution

Evaluate $\displaystyle\int_{0}^{2} x(x^{2} + 1)^{3}\,dx$

Step 1: Choose $u$ , compute $du$ , and solve for $x\,dx$

The inner function $(x^2 + 1)$ is the natural substitution. Set $u = x^2 + 1$ , so $du = 2x\,dx$ , which gives $x\,dx = \tfrac{1}{2}du$ . The factor $x\,dx$ appears in the integrand, so the substitution removes all $x$ from the integral.

Step 2: Change the limits to $u$ -values

Converting the limits avoids the need to substitute back at the end, making the definite integral cleaner to evaluate. When $x = 0$ , $u = 0^2 + 1 = 1$ ; when $x = 2$ , $u = 2^2 + 1 = 5$ .

Step 3: Substitute and evaluate the integral in $u$

The integral becomes a straightforward power rule, evaluated between the new limits.

\int_{0}^{2} x(x^2+1)^3\,dx = \int_{1}^{5} u^{3}\cdot\tfrac{1}{2}\,du = \frac{1}{2}\left[\frac{u^{4}}{4}\right]_{1}^{5} = \frac{1}{8}\left(5^4 - 1^4\right).

Now $5^4 = 625$ , so

= \frac{1}{8}(625 - 1) = \frac{624}{8} = 78.

Verify by differentiating

For an indefinite integral you can always check your answer by differentiating it - the chain rule should return the original integrand. This catches missing constant factors immediately.

Trigonometric substitutions

A frequent SACE pattern uses a trigonometric inside function. To integrate something like $\int \sin^3 x\,dx$ , first rewrite using an identity such as $\sin^3 x = \sin x(1 - \cos^2 x)$ , then substitute $u = \cos x$ so that $du = -\sin x\,dx$ . The leftover $\sin x\,dx$ becomes $-du$ , and the integral turns into a simple polynomial in $u$ . The general lesson is that an odd power of sine or cosine can always be split so that one factor pairs with $dx$ to form the differential of the other function. Recognising this pairing is the key skill the technique tests.

Common errors

Forgetting to change $dx$ to $du$: You must replace $dx$ entirely; leaving a stray $dx$ or $x$ after substituting means the substitution was not completed correctly.
Not changing the limits for a definite integral: If you switch to $u$ but keep the original $x$ -limits, the answer is wrong. Either convert the limits to $u$ , or convert the antiderivative back to $x$ first.
Mishandling the constant factor: When $du = k\,dx$ , you must divide by $k$ ; dropping this factor scales the whole answer.
Choosing a $u$ whose derivative is absent: If $g'(x)$ does not appear (even up to a constant) in the integrand, the substitution will not simplify it - try a different method.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2023-style3 marks

Calculator-free. Given that

\sin^3 x = \sin x - \sin x\cos^2 x

, show that

\displaystyle\int_0^{\pi/3} \sin^3 x\,dx = \dfrac{5}{24}

Show worked answer →

Rewrite using the identity: $\int_0^{\pi/3} (\sin x - \sin x\cos^2 x)\,dx$ .

Let $u = \cos x$ , so $du = -\sin x\,dx$ , i.e. $\sin x\,dx = -du$ . Change limits: $x = 0 \Rightarrow u = 1$ ; $x = \tfrac{\pi}{3} \Rightarrow u = \tfrac{1}{2}$ . [1 mark]

The antiderivative is $-\cos x + \dfrac{\cos^3 x}{3}$ (since $\int \sin x\cos^2 x\,dx = \int -u^2\,du = -\tfrac{u^3}{3}$ ). [1 mark]

Evaluate: at $x = \tfrac{\pi}{3}$ , $-\tfrac{1}{2} + \tfrac{1}{24} = -\tfrac{11}{24}$ ; at $x = 0$ , $-1 + \tfrac{1}{3} = -\tfrac{16}{24}$ . Difference $= -\tfrac{11}{24} + \tfrac{16}{24} = \tfrac{5}{24}$ , as required. [1 mark]

SACE 2022-style3 marks

Calculator-free. Evaluate

\displaystyle\int_0^2 x(x^2 + 1)^3\,dx

using a substitution.

Show worked answer →

Let $u = x^2 + 1$ , so $du = 2x\,dx$ and $x\,dx = \tfrac{1}{2}du$ . Change limits: $x = 0 \Rightarrow u = 1$ ; $x = 2 \Rightarrow u = 5$ .

\int_0^2 x(x^2+1)^3\,dx = \int_1^5 u^3\cdot\tfrac{1}{2}\,du = \frac{1}{2}\Big[\frac{u^4}{4}\Big]_1^5 = \frac{1}{8}(625 - 1) = 78.

Marks: one for the substitution and limit change, one for the antiderivative, one for the value $78$ .

SACE 2021-style2 marks

Calculator-free. Find

\displaystyle\int x e^{x^2}\,dx

Show worked answer →

Let $u = x^2$ , so $du = 2x\,dx$ and $x\,dx = \tfrac{1}{2}du$ :

\int x e^{x^2}\,dx = \int e^u\cdot\tfrac{1}{2}\,du = \tfrac{1}{2}e^u + C = \tfrac{1}{2}e^{x^2} + C.

Marks: one for the substitution with the constant factor, one for the answer including $+C$ .

Practise this

Sources & how we know this

Stage 2 Specialist Mathematics subject outline (2025 onwards) — SACE Board
Stage 2 Specialist Mathematics: external assessment — SACE Board

What this dot point is asking

The method

A worked indefinite integral

Find ∫2xx2+1 dx\displaystyle\int 2x\sqrt{x^{2} + 1}\,dx∫2xx2+1​dx

Step 1: Choose uuu, compute dududu, and note it matches the integrand

Step 2: Substitute and integrate in terms of uuu

Step 3: Substitute back to express the answer in terms of xxx

Why substitution works

Handling a constant factor

Definite integrals: change the limits

Evaluate ∫02x(x2+1)3 dx\displaystyle\int_{0}^{2} x(x^{2} + 1)^{3}\,dx∫02​x(x2+1)3dx

Step 1: Choose uuu, compute dududu, and solve for x dxx\,dxxdx

Step 2: Change the limits to uuu-values

Step 3: Substitute and evaluate the integral in uuu

Trigonometric substitutions

Exam-style questions

Practise this

Sources & how we know this

Find $\displaystyle\int 2x\sqrt{x^{2} + 1}\,dx$

Step 1: Choose $u$ , compute $du$ , and note it matches the integrand

Step 2: Substitute and integrate in terms of $u$

Step 3: Substitute back to express the answer in terms of $x$

Evaluate $\displaystyle\int_{0}^{2} x(x^{2} + 1)^{3}\,dx$

Step 1: Choose $u$ , compute $du$ , and solve for $x\,dx$

Step 2: Change the limits to $u$ -values

Step 3: Substitute and evaluate the integral in $u$