What is 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.

SACE Board SACE 2023-style practice: Calculator-free. Given that sin^3 x = sin x - sin xcos^2 x, show that _0^π/3 sin^3 x\,dx = 524.

Rewrite using the identity: _0^π/3 (sin x - sin xcos^2 x)\,dx. Let u = cos x, so du = -sin x\,dx, i.e. sin x\,dx = -du. Change limits: x = 0 u = 1; x = π3 u = 12. [1 mark] The antiderivative is -cos x + cos^3 x3 (since int sin xcos^2 x\,dx = int -u^2\,du = -u^33). [1 mark] Evaluate: at x = π3, -12 + 124 = -1124; at x = 0, -1 + 13 = -1624. Difference = -1124 + 1624 = 524, as required. [1 mark]

SACE Board SACE 2022-style practice: Calculator-free. Evaluate _0^2 x(x^2 + 1)^3\,dx using a substitution.

Let u = x^2 + 1, so du = 2x\,dx and x\,dx = 12du. Change limits: x = 0 u = 1; x = 2 u = 5. $_0^2 x(x^2+1)^3\,dx = _1^5 u^3·12\,du = (1)/(2)[(u^4)/(4)]_1^5 = (1)/(8)(625 - 1) = 78.$ Marks: one for the substitution and limit change, one for the antiderivative, one for the value 78.

SACE Board SACE 2021-style practice: Calculator-free. Find x e^x^2\,dx.

Let u = x^2, so du = 2x\,dx and x\,dx = 12du: $int x e^x^2\,dx = int e^u·12\,du = 12e^u + C = 12e^x^2 + C.$ Marks: one for the substitution with the constant factor, one for the answer including +C.

§-Syllabus dot point

SASpecialist MathematicsSyllabus dot point

How does reversing the chain rule let us integrate composite functions?

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

Using substitution to reverse the chain rule, choosing u and replacing dx with du, handling definite integrals by changing the limits, and recognising when substitution applies.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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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

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.

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.

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.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20233 marksCalculator-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 20223 marksCalculator-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 20212 marksCalculator-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 $+C$ .