NSWMaths Extension 1Syllabus dot point

How do we use the substitution method to evaluate integrals that arise from the reverse chain rule?

Apply integration by substitution to evaluate definite and indefinite integrals, including reverse chain rule cases

A focused answer to the HSC Maths Extension 1 dot point on integration by substitution. Choosing the right substitution, transforming the integrand and differential, changing limits for definite integrals, and standard reverse chain rule patterns, with worked examples.

Generated by Claude OpusReviewed by Better Tuition Academy9 min answerUpdated 2026-05-20

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What this dot point is asking

NESA wants you to evaluate integrals where the integrand is a composition or contains a factor that is (almost) the derivative of another factor, by choosing a substitution $u = g(x)$ that undoes the chain rule. You should be fluent both with the rote mechanics and with recognising patterns that signal a substitution is appropriate.

The answer

The general method

For an integral $\int f(g(x)) g'(x) \, dx$ :

Identify the inner function $g(x)$ and let $u = g(x)$ .
Compute $du = g'(x) \, dx$ .
Rewrite the integral entirely in terms of $u$ : $\int f(u) \, du$ .
Evaluate the integral in $u$ .
For an indefinite integral, substitute back to $x$ . For a definite integral, either substitute back and use the original limits, or change the limits to $u$ -values and skip the back-substitution.

Choosing a good IMATH_17

Look for one of these patterns in the integrand:

A function inside another function ( $(x^2 + 1)^5$ , $\sqrt{x + 4}$ , $e^{x^3}$ ). Set $u$ equal to the inner.
A function and its derivative appearing as a product. Set $u$ equal to the function whose derivative is present.
A linear inside argument: $\sin(2 x + 1)$ , $e^{3 x - 5}$ . Substitute or use the linear-argument shortcut.

Changing limits

For $\int_a^b f(g(x)) g'(x) \, dx$ , when $u = g(x)$ , the new limits are $u = g(a)$ at the bottom and $u = g(b)$ at the top.

After substitution, the integral becomes $\int_{g(a)}^{g(b)} f(u) \, du$ , and no back-substitution to $x$ is needed.

Linear inside argument

For $\int f(a x + b) \, dx$ , the shortcut is $\int f(a x + b) \, dx = \frac{1}{a} F(a x + b) + C$ , where $F$ is the antiderivative of $f$ .

For example, $\int \cos(3 x + 2) \, dx = \frac{1}{3} \sin(3 x + 2) + C$ .

Reverse chain rule patterns

Memorise these common patterns:

\int \frac{f'(x)}{f(x)} \, dx = \ln |f(x)| + C,

\int f'(x) e^{f(x)} \, dx = e^{f(x)} + C,

\int f'(x) [f(x)]^n \, dx = \frac{[f(x)]^{n + 1}}{n + 1} + C \quad (n \neq -1),

\int f'(x) \cos f(x) \, dx = \sin f(x) + C,

\int f'(x) \sin f(x) \, dx = -\cos f(x) + C.

Worked example

Polynomial inside

Evaluate $\int 2 x (x^2 + 1)^4 \, dx$ .

$u = x^2 + 1$ , $du = 2 x \, dx$ .

$\int u^4 \, du = \frac{u^5}{5} + C = \frac{(x^2 + 1)^5}{5} + C$ .

Logarithm pattern

Evaluate $\int \frac{2 x}{x^2 + 1} \, dx$ .

Numerator is the derivative of the denominator. $\int \frac{f'(x)}{f(x)} \, dx = \ln |f(x)| + C$ .

$\int \frac{2 x}{x^2 + 1} \, dx = \ln(x^2 + 1) + C$ (no absolute value needed since $x^2 + 1 > 0$ ).

Trig inside

Evaluate $\int \sin^2 x \cos x \, dx$ .

$u = \sin x$ , $du = \cos x \, dx$ .

$\int u^2 \, du = \frac{u^3}{3} + C = \frac{\sin^3 x}{3} + C$ .

Definite integral with limit change

Evaluate $\int_0^2 x e^{x^2} \, dx$ .

$u = x^2$ , $du = 2 x \, dx$ , so $x \, dx = \frac{1}{2} \, du$ .

Limits: $x = 0 \implies u = 0$ ; $x = 2 \implies u = 4$ .

$\int_0^2 x e^{x^2} \, dx = \frac{1}{2} \int_0^4 e^u \, du = \frac{1}{2} (e^4 - 1)$ .

Mixed substitution

Evaluate $\int x \sqrt{1 - x} \, dx$ .

$u = 1 - x$ , so $x = 1 - u$ and $du = -dx$ , i.e. $dx = -du$ .

$\int (1 - u) \sqrt{u} \cdot (-du) = -\int (1 - u) u^{1/2} \, du = -\int u^{1/2} \, du + \int u^{3/2} \, du$

$= -\frac{2}{3} u^{3/2} + \frac{2}{5} u^{5/2} + C = -\frac{2}{3} (1 - x)^{3/2} + \frac{2}{5} (1 - x)^{5/2} + C$ .

Linear inside argument shortcut

Evaluate $\int e^{2 x + 1} \, dx$ .

$\frac{1}{2} e^{2 x + 1} + C$ (the coefficient of $x$ inside is $2$ , so divide by $2$ ).

Common traps

Not transforming $dx$: Substitution rewrites every $x$ in the integrand including the $dx$ . Skipping the $du = g'(x) \, dx$ step is the most common error.
Forgetting to change the limits: Either change limits to $u$ -values, or substitute back to $x$ before evaluating. Do not mix the two; substituting $u$ -values into the back-substituted $x$ expression gives wrong answers.
Picking the wrong $u$: If your chosen $u$ does not eliminate $x$ from the integral, the substitution failed. Try $u =$ a different inner function.
Dropping a constant when $du = k \, dx$: IMATH_13 means $x \, dx = \frac{1}{2} du$ , so the integral picks up a $\frac{1}{2}$ .
Missing the absolute value in $\ln$: IMATH_17 in general; you can drop the absolute value only when $x > 0$ or $x < 0$ throughout the interval of integration.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2022 HSC Q153 marksUse the substitution

u = x^2 + 3

to evaluate

\int_0^1 x \sqrt{x^2 + 3} \, dx

Show worked answer →

With $u = x^2 + 3$ , $\frac{du}{dx} = 2 x$ , so $x \, dx = \frac{1}{2} du$ .

Change limits: $x = 0 \implies u = 3$ ; $x = 1 \implies u = 4$ .

$\int_0^1 x \sqrt{x^2 + 3} \, dx = \int_3^4 \sqrt{u} \cdot \frac{1}{2} \, du = \frac{1}{2} \int_3^4 u^{1/2} \, du = \frac{1}{2} \cdot \frac{2}{3} [u^{3/2}]_3^4 = \frac{1}{3} (4^{3/2} - 3^{3/2})$ .

$= \frac{1}{3} (8 - 3 \sqrt{3})$ .

Markers reward the substitution statement, the differential transformation, the limit change, and a clean exact answer.

2021 HSC Q184 marksUse the substitution

u = \sin x

to evaluate

\int_0^{\pi/2} \sin^3 x \cos x \, dx

Show worked answer →

$u = \sin x \implies du = \cos x \, dx$ .

Limits: $x = 0 \implies u = 0$ ; $x = \frac{\pi}{2} \implies u = 1$ .

$\int_0^{\pi/2} \sin^3 x \cos x \, dx = \int_0^1 u^3 \, du = \left[ \frac{u^4}{4} \right]_0^1 = \frac{1}{4}$ .

Markers reward the substitution choice, the differential, the limit change, and a clean evaluation. An alternative is to recognise $\sin^3 x \cos x = \frac{d}{dx}\left( \frac{\sin^4 x}{4} \right)$ directly.