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

A root-in-the-denominator substitution. Let u = x^2 + 1, so dudx = 2x, giving x\,dx = 12\,du. Change the limits: x = 0 u = 1; x = sqrt(3) u = 4. The integral becomes 12_1^4 u^-1/2\,du = 12[2u^1/2]_1^4 = [sqrt(u)]_1^4 = sqrt(4) - sqrt(1) = 2 - 1 = 1. Markers reward the substitution, x\,dx = 12\,du, changing the limits to u-values, and the answer 1.

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WASpecialist MathematicsSyllabus dot point

How does reversing the chain rule let us integrate functions built by composition?

Integrate by substitution, changing the variable and the limits to evaluate definite and indefinite integrals

WACE Specialist Unit 4 integration by substitution: choosing u, replacing dx with du over the derivative, changing limits for definite integrals, and the reverse chain rule, with a worked example.

Generated by Claude Opus 4.87 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
Choosing the substitution
Why substitution reverses the chain rule
Recognition versus full working
Definite integrals: change the limits
Substitutions that simplify rather than match
A reliable checking habit

What this dot point is asking

SCSA wants confident use of substitution for both indefinite and definite integrals, including recognising when the integrand contains a function and (a multiple of) its derivative.

The method

Choose $u = g(x)$ so that the integrand simplifies. Differentiate to get $du = g'(x)\,dx$ , and replace every $x$ -expression, including $dx$ , by its $u$ -equivalent. The integral must end up entirely in $u$ with no stray $x$ remaining. Integrate, then for an indefinite integral substitute $x$ back.

Choosing the substitution

Look for an inner function whose derivative is also present (up to a constant factor). Common choices: the expression under a root, the exponent of $e$ , the argument of a trig function, or the denominator. If the derivative is present only up to a constant, adjust by that constant.

Why substitution reverses the chain rule

The substitution rule is the chain rule read in reverse. If $F$ is an antiderivative of $f$ , then by the chain rule $\dfrac{d}{dx}F(g(x)) = f(g(x))\,g'(x)$ . Integrating both sides recovers $\displaystyle\int f(g(x))\,g'(x)\,dx = F(g(x)) + C$ , which is exactly what the substitution $u = g(x)$ , $du = g'(x)\,dx$ computes when it turns the integral into $\displaystyle\int f(u)\,du = F(u) + C$ . Understanding the rule this way makes it clear why the integrand must contain the derivative $g'(x)$ of the inner function: that factor is the leftover from differentiating the composite, and without it (or a constant multiple of it) the substitution will not close.

Recognition versus full working

With practice, many substitutions can be done by recognition, writing the answer directly. An integral of the form $\displaystyle\int g'(x)\,[g(x)]^n\,dx$ is immediately $\dfrac{[g(x)]^{n+1}}{n+1} + C$ , and $\displaystyle\int \dfrac{g'(x)}{g(x)}\,dx$ is immediately $\ln|g(x)| + C$ . SCSA accepts recognition, but in the calculator-free section it is safer to show the substitution explicitly for any non-trivial case, because a single dropped constant factor loses the answer. A good habit is to differentiate your proposed antiderivative mentally as a check: it should return the original integrand exactly.

Definite integrals: change the limits

This is cleaner and avoids errors from reverting. If you do keep $x$ -limits, you must back-substitute before applying them.

Substitutions that simplify rather than match

Not every substitution exploits a derivative already present; some are chosen simply to simplify an awkward expression. A linear substitution $u = ax + b$ handles integrands like $\sqrt{2x + 1}$ or $(3x - 4)^7$ , where $du = a\,dx$ supplies a constant factor. A trigonometric substitution such as $x = a\sin\theta$ converts $\sqrt{a^2 - x^2}$ into $a\cos\theta$ via the Pythagorean identity, turning a root into a clean trig integrand; this is how integrals leading to $\arcsin$ are handled. The guiding principle is to pick the substitution that removes the structural obstacle, whether that is a composition, a root, or a sum of squares.

A reliable checking habit

Because substitution has several places to slip, build in a check. After finding an antiderivative, differentiate it mentally and confirm it returns the original integrand, including any constant factor. For a definite integral, sanity-check the sign and rough size of the answer against the graph: an integral of a positive function over an interval must be positive, and its magnitude should be comparable to the area under the curve. These quick checks catch the most common errors, a dropped constant or an unconverted limit, before they cost marks.

Worked example

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

Step 1: Choose a substitution

The integrand contains $e^{x^2}$ , whose exponent is a composite function. Recognising that $x$ in the integrand is proportional to the derivative of $x^2$ signals that the substitution $u = x^2$ will collapse the composition.

Let $u = x^2$ , so $\dfrac{du}{dx} = 2x$ , which rearranges to $x\,dx = \tfrac{1}{2}\,du$ .

Step 2: Convert the limits

Because this is a definite integral, converting the limits to $u$ -values now means we never need to revert to $x$ . Apply $u = x^2$ to each original limit.

When $x = 0$ : $u = 0^2 = 0$ . When $x = 2$ : $u = 2^2 = 4$ .

Step 3: Rewrite and evaluate the integral

Replacing $x\,dx$ with $\tfrac{1}{2}\,du$ and $e^{x^2}$ with $e^u$ transforms the integral entirely into $u$ . The antiderivative of $e^u$ is simply $e^u$ , so evaluation at the new limits is direct.

\int_0^2 x\,e^{x^2}\,dx = \frac{1}{2}\int_0^4 e^{u}\,du = \frac{1}{2}\Big[e^{u}\Big]_0^4 = \frac{1}{2}(e^4 - 1)

Final answer: $\dfrac{1}{2}(e^4 - 1)$ .

Exam-style practice questions

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

WACE 20225 marksCalculator-free. Evaluate

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

using a substitution.

Show worked answer →

A root-in-the-denominator substitution.

Let $u = x^2 + 1$ , so $\dfrac{du}{dx} = 2x$ , giving $x\,dx = \dfrac{1}{2}\,du$ . Change the limits: $x = 0 \Rightarrow u = 1$ ; $x = \sqrt{3} \Rightarrow u = 4$ .

The integral becomes $\dfrac{1}{2}\displaystyle\int_1^4 u^{-1/2}\,du = \dfrac{1}{2}\Big[2u^{1/2}\Big]_1^4 = \Big[\sqrt{u}\Big]_1^4 = \sqrt{4} - \sqrt{1} = 2 - 1 = 1$ .

Markers reward the substitution, $x\,dx = \dfrac{1}{2}\,du$ , changing the limits to $u$ -values, and the answer $1$ .

WACE 20246 marksCalculator-assumed. Find

\displaystyle\int (2x - 1)(x^2 - x + 4)^5\,dx

Show worked answer →

A reverse-chain-rule substitution.

Let $u = x^2 - x + 4$ , so $\dfrac{du}{dx} = 2x - 1$ , giving $du = (2x - 1)\,dx$ . The factor $(2x - 1)\,dx$ in the integrand is exactly $du$ .

The integral becomes $\displaystyle\int u^5\,du = \dfrac{u^6}{6} + C = \dfrac{(x^2 - x + 4)^6}{6} + C$ .

Markers reward spotting that $2x - 1$ is the derivative of the inner function, the clean substitution, and substituting back to $x$ with the constant. This is an indefinite integral, so reverting to $x$ is required.