Integration by substitution in HSC Maths Extension 1: choosing $u$, transforming the integral and changing limits

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 the patterns that signal a substitution is appropriate.

The reason substitution works is that integration is differentiation run backwards, and the chain rule is the differentiation rule for compositions. So when an integrand looks like "a function of something, times the derivative of that something", a substitution collapses it back to the simple integral it came from. Seeing that structure quickly is what separates a fast, clean solution from a stuck one.

The answer

The general method

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

1. Identify the inner function $g(x)$ and let $u = g(x)$.
2. Compute $du = g'(x) \, dx$.
3. Rewrite the integral entirely in terms of $u$: $\int f(u) \, du$. Nothing in $x$ may remain.
4. Evaluate the integral in $u$.
5. 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 $u$

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.

The quick test: after substituting, does every $x$ disappear? If a stray $x$ survives that you cannot express through $u$, the choice was wrong - try a different inner function.

Changing limits, stage by stage

The mechanics that students most often slip on is the definite integral: choosing $u$, transforming the differential, and changing the limits in step. The figures below walk through the HSC-style integral $\displaystyle\int_0^1 x\sqrt{x^2+3}\,dx$ to show how the substitution moves the whole problem from $x$-space to $u$-space while preserving the area.

Stage 1, the integral in $x$. The integral is the area under $y = x\sqrt{x^2+3}$ from $x = 0$ to $x = 1$. We will not compute this area directly; we will transform it.

Stage 2, choose the substitution. The inner function is $x^2 + 3$, and the spare factor $x$ is (up to a constant) its derivative. Let $u = x^2 + 3$; then $\dfrac{du}{dx} = 2x$, so $x\,dx = \tfrac12\,du$. That is exactly the leftover piece of the integrand, which is the sign of a good choice.

Stage 3, change the limits. Because the variable has changed, the limits must change with it: feed each $x$-limit through $u = x^2 + 3$. So $x = 0$ gives $u = 3$, and $x = 1$ gives $u = 4$. The integral now runs from $u = 3$ to $u = 4$.

Stage 4, the integral in $u$. Replacing $x\sqrt{x^2+3}\,dx$ with $\sqrt{u}\cdot\tfrac12\,du$ turns the problem into $\tfrac12\displaystyle\int_3^4 \sqrt{u}\,du$, a one-line power integral. The shaded area is identical to the one in Stage 1, so no back-substitution to $x$ is needed - the $u$-limits already carry the right interval. Evaluating gives $\tfrac13\big(4^{3/2} - 3^{3/2}\big) = \tfrac13(8 - 3\sqrt 3)$.

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, $\int \cos(3 x + 2) \, dx = \frac{1}{3} \sin(3 x + 2) + C$. The $\frac{1}{a}$ is the substitution $u = ax + b$ done in your head; the constant slope means $du = a\,dx$ contributes a single factor.

Reverse chain rule patterns

Memorise these common patterns; recognising them on sight saves setting up a full substitution:

When the substitution leaves a stray $x$

Sometimes the natural substitution does not clear every $x$ at once. For $\int x\sqrt{1-x}\,dx$, choosing $u = 1 - x$ handles the root but leaves an $x$; the trick is that $u = 1-x$ also gives $x = 1 - u$, so the stray $x$ is itself expressible in $u$. Rearranging the substitution to express the leftover factor is a standard Extension 1 move, shown in the worked examples below.

How exam questions ask about integration by substitution

• "Use the substitution $u = \ldots$ to evaluate ..." The substitution is handed to you; show $du$, the transformed integrand, the limit change (if definite), and a clean exact answer.
• "Evaluate $\int \ldots$" (no substitution named). You must spot the structure yourself. Look for an inner function whose derivative is present as a factor.
• "Find $\int \frac{f'(x)}{f(x)}\,dx$"-shaped integrands. Recognise the log pattern; the answer is $\ln|f(x)| + C$.
• "Hence find the exact value ..." Definite integral: change limits to $u$-values, integrate, evaluate. "Exact" means leave surds, $e$, $\ln$ and $\pi$ in - no decimals.
• A square root times a linear factor. Substitute $u =$ the inside of the root and re-express the stray factor through $u$.
• A linear inside a standard function ($\sin(2x+1)$, $e^{3x}$). Use the $\frac{1}{a}$ shortcut.