Integration techniques: VCE Specialist Mathematics Unit 4

Syllabus dot point

“Antidifferentiation techniques including integration by substitution, the use of partial fractions, trigonometric identities and inverse-trigonometric standard forms, and the evaluation of definite integrals using these techniques”

8 min readUpdated 2 Jun 2026

Quick answer

Integration by substitution reverses the chain rule: set $u = g(x)$ , replace $\mathrm{d}x$ with $\frac{\mathrm{d}u}{g'(x)}$ , and change the terminals to $u$ -values for a definite integral. Partial fractions split a rational integrand into simpler pieces that integrate to logarithms or inverse trig functions. Trigonometric identities (such as $\cos^2 x = \frac{1 + \cos 2x}{2}$ ) convert products and even powers into integrable sums. The inverse-trig standard forms give $\int \frac{\mathrm{d}x}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac{x}{a} + c$ and $\int \frac{\mathrm{d}x}{a^2 + x^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + c$ .

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What this dot point is asking
Integration by substitution
Partial fractions
Trigonometric identities
Inverse-trigonometric standard forms
Choosing the right technique
Examples in context
Try this
Exam-style questions

What this dot point is asking

VCAA wants you to extend antidifferentiation beyond the basic rules by using substitution, partial fractions, trigonometric identities and the inverse-trigonometric standard forms, and then to evaluate definite integrals with these techniques. The marker is testing whether you can recognise which method a given integrand calls for and execute it cleanly, including adjusting the terminals when you substitute.

Integration by substitution

Substitution undoes the chain rule. To integrate $\int f(g(x))\,g'(x)\,\mathrm{d}x$ , let $u = g(x)$ so $\mathrm{d}u = g'(x)\,\mathrm{d}x$ , turning the integral into $\int f(u)\,\mathrm{d}u$ . The aim is to choose $u$ so that the rest of the integrand is (a constant multiple of) $\frac{\mathrm{d}u}{\mathrm{d}x}$ .

For a definite integral, change the terminals to the corresponding $u$ -values rather than substituting back. If the original terminals are $x = a$ and $x = b$ , the new terminals are $u = g(a)$ and $u = g(b)$ :

\int_a^b f(g(x))\,g'(x)\,\mathrm{d}x = \int_{g(a)}^{g(b)} f(u)\,\mathrm{d}u.

Partial fractions

A proper rational function (numerator degree less than denominator degree) whose denominator factorises can be split into a sum of simpler fractions. For distinct linear factors,

\frac{1}{(x - a)(x - b)} = \frac{A}{x - a} + \frac{B}{x - b},

and each piece integrates to a logarithm: $\int \frac{A}{x - a}\,\mathrm{d}x = A\ln|x - a| + c$ . If the rational function is improper, divide first to extract a polynomial part. Repeated and irreducible quadratic factors need the appropriate forms, but the linear-factor case is the staple.

Trigonometric identities

Integrands involving powers or products of trig functions are rewritten with identities into a form that integrates term by term. The most used are the double-angle rearrangements

\cos^2 x = \frac{1 + \cos 2x}{2}, \qquad \sin^2 x = \frac{1 - \cos 2x}{2},

which turn even powers into linear combinations of $1$ and $\cos 2x$ . Products such as $\sin 3x \cos x$ are handled with product-to-sum identities, and $\tan^2 x = \sec^2 x - 1$ converts a tangent squared into something with a known antiderivative.

Inverse-trigonometric standard forms

Two standard forms appear constantly and are worth memorising:

\int \frac{1}{\sqrt{a^2 - x^2}}\,\mathrm{d}x = \sin^{-1}\!\frac{x}{a} + c, \qquad \int \frac{1}{a^2 + x^2}\,\mathrm{d}x = \frac{1}{a}\tan^{-1}\!\frac{x}{a} + c.

Recognising a denominator of the form $a^2 - x^2$ under a square root, or $a^2 + x^2$ , signals an inverse-trig answer. Completing the square first often reveals one of these forms in a disguised integrand.

Worked example

A definite integral by substitution

Evaluate $\displaystyle\int_0^2 x\sqrt{x^2 + 1}\,\mathrm{d}x$ .

Step 1: Choose the substitution

The factor $x^2 + 1$ is under the square root and its derivative $2x$ appears alongside it as a factor, so setting $u = x^2 + 1$ will convert the integrand into a simple power of $u$ . Let $u = x^2 + 1$ . Then $\dfrac{\mathrm{d}u}{\mathrm{d}x} = 2x$ , so $x\,\mathrm{d}x = \tfrac{1}{2}\,\mathrm{d}u$ . The integrand $x\sqrt{x^2 + 1}\,\mathrm{d}x$ becomes $\sqrt{u}\cdot\tfrac12\,\mathrm{d}u$ .

Step 2: Change the terminals

Because the integral is definite, we convert the $x$ -limits to $u$ -values so we can evaluate directly in $u$ without substituting back. When $x = 0$ , $u = 0^2 + 1 = 1$ . When $x = 2$ , $u = 2^2 + 1 = 5$ . So

\int_0^2 x\sqrt{x^2 + 1}\,\mathrm{d}x = \frac{1}{2}\int_1^5 u^{1/2}\,\mathrm{d}u.

Step 3: Integrate

Apply the power rule for $u^{1/2}$ , increasing the exponent by one and dividing by the new exponent. This converts the integral into a closed-form expression to be evaluated at the terminals.

\frac{1}{2}\int_1^5 u^{1/2}\,\mathrm{d}u = \frac{1}{2}\cdot\frac{2}{3}\Big[u^{3/2}\Big]_1^5 = \frac{1}{3}\Big(5^{3/2} - 1^{3/2}\Big).

Step 4: Evaluate

Simplify $5^{3/2} = 5\sqrt{5}$ and $1^{3/2} = 1$ , then subtract to obtain the exact value.

\frac{1}{3}\Big(5^{3/2} - 1\Big) = \frac{5\sqrt{5} - 1}{3}.

Numerically $5\sqrt{5} \approx 11.180$ , so the integral is about $\frac{10.180}{3} \approx 3.393$ , which is positive as expected since the integrand is positive on $[0, 2]$ .

Final answer: $\dfrac{5\sqrt{5} - 1}{3}$ .

Choosing the right technique

The skill is recognition. A factor that is (a multiple of) the derivative of another part suggests substitution. A rational integrand with a factorable denominator suggests partial fractions. Even powers or products of sines and cosines suggest trig identities. A square root of $a^2 - x^2$ , or a sum $a^2 + x^2$ in a denominator, suggests an inverse-trig standard form, possibly after completing the square. Scan the integrand for these signatures before committing to a method.

Common exam traps

Not changing the terminals: After substituting $u = g(x)$ in a definite integral you must either change the limits to $u$ -values or substitute back to $x$ before applying the original limits. Mixing $x$ -limits with a $u$ -integrand is wrong.
Forgetting the constant of integration: For indefinite integrals always add $+ c$ .
Dropping the absolute value in log integrals: $\int \frac{1}{x - a}\,\mathrm{d}x = \ln|x - a| + c$ . The absolute value keeps the antiderivative valid on both sides of $x = a$ .
Mismatching the inverse-trig form: $\frac{1}{\sqrt{a^2 - x^2}}$ gives $\sin^{-1}$ , while $\frac{1}{a^2 + x^2}$ gives $\frac{1}{a}\tan^{-1}$ . Confusing the two, or forgetting the $\frac{1}{a}$ factor, loses marks.

Examples in context

Example 1. Partial fractions. $\dfrac{1}{(x-1)(x+1)} = \dfrac{1/2}{x-1} - \dfrac{1/2}{x+1}$ , so $\int \dfrac{\mathrm{d}x}{x^2 - 1} = \tfrac12\ln|x-1| - \tfrac12\ln|x+1| + c = \tfrac12\ln\left|\dfrac{x-1}{x+1}\right| + c$ .

Example 2. Inverse-trig form. $\int \dfrac{1}{9 + x^2}\,\mathrm{d}x = \tfrac13\tan^{-1}\dfrac{x}{3} + c$ , taking $a = 3$ .

Try this

Q1. Evaluate $\displaystyle\int 2x\,e^{x^2}\,\mathrm{d}x$ . [2 marks]

Cue. Let $u = x^2$ , $\mathrm{d}u = 2x\,\mathrm{d}x$ : $\int e^u\,\mathrm{d}u = e^{x^2} + c$ .

Q2. Evaluate $\displaystyle\int_0^{1} \frac{1}{1 + x^2}\,\mathrm{d}x$ . [2 marks]

Cue. $[\tan^{-1}x]_0^1 = \frac{\pi}{4} - 0 = \frac{\pi}{4}$ .

Q3. Evaluate $\displaystyle\int \cos^2 x\,\mathrm{d}x$ . [2 marks]

Cue. Use $\cos^2 x = \frac{1 + \cos 2x}{2}$ : $\frac{x}{2} + \frac{\sin 2x}{4} + c$ .

Exam-style questions

Questions in the style of VCAA 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.

Practice4 marks

Evaluate

\displaystyle\int_0^{1} \dfrac{2x}{x^2 + 1}\,dx

exactly, and state the technique used.

Show worked answer →

The numerator $2x$ is exactly the derivative of the denominator $x^2 + 1$ , so use substitution $u = x^2 + 1$ , $du = 2x\,dx$ .

Change the terminals: $x = 0 \Rightarrow u = 1$ ; $x = 1 \Rightarrow u = 2$ .

$\displaystyle\int_0^1 \dfrac{2x}{x^2 + 1}\,dx = \int_1^2 \dfrac{1}{u}\,du = \big[\ln u\big]_1^2 = \ln 2 - \ln 1 = \ln 2$ .

Markers reward recognising the $\frac{f'}{f}$ form (substitution), changing the terminals, and the exact value $\ln 2$ .

Practice5 marks

(a) Express

\dfrac{5}{(x - 1)(x + 4)}

in partial fractions. (b) Hence evaluate

\displaystyle\int_2^{6} \dfrac{5}{(x - 1)(x + 4)}\,dx

, giving the exact answer.

Show worked answer →

(a) Write $\dfrac{5}{(x - 1)(x + 4)} = \dfrac{A}{x - 1} + \dfrac{B}{x + 4}$ , so $5 = A(x + 4) + B(x - 1)$ .

Let $x = 1$ : $5 = 5A$ , so $A = 1$ . Let $x = -4$ : $5 = -5B$ , so $B = -1$ . Thus $\dfrac{5}{(x - 1)(x + 4)} = \dfrac{1}{x - 1} - \dfrac{1}{x + 4}$ .

(b) $\displaystyle\int_2^6 \left(\dfrac{1}{x - 1} - \dfrac{1}{x + 4}\right)dx = \big[\ln|x - 1| - \ln|x + 4|\big]_2^6 = \left(\ln 5 - \ln 10\right) - \left(\ln 1 - \ln 6\right)$ .

$= \ln\dfrac{5}{10} - \ln\dfrac{1}{6} = \ln\dfrac{1}{2} + \ln 6 = \ln 3$ .

Markers reward the partial-fraction constants, integrating to logarithms, and the exact value $\ln 3$ .