Integration techniques and applications (WACE Year 12 Mathematics Specialist Unit 4)

Syllabus dot point

“Integrate using substitution, partial fractions and trig identities, and apply integration to volumes and differential equations”

10 min readUpdated 2 Jun 2026

Quick answer

Integrate harder functions using $u$ -substitution, partial fractions (splitting a rational function into simpler pieces), and trigonometric identities such as the double-angle and Pythagorean forms; then apply definite integrals to find volumes of revolution $V = \pi\int_a^b y^2\,dx$ and to solve separable differential equations $\frac{dy}{dx} = f(x)g(y)$ by separating variables and integrating.

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What this dot point is asking
Substitution
Partial fractions
Trigonometric integrals
Volumes of revolution
Differential equations
Choosing the technique from the integrand
Definite integrals and applications
Exam-style questions

What this dot point is asking

SCSA Unit 4 brings together a toolkit of integration methods and their applications. You choose the technique from the structure of the integrand, evaluate definite integrals, find volumes generated by rotating a region about an axis, and solve first-order separable differential equations including growth, decay and logistic models.

Substitution

Reverse the chain rule. Choose $u = g(x)$ so that $du = g'(x)\,dx$ appears (up to a constant) in the integrand, rewrite everything in terms of $u$ , integrate, and for a definite integral either change the limits to $u$ -values or convert back to $x$ .

Partial fractions

A proper rational function with a factorable denominator splits into simpler fractions. For distinct linear factors,

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

and each piece integrates to a logarithm. If the numerator degree is greater than or equal to the denominator degree, perform polynomial division first.

Partial fractions

Evaluate $\displaystyle\int \frac{5x - 1}{(x - 1)(x + 2)}\,dx$ .

Step 1: Set up the partial-fraction decomposition

The integrand is a proper rational function whose denominator factors into two distinct linear terms. Expressing it as a sum of simpler fractions, each with one linear factor in the denominator, converts the integral into two standard logarithmic integrals. Multiplying through by $(x - 1)(x + 2)$ gives a polynomial identity.

\frac{5x - 1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2} \implies 5x - 1 = A(x+2) + B(x-1)

Step 2: Find the constants by substitution

Setting $x$ equal to each factor's root eliminates one constant at a time, giving a direct equation for the other. This substitution method is faster and less error-prone than expanding and equating coefficients.

Set $x = 1$ : $5(1) - 1 = A(1 + 2)$ , so $4 = 3A$ and $A = \tfrac{4}{3}$ .

Set $x = -2$ : $5(-2) - 1 = B(-2 - 1)$ , so $-11 = -3B$ and $B = \tfrac{11}{3}$ .

Step 3: Integrate each partial fraction

Each piece now has the form $\tfrac{k}{x - a}$ , whose antiderivative is $k\ln|x - a|$ . Combining the two results and adding the constant of integration gives the final answer.

\int \frac{5x - 1}{(x-1)(x+2)}\,dx = \frac{4}{3}\ln|x - 1| + \frac{11}{3}\ln|x + 2| + C

Final answer: $\dfrac{4}{3}\ln|x - 1| + \dfrac{11}{3}\ln|x + 2| + C$ .

Trigonometric integrals

Use identities to rewrite integrands into directly integrable forms. The key ones are the double-angle identities for $\cos^2$ and $\sin^2$ :

Power-reduction identities

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

For example $\displaystyle\int \cos^2 x\,dx = \int \frac{1 + \cos 2x}{2}\,dx = \frac{x}{2} + \frac{\sin 2x}{4} + C$ .

Volumes of revolution

Rotating the region under $y = f(x)$ between $x = a$ and $x = b$ about the $x$ -axis generates a solid of volume

V = \pi \int_a^b y^2\,dx.

Rotation about the $y$ -axis uses $V = \pi\displaystyle\int_c^d x^2\,dy$ with $x$ expressed in terms of $y$ .

Differential equations

A separable equation $\dfrac{dy}{dx} = f(x)g(y)$ is solved by separating variables and integrating both sides:

\int \frac{1}{g(y)}\,dy = \int f(x)\,dx.

This produces a general solution with one constant; an initial condition fixes the constant. Exponential growth and decay ( $\frac{dy}{dt} = ky$ ) and logistic models appear in context.

Choosing the technique from the integrand

The first move on any integral is to read its structure and pick the matching technique. If the integrand contains a function and a multiple of its derivative, use substitution. If it is a proper rational function with a factorable denominator, use partial fractions; if it is improper, divide first. If it involves powers or products of sine and cosine, use the trigonometric identities, with power reduction for even powers and split-and-substitute for odd powers. If it is of the standard form $\dfrac{1}{a^2 + x^2}$ or $\dfrac{1}{\sqrt{a^2 - x^2}}$ , recognise the inverse-trig antiderivative directly. Building this diagnostic habit means you spend the exam executing a chosen method rather than guessing among them.

Definite integrals and applications

A definite integral evaluates to a number and is the workhorse for the applications in this dot point. For an area between curves it is top-minus-bottom integrated over the interval; for a volume of revolution it is $\pi\int r^2$ over the axis range; for a distance travelled it is the integral of speed. In each case the technique above is used to find the antiderivative, then the fundamental theorem of calculus evaluates it at the limits. Keeping the application separate from the integration technique, decide what integral the problem requires, then choose the method to compute it, keeps multi-step questions organised.

Solving a separable differential equation

Solve $\dfrac{dy}{dx} = 2xy$ with $y(0) = 3$ .

Step 1: Separate the variables

The equation can be rearranged so that all $y$ terms are on the left and all $x$ terms are on the right. This is possible because the right-hand side is a product of a function of $x$ and a function of $y$ , which is the defining feature of a separable equation.

\frac{1}{y}\,dy = 2x\,dx

Step 2: Integrate both sides

Integrating $\tfrac{1}{y}$ with respect to $y$ gives $\ln|y|$ , and integrating $2x$ with respect to $x$ gives $x^2$ . Only one constant of integration is needed; it appears on the right-hand side.

\ln|y| = x^2 + C

Step 3: Solve for $y$

Exponentiating both sides removes the logarithm. The constant $e^C$ is positive but its sign is absorbed by writing $A = \pm e^C$ , allowing $A$ to take any non-zero real value.

y = A e^{x^2} \quad \text{where } A = \pm e^{C}

Step 4: Apply the initial condition

Substituting $x = 0$ and $y = 3$ into the general solution determines the value of $A$ . Since $e^0 = 1$ , the equation simplifies immediately.

3 = A e^{0} = A \implies A = 3

y = 3 e^{x^2}

Final answer: $y = 3e^{x^2}$ .

Common traps

Not changing the limits after substitution: In a definite integral, either convert the limits to $u$ -values or convert back to $x$ before substituting the original limits. Mixing them gives a wrong number.
Squaring incorrectly in volumes: $V = \pi\int y^2\,dx$ uses $y^2$ , so $(2x + 1)^2$ must be fully expanded before integrating, not integrated and then squared.
Forgetting partial-fraction division: If the integrand is improper (numerator degree at least the denominator degree), divide first; otherwise the partial fractions are wrong.
Losing the constant in a differential equation: Integrate, include $+C$ , then apply the initial condition. Applying the condition before integrating, or dropping $C$ , breaks the solution.

Exam-style questions

Questions in the style of SCSA 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.

WACE 2023-style8 marks

Calculator-assumed. The region bounded by

y = \sqrt{x}

, the

x

-axis and the line

x = 4

is rotated about the

x

-axis. (a) Find the volume of the solid generated. (b) Hence find

\displaystyle\int_0^4 \pi x\,dx

and confirm your answer.

Show worked answer →

A volume of revolution drawing on several techniques.

(a) $V = \pi\displaystyle\int_0^4 y^2\,dx = \pi\displaystyle\int_0^4 (\sqrt{x})^2\,dx = \pi\displaystyle\int_0^4 x\,dx = \pi\left[\dfrac{x^2}{2}\right]_0^4 = \pi \times 8 = 8\pi$ .

(b) The integral $\displaystyle\int_0^4 \pi x\,dx = \pi\left[\dfrac{x^2}{2}\right]_0^4 = 8\pi$ , the same value, confirming the volume.

Markers reward $V = \pi\int y^2\,dx$ , squaring $\sqrt{x}$ to $x$ , the antiderivative, and the volume $8\pi$ cubic units.

WACE 2021-style6 marks

Calculator-free. Solve the differential equation

\dfrac{dy}{dx} = \dfrac{x}{y}

given that

y = 3

when

x = 0

Show worked answer →

A separable differential equation.

Separate: $y\,dy = x\,dx$ . Integrate both sides: $\dfrac{y^2}{2} = \dfrac{x^2}{2} + C$ , so $y^2 = x^2 + 2C$ . Write $2C = K$ : $y^2 = x^2 + K$ .

Apply $y = 3$ at $x = 0$ : $9 = 0 + K$ , so $K = 9$ . Thus $y^2 = x^2 + 9$ , and since $y = 3 > 0$ , $y = \sqrt{x^2 + 9}$ .

Markers reward separating the variables, integrating both sides, applying the initial condition, and choosing the positive root.

Practise this

Sources & how we know this

Mathematics Specialist ATAR course: syllabus and support materials — SCSA
Past ATAR course examinations — SCSA

What this dot point is asking

Substitution

Partial fractions

Step 1: Set up the partial-fraction decomposition

Step 2: Find the constants by substitution

Step 3: Integrate each partial fraction

Trigonometric integrals

Volumes of revolution

Differential equations

Choosing the technique from the integrand

Definite integrals and applications

Step 1: Separate the variables

Step 2: Integrate both sides

Step 3: Solve for yyy

Step 4: Apply the initial condition

Exam-style questions

Practise this

Sources & how we know this

Step 3: Solve for $y$