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Which techniques unlock the integrals and applications of Specialist Unit 4?

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

WACE Specialist Unit 4 integration: substitution, partial fractions, trigonometric identities and the double-angle method, definite integrals, volumes of revolution, and solving separable differential equations including exponential and logistic models, with worked examples.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-02

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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

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 $\sin^2$ :

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

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 20238 marksCalculator-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 20216 marksCalculator-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.