Integration techniques: substitution and partial fractions (QCE Specialist Mathematics Unit 4)

What this dot point is asking

QCAA wants you to evaluate integrals that go beyond the standard antiderivatives by choosing and applying a technique: substitution, partial fractions, or a trigonometric simplification. You then apply these to find areas and volumes of solids of revolution. Integration techniques are core to IA3 and the external assessment, where selecting the right method quickly is the difference between a full and partial response.

The answer

Integration by substitution

Substitution reverses the chain rule. If the integrand contains a function and (a multiple of) its derivative, let $u$ be the inner function. With $u = g(x)$ and $du = g'(x)\,dx$,

For a definite integral, change the limits to $u$-values rather than converting back to $x$:

Changing the limits removes the need to back-substitute and is the cleaner approach.

Integration by partial fractions

A proper rational function with a factorisable denominator can be split into simpler fractions. For distinct linear factors,

and each piece integrates to a logarithm:

Find $A$ and $B$ by multiplying out and equating coefficients, or by substituting the values $x = \alpha$ and $x = \beta$. If the numerator degree is not less than the denominator degree, divide first.

Trigonometric integrals

Use identities to rewrite products and powers into integrable forms. The double-angle identity $\cos^2 x = \tfrac{1}{2}(1 + \cos 2x)$ handles even powers, and $\sin^2 x = \tfrac{1}{2}(1 - \cos 2x)$ similarly. For odd powers of sine or cosine, peel off one factor and substitute. The Pythagorean identity $\sin^2 x + \cos^2 x = 1$ converts between the two functions.

Volumes of solids of revolution

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

Each thin disc has radius $f(x)$ and thickness $dx$, so its volume is $\pi[f(x)]^2\,dx$; integrating sums the discs. Rotation about the $y$-axis uses $V = \pi\int [g(y)]^2\,dy$ with $x = g(y)$.

Repeated and quadratic factors in partial fractions

Distinct linear factors are the simplest case, but denominators can be more involved. A repeated linear factor needs a term for each power, so $\dfrac{1}{(x - 1)^2}$ contributes $\dfrac{A}{x - 1} + \dfrac{B}{(x - 1)^2}$. An irreducible quadratic factor $x^2 + a^2$ contributes a numerator linear in $x$, namely $\dfrac{Cx + D}{x^2 + a^2}$, which integrates to a logarithm (from the $Cx$ part) plus an inverse tangent (from the $D$ part). Setting up the right form before solving for the constants is half the work, and a wrong template makes the system unsolvable.

Choosing the substitution

A good substitution makes the derivative of the inner function appear (up to a constant) elsewhere in the integrand. Telltale signs are a composite function with its inner derivative present, such as $6x$ alongside $3x^2 + 1$, or an expression of the form $\dfrac{g'(x)}{g(x)}$, whose integral is $\ln|g(x)|$. When the integrand is a single awkward function with no obvious inner derivative, a trigonometric substitution (for example $x = a\sin\theta$ to handle $\sqrt{a^2 - x^2}$) may be the intended route.

Standard integrals to recognise

Fluent integration depends on recognising a small library of standard forms: $\displaystyle\int \dfrac{1}{x}\,dx = \ln|x| + C$, $\displaystyle\int \dfrac{1}{a^2 + x^2}\,dx = \dfrac{1}{a}\arctan\dfrac{x}{a} + C$, $\displaystyle\int \dfrac{1}{\sqrt{a^2 - x^2}}\,dx = \arcsin\dfrac{x}{a} + C$, and $\displaystyle\int e^{kx}\,dx = \dfrac{1}{k}e^{kx} + C$. Many integration questions reduce, after a substitution or a partial-fraction split, to one of these forms, so the strategy is always to manipulate the integrand until it matches a standard result.