Apply integration techniques including substitution, integration by partial fractions, and trigonometric integrals, and use them to evaluate definite integrals and compute areas and volumes of solids of revolution

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 $x = g(y)$.