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

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 $u = g(b)$:

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,

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

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:

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.

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.

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