In VCE Specialist Mathematics, integration extends beyond the basic power rule to include more complex functions and techniques. This involves mastering the integration of rational, trigonometric, exponential, and logarithmic functions using advanced methods such as substitution and integration by parts.
Before applying advanced techniques, you must be fluent in the standard forms derived from Mathematical Methods and extended in Specialist Mathematics.
| Function $f(x)$ | Antiderivative $\int f(x) \, dx$ |
|---|---|
| $x^n$ | $\frac{x^{n+1}}{n+1} + c, \quad n \neq -1$ |
| $\frac{1}{ax+b}$ | $\frac{1}{a} \log_e |
| $e^{ax+b}$ | $\frac{1}{a} e^{ax+b} + c$ |
| $\sin(ax+b)$ | $-\frac{1}{a} \cos(ax+b) + c$ |
| $\cos(ax+b)$ | $\frac{1}{a} \sin(ax+b) + c$ |
| $\sec^2(ax+b)$ | $\frac{1}{a} \tan(ax+b) + c$ |
REMEMBER: Always include the constant of integration $+c$ for indefinite integrals. For logarithmic results, the absolute value $|ax+b|$ is technically required to ensure the argument is positive, though VCAA often defines the domain such that $ax+b > 0$.
Rational functions are ratios of polynomials. The technique used depends on the degree of the numerator and denominator.
If the integrand is of the form $\frac{g’(x)}{g(x)}$, the result is a natural logarithm:
$$\int \frac{g’(x)}{g(x)} \, dx = \log_e|g(x)| + c$$
When the denominator is a product of linear factors, use partial fractions to decompose the integrand.
* Form: $\frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$
* Method: Equate numerators and solve for $A$ and $B$ by substituting $x=a$ and $x=b$.
If the degree of the numerator is $\ge$ the degree of the denominator, perform polynomial long division first.
$$\text{Integrand} = \text{Quotient} + \frac{\text{Remainder}}{\text{Divisor}}$$
EXAM TIP: If you see a quadratic in the denominator that cannot be factored, consider completing the square to use inverse trigonometric forms (e.g., $\arctan$ or $\arcsin$).
Advanced trigonometric integration relies heavily on identities to transform the integrand into a integrable form.
KEY TAKEAWAY: For even powers of $\sin(x)$ or $\cos(x)$, you must use double-angle formulas to reduce the degree of the expression before integrating.
Substitution is the “reverse chain rule.” It is used when the integrand contains a function and its derivative.
If $u = g(x)$, then $du = g’(x) \, dx$:
$$\int f(g(x))g’(x) \, dx = \int f(u) \, du$$
When performing substitution on a definite integral, you must change the terminals (limits of integration).
$$\int_{a}^{b} f(g(x))g’(x) \, dx = \int_{g(a)}^{g(b)} f(u) \, du$$
COMMON MISTAKE: Forgetting to change the limits of integration in a definite integral. If you change the variable to $u$, the boundaries $a$ and $b$ must also be converted using the substitution formula $u = g(x)$.
Integration by parts is the “reverse product rule.” It is used when the integrand is a product of two unrelated functions (e.g., $x e^x$ or $x \sin(x)$).
$$\int u \frac{dv}{dx} \, dx = uv - \int v \frac{du}{dx} \, dx$$
To choose which part of the integrand should be $u$, follow the LIATE priority:
1. Logarithmic functions ($\log_e x$)
2. Inverse trigonometric functions ($\arcsin x$, $\arctan x$)
3. Algebraic functions ($x^n$, polynomials)
4. Trigonometric functions ($\sin x$, $\cos x$)
5. Exponential functions ($e^x$)
Pick $u$ based on which function appears higher in the list.
VCAA FOCUS: Integration by parts is frequently tested in Section B (Extended Response) of Exam 2, often requiring you to show the intermediate steps of the formula application.
| Integrand Type | Recommended Technique |
|---|---|
| $\frac{\text{linear}}{\text{quadratic (factorable)}}$ | Partial Fractions |
| $\sin^n(x)$ where $n$ is odd | Substitution ($u = \cos x$) |
| $\sin^n(x)$ where $n$ is even | Double-angle Identities |
| $f(g(x)) \cdot g’(x)$ | Substitution ($u = g(x)$) |
| $x^n \cdot e^x$ or $x^n \cdot \sin(x)$ | Integration by Parts |
| $\log_e(x)$ | Integration by Parts (let $u = \log_e(x), dv = 1$) |
STUDY HINT: Practice recognizing “patterns.” If you see a function and its derivative multiplied together, think substitution. If you see two different types of functions multiplied, think integration by parts.