Station 01 · Trigonometric products as sums
Products to sums
Turn a product like sin 3x cos x into a sum, and it becomes easy to integrate and to solve.
- derive the four identities
- ∫ sin mx cos nx dx and friends
- solving equations over restricted domains
Convert and integrate
Deriving the identities
Solving equations
Station 02 · The t-formulas
t = tan(A/2)
The line from (−1, 0) to the point at angle A on the unit circle has gradient t = tan(A/2), because the angle at the circumference is half the angle at the centre. Its y-intercept is t.
- sin A = 2t/(1 + t²)
- cos A = (1 − t²)/(1 + t²)
- tan A = 2t/(1 − t²)
- solving trig equations
Deriving the t-formulas
Solve a sin x + b cos x = c for 0 ≤ x ≤ 2π
Station 03 · Integration by substitution
Substitution
Choose u so that its derivative appears as a factor. For definite integrals, change the limits as well.
- given substitutions
- choosing a substitution
- changing limits
- trig and t substitutions
Pick the substitution
Check any antiderivative
Differentiates your answer numerically and compares it with the integrand.
Station 04 · Partial fractions
Partial fractions
Split a rational function into simpler fractions you can integrate term by term. Change the numbers to build your own.
- distinct linear factors
- repeated (perfect square) factors
- irreducible quadratic factors
- integrating the result
Numerator coefficients from the highest power down: 5, 1 means 5x + 1; 1, 0, 2 means x² + 2. Its degree must be less than the denominator's.
Decompose, then integrate
Station 05 · Rational functions and algebraic manipulation
Completing the square, division and rearranging
An irreducible quadratic denominator leads to ln and tan⁻¹. A numerator of equal or higher degree needs division first.
- completing the square
- degree of numerator ≥ degree of denominator
- algebraic manipulation
∫ (px + q) / (x² + bx + c) dx
Divide first: ∫ N(x) / D(x) dx
2, 3, 0, 1 means 2x³ + 3x² + 1. D may be linear or quadratic.
Algebraic manipulation
Station 06 · Integration by parts
Integration by parts
∫ u dv = uv − ∫ v du. The choice of u decides whether the new integral is easier or harder.
- derive the method
- choosing u
- repeated application
- cyclic integrals
- definite integrals
Choose u
Deriving the formula
Repeated parts: the tabular (DI) method
Station 07 · Recurrence relations
Reduction formulas
Integration by parts turns Iₙ into an expression involving Iₙ₋₁ or Iₙ₋₂. Then climb the ladder from I₀ or I₁.
- derive recurrence relations by parts
- use them to evaluate integrals
Climbing the ladder
The integrands
Darker curves are higher n.
Station 08 · Practice
Practice and HSC-style questions
Exact answers can be typed with ln, pi, sqrt, e and arctan, for example ln(3/2) or pi/8.