MTH1020 · Analysis of Change
Integration Techniques
Week 11 of Monash MTH1020 Analysis of Change assembles the toolkit for evaluating integrals that the standard antiderivatives cannot handle directly. It covers substitution (the reverse chain rule), partial fractions for rational functions, integration by parts (the reverse product rule), and trigonometric integrals and substitutions. Recognising which technique a given integral needs is itself the examinable skill, and these methods are staple final-exam questions that feed directly into the Week-12 applications of integration.
What this chapter covers
- 01Substitution (reverse chain rule): spot (inner function)·(its derivative), set u = inner, replace dx via du
- 02Definite integrals by substitution: convert the terminals to the new variable u
- 03Partial fractions: split a proper rational function; do polynomial division first if improper
- 04Partial-fraction forms: linear A/(ax+b), repeated (ax+b)², irreducible quadratic (Dx+E)/(ax²+bx+c)
- 05Integration by parts: ∫ f g′ dx = f g − ∫ f′ g dx (reverse product rule); choosing f and g′
- 06Repeated by-parts and rearrangement (e.g. ∫eˣcos x dx solved after two applications)
- 07Trig integrals via double-angle: cos 2x = 1 − 2sin²x = 2cos²x − 1 to reduce even powers
- 08Trig substitution: √(a²−x²) via x = a sin u; standard forms such as ∫1/(x²+a²) dx = (1/a)Tan⁻¹(x/a) + C
Choosing the technique: substitution and integration by parts
- +1(a) Choose the substitution. The integrand is (function of x²)·(2x), and 2x is the derivative of the inner function x². Let u = x², so du = 2x dx, which replaces 2x dx exactly.
- +1(a) Integrate in u and back-substitute. ∫ 2x cos(x²) dx = ∫ cos u du = sin u + C = sin(x²) + C. (Check: d/dx[sin(x²)] = cos(x²)·2x ✓.)
- +1(b) Set up integration by parts. Use ∫ f g′ dx = f g − ∫ f′ g dx. Choose f = x (so f′ = 1, which simplifies) and g′ = eˣ (so g = eˣ).
- +1(b) Apply the formula and finish. ∫ x eˣ dx = x eˣ − ∫ 1·eˣ dx = x eˣ − eˣ + C = eˣ(x − 1) + C. (Check: d/dx[eˣ(x−1)] = eˣ(x−1) + eˣ = x eˣ ✓.)
Key terms
- Substitution (reverse chain rule)
- Setting u = inner function when the integrand is (function of inner)·(derivative of inner), replacing dx via du = (du/dx) dx, integrating in u, then back-substituting. Convert the limits for definite integrals.
- Partial fractions
- Splitting a proper rational function into simpler fractions (over linear, repeated and irreducible-quadratic factors) that integrate individually. Do polynomial division first if the fraction is improper.
- Integration by parts
- ∫ f g′ dx = f g − ∫ f′ g dx, the reverse of the product rule. Choose f to differentiate (getting simpler) and g′ to integrate.
- Improper rational function
- A quotient g/h with deg g ≥ deg h. Reduce it to a polynomial plus a proper remainder r/h by polynomial division before applying partial fractions.
- Trigonometric substitution
- Replacing x with a trig function to simplify a root; e.g. x = a sin u turns √(a²−x²) into a cos u. Standard companion result: ∫1/(x²+a²) dx = (1/a)Tan⁻¹(x/a) + C.
- Double-angle reduction
- Using cos 2x = 1 − 2sin²x = 2cos²x − 1 to rewrite even powers of sine or cosine into a form that integrates directly.
Integration Techniques FAQ
How do I know which integration technique to use?
Read the structure of the integrand. If you see a composite function multiplied by (something close to) the derivative of its inner part, use substitution. If you see a product of two unlike functions — a polynomial times an exponential or trig, say — use integration by parts. If you have a rational function (a ratio of polynomials), use partial fractions, after polynomial division if it is improper. Roots like √(a²−x²) point to a trig substitution. Recognising the type is itself examined, so practise classifying before integrating.
How do I choose f and g′ for integration by parts?
Pick f to be the factor that becomes simpler when differentiated, and g′ to be the factor you can integrate. The LIATE guide (Logarithmic, Inverse-trig, Algebraic, Trigonometric, Exponential) suggests which to call f, earliest first. For ∫ x eˣ dx you take f = x (its derivative 1 is simpler) and g′ = eˣ, so the remaining integral ∫ eˣ dx is easy. A poor choice makes the new integral harder, not easier — a signal to swap.
Why does substitution need the derivative of the inner function present?
Substitution is the chain rule run backwards, and the chain rule produces exactly that inner derivative as a factor. When you set u = inner, du = (inner)′ dx, so you can only cleanly replace dx if (inner)′ already appears (up to a constant) in the integrand. In ∫ 2x cos(x²) dx the factor 2x is precisely d/dx[x²], which is why the substitution closes perfectly to ∫ cos u du.
Can Sia help me pick and apply integration techniques?
Yes. Sia can help you classify an integral, choose the substitution or the by-parts split, and carry the calculation step by step — then check your antiderivative by differentiating it back with you. It teaches the method and checks your reasoning on your own practice questions; it does not do graded assessment for you, and Monash academic-integrity rules apply.
Exam move
The examinable skill here is recognition, so drill classification before calculation: given twenty mixed integrals, first label each as substitution, parts, partial fractions or trig substitution, then evaluate. Keep a decision card — composite-with-inner-derivative → substitution; product of unlike functions → by parts; ratio of polynomials → partial fractions; root √(a²−x²) → trig substitution. Always verify an antiderivative by differentiating it back, and never drop the + C or, for definite integrals, forget to convert the limits. These techniques are staple final-exam questions and are exactly what Week 12 uses to compute areas and volumes, so over-learn them. Ask Sia to generate a mixed set and check both your classification and your working.
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