Advanced Integration and Inverse Trigonometric Functions
Integrating through partial fractions including quadratic factors, differentiating inverse trigonometric functions and choosing appropriate trigonometric substitutions.
How to study A-level Further Mathematics
Define the objects and conditions, select a representation, carry out exact mathematics, then validate, interpret and communicate the result.
Core concepts
Concept 1
Partial fractions decompose a rational function according to repeated linear and irreducible quadratic factors.
Exam cue: Compare polynomial degrees and divide first when the rational function is improper.
Concept 2
Inverse trigonometric derivatives depend on both formula and domain.
Exam cue: Choose numerator forms Ax + B for irreducible quadratic factors.
Concept 3
Substitutions such as x = a sin theta or x = a tan theta simplify characteristic square-root or quadratic forms.
Exam cue: Set the substitution range so the back-substituted signs are justified.
Risk pitfalls and guardrails
Using a constant numerator above a quadratic factor.
Guardrail: Do not replace proof with examples, exact reasoning with unverified calculator output, or a valid awarding-body route with an invented mix of options.
Dropping the derivative of the inner function.
Guardrail: Do not replace proof with examples, exact reasoning with unverified calculator output, or a valid awarding-body route with an invented mix of options.
Back-substituting from a triangle without checking the angle range.
Guardrail: Do not replace proof with examples, exact reasoning with unverified calculator output, or a valid awarding-body route with an invented mix of options.
Memory anchors
Improper Rational Function
An improper rational function has numerator degree at least as large as denominator degree.
Quadratic Partial Fraction
An irreducible quadratic factor requires a linear numerator.
Inverse Trigonometric Function
An inverse trigonometric function returns an angle on a specified principal range.
Trigonometric Substitution
A trigonometric substitution uses an identity to simplify an algebraic form.
Back-substitution
Back-substitution returns the antiderivative to the original variable with valid signs and domains.
Checkpoint rule
Do the check-up only after you can summarize each concept in one sentence and identify one dangerous pitfall from memory.
Knowledge Check (after reading)
Short check-up to confirm understanding of this module.
Check-up Questions
Decompose (5x+1)/[(x−1)(x+2)].
Which partial-fraction form is required for 1/[(x−1)²(x²+4)]?
Answer all questions to submit.
Next step personalized recommendations
Continue learning
Move forward only after this module is stable.
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