Technique Masterclasses

Pure Mathematics 3

Deep Dive: Integration By Recognition

Provenance (2026-09-29): read research/veridian/WMA13-verified-facts.md in full and grepped it specifically for spec 5.2 (§1 spec table, §3 formula-booklet/memorise-list split, §4 mark-scheme conventions, §5 Integration, §8 flagged issues, §9 lesson-audit log for wma13-integration-by-recognition); read content/veridian/curriculum/wma13/integration-by-recognition.ts in full (897 lines, including its own header's sourcing notes); read all three existing technique-hub documents in content/veridian/technique-hub/wma13/ (codex-formula-and-method-bank.md, forge-exam-technique-guide.md, signal-frequency-and-boundaries.md) end to end, and confirmed content/veridian/question-bank/wma13.ts contains no spec-5.2 material to duplicate. Beyond those, I went back to the two primary-source-derived consolidated files the technique-hub documents themselves cite — ~/Documents/SOL/Papers/IAL_Paper3_Complete/Master_Consolidated_Docs/Mathematics_Paper3_Mark_Schemes.md and Mathematics_Paper3_Examiner_Reports.md — and located and read, in full, the October 2021 Q5 and January 2022 Q3 examiner-report prose and the June 2023 Q9(c) mark-scheme cell, none of which had been quoted at this level of detail anywhere in this course before. Every claim below is tagged to one of these places; nothing is carried over from memory of Pearson content generally.

CODEX Deep-Dive: Integration by Recognition (WMA13, spec 5.2)

Why this document exists

integration-by-recognition.ts already teaches the two named formulae, derives them, and drills them with a worked chain, a chain drill, a method-comparison and a marked solution. codex-formula-and-method-bank.md already gives the compressed formula-and-mark-skeleton version. This document does neither of those jobs again. It does three things that don't exist anywhere else in this course yet:

  1. States the actual theoretical structure spec 5.2 covers — not two formulae, but three genuinely distinct recognition mechanisms plus one look-alike that isn't recognition at all — and shows precisely why the spec only bothers to name two of the three as formulae.
  2. Pulls fuller primary-source detail than has been quoted before on this paper's own real spec-5.2 questions (Oct 2021 Q5, Jan 2022 Q3, Jun 2023 Q9(c)) directly from the underlying examiner-report and mark-scheme text, surfacing several genuinely new, verbatim-sourced error modes.
  3. Independently reconfirms, from a second primary-source location, a correction codex-formula-and-method-bank.md had already made to WMA13-verified-facts.md's own account of the June 2023 Q9(c) question.

Spec 5.2 is real but narrow evidence: signal-frequency-and-boundaries.md's own frequency table records integration (5.1–5.2 combined) as seen in 4 of the 13 series checked for this paper (Oct 2020, Oct 2021, Jan 2022, Jun 2023) — one of the paper's less frequently and less richly evidenced spec items, not absent, but not in the same league as functions (1.2) or numerical methods (6.1–6.2) at 9 of 13 each. This document's length is scoped to match that: deep on structure, honest about how thin the past-paper anchor set actually is.


1. The real theoretical structure

1.1 What the spec actually asks for

Verbatim, spec 5.2 (WMA13-verified-facts.md §1): "Integration by recognition of known derivatives to include integrals of the form ∫f′(x)/f(x)dx = ln(f(x))+c and ∫f′(x)[f(x)]ⁿdx = [f(x)]^(n+1)/(n+1)+c." The guidance column adds, verbatim: "E.g. to include integration of tan x, sec²2x; students expected to use trigonometric identities to integrate, e.g., sin²x, tan²x, cos²3x."

Read literally, that guidance line names three things the spec expects, not one:

The lesson file and the technique-hub bank both teach the two formulae and the identity-rewrite skill. Neither document stops to say explicitly why tan x and sec²2x sit in the guidance line at all, given that neither is literally f′(x)/f(x)f'(x)/f(x) or f′(x)[f(x)]nf'(x)[f(x)]^n for any natural choice of ff. That gap is this section's first job.

1.2 The one general principle behind everything in this topic

Both named formulae are specialisations of a single line the spec's own "must memorise" list states once (WMA13-verified-facts.md §3, integration): ∫f′(g(x)) g′(x) dx=f(g(x))+c\int f'(g(x))\,g'(x)\,dx = f(g(x)) + c — the chain rule, run backwards, for an arbitrary outer function ff. Recognition is the skill of spotting that an integrand already has this shape, for some outer function, disguised.

Differentiate the two candidate antiderivatives directly to see where the two named formulae come from as special cases of that one line:

That n≠−1n \neq -1 condition is not a footnote — it is why the log form exists as its own named formula at all. The ordinary power rule for integration, ∫un du=un+1n+1+c\int u^n\,du = \frac{u^{n+1}}{n+1}+c, has a removable-looking but genuinely fatal discontinuity at n=−1n=-1: the formula divides by zero. ln⁡∣u∣\ln|u| is exactly the patch for that one missing case. So the "two formulae" of spec 5.2 are not two unrelated facts to memorise — they are the single power-rule family for integrating [g(x)]ng′(x)[g(x)]^n g'(x), split at the one value of nn where that family's own formula breaks. (This single-formula-family framing is a direct mathematical consequence of the two mechanism derivations already in integration-by-recognition.ts; it is not itself sourced to a Pearson document — no exam board document states the two formulae are "the same family" — but it is checkable by direct differentiation, and it is a real structural fact about why the spec's own guidance groups these two formulae together rather than in separate sections.)

1.3 The third real sub-case the spec names but doesn't formula-ise: direct reverse-lookup

tan x and sec²2x, the guidance line's own named examples, are recognition integrals for a third reason: their outer function's reversed form is already sitting on the formula sheet or in the P3-memorised differentiation list, so no new formula needs deriving at all — the recognition is simply "this integrand is the derivative of something I already have, read backwards."

Checked directly against the booklet contents transcribed in WMA13-verified-facts.md §3:

This is structurally different from the log-form and power-form cases in §1.2: there, the spec supplies a general formula for an arbitrary f(x)f(x), because ln⁡∣u∣\ln|u| and un+1/(n+1)u^{n+1}/(n+1) are not themselves printed as reverse-differentiation entries anywhere on the sheet — the student has to build the antiderivative from the chain rule directly. Here, the outer function (tan⁡\tan, csc⁡\csc, or the plain power built into sec⁡2\sec^2) already has its own printed or memorised reverse-lookup entry, so "recognition" collapses to noticing a match against a table the student already has, not deriving anything new. Both are legitimately spec 5.2, and both require the same underlying skill — spot a disguised derivative — but they are not the same mechanism, and a document that only teaches the two named formulae is quietly leaving out the guidance line's own two worked examples as a distinct category.

1.4 The look-alike that is not recognition: squared trig functions

sin⁡2x\sin^2 x, cos⁡2x\cos^2 x, tan⁡2x\tan^2 x are the guidance line's third component, and they are explicitly not an instance of any of the three mechanisms above. integration-by-recognition.ts's own teach block gets the reason exactly right and it is worth restating precisely, because it is the single most checkable fact in this whole sub-topic: the power-form pattern needs f′(x)f'(x) present as a genuine multiplying factor alongside [f(x)]n[f(x)]^n. sin⁡2x\sin^2 x is [sin⁡x]2[\sin x]^2 with no accompanying cos⁡x\cos x factor anywhere in the integrand — there is nothing playing the role of f′(x)f'(x), and no algebraic manipulation of sin⁡2x\sin^2 x alone conjures one into existence. The fix is a substitution of technique, not a disguised application of the same one: rewrite the square away entirely, using the double-angle identities from spec 2.3 (memorised, not on the sheet — WMA13-verified-facts.md §3), and only then integrate two ordinary spec-5.1 terms with no recognition step left to do.

1.5 The modulus condition, stated precisely

∫f′(x)f(x) dx=ln⁡∣f(x)∣+c\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)|+c holds for every xx with f(x)≠0f(x)\neq 0, negative values included, because ln⁡∣u∣\ln|u| is defined and differentiates to 1/u1/u on the whole domain u≠0u\neq0, while ln⁡(u)\ln(u) alone is undefined the instant u<0u<0. Whether the modulus can be safely dropped in a specific question is a domain question about that specific f(x)f(x), not a stylistic choice — integration-by-recognition.ts's own MCQ bank already tests this correctly (three functions provably always positive, one provably negative on the stated interval). Nothing to add there theoretically; §3 below adds the real, quotable exam evidence for how often this specific step is dropped.


2. Micro-worked examples, one per real sub-case

Short illustrations only — the full-length worked chain, chain drill, method-comparison and marked solution for these already exist in integration-by-recognition.ts and are not reproduced here.

Log form, coefficient mismatch. ∫6x3x2+1 dx\int \frac{6x}{3x^2+1}\,dx: f(x)=3x2+1f(x)=3x^2+1, f′(x)=6xf'(x)=6x — numerator matches f′(x)f'(x) exactly here (no adjustment needed), so the integral is ln⁡∣3x2+1∣+c\ln|3x^2+1|+c directly. Check: ddx[ln⁡∣3x2+1∣]=6x3x2+1\frac{d}{dx}[\ln|3x^2+1|] = \frac{6x}{3x^2+1}. ✓.

Power form, positive integer nn. ∫10x(x2−4)4 dx\int 10x(x^2-4)^4\,dx: f(x)=x2−4f(x)=x^2-4, f′(x)=2xf'(x)=2x, so 10x(x2−4)4=5⋅f′(x)[f(x)]410x(x^2-4)^4 = 5\cdot f'(x)[f(x)]^4, integrating to 5⋅[f(x)]55+c=(x2−4)5+c5\cdot\frac{[f(x)]^5}{5}+c=(x^2-4)^5+c. Check: ddx[(x2−4)5]=5(x2−4)4⋅2x=10x(x2−4)4\frac{d}{dx}[(x^2-4)^5]=5(x^2-4)^4\cdot2x=10x(x^2-4)^4. ✓.

Power form, negative integer nn (the disguised-reciprocal case). ∫6(3x+1)2 dx=∫6(3x+1)−2 dx\int \frac{6}{(3x+1)^2}\,dx = \int 6(3x+1)^{-2}\,dx: f(x)=3x+1f(x)=3x+1, f′(x)=3f'(x)=3, so 6(3x+1)−2=2⋅f′(x)[f(x)]−26(3x+1)^{-2}=2\cdot f'(x)[f(x)]^{-2}, integrating to 2⋅[f(x)]−1−1+c=−23x+1+c2\cdot\frac{[f(x)]^{-1}}{-1}+c=\frac{-2}{3x+1}+c. Check: ddx[−23x+1]=6(3x+1)2\frac{d}{dx}\left[\frac{-2}{3x+1}\right] = \frac{6}{(3x+1)^2}. ✓. This is the exact structural shape of the real Oct 2021 Q5(i) question (§3.2 below), simplified to remove its definite-integral limits step.

Direct reverse-lookup (§1.3's third mechanism). ∫csc⁡23x dx\int \csc^2 3x\,dx: the printed differentiation-table result tan⁡kx→ksec⁡2kx\tan kx \to k\sec^2 kx has no direct csc⁡2\csc^2 counterpart on the P3 sheet, but cot⁡kx→−kcsc⁡2kx\cot kx \to -k\csc^2 kx is printed (WMA13-verified-facts.md §3), so ∫csc⁡23x dx=−13cot⁡3x+c\int \csc^2 3x\,dx = -\frac{1}{3}\cot 3x+c by the same reverse-lookup logic as ∫sec⁡22x dx=12tan⁡2x+c\int\sec^2 2x\,dx=\frac12\tan2x+c or ∫csc⁡xcot⁡x dx=−csc⁡x+c\int\csc x\cot x\,dx=-\csc x+c. Check: ddx[−13cot⁡3x]=−13×(−3csc⁡23x)=csc⁡23x\frac{d}{dx}\left[-\frac13\cot 3x\right] = -\frac13\times(-3\csc^23x)=\csc^23x. ✓.

Identity-rewrite, not recognition. ∫cos⁡25x dx\int \cos^2 5x\,dx: no sin⁡5x\sin 5x factor present, so this is not the power form. Using cos⁡2A≡2cos⁡2A−1⇒cos⁡2A≡1+cos⁡2A2\cos 2A \equiv 2\cos^2A-1 \Rightarrow \cos^2A\equiv\frac{1+\cos2A}{2} with A=5xA=5x: cos⁡25x=1+cos⁡10x2\cos^25x=\frac{1+\cos10x}{2}, integrating to x2+120sin⁡10x+c\frac{x}{2}+\frac{1}{20}\sin10x+c. Check: ddx[x2+120sin⁡10x]=12+12cos⁡10x=cos⁡25x\frac{d}{dx}\left[\frac x2+\frac1{20}\sin10x\right]=\frac12+\frac12\cos10x = \cos^25x by the same identity, run forward. ✓.


3. Every real exam angle, synthesised from primary sources

3.1 Real frequency and the honest scope of the evidence

signal-frequency-and-boundaries.md: integration (5.1–5.2 combined, not tallied separately in the source material) is seen in 4 of the 13 series checked for this paper — Oct 2020, Oct 2021, Jan 2022, Jun 2023 — and is not listed among WMA13-verified-facts.md §7's topics judged to have "enough verified material for a genuine standalone lesson" (that list names functions, numerical methods, exponential modelling, trig identities/harmonic form, and product/quotient/chain rule; integration by recognition is absent from it). The real past-paper anchor set for spec 5.2 specifically, across those 4 series, is: Oct 2020 Q9(c) (modulus-drop trap only — the full question is not otherwise transcribed in the research file), Oct 2021 Q5(i) and Q5(ii), Jan 2022 Q3(i) and Q3(ii), and Jun 2023 Q9(c). That is five real questions across four series. Everything below is drawn from those five, plus the formula-booklet/spec text already covered in §1.

3.2 The real mark-scheme architecture — verified across three of the five real questions

WMA13-verified-facts.md §9's own audit of this exact lesson (dated 2026-09) pulled Oct 2021 Q5(i)/(ii) and Jan 2022 Q3(i)/(ii) verbatim and states the finding precisely: for the antiderivative step itself, real WMA13 schemes always split it as one M1 (reach the correct form — power raised by one, or log of the bracket — with any constant, right or wrong) followed by one A1 (the exact constant, together with the +c+c on an indefinite integral, as a single combined requirement). dM1 is never used to separate "spot the pattern" from "get the constant exactly right" within one antiderivative; a real dM1 in this topic is reserved for the genuinely later, separate step of substituting definite-integral limits into an antiderivative already scored. Verbatim anchors: Oct 2021 Q5(ii) M1 — "Achieves k(x2+3)8k(x^2+3)^8 ... where kk is a constant"; its A1 — "116(x2+3)8+c\frac{1}{16}(x^2+3)^8+c. Must be in terms of xx and the +c+c must be present"; Jan 2022 Q3(ii) M1 — "Achieves bln⁡(1+2cos⁡x)b\ln(1+2\cos x) ... where bb is a constant"; its A1 — "Achieves −2ln⁡(1+2cos⁡x)-2\ln(1+2\cos x) ... There is no need for +c+c" (a definite integral, correctly carrying no +c+c requirement) followed by its own separate dM1/A1 pair for the limit substitution.

The same M1(form)/A1(exact) shape holds a third time, independently, in the question this document adds fuller detail on below (Jun 2023 Q9(c)): M1 "For using part (a) and achieving kcsc⁡xk\csc x oe for the integral (limits not required for this mark. May arise from longer methods, but must achieve the correct form", A1 for the exact evaluated value, with the explicit note "Must have scored the M — answer only with no integration shown is M0A0" (Mathematics_Paper3_Mark_Schemes.md, June 2023 section, Q9 notes). Three independent real questions, three independent confirmations of the identical generous-M1/exact-A1 shape — this is now the best-evidenced single fact in this sub-topic.

3.3 A genuinely fuller reading of Oct 2021 Q5 — new detail beyond what any existing document quotes

Mathematics_Paper3_Examiner_Reports.md (October 2021 section, Question 5) gives real detail beyond the single sentence already quoted in WMA13-verified-facts.md and the lesson file:

3.4 A genuinely fuller reading of Jan 2022 Q3 — three new, single-instance misconceptions

Mathematics_Paper3_Examiner_Reports.md (January 2022 section, Question 3) on part (ii), ∫0π/34sin⁡x1+2cos⁡x dx\int_0^{\pi/3}\frac{4\sin x}{1+2\cos x}\,dx: "Some candidates seemed at a loss... Some integrated numerator and denominator separately, others attempted to use a 'quotient rule' of sorts and others incorrectly attempted to split the integrand into asin⁡x+btan⁡xa\sin x+b\tan x — all such approaches which were given no credit." Three distinct, real, zero-credit misconceptions on the log-form pattern, none previously quoted in this course's WMA13 documents: (a) integrating the numerator and denominator as two separate integrals (a nonsensical operation for a quotient, but a real one candidates attempted); (b) reaching for differentiation's quotient rule as though it had an integration analogue (it does not — there is no general "reverse quotient rule" for integration); (c) trying to force the specific trig integrand into a sum of two standard spec-5.1 terms it does not actually decompose into. The same report also names a specific, real sign slip distinct from the modulus/constant-of-integration traps already documented: "It was fairly common though to see a multiplier of +2+2 rather than −2-2 in the result" — because f(x)=1+2cos⁡xf(x)=1+2\cos x differentiates to f′(x)=−2sin⁡xf'(x)=-2\sin x, and the numerator 4sin⁡x4\sin x is −2×(−2sin⁡x)-2\times(-2\sin x); missing the sign on f′(x)f'(x) itself, not just the magnitude, inverts the final coefficient.

On part (i)'s power-form pattern, ∫(2x−5)7 dx\int(2x-5)^7\,dx, the same report adds one real, quotable detail on where the coefficient slip actually lands: "It was not uncommon for slips in the calculation of the correct coefficient (1/16) and weaker candidates often obtained 1/8 instead" — a factor-of-2 slip specifically in dividing by (n+1)=8(n+1)=8 and by the constant f′(x)=2f'(x)=2 together, landing on half the correct coefficient rather than a randomly wrong one. It also independently reconfirms this document's §1.2 framing that most candidates default to direct recognition over substitution on the power-form pattern specifically: "Most candidates integrated here by recognition rather than by substitution. Although candidates who employed substitution here and again in part (ii) were usually successful" — and, for those who did substitute on part (ii): "Most candidates using substitution kept their limits in terms of xx, undoing their substitution before applying the limits... and subtracting" — i.e. the safe order of operations for a substituted definite integral is to convert back to xx before substituting the original limits, not to convert the limits into uu and risk a slip there instead. Both are real, viable orderings; the report is simply recording which one candidates who substituted actually used successfully.

3.5 The cosec x / cosec x cot x correction — independently reconfirmed from a second primary source

codex-formula-and-method-bank.md (§1.5 of that document) had already corrected WMA13-verified-facts.md's own account of June 2023 Q9(c): the research file's original write-up treated that question as testing ∫csc⁡x dx\int\csc x\,dx (a harder P4-booklet result whose presence on a P3 paper would be a genuine boundary question), but a direct re-check of the real mark scheme showed the question is actually ∫csc⁡xcot⁡x dx\int\csc x\cot x\,dx — the direct reverse-lookup case from §1.3 above, using the P3-printed derivative csc⁡x→−csc⁡xcot⁡x\csc x\to-\csc x\cot x, not the harder standalone csc⁡x\csc x antiderivative at all.

This document independently reconfirms that correction from the mark-scheme cell itself (Mathematics_Paper3_Mark_Schemes.md, June 2023 section, Q9(c)): "M1: For using part (a) and achieving kcsc⁡xk\csc x oe for the integral" — the target antiderivative is csc⁡x\csc x alone (from csc⁡xcot⁡x\csc x\cot x integrating to −csc⁡x-\csc x), not a csc⁡x\csc x-and-cot⁡x\cot x log expression, which is what a genuine ∫csc⁡x dx\int\csc x\,dx antiderivative would look like. The matching examiner-report prose (Mathematics_Paper3_Examiner_Reports.md, June 2023 section, Question 9) independently confirms the same reading in its own words: "showing an understanding to again use part (a) to replace the integrand and use the reverse differentiation of cosec x" and "Most students achieved the integral directly from the known result in the formula book" — "the reverse differentiation of cosec x" is precisely csc⁡xcot⁡x→−csc⁡x\csc x\cot x\to-\csc x, read backwards, not a derivation of ∫csc⁡x dx\int\csc x\,dx from scratch. The exact numeric value of the evaluated definite integral is garbled in this document's extraction of that mark-scheme cell (a fraction/root rendering casualty of pdftotext, the same class of extraction failure WMA13-verified-facts.md §2's own Summer-2026 audit pass records and works around by re-rendering the page as an image) and is left unverified here rather than guessed at — the mark-scheme structure and the identity of the integral being tested are the load-bearing, independently-confirmed facts; the specific evaluated number is not asserted.

One further real detail from the same examiner-report passage, not previously quoted anywhere in this course: "some students did not realise the result could be found from their Formula Booklet, and instead used a substitution and changed the limits, which were not always correct, and if the substitution was not undone lost the final mark. The substitution approach often gave the integral as −1sin⁡x-\frac{1}{\sin x}, which is of course also correct." Two real facts here: substitution is confirmed, a third time in this sub-topic (after tan x in the lesson's own method-comparison, and Q5/Q3 above), as a fully creditable alternative route on a recognition integral — and −1sin⁡x-\frac{1}{\sin x} is confirmed algebraically equivalent to −csc⁡x-\csc x, so a student reaching that form via substitution loses nothing for the alternative notation, only for failing to undo the substitution before applying limits. There is also a distinct, real zero-credit trap named in the same passage — "numerous students who did not show any form for the integral and simply used their calculators to evaluate directly scoring M0A0" — which is the same "answer only, no integration shown" boundary already stated as a mark-scheme rule in §3.2 above, now with its real consequence attached: a numerically correct final value earns nothing at all without the antiderivative line actually written down.

3.6 Synthesised trap taxonomy, with real counts

Every trap below is tagged with how many of the five real questions located for this sub-topic actually exhibit it — not a probability, a count.

TrapReal anchor(s)Count
Constant of integration omitted on an indefinite integralOct 2021 Q5(ii); Jan 2022 Q3(i)2 of 5
Modulus dropped in log-form integration, leading to ln⁡\ln(negative)Oct 2020 Q9(c); Jun 2023 Q9(c) (structurally, via the "answer only" route, not a modulus drop specifically — see below)2 of 5, but see caveat
Formula-booklet/memorised result not recognised, substitution used instead (valid, but slower and riskier)Jan 2022 Q3(i)/(ii) (substitution confirmed used successfully); Jun 2023 Q9(c)2 of 5
Answer given with no antiderivative line shown (calculator-only), scoring nothing regardless of correctnessJun 2023 Q9(c)1 of 5, single instance
Log-form and power-form patterns confused with each other (multiplying out, or reaching for a log where none is needed)Oct 2021 Q5(i)1 of 5, single instance
Differentiation applied where integration was required (power moves the wrong way)Oct 2021 Q5(i)1 of 5, single instance
Product treated as a sum — factors integrated separately instead of recognised as one unitOct 2021 Q5(ii)1 of 5, single instance
A non-existent "reverse quotient rule" or term-splitting attempted on a log-form quotientJan 2022 Q3(ii)1 of 5, single instance
Sign of f′(x)f'(x) lost, inverting the final coefficient (not just its magnitude)Jan 2022 Q3(ii)1 of 5, single instance
Coefficient miscalculated via a factor-of-2 slip on the (n+1)(n{+}1)-and-f′(x)f'(x) combinationJan 2022 Q3(i)1 of 5, single instance
Substituted definite integral: limits not undone/converted back correctly before applyingJun 2023 Q9(c)1 of 5, single instance

The caveat on the modulus row: Oct 2020 Q9(c) is a genuine, independently-verified modulus-drop (WMA13-verified-facts.md §5: "many did not have the modulus and left the values as 5ln(-2) and 5ln(-4) not realising these are undefined"); Jun 2023 Q9(c)'s failure mode, on closer reading via §3.5 above, is really the separate "answer only" trap, not a modulus drop specifically — it is listed against modulus in this course's existing documents only because both are integration-context errors on questions in the same "Q9(c)" slot across different series, not because they are the same mechanism. This document's count above treats them as the two separate things they actually are, per §3.5's correction.

Everything in this table beyond the first row is either new to this course's documentation (rows 4, 6, 7, 8, 9, 10, 11) or previously stated only as a single fused sentence without the distinct mechanism separated out (row 2's caveat). Nothing in the table is asserted beyond a plain count of the five located real anchors — no likelihood, frequency-in-general, or grade-boundary claim is attached to any row.


4. What remains genuinely unverified


Sources