Integration by Recognising a Known Derivative

Before you read on

Two or three questions on exactly what this lesson teaches. Being wrong here is fine — it's the fastest way to find out what to pay attention to next.

What this lesson assumes — recapped, not retaught

Spec 5.1 gives four standard integrals, used without comment from here on: $\int e^{kx}\,dx = \frac{1}{k}e^{kx}+c$ ; $\int \frac{1}{x}\,dx = \ln|x|+c$ (for $x \neq 0$ ); $\int \sin kx\,dx = -\frac{1}{k}\cos kx+c$ ; and $\int \cos kx\,dx = \frac{1}{k}\sin kx+c$ . All four sit on the spec's own 'must memorise' list, not the formula booklet — reversing a derivative already known is the whole method, so there is nothing to look up.

The is used just as constantly, in reverse. Differentiating a composite function $f(g(x))$ gives $f'(g(x)) \cdot g'(x)$ : the outer function's derivative, evaluated at the inner function, multiplied by the inner function's own derivative. Two specific instances of it matter for this whole lesson: $\frac{d}{dx}[\ln u] = \frac{1}{u}\cdot\frac{du}{dx}$ and $\frac{d}{dx}[u^n] = nu^{n-1}\cdot\frac{du}{dx}$ , both with $u$ standing for some function of $x$ . Recognising an integral of either type covered here is nothing but spotting one of these two shapes already sitting in the integrand — in reverse, and often in disguise.

The identity this lesson also leans on — memorised, not looked up

Spec 2.3's double-angle formulae get used later in this lesson to rewrite squared trig functions before integrating them, and — like the standard integrals above — none of the following are printed on the formula sheet; the spec states plainly that this list 'will not appear in the booklet.' $\cos 2A$ has three equivalent forms, each useful for a different target: $\cos 2A \equiv \cos^2 A - \sin^2 A \equiv 2\cos^2 A - 1 \equiv 1 - 2\sin^2 A$ .

One more earns its place here for the same reason: $\sec^2 A \equiv 1 + \tan^2 A$ , rearranged to $\tan^2 A \equiv \sec^2 A - 1$ . $\tan^2 x$ has no elementary to reach for directly, but $\sec^2 x - 1$ is a sum of two things already integrable: $\sec^2 kx$ is printed directly on the formula sheet ( $\int \sec^2 kx\,dx = \frac{1}{k}\tan kx + c$ ), and $\int 1\,dx = x + c$ needs no formula at all.

Why "5 ÷ the bracket" integrates to a log of the bracket — before the letters

In plain terms

Before any letters appear, check the central claim of this lesson with one plain bracket and real numbers only — no f, no x as an abstract symbol, just a rule in words and the arithmetic it produces. The bracket's rule: take a chosen whole number, multiply it by 5, then add 2. Choose the number 1: the bracket comes out at 5 × 1 + 2 = 7. Choose the number 4 instead: the bracket comes out at 5 × 4 + 2 = 22. Whichever number is chosen, the bracket's own rate of change is a flat 5 — every time the chosen number goes up by 1, the bracket goes up by exactly 5, because that's what "multiply by 5" means; that flat 5 is the number this lesson will later call the bracket's derivative. Now check the actual claim: that adding up "5 ÷ the bracket" as the chosen number climbs gives back a log of the bracket, plus a constant. Test a claim about undoing a derivative the way any such claim gets tested — differentiate the proposed answer and see whether it lands back on the original question. Differentiating a log of a bracket means: take the bracket's rate of change, and divide it by the bracket's own current value. At the number 1: rate of change (5) ÷ bracket's value (7) = 5/7 — compare that with the original expression, "5 ÷ the bracket", at the number 1: 5 ÷ 7. Exact match. At the number 4: rate of change (5) ÷ bracket's value (22) = 5/22 — original expression at the number 4: 5 ÷ 22. Exact match again. Two different numbers, the same match both times — not a coincidence. Differentiating a log of the bracket really does hand back "5 ÷ the bracket", for any number chosen, which is exactly why running that process in reverse — integrating "5 ÷ the bracket" — has to return a log of the bracket. Nothing here was taken on trust; it was checked, twice, with ordinary arithmetic.

Name what was just checked. "The bracket" is what the rest of this lesson calls $f(x)$ — some expression built from $x$ , treated as one object. "The bracket's rate of change" is $f'(x)$ , its derivative. The rule "multiply the chosen number by 5, then add 2" was the concrete bracket $5x+2$ , with $f'(x) = 5$ for every $x$ , since a straight-line rule always has a constant rate of change. The claim tested twice above — differentiate a log of the bracket, get the bracket's rate of change divided by the bracket's own value — is the general pattern this lesson is named for: $\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + c$ , true for any $f(x)$ , not just the one bracket tested here.

Formally

For any differentiable $f(x)$ with $f(x) \neq 0$ on the interval considered, $\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + c$ . This is exactly what was checked twice above, generalised: differentiating $y = \ln|f(x)|$ via the chain rule (outer function $\ln|u|$ , inner function $u=f(x)$ ) gives $\frac{dy}{dx} = \frac{1}{f(x)}\cdot f'(x) = \frac{f'(x)}{f(x)}$ for every valid $x$ , not just the two numbers tested above — so integrating $\frac{f'(x)}{f(x)}$ must return $\ln|f(x)|+c$ . The mechanism block immediately below carries out this exact derivation in full, including why the modulus is not optional.

Mechanism

Where $\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + c$ actually comes from

Start from the claimed answer and differentiate it, because that is the fastest way to see why the result is true rather than just that it is. Take $y = \ln|f(x)|$ and apply the chain rule: the outer function is $\ln|u|$ , the inner function is $u = f(x)$ . $\frac{d}{du}[\ln|u|] = \frac{1}{u}$ is itself one of the standard results memorised from spec 5.1 (it is on the 'must memorise' list, not the sheet), so $\frac{dy}{dx} = \frac{1}{f(x)} \cdot f'(x) = \frac{f'(x)}{f(x)}$ . That is the entire theorem: differentiating $\ln|f(x)|$ , using nothing beyond the chain rule already known before this lesson, produces exactly $\frac{f'(x)}{f(x)}$ — so integrating $\frac{f'(x)}{f(x)}$ has to return $\ln|f(x)| + c$ , because integration is defined as the operation that undoes differentiation, not as a separate technique that happens to agree with it. The modulus is not decoration: $\ln|u|$ is defined and differentiates to $\frac{1}{u}$ for every $u \neq 0$ , negative values included, while $\ln(u)$ alone is undefined the moment $u$ turns negative. The modulus is what keeps the result true on the whole domain where $f(x)$ can be negative — not a convention layered on top of the theorem, but part of what makes the theorem correct in the first place.

Mechanism

Where $\int f'(x)[f(x)]^n\,dx = \frac{[f(x)]^{n+1}}{n+1} + c$ actually comes from

Run the identical move on the other pattern. Differentiate $y = [f(x)]^{n+1}$ : outer function $u^{n+1}$ , inner function $u = f(x)$ . $\frac{d}{du}[u^{n+1}] = (n+1)u^n$ , so $\frac{dy}{dx} = (n+1)[f(x)]^n \cdot f'(x)$ . Divide both sides by the constant $(n+1)$ — differentiation is linear, so a constant factor carries straight through the derivative — and $\frac{d}{dx}\left[\frac{[f(x)]^{n+1}}{n+1}\right] = f'(x)[f(x)]^n$ . Integrating $f'(x)[f(x)]^n$ must therefore return $\frac{[f(x)]^{n+1}}{n+1} + c$ . The two results in this lesson are not two separate facts to store — they are the same one-line chain-rule reversal, carried out for two different outer functions, $\ln|u|$ and $u^{n+1}$ . The spec's own 'must memorise' list, verified directly against the specification text, states the general pattern once, as $\int f'(g(x))\,g'(x)\,dx = f(g(x)) + c$ : this lesson's two named formulae are that single line, specialised to $f(u) = \ln|u|$ and $f(u) = \frac{u^{n+1}}{n+1}$ .

The other half of 5.2: squared trig functions fit neither pattern

$\sin^2 x$ , $\cos^2 x$ and $\tan^2 x$ look like candidates for the power form above — $f(x) = \sin x$ , $n = 2$ — but they are not, and seeing exactly why matters more than memorising the exception. The power form needs $f'(x)$ sitting alongside $[f(x)]^n$ as a genuine factor: $f'(x)[f(x)]^n$ . $\sin^2 x$ is just $[\sin x]^2$ on its own, with no accompanying $\cos x$ factor anywhere — nothing is playing the role of $f'(x)$ , and multiplying by an invisible 1 does not conjure one into existence.

So a different tool is needed: a trig identity that rewrites the squared function as something with no square in it at all. $\cos 2A \equiv 1 - 2\sin^2 A$ rearranges to $\sin^2 A \equiv \frac{1 - \cos 2A}{2}$ , turning the integral into a sum of two ordinary spec 5.1 integrals — $\frac{1}{2}$ and $-\frac{1}{2}\cos 2A$ — with no recognition needed at all. $\cos^2 A \equiv \frac{1 + \cos 2A}{2}$ comes the same way from $\cos 2A \equiv 2\cos^2 A - 1$ ; and $\tan^2 A \equiv \sec^2 A - 1$ comes from $\sec^2 A \equiv 1 + \tan^2 A$ , integrating using the $\sec^2 kx$ result printed directly on the formula sheet. All three source identities are on the spec's memorise list, not the sheet — so getting the rearrangement wrong under pressure, with nothing printed to check it against, is exactly the failure mode real examiner reports record (see the trap taxonomy below).

Worked, in full

Find $\int 3\cos 2x \,/\, (5 + 3\sin 2x)\,dx$ — spotting a coefficient mismatch, not just the pattern

01
Check whether the denominator's derivative is hiding in the numerator. Let $f(x) = 5 + 3\sin 2x$ . Differentiating: $f'(x) = 3 \times 2\cos 2x = 6\cos 2x$ (the constant 3 stays, the chain rule contributes the extra factor of 2 from $\sin 2x$ 's inner function). The numerator $3\cos 2x$ is not $f'(x)$ itself, but it is a constant multiple of it — reaching an expression of the form $k\ln|5+3\sin 2x|$ for *some* constant $k$ , correct or not, is the recognition move this stage is for.
Earns: M1 — achieves $k\ln|5+3\sin 2x|$ (or the equivalent $k\ln|u|$ with $u=5+3\sin 2x$ ) for any constant $k$ . Verified against a real WMA13 recognition-integral mark scheme testing the identical $f'(x)/f(x)$ -with-coefficient-mismatch pattern (Jan 2022 Q3(ii), $\int \frac{4\sin x}{1+2\cos x}\,dx$ ): the mark scheme's M1 reads "Achieves $b\ln(1+2\cos x)$ ... where $b$ is a constant" — the exact value of the constant is explicitly NOT required for this mark, only the correct log-of-the-bracket form. A candidate who writes down any constant times $\ln|5+3\sin 2x|$ — even a wrong one — banks this mark.
02
Now get the constant exactly right. $3\cos 2x = \frac{1}{2} \times 6\cos 2x = \frac{1}{2}f'(x)$ , so $\frac{1}{2}\int \frac{f'(x)}{f(x)}\,dx = \frac{1}{2}\ln|f(x)| = \frac{1}{2}\ln|5+3\sin 2x|$ — and, because this integral is indefinite, the constant of integration belongs on this same line, not a later one: $\frac{1}{2}\ln|5+3\sin 2x| + c$ .
Earns: A1 — the exact constant AND the constant of integration together, cao. These are ONE combined mark in real WMA13 schemes, not two separate ones: verified against Oct 2021 Q5(ii), a real power-form recognition integral ( $\int x(x^2+3)^7\,dx$ ), where the matching A1 reads "Achieves $\frac{1}{16}(x^2+3)^8+c$ ... the $+c$ must be present" as a single all-or-nothing requirement. A correct coefficient with a missing $+c$ loses this whole mark, not half of it — exactly why the examiner report below records candidates losing "the final mark" outright, on an answer that was otherwise entirely correct.
03
Check whether the modulus is actually load-bearing here. $\sin 2x$ ranges over $[-1, 1]$ , so $3\sin 2x$ ranges over $[-3, 3]$ , and $5 + 3\sin 2x$ ranges over $[2, 8]$ — strictly positive for every real $x$ . The modulus is never wrong to write, but on this particular domain it is not essential: $5+3\sin 2x$ can never be the negative value that would make it matter.
Earns: Nothing — not a markable step on its own, but the check that catches a dropped modulus before it costs a real mark on a harder version where $f(x)$ does change sign (see the trap taxonomy below).
04
State the final answer: $\frac{1}{2}\ln|5+3\sin 2x| + c$ .
Earns: No further mark here — this restates the answer that stage 2's single A1 already covered in full, algebra and $+c$ together. It is not a second, smaller mark for the constant of integration alone; a real mark scheme has already run out by this point.

Source — Examiner report, Oct 2021

"A fair number of candidates could not be awarded the final mark, despite having obtained the correct algebraic expression, as they had forgotten to include the constant of integration"

Complete it yourself

Complete the chain — find $\int 12\sin^3 x \cos x\,dx$

01
Identify the structure. $\cos x$ is exactly the derivative of $\sin x$ , and it is multiplying a power of $\sin x$ — so this is the $\int f'(x)[f(x)]^n\,dx$ pattern, with $f(x) = \sin x$ , $f'(x) = \cos x$ , $n = 3$ .
02
Separate the constant. $12\sin^3 x \cos x = 12 \cdot f'(x)[f(x)]^3$ , so the integral is $12\int f'(x)[f(x)]^3\,dx$ .

Same question, every valid method

Find $\int \tan x \, dx$ , given that $\tan x \equiv \frac{\sin x}{\cos x}$ . (VERIDIAN-original question. The given identity mirrors the spec's own guidance that 'students [are] expected to use trigonometric identities to integrate' this exact function — not a reproduction of any past-paper question.)

3 valid methods · every one reaches $\ln|\sec x| + c$ (equivalently $-\ln|\cos x| + c$ ) · 3 marks available

01
$\tan x = \frac{\sin x}{\cos x} = -\frac{-\sin x}{\cos x}$ , recognising the numerator, up to sign, as the derivative of the denominator: $f(x) = \cos x$ , $f'(x) = -\sin x$ .
Method mark for identifying the f'(x)/f(x) structure with the sign correctly accounted for. The minus sign is the entire difficulty here — cos x's derivative is −sin x, not sin x.
M1
02
$\int \tan x\,dx = -\int \frac{-\sin x}{\cos x}\,dx = -\ln|\cos x| + c$
Accuracy mark for correctly applying the recognition result to the signed form from the line above.
A1
03
$= \ln|\cos x|^{-1} + c = \ln|\sec x| + c$
Accuracy mark for the equivalent final form matching the one printed on the formula sheet, using $-\ln|u| = \ln|u^{-1}|$ .
A1

Fastest once the pattern is automatic — no substitution variable to introduce or convert back from afterwards. Its one real risk is the sign: cos x sits on the bottom, but its derivative is −sin x, not sin x, and dropping that minus sign is the single easiest way to lose the method mark on this exact question.

Marked, line by line

$f(x) = \ln(2 + \sin x)$ . (a) Find $f'(x)$ . (2) (b) Hence find $\int \frac{\cos x}{2+\sin x}\,dx$ . (2) (c) Using the identity $\cos 6x \equiv 1 - 2\sin^2 3x$ , find $\int \sin^2 3x\,dx$ . (3) (d) Hence show that $\int_0^{\pi/6} \sin^2 3x\,dx = \frac{\pi}{12}$ . (2) — VERIDIAN-original question, inspired by the real WMA13 pattern of a 'find the derivative, hence integrate' opening part followed by an identity-based part (spec 5.2 and 2.3 in one question, matching the spec's own guidance that both techniques belong together). Not a reproduction of any past-paper question, and the per-line mark allocations below are modelled on verified mark-scheme conventions rather than copied from a real scheme.

9 marks available

(a) — 2 marks

01
$f'(x) = \frac{d}{dx}[2+\sin x] \div (2+\sin x)$ , i.e. the chain-rule structure $\frac{u'}{u}$ with $u = 2+\sin x$ .
Method mark for the correct chain-rule structure for differentiating a log — the derivative of the bracket, over the bracket.
M1
02
$f'(x) = \frac{\cos x}{2+\sin x}$
Accuracy mark, correct answer only, dependent on the method mark above.
A1

(b) — 2 marks

101
"Hence" points back to (a): the expression $\frac{\cos x}{2+\sin x}$ IS $f'(x)$ , so this integral is asking for the function that differentiates to it — which part (a) has just supplied.
Method mark for recognising the reversal — using the given result from (a) rather than re-deriving the integral from scratch, which is what 'hence' specifically instructs.
M1
102
$\int \frac{\cos x}{2+\sin x}\,dx = \ln(2+\sin x) + c$
Accuracy mark for the correct result with the constant of integration present. The modulus can be safely dropped here specifically: $\sin x \in [-1,1]$ , so $2+\sin x \in [1,3]$ is always strictly positive — writing $\ln|2+\sin x|+c$ is also accepted, since it is never wrong, just not essential on this domain.
A1

(c) — 3 marks

201
Rearranging the given identity: $2\sin^2 3x = 1 - \cos 6x$ , so $\sin^2 3x = \frac{1-\cos 6x}{2}$ .
Independent accuracy mark for correctly rearranging the given identity — no method beyond algebra is needed, which is exactly what a B mark rewards.
B1
202
$\int \sin^2 3x\,dx = \int \left(\frac{1}{2} - \frac{1}{2}\cos 6x\right)dx$
Method mark for integrating the rearranged expression term by term, using the standard spec 5.1 result for $\cos kx$ .
M1
203
$= \frac{x}{2} - \frac{1}{12}\sin 6x + c$
Accuracy mark for the correct simplified result. The $\frac{1}{2} \times \frac{1}{6}$ combining to $\frac{1}{12}$ is where a coefficient slip most often happens.
A1

(d) — 2 marks

301
Substitute both limits into the antiderivative from (c): $\left[\frac{\pi/6}{2} - \frac{1}{12}\sin\left(6 \times \frac{\pi}{6}\right)\right] - \left[0 - \frac{1}{12}\sin 0\right] = \frac{\pi}{12} - \frac{1}{12}\sin\pi - 0$
Method mark for substituting both limits into their own antiderivative from (c) and evaluating $\sin\pi$ correctly as 0 — the step candidates most often skip is showing this line at all rather than jumping to the given answer.
M1
302
$= \frac{\pi}{12} - 0 = \frac{\pi}{12}$ , as required.
Accuracy mark, correct solution only. This is a 'show that' with the target printed in the question, so cso applies in full: the working must reach exactly $\frac{\pi}{12}$ , not a numerically close or unsimplified equivalent.
A1

In your own words

In one sentence: why do $\int \frac{f'(x)}{f(x)}\,dx$ and $\int f'(x)[f(x)]^n\,dx$ count as one technique rather than two separate formulae to memorise?

Named traps

modulus-dropped-in-log-integration: Confirmed directly on a real recognition-integration question involving negative values: "many did not have the modulus and left the values as 5ln(-2) and 5ln(-4) not realising these are undefined" (Oct 2020, Q9(c)); independently confirmed in Jun 2023, Q9(c). The mechanism this lesson's own derivation makes explicit: $\ln|f(x)|$ is defined and differentiates correctly for every $f(x) \neq 0$ , including negative values, while $\ln(f(x))$ alone breaks the moment $f(x)$ turns negative. Whether the modulus is essential or just safe depends entirely on whether $f(x)$ can actually be negative on the domain in question — check that before deciding it can be dropped, never assume.
constant-of-integration-omitted: Confirmed across at least two series on this exact topic: "A fair number of candidates could not be awarded the final mark, despite having obtained the correct algebraic expression, as they had forgotten to include the constant of integration" (Oct 2021, Q5(ii)); independently confirmed, "a substantial number of candidates neglected to include the constant of integration and were penalised" (Jan 2022, Q3(i)). It costs one mark on an otherwise perfect answer, on an indefinite integral specifically — a definite integral (like part (d) of the marked solution above) has no constant to lose, which is worth noticing precisely so the two cases are not confused.
double-angle-form-confused: A real double-angle "show that" question records candidates who "commonly wrote cos 2x as cos²x − 1 or 1 − cos²x" (Oct 2020, Q1) — misapplying or confusing the three equivalent double-angle-for-cosine forms with each other, or with the plain Pythagorean identity. This lesson's identity-rewrite technique for $\sin^2 x$ , $\cos^2 x$ and $\tan^2 x$ depends entirely on quoting the RIGHT rearranged form with the RIGHT coefficient — get the identity wrong (as in the marked solution's own common wrong path above) and the integration method that follows can be flawless and still reach the wrong answer.
intermediate-lines-omitted-on-hence-or-show-that: Confirmed as a general pattern across multiple series, not tied to one question: "many candidates omitted important lines when proceeding to the given solution resulting in the loss of some vital marks" (Jan 2022, general summary); the same report separately records "many good candidates lost marks here for merely writing down the given answer from a correct dx/dy without any intermediate lines." A "hence, show that" part with a printed target answer — exactly the shape of part (d) in the marked solution above — is where this costs the most: every line from the antiderivative to the printed target has to actually appear.
booklet-result-not-recognised: Confirmed on a real recognition-type integral: "some students did not realise the result could be found from their Formula Booklet, and instead used a substitution" (Jun 2023, Q9(c), on ∫cosec x dx). Substitution is a fully valid, mark-scheme-credited alternative route — the method-comparison block above shows exactly that — but re-deriving from scratch under time pressure, when a result is already printed and the question does not demand a derivation, spends time and adds a line where a sign or arithmetic slip can happen that a direct quote could not have introduced.

Choosing the right move under exam pressure

Three checks cover almost every 5.2 question, worth running in this order rather than scanning the whole expression at once. First: is there a squared trig function to rewrite before anything else can apply — $\sin^2$ , $\cos^2$ or $\tan^2$ of something, with no accompanying derivative factor sitting beside it? If so, the identity-rewrite move from earlier in this lesson comes first, always, because neither recognition formula applies to a bare power of a trig function. Second, once any squared trig terms are gone: does what remains fit $\frac{f'(x)}{f(x)}$ or $f'(x)[f(x)]^n$ — a function's derivative sitting next to a function of that same function? Third, and easiest to rush past: does the numerator match $f'(x)$ exactly, or only up to a constant multiple that needs pulling out first, the way the worked chain and chain drill above both needed?

Every route this lesson has shown — direct recognition, substitution, quoting the formula sheet — reduces to the same underlying check: differentiate a candidate answer and see whether it reproduces the original integrand. When an integral in the exam doesn't obviously match either named formula on sight, that check is still the fastest way in, not a fallback to reach for only once the named patterns have failed. It's exactly how both formulae were derived from scratch earlier in this lesson, and it's the one method that works even on a version of the pattern dressed up enough to not be recognised immediately.

Retrieval — with feedback on every choice

Question 1

2 marks

Find $\int \frac{\cos 3x}{2+\sin 3x}\,dx$ .

Question 2

3 marks

Find $\int 8x(x^2+3)^3\,dx$ .

Question 3

3 marks

Given that $\cos 4x \equiv 1 - 2\sin^2 2x$ .

Find $\int \sin^2 2x\,dx$ .

Question 4

2 marks

In which of these is it essential — not just safe — to keep the modulus in $\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)|+c$ ?

Question 5

1 mark

A student correctly establishes that $\frac{d}{dx}[\ln(\sec 3x)] = 3\tan 3x$ , for a domain restricted so that $\sec 3x > 0$ throughout, then writes: ' $\int \tan 3x\,dx = \frac{1}{3}\ln(\sec 3x)$ .'

Which single change would a real mark scheme most likely require before awarding full marks?

Reference — not a study method, a lookup

f'(x)/f(x) → ln|f(x)|+c. f'(x)[f(x)]ⁿ → [f(x)]ⁿ⁺¹/(n+1)+c. Both are the chain rule, reversed.
Check: does the integrand contain something's derivative sitting beside a function of that same something?
Coefficient mismatch: pull out or introduce a constant first, so the numerator becomes exactly f'(x).
sin²A, cos²A, tan²A: rewrite first — cos2A ≡ 1−2sin²A ≡ 2cos²A−1 ≡ cos²A−sin²A, or sec²A ≡ 1+tan²A. Never integrate a squared trig function directly.
Keep the modulus unless f(x) provably never changes sign. Always add +c.

Not affiliated with or endorsed by Pearson Edexcel. Every quotation and figure attributed to a mark scheme or examiner report in this lesson was independently verified against the primary Pearson document, not carried over from prior course material. Every question in this lesson — prequestion, worked chain, chain drill, method comparison, marked solution and MCQ alike — is VERIDIAN-original wording, inspired by confirmed real question types, never a reproduction of a real Pearson question; and because the questions are original, the per-line mark allocations attached to them are modelled on verified mark-scheme conventions (what M, A and B marks mean, when follow-through applies, when cso applies to a printed target answer) rather than transcribed from a real mark scheme, which for an original question does not exist.

Question 12 marks

Find $\int \frac{\cos 3x}{2+\sin 3x}\,dx$ .

$\frac{1}{3}\ln|2+\sin 3x| + c$
Correct. With $f(x)=2+\sin 3x$ , $f'(x) = 3\cos 3x$ — but the numerator here is only $\cos 3x$ , one third of $f'(x)$ . Pulling out the missing factor of $\frac{1}{3}$ first, then applying $\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)|+c$ , gives exactly this.
B $\ln|2+\sin 3x| + c$
This is the answer for $\int \frac{3\cos 3x}{2+\sin 3x}\,dx$ — i.e. it treats the numerator as already matching $f'(x) = 3\cos 3x$ exactly, when it is actually $\cos 3x$ , a third of that. The coefficient check is not optional; it decides whether any adjustment is needed at all.
C $3\ln|2+\sin 3x| + c$
The adjustment has been applied in the wrong direction — multiplying by 3 instead of dividing by it. Differentiating this answer gives $\frac{9\cos 3x}{2+\sin 3x}$ , nine times too large, not the original integrand.
D $\frac{1}{3}\ln|2+\sin 3x|$
The expression itself is correct and the constant of integration is missing — the same omission that real mark schemes withhold the final accuracy mark for, confirmed across multiple series on this exact topic.

Traps tested: Coefficient mismatch ignored · Coefficient adjustment inverted · Constant of integration dropped

Question 23 marks

Find $\int 8x(x^2+3)^3\,dx$ .

$(x^2+3)^4 + c$
Correct. With $f(x)=x^2+3$ , $f'(x)=2x$ , and $n=3$ : $8x(x^2+3)^3 = 4 \times 2x(x^2+3)^3 = 4f'(x)[f(x)]^3$ , so the integral is $4 \times \frac{[f(x)]^4}{4} + c = [f(x)]^4 + c$ . The 4s cancel exactly — check by differentiating: $\frac{d}{dx}[(x^2+3)^4] = 4(x^2+3)^3 \times 2x = 8x(x^2+3)^3$ . ✓
B $4(x^2+3)^4 + c$
The constant 4 was pulled out correctly but never divided by $(n+1)=4$ afterwards, so it was applied once instead of cancelling. Differentiating this gives $32x(x^2+3)^3$ , four times too large.
C $\frac{(x^2+3)^4}{4} + c$
This divides by $(n+1)=4$ a second time, on top of the division that already happened when the coefficient 4 was extracted. Differentiating this gives $2x(x^2+3)^3$ , four times too small.
D $2(x^2+3)^3 + c$
The power has gone down, not up — this is what differentiating $(x^2+3)^4$ produces before the chain-rule factor is applied, not what integrating $8x(x^2+3)^3$ produces. Integration by recognition always raises the power by one; it never lowers it.

Traps tested: Power rule denominator omitted · Power rule coefficient divided twice · Differentiation rule applied instead of integration

Question 33 marks

Given that $\cos 4x \equiv 1 - 2\sin^2 2x$ .

Find $\int \sin^2 2x\,dx$ .

$\frac{x}{2} - \frac{1}{8}\sin 4x + c$
Correct. Rearranging: $\sin^2 2x = \frac{1-\cos 4x}{2}$ , so the integral is $\int \left(\frac{1}{2} - \frac{1}{2}\cos 4x\right)dx = \frac{x}{2} - \frac{1}{2}\times\frac{1}{4}\sin 4x + c = \frac{x}{2} - \frac{1}{8}\sin 4x + c$ .
B $x - \frac{1}{4}\sin 4x + c$
This comes from rearranging the given identity as $\sin^2 2x = 1 - \cos 4x$ — dropping the coefficient of 2 that the identity actually has, exactly the double-angle-form confusion that real examiner reports record on this technique. The result is precisely double the correct antiderivative.
C $\frac{x}{2} + \frac{1}{8}\sin 4x + c$
The sign on the $\cos 4x$ term has flipped somewhere in the integration. $\int -\frac{1}{2}\cos 4x\,dx = -\frac{1}{8}\sin 4x$ , not $+\frac{1}{8}\sin 4x$ — check by differentiating this option: it returns $\frac{1}{2}-\frac{1}{2}\cos 4x$ , which is $\cos^2 2x$ , not $\sin^2 2x$ .
D $2\sin 4x + c$
This is what DIFFERENTIATING $\sin^2 2x$ produces (via the chain rule: $2\sin 2x \times \cos 2x \times 2 = 2\sin 4x$ , using $\sin 4x \equiv 2\sin 2x\cos 2x$ ), not integrating it. The question asks for the antiderivative, the reverse operation.

Traps tested: Double angle coefficient dropped · Sign lost integrating the cosine term · Differentiated instead of integrated

Question 42 marks

In which of these is it essential — not just safe — to keep the modulus in $\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)|+c$ ?

$\int \frac{2x}{x^2-9}\,dx$ , for $x$ between $-2$ and $0$
Correct. $f(x) = x^2-9$ on this interval ranges from $-9$ (at $x=0$ ) to $-5$ (at $x=-2$ ) — strictly negative throughout. $\ln(f(x))$ is undefined for every value in this range; only $\ln|f(x)|$ gives a real, differentiable result here, which is exactly the failure mode a real examiner report records: candidates left with expressions like $\ln(-4)$ , 'not realising these are undefined.'
B $\int \frac{\cos x}{2+\sin x}\,dx$ , for any real $x$
$2+\sin x$ ranges over $[1,3]$ — always strictly positive, since $\sin x \in [-1,1]$ . The modulus is never wrong to include, but it is not essential here; this is the exact function from part (b) of the marked solution above.
C $\int \frac{4x^3}{x^4+1}\,dx$ , for any real $x$
$x^4+1 \geq 1$ for every real $x$ , since $x^4 \geq 0$ always — strictly positive, so the modulus is safe to include but not essential.
D $\int \frac{6\cos 2x}{5+3\sin 2x}\,dx$ , for any real $x$
$5+3\sin 2x$ ranges over $[2,8]$ — always strictly positive, exactly as established in the worked chain earlier in this lesson. The modulus is safe here too, not essential.

Traps tested: Modulus treated as always required

Question 51 mark

A student correctly establishes that $\frac{d}{dx}[\ln(\sec 3x)] = 3\tan 3x$ , for a domain restricted so that $\sec 3x > 0$ throughout, then writes: ' $\int \tan 3x\,dx = \frac{1}{3}\ln(\sec 3x)$ .'

Which single change would a real mark scheme most likely require before awarding full marks?

Add the constant of integration, giving $\frac{1}{3}\ln(\sec 3x) + c$
Correct. The algebra and the coefficient are both right, and the domain restriction means the modulus genuinely is not needed here — the one thing missing is the $+c$ , the exact omission real examiner reports record costing the final accuracy mark on an otherwise fully correct indefinite integral.
BNothing — this is already the accepted final form
An indefinite integral without a constant of integration is not a complete answer; real mark schemes withhold the final accuracy mark for this specific omission, confirmed across multiple series on this exact topic.
CMultiply the whole answer by 3
This would make the answer wrong. Differentiating $\ln(\sec 3x)$ gives $3\tan 3x$ , so integrating $\tan 3x$ (without the 3) needs the $\frac{1}{3}$ that is already correctly present — multiplying it away reintroduces the mismatch the $\frac{1}{3}$ exists to fix.
DChange $\frac{1}{3}\ln(\sec 3x)$ to $3\ln(\sec 3x)$
This inverts the correct coefficient rather than fixing anything. $\frac{1}{3}$ is the constant that makes $\frac{d}{dx}\left[\frac{1}{3}\ln(\sec 3x)\right] = \tan 3x$ true; replacing it with 3 breaks that equality by a factor of 9.

Traps tested: Constant of integration treated as optional · Coefficient adjustment misapplied · Reciprocal confused with coefficient

Practice this for real

This site teaches the mechanism; the exam is sat on Pearson's own real questions. Go find and attempt these yourself — nothing here substitutes for actually sitting a timed paper.

Examiner report: Oct 2021 · Q5(ii) — cited directly in this lesson

Pearson's official past-papers portal

Select International Advanced Level → Mathematics → any series, then look for WMA13.

Integration by Recognising a Known Derivative

What this lesson assumes — recapped, not retaught

The identity this lesson also leans on — memorised, not looked up

Why "5 ÷ the bracket" integrates to a log of the bracket — before the letters

Where ∫f′(x)f(x) dx=ln⁡∣f(x)∣+c\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + c∫f(x)f′(x)​dx=ln∣f(x)∣+c actually comes from

Where ∫f′(x)[f(x)]n dx=[f(x)]n+1n+1+c\int f'(x)[f(x)]^n\,dx = \frac{[f(x)]^{n+1}}{n+1} + c∫f′(x)[f(x)]ndx=n+1[f(x)]n+1​+c actually comes from