Integration by Parts and by Substitution

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

P4.2 states plainly that "a knowledge of the specifications for P1, P2 and P3, their prerequisites and associated formulae, is assumed and may be tested" — P4 is cumulative on the whole Pure strand, and this lesson leans on two results from that strand without re-deriving either: the , $\frac{d}{dx}[uv] = u\frac{dv}{dx} + v\frac{du}{dx}$ , and the , $\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}$ . Both are on the specification's own "must memorise, will not appear in the booklet" list, and both were derived from first principles in the prerequisite differentiation-rules lesson — this one uses them, run in reverse, rather than proving them again.

The standard integrals from earlier Pure units are also used freely from here on: $\int e^{kx}\,dx = \frac{1}{k}e^{kx}+c$ , $\int \frac{1}{x}\,dx = \ln|x|+c$ , $\int \sin kx\,dx = -\frac{1}{k}\cos kx+c$ , $\int \cos kx\,dx = \frac{1}{k}\sin kx+c$ . None of these are new; what is new is that they now get reused repeatedly, inside a single question, as the smaller integral that a first application of parts or substitution reduces the original problem down to.

Mechanism

Where the integration by parts formula actually comes from

Start from the product rule, exactly as stated above: $\frac{d}{dx}[uv] = u\frac{dv}{dx} + v\frac{du}{dx}$ . Integrate both sides with respect to $x$ . The left-hand side integrates back to exactly what was differentiated — integration undoes differentiation, so $\int \frac{d}{dx}[uv]\,dx = uv$ . The right-hand side splits, since integration distributes over a sum: $\int u\frac{dv}{dx}\,dx + \int v\frac{du}{dx}\,dx$ . So $uv = \int u\frac{dv}{dx}\,dx + \int v\frac{du}{dx}\,dx$ , and rearranging for the first integral gives $\int u\frac{dv}{dx}\,dx = uv - \int v\frac{du}{dx}\,dx$ — exactly the formula printed in the Mathematical Formulae and Statistical Tables booklet, confirmed verbatim against the booklet's own P4 section. This is precisely what the specification means by describing the two techniques in this lesson as "the reverse processes of the chain and product rules respectively": nothing here is a new rule invented for integration, it is the product rule, read backwards. Because the formula is printed, there is nothing in it to memorise — the entire content of the skill lives in the two choices the formula itself is silent on: which factor becomes $u$ , and which becomes $\frac{dv}{dx}$ .

Choosing u and dv/dx — the one thing the formula can't do for you

Two questions decide the choice, and they matter in this order. First: which factor gets SIMPLER when differentiated? A polynomial's degree drops by one every time ( $x^2\to 2x\to 2\to 0$ ); $\ln x$ differentiates to the algebraic fraction $\frac{1}{x}$ , simpler in kind even though it isn't a polynomial. That factor is $u$ . Second: which factor can actually be integrated, and stays no harder to integrate a second time? $e^{kx}$ , $\sin kx$ and $\cos kx$ all reproduce themselves in kind under integration — differentiating or integrating any of them again is no harder than the first time. That factor is $\frac{dv}{dx}$ .

One consequence of the first question is absolute, not just a preference: a factor with no simple to write down can never be $\frac{dv}{dx}$ , because $v$ has to exist before the formula can even be applied. $\ln x$ is exactly this case — there is no elementary $v$ such that $\frac{dv}{dx}=\ln x$ that this course has met — so $\ln x$ must always be differentiated (as $u$ ), never integrated. That single restriction is what makes the very next worked example a genuinely special case rather than a routine application.

Worked, in full

Find $\int \ln x\,dx$ — the special case where dv/dx = 1 is invisible until you need it

01
$\ln x$ is a single factor, not a visible product — there is nothing obvious to split into $u$ and $\frac{dv}{dx}$ . Spec 6.2's own guidance names this exact integral as required, and the trick is to treat it as a product with an invisible second factor: $\ln x \times 1$ . Set $u = \ln x$ , $\frac{dv}{dx} = 1$ .
Earns: M1 — writes ln x as (ln x)×1 and attempts parts with u=ln x, dv/dx=1. The mark is for spotting the disguised product, which is the entire difficulty of this special case.
02
Differentiate and integrate the two chosen pieces, both already known from earlier Pure units: $\frac{du}{dx} = \frac{1}{x}$ , and $v = \int 1\,dx = x$ .
Earns: B1 — du/dx=1/x and v=x both correctly stated; both come from results already known, not from anything new in this lesson.
03
Substitute into the formula: $\int \ln x\,dx = x\ln x - \int x\cdot\frac{1}{x}\,dx = x\ln x - \int 1\,dx$ .
Earns: M1 — correct substitution into u(dv/dx)dx = uv − ∫v(du/dx)dx, with the x and 1/x inside the remaining integral correctly cancelling to 1. That cancellation is the entire payoff of the u=ln x, dv/dx=1 choice: a single, intractable-looking integral collapses to the simplest possible one.
04
Complete it: $\int 1\,dx = x + c$ , so $\int \ln x\,dx = x\ln x - x + c$ .
Earns: A1 — correct final result, including the constant of integration.
05
Check by differentiating: $\frac{d}{dx}[x\ln x - x] = \ln x + x\cdot\frac{1}{x} - 1 = \ln x + 1 - 1 = \ln x$ . ✓
Earns: Nothing — not a markable step, but the one check that confirms the invisible-factor trick actually worked, in one line.

Recognition, substitution, or parts — deciding before you start

The specification's own framing of this lesson's two techniques — "understand these methods as the reverse processes of the chain and product rules respectively" — sits alongside a third reverse process this course already covers: (the prerequisite lesson, spec 5.2), which reverses the chain rule in its OWN way, spotting $\int \frac{f'(x)}{f(x)}\,dx$ or $\int f'(x)[f(x)]^n\,dx$ already sitting in an integrand with no new variable needed at all. All three techniques on this course are "differentiation, done backwards" — they differ only in which structural shape they're built to reverse, and picking the wrong one is a real, examined failure mode, not just a slower route to the same answer.

A working checklist, in this order. First: does the integrand already contain a function's own derivative — exactly, or up to a constant multiple — sitting beside a function OF that function? If so, that is recognition: no new variable, no formula, done in one line, and reaching for parts or substitution instead is unnecessary work. Second: does the integrand's structure suggest a direct replacement for part of itself — and, at this level, has a substitution actually been given, since spec 6.2's guidance states the substitution will be handed to you in more complicated integrals? If so, substitution. Third: is the integrand a product of two structurally DIFFERENT kinds of function — algebraic × exponential, algebraic × trig, a log sitting on its own — with no derivative-match anywhere for recognition to grab onto? That is parts.

Getting this choice wrong is real and documented, even outside spec 6.2 itself. A real examiner report on a volume-of-revolution question (spec 6.1, not this lesson's own spec point, but the failure it names generalises directly) records: "The major stumbling block for the majority of candidates was a failure to choose a correct approach to the integration... few candidates made this choice and wasted much time pursuing incorrect methods which included... using integration by parts" (Oct 2022, Q5) — where the efficient route was recognition or substitution instead. Integration by parts is a real, always-available-looking tool, which is exactly why it gets reached for out of habit even when it is the wrong one.

Mechanism

Where integration by substitution actually comes from

Suppose $x = g(u)$ , and let $F(x)$ be any antiderivative of $f(x)$ , so $F'(x) = f(x)$ by definition. Define a new function of $u$ : $G(u) = F(g(u))$ — the composition of $F$ with the substitution itself. Differentiate $G$ with respect to $u$ using the chain rule, exactly as stated in this lesson's own recap: $\frac{dG}{du} = F'(g(u))\cdot g'(u) = f(g(u))\cdot\frac{dx}{du}$ , writing $\frac{dx}{du}$ for $g'(u)$ . Because $\frac{dG}{du} = f(g(u))\cdot\frac{dx}{du}$ , integrating both sides with respect to $u$ gives $G(u) = \int f(g(u))\cdot\frac{dx}{du}\,du$ . But $G(u) = F(g(u)) = F(x)$ , and $F(x)$ is exactly $\int f(x)\,dx$ by definition of an antiderivative. So $\int f(x)\,dx = \int f(g(u))\cdot\frac{dx}{du}\,du$ — this IS the substitution rule, and it fell straight out of the chain rule applied to a composite antiderivative, exactly matching the spec's own description of substitution as the chain rule reversed. Read the middle of that derivation again, because it is the entire mechanics of the technique used in practice below: $dx$ is never simply crossed out and relabelled $du$ . It is replaced by the full product $\frac{dx}{du}\,du$ — and that extra factor is not decoration, it is the chain-rule derivative that makes the whole substitution valid in the first place.

Substitution in practice — three things change together, or none of them are safe

Three things change when a substitution is applied, and all three have to change, every time, or the result is wrong. The variable itself: every occurrence of $x$ in the original integrand — not just the one that looks like it prompted the substitution — has to be rewritten in terms of $u$ . The differential: $dx$ becomes the FULL expression $\frac{dx}{du}\,du$ , found by differentiating the substitution, exactly as the mechanism above derives — never $du$ on its own. And, on a only: the two numbers at the top and bottom of the integral sign describe values of $x$ before the substitution and have to be converted to the corresponding values of $u$ afterwards, using the substitution itself.

One rule closes the process off cleanly: once every $x$ has been replaced by $u$ , the whole integral is a function of $u$ and nothing else — evaluate it entirely in $u$ (using converted limits, for a definite integral) or convert the final antiderivative back to $x$ (for an indefinite one, or a definite one where converting back is easier). Never mix the two: an expression left in $u$ should never be evaluated at values of $x$ , and an expression already converted back to $x$ needs the original $x$ -limits, not converted $u$ -limits. Both worked examples below are built to make this rule concrete.

Worked, in full

Find $\int x\sqrt{x-2}\,dx$ — the spec's own named example of a substitution question

01
With $\sqrt{x-2}$ sitting inside the integrand, let $u = x-2$ , so $x = u+2$ and $\frac{du}{dx}=1$ , i.e. $dx = du$ . Every occurrence of $x$ needs replacing — both the $x$ multiplying the surd, and the $x-2$ inside it.
Earns: M1 — sets up the substitution and correctly finds dx in terms of du (here, simply dx=du, since the substitution is a shift by a constant).
02
Rewrite the WHOLE integral in $u$ , with no $x$ left anywhere: $\int x\sqrt{x-2}\,dx = \int (u+2)\sqrt{u}\,du = \int (u+2)u^{1/2}\,du$ .
Earns: M1 — every instance of x replaced, including the one multiplying the surd, leaving an integral entirely in u. A partial substitution, with an x still sitting somewhere in the expression, cannot be integrated in u at all — see the trap taxonomy below.
03
Expand and integrate term by term: $\int (u^{3/2}+2u^{1/2})\,du = \frac{2}{5}u^{5/2} + \frac{4}{3}u^{3/2} + c$ .
Earns: A1 — correct expansion and integration, each power raised by one and divided by the new power.
04
Substitute back to $x$ , since this is an indefinite integral with no limits to convert instead: $\frac{2}{5}(x-2)^{5/2} + \frac{4}{3}(x-2)^{3/2} + c$ .
Earns: A1 — correct final expression, back in terms of x.
05
Check by differentiating: $\frac{d}{dx}\left[\frac{2}{5}(x-2)^{5/2}\right] = (x-2)^{3/2}$ and $\frac{d}{dx}\left[\frac{4}{3}(x-2)^{3/2}\right] = 2(x-2)^{1/2}$ , summing to $(x-2)^{1/2}\left[(x-2)+2\right] = x\sqrt{x-2}$ — exactly the original integrand.
Earns: Nothing — the mark scheme has run out, but this is the check that catches a wrong power or a dropped factor before it costs a mark.

Source — Examiner report, Oct 2021

"candidates who ignored the dx and simply wrote it as du"

Marked, line by line

Using the substitution $u = x - 2$ : (a) Find $\int x\sqrt{x-2}\,dx$ . (4) (b) Hence find the exact value of $\int_3^6 x\sqrt{x-2}\,dx$ . (3) — VERIDIAN-original numeric scenario built around the spec's own named substitution example (spec 6.2 guidance: 'Students will be expected to use a substitution to find, e.g. ∫x√(x−2) dx'); the bounds 3 and 6 and the per-line mark allocations below are VERIDIAN-original, modelled on verified mark-scheme conventions rather than transcribed from a real mark scheme, which for an original question does not exist.

7 marks available

(a) — 4 marks

01
$u=x-2$ , so $x=u+2$ and $\frac{dx}{du}=1$ , i.e. $dx=du$ .
Method mark for a correctly set-up substitution, with dx correctly found in terms of du.
M1
02
$\int x\sqrt{x-2}\,dx = \int (u+2)u^{1/2}\,du$
Method mark for rewriting the whole integral in u, with no x remaining anywhere in the expression.
M1
03
$= \frac{2}{5}u^{5/2}+\frac{4}{3}u^{3/2}+c$
Accuracy mark for correct term-by-term integration.
A1
04
$= \frac{2}{5}(x-2)^{5/2}+\frac{4}{3}(x-2)^{3/2}+c$
Accuracy mark for correctly substituting back to x, dependent on the method marks above. This is the expression part (b) needs, whether evaluated directly in x or left in u with the limits converted instead.
A1

(b) — 3 marks

101
Convert the limits using the same substitution: when $x=3$ , $u=1$ ; when $x=6$ , $u=4$ .
Method mark for converting the original x-limits into u-limits via the substitution actually used — the exact step a real mark scheme withholds credit for skipping (see the trap taxonomy below).
M1
102
$\left(\frac{2}{5}(4)^{5/2}+\frac{4}{3}(4)^{3/2}\right) - \left(\frac{2}{5}(1)^{5/2}+\frac{4}{3}(1)^{3/2}\right)$
Dependent method mark for substituting both converted u-limits into the u-form expression they came from, dependent on the limit-conversion mark above.
dM1
103
$= \left(\frac{64}{5}+\frac{32}{3}\right) - \left(\frac{2}{5}+\frac{4}{3}\right) = \frac{352}{15} - \frac{26}{15} = \frac{326}{15}$
Accuracy mark, exact value. A decimal approximation would not be safe here without full working shown — the general principles state exact answers are expected wherever an exact value is asked for.
A1

Same question, every valid method

Find $\int x^2e^x\,dx$ . (VERIDIAN-original question wording, built around the integral spec 6.2's own guidance names directly as needing more than one application of parts: 'More than one application of integration by parts may be required, for example, ∫x²eˣ dx'.)

2 valid methods · every one reaches $e^x(x^2-2x+2)+c$ · 4 marks available

01
First application: $u=x^2$ , $\frac{dv}{dx}=e^x$ , so $\frac{du}{dx}=2x$ , $v=e^x$ . $\int x^2e^x\,dx = x^2e^x - \int 2xe^x\,dx = x^2e^x - 2\int xe^x\,dx$ .
Method mark for the first application, with u chosen as the algebraic factor (which gets simpler when differentiated) and dv/dx as the exponential factor (which reproduces itself when integrated).
M1
02
Second application, on the new integral $\int xe^x\,dx$ : $u=x$ , $\frac{dv}{dx}=e^x$ , so $\frac{du}{dx}=1$ , $v=e^x$ . $\int xe^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + c$ .
Dependent method mark for a second application of parts to the integral produced by the first, awarded only if the first application's M1 was earned — a genuinely different integral needing its own u and dv/dx chosen from scratch, but the mark itself is gated on the first step's success, not independent of it.
dM1
03
Substitute back into the first line's result, distributing the −2 across BOTH terms: $x^2e^x - 2(xe^x - e^x) + c = x^2e^x - 2xe^x + 2e^x + c$ .
Accuracy mark for correctly distributing the −2 across both terms of the second application's result. This single distribution step is where a coefficient most often survives one term and vanishes from the other.
A1
04
$= e^x(x^2-2x+2)+c$
Accuracy mark for the factorised final form, dependent on both method marks above.
A1

Shows every intermediate integral explicitly, which is safer when coefficient tracking is the real risk — each application is a self-contained, checkable step on its own. The cost is length: two full applications, each needing u, dv/dx, du/dx and v written out in full, is a lot of places for a sign or a factor to go missing.

Worked, in full

Find $\int e^x\sin x\,dx$ — the case where parts never terminates, so you solve for the integral instead

01
First application: $u=\sin x$ (differentiates to $\cos x$ — still a trig function, never reaching 0 the way a polynomial factor would), $\frac{dv}{dx}=e^x$ (integrates to $e^x$ , unchanged). $\frac{du}{dx}=\cos x$ , $v=e^x$ . Writing $I=\int e^x\sin x\,dx$ : $I = e^x\sin x - \int e^x\cos x\,dx$ .
Earns: M1 — first application of parts, correctly identified: u=sin x, dv/dx=eˣ.
02
Second application, applied to $\int e^x\cos x\,dx$ , choosing the SAME kind of factor to differentiate as before — the trig one, again: $u=\cos x$ , $\frac{dv}{dx}=e^x$ , $\frac{du}{dx}=-\sin x$ , $v=e^x$ . $\int e^x\cos x\,dx = e^x\cos x - \int e^x(-\sin x)\,dx = e^x\cos x + \int e^x\sin x\,dx = e^x\cos x + I$ .
Earns: dM1 — a second, dependent application (earned only if the first application's M1 was scored), with the SAME factor (the trig one) chosen to differentiate both times. This consistency is not a stylistic preference: switching which factor is differentiated on the second application undoes the first and produces no new information at all (see stage 5).
03
Substitute the second result back into the first: $I = e^x\sin x - (e^x\cos x + I) = e^x\sin x - e^x\cos x - I$ .
Earns: A1 — correct substitution, with the sign of the trailing −I handled correctly through the outer bracket.
04
This is not stuck — it is a genuine equation in $I$ , and $I$ appears on both sides of it. Collect the $I$ terms exactly as with any other equation: $I + I = e^x\sin x - e^x\cos x$ , so $2I = e^x(\sin x - \cos x)$ , and $I = \frac{1}{2}e^x(\sin x - \cos x) + c$ .
Earns: A1 — solves the equation in I correctly, including recognising that a +c belongs on the final answer even though no explicit integral sign remains in the last line — the constant was implicitly present in the original ∫eˣsin x dx all along.
05
Check what the INCONSISTENT choice would have done, on the second application: choosing $u=e^x$ instead of $u=\cos x$ this time gives $\int e^x\cos x\,dx = e^x\sin x - \int e^x\sin x\,dx = e^x\sin x - I$ . Substituting back: $I = e^x\sin x - (e^x\sin x - I) = I$ — a true statement carrying no information at all. Both applications have to make the same choice of which factor to differentiate, or the second one simply reverses the first and the working goes in a circle instead of closing.
Earns: Nothing — not a markable step, and this exact self-cancelling outcome is not documented in any source used for this lesson; it is included because it is real, checkable in one line, and explains why the consistency in stage 2 is a genuine requirement, not a convention.
06
Check by differentiating the final answer: $\frac{d}{dx}\left[\frac{1}{2}e^x(\sin x-\cos x)\right] = \frac{1}{2}\left[e^x(\sin x-\cos x)+e^x(\cos x+\sin x)\right] = \frac{1}{2}e^x(2\sin x) = e^x\sin x$ . ✓
Earns: Nothing — the mark scheme has run out, but this is the check that confirms the whole cyclic argument actually closed correctly.

Complete it yourself

Complete the chain — find $\int x(2x-1)^4\,dx$ , using the substitution $u=2x-1$

01
The substitution is given: $u=2x-1$ , so $x=\frac{u+1}{2}$ , and $\frac{du}{dx}=2$ , i.e. $dx=\frac{du}{2}$ . Both pieces of the original integrand need replacing — the $x$ outside the bracket, and the $(2x-1)^4$ itself.
02
Rewrite the whole integral in $u$ , with no $x$ left anywhere: $\int x(2x-1)^4\,dx = \int \frac{u+1}{2}\cdot u^4\cdot\frac{du}{2} = \frac{1}{4}\int(u^5+u^4)\,du$ .

In your own words

In one sentence: why does solving $2I = e^x(\sin x - \cos x)$ for $I$ in the cyclic case above still count as integration by parts, even though the last step is pure algebra with no integral sign left in it?

Named traps

ln-x-squared-treated-as-interchangeable-with-2-ln-x-mid-derivation: Confirmed on a real repeated-parts question that reused an earlier part's result: "many candidates assumed that ln(x²) was identical to 2 ln x and failed to score" (Oct 2021, Q8(b)). The identity itself is true — ln(x²) ≡ 2ln|x| — so the failure is not the algebra, it is using it as an unannounced shortcut in the middle of a derivation the mark scheme is following line by line; a substitution the examiner cannot see the justification for breaks the chain of reasoning being credited, even when the number it produces is correct. Write out any such rewrite as its own explicit line, not an invisible mental step.
coefficient-and-differentiation-slips-in-a-second-parts-application: A real second application of parts, building on an earlier part of the same question, records two separate documented failures at once: "A common error was then to integrate/differentiate the term cos2x incorrectly, losing the accuracy mark," and, further into the same question, "Some candidates failed to realise they required a factor of 3 within the final integral" (both Oct 2022, Q7(ii)). Together they describe exactly the risk the traditional method in the method comparison above is built to make visible: a SECOND application of the formula has to re-differentiate or re-integrate correctly all over again, and any constant carried from the first application has to survive being distributed across every term the second application produces — not just the first one it touches.
inverse-function-misapplied-to-a-substitution-term: Confirmed on a real substitution question: examiners record "candidates who changed sin 2x to sin⁻¹((u−3)/4)" (Oct 2021, Q6) — applying an inverse trig function to part of the substitution instead of substituting directly for the variable. A substitution replaces x, and every function of x, with an expression in u by direct algebraic rearrangement of the substitution itself; nothing about the technique ever calls for an inverse function, and reaching for one is a sign the substitution's own rearrangement — solving it for x, not for some other quantity — was skipped.
dx-relabelled-as-du-without-being-converted: Confirmed on the same question: "candidates who ignored the dx and simply wrote it as du" (Oct 2021, Q6) — one of the most fundamental substitution failures in this facts bank, and the exact failure the ∫x√(x−2)dx worked chain above is built to demonstrate the opposite of. dx is never simply crossed out and replaced with du; it is replaced by the FULL expression dx/du × du, found by differentiating the substitution. Dropping that factor is not a rounding error — it deletes the entire chain-rule content of the technique, leaving an expression that only coincidentally resembles the correct one when the factor happens to equal 1.
limits-of-integration-not-converted-to-the-new-variable: Confirmed on a real definite-integral substitution question: examiners record marks lost to "use of the x limits, ln7 and ln5 instead of the u limits of 4 and 2" (Oct 2022, Q7(i)) — evaluating the u-form antiderivative at the original x-values instead of converting them first. The marked solution above is built around exactly this failure: once the variable of integration changes, the two numbers at the top and bottom of the integral sign describe values of the new variable or the old one, never a mix of both.
a-numerical-factor-lost-within-the-substituted-expression: Confirmed on the same question, as a separate failure from the limits error above: "losing the factor 4 in the expression" (Oct 2022, Q7(i)). This is the coefficient-tracking risk from earlier in this lesson, arriving from the substitution side rather than the repeated-parts side: whenever dx/du is anything other than 1, that factor has to be carried through every remaining line of working, not dropped the moment it stops being the newest thing written down.

Retrieval — with feedback on every choice

Question 1

2 marks

Which method most efficiently evaluates $\int 4x^3e^{x^4}\,dx$ ?

Question 2

3 marks

Find $\int x\cos 3x\,dx$ .

Question 3

2 marks

Using the substitution $u=3x-1$ , find $\int (3x-1)^4\,dx$ .

Question 4

2 marks

$\int_1^4 f(x)\,dx$ is to be evaluated using the substitution $u=2x-1$ . What are the correct limits for the resulting integral in u?

Question 5

3 marks

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

Reference — not a study method, a lookup

∫u(dv/dx)dx = uv − ∫v(du/dx)dx. Given in the booklet — the job is choosing u and dv/dx, not recalling the formula.
u gets simpler when differentiated (polynomials, ln x). dv/dx must be integrable and stay 'the same kind' (eˣ, sin, cos).
∫ln x dx: treat as (ln x)×1, u=ln x, dv/dx=1 → x ln x − x + c.
Repeated parts: track every constant through every term. Cyclic case (eˣsinx-type): differentiate the SAME factor both times, then solve the resulting equation for the integral.
Substitution replaces dx with (dx/du)du — never just relabels it. On a definite integral, convert the LIMITS too, via the same substitution applied to the original x-limits.

Not affiliated with or endorsed by Pearson Edexcel. Every quotation and figure attributed to the specification, a mark scheme or an examiner report in this lesson was independently verified against the primary Pearson document, not carried over from prior course material — including every spec-guidance quote (the required ∫ln x dx, the named ∫x²eˣ dx and ∫eˣ sin x dx repeated-parts examples, and the ∫x√(x−2) dx substitution example), which come from WMA14-verified-facts.md §1's verbatim transcription of the spec's own 6.2 guidance column. Every question in this lesson — prequestion, worked chain, chain drill, method comparison, marked solution and MCQ alike — is VERIDIAN-original wording built around those spec-named integrands; none reproduces a real Pearson question, and because the specific numeric bounds and per-line mark allocations attached to them are original, they are modelled on verified mark-scheme conventions (what M, A and B marks mean, when a mark is dependent on another, when cao applies) rather than transcribed from a real mark scheme, which for an original question does not exist. This lesson draws on three examiner-report series specifically for its spec 6.2 content — October 2021, October 2022 and January 2024 — out of the four full examiner reports reviewed for this unit; a fourth (January 2021) contributed no 6.2-specific finding here and is not cited as though it had. The one genuinely novel mathematical observation in this lesson beyond the facts bank — that an inconsistent choice of which factor to differentiate on the second application of parts to ∫eˣsin x dx collapses to the tautology I = I — is checked, original content, not a transcribed examiner finding, and is labelled as such at the point it appears.

Question 12 marks

Which method most efficiently evaluates $\int 4x^3e^{x^4}\,dx$ ?

Recognition: $4x^3$ is exactly the derivative of $x^4$ , so $\int 4x^3e^{x^4}\,dx = e^{x^4}+c$ in one line
Correct. This is already f'(x)e^{f(x)} in the reverse-chain-rule pattern from the prerequisite lesson — the exponent's own derivative sits in the integrand as an exact factor, so no new variable and no parts formula are needed at all.
BIntegration by parts, with $u=e^{x^4}$ and $\frac{dv}{dx}=4x^3$
This leads nowhere useful: differentiating $e^{x^4}$ reintroduces the same exponential multiplied by an extra power of x ( $4x^3e^{x^4}$ ), so the new integral produced is not simpler than the one just started with. Parts is the wrong tool exactly because nothing about this integrand gets simpler under either operation the formula offers.
C $e^{4x^3}+c$ , treating the coefficient $4x^3$ as though it belonged inside the exponent
This differentiates the wrong target: $\frac{d}{dx}\left[e^{4x^3}\right]=12x^2e^{4x^3}$ , nothing like the original integrand. The coefficient $4x^3$ is the derivative of the exponent $x^4$ , not part of a different exponent.
DA substitution is required first, since the exponent $x^4$ is not linear
Untrue as a general rule: the condition this technique needs is that the exponent's own derivative sits in the integrand as a factor (or a constant multiple of it), whatever shape the exponent takes — already satisfied here with no new variable introduced. A substitution u=x⁴ would also reach the right answer, just in more lines, not because it is required.

Traps tested: By parts chosen where recognition applies · Coefficient and exponent confused · Assumes nonlinear exponent always needs explicit substitution

Question 23 marks

Find $\int x\cos 3x\,dx$ .

$\frac{x}{3}\sin 3x + \frac{1}{9}\cos 3x + c$
Correct. With u=x, dv/dx=cos3x: du/dx=1, v=(1/3)sin3x. $x\cdot\frac{1}{3}\sin 3x - \int\frac{1}{3}\sin 3x\,dx = \frac{x}{3}\sin 3x - \left(-\frac{1}{9}\cos 3x\right) = \frac{x}{3}\sin 3x + \frac{1}{9}\cos 3x + c$ .
B $\frac{x}{3}\sin 3x - \frac{1}{9}\cos 3x + c$
The sign on the second term is wrong — this comes from adding $\int \frac{1}{3}\sin 3x\,dx$ instead of subtracting it, or equivalently from a sign lost while integrating $\frac{1}{3}\sin 3x$ itself. The formula's own minus sign has to survive through to the very last line.
C $\frac{x}{3}\sin 3x + \frac{1}{3}\cos 3x + c$
The coefficient on the second term has only been divided by 3 once, not twice. v = (1/3)sin3x is correct on its own — but integrating (1/3)sin3x needs a FURTHER factor of 1/3 from integrating sin3x itself, giving −1/9 cos3x overall, not −1/3 cos3x.
D $\frac{x}{3}\sin 3x + \frac{1}{9}\cos 3x$
The expression itself is correct and the constant of integration is missing — real mark schemes withhold the final accuracy mark for exactly this omission, even when every other line is right.

Traps tested: Sign lost in the parts formula · Coefficient only divided once not twice · Constant of integration dropped

Question 32 marks

Using the substitution $u=3x-1$ , find $\int (3x-1)^4\,dx$ .

$\frac{1}{15}(3x-1)^5 + c$
Correct. $\frac{du}{dx}=3$ , so $dx=\frac{du}{3}$ : $\int u^4\cdot\frac{du}{3} = \frac{1}{3}\times\frac{u^5}{5}+c = \frac{u^5}{15}+c$ , back in x giving $\frac{(3x-1)^5}{15}+c$ . Check by differentiating: $5(3x-1)^4\times 3 \div 15 = (3x-1)^4$ . ✓
B $\frac{1}{5}(3x-1)^5 + c$
The factor 1/3 from dx=du/3 has been dropped entirely — this treats dx as though it equalled du directly. Differentiating this answer gives $3(3x-1)^4$ , three times too large.
C $\frac{3}{5}(3x-1)^5 + c$
The adjustment has been applied in the wrong direction — multiplying by 3 instead of dividing by it. Differentiating this gives $9(3x-1)^4$ , nine times too large.
D $\frac{1}{15}(3x-1)^5$
The expression itself is correct and the constant of integration is missing.

Traps tested: Substitution factor dropped from dx · Coefficient adjustment inverted · Constant of integration dropped

Question 42 marks

$\int_1^4 f(x)\,dx$ is to be evaluated using the substitution $u=2x-1$ . What are the correct limits for the resulting integral in u?

$u=1$ to $u=7$
Correct. $u(1)=2(1)-1=1$ and $u(4)=2(4)-1=7$ — each original x-limit converted separately, using the substitution itself.
B $u=1$ to $u=4$
This reuses the original x-limits directly, with no conversion attempted at all — exactly the real, documented failure this lesson's marked solution is built around: evaluating a u-form expression at x-values rather than at the u-values those x-values actually correspond to.
C $u=7$ to $u=1$
The individual values are right and the order is reversed — the lower x-limit (x=1) converts to the LOWER u-limit (u=1), and the upper x-limit (x=4) converts to the upper u-limit (u=7). Swapping the order flips the sign of the final answer.
D $u=2$ to $u=8$
This computes $u=2x$ , dropping the $-1$ from the substitution entirely. The substitution is $u=2x-1$ , not $u=2x$ — every term has to be applied, including the constant.

Traps tested: Limits not converted to the new variable · Converted limits reversed in order · Minus one term dropped in conversion

Question 53 marks

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

$x^2\sin x + 2x\cos x - 2\sin x + c$
Correct. First application (u=x², dv/dx=cosx): x²sinx − ∫2xsinx dx. Second application (u=x, dv/dx=sinx): ∫xsinx dx = −xcosx + sinx + c. Distributing the −2 across BOTH terms: x²sinx − 2(−xcosx+sinx) = x²sinx+2xcosx−2sinx+c.
B $x^2\sin x + 2x\cos x + \sin x + c$
The −2 has been distributed onto the first term (−xcosx → +2xcosx, correctly) but not onto the second (sinx should become −2sinx, not stay as +sinx). A coefficient produced by a first application has to reach every term the second application produces, not just the one closest to it.
C $x^2\sin x - 2x\cos x - 2\sin x + c$
The sign on the 2xcosx term is wrong. −2 × (−xcosx) = +2xcosx, not −2xcosx — check by differentiating this option: it does not return x²cosx.
D $x^2\sin x + 2x\cos x - 2\sin x$
The expression is correct and the constant of integration is missing.

Traps tested: Coefficient not distributed across both terms · Sign lost in the parts formula · Constant of integration dropped

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 · Q6 — cited directly in this lesson

Pearson's official past-papers portal

Select International Advanced Level → Mathematics → any series, then look for WMA14.

Integration by Parts and by Substitution

What this lesson assumes — recapped, not retaught

Where the integration by parts formula actually comes from

Choosing u and dv/dx — the one thing the formula can't do for you

Find ∫ln⁡x dx\int \ln x\,dx∫lnxdx — the special case where dv/dx = 1 is invisible until you need it

Recognition, substitution, or parts — deciding before you start

Where integration by substitution actually comes from

Substitution in practice — three things change together, or none of them are safe

Find ∫xx−2 dx\int x\sqrt{x-2}\,dx∫xx−2​dx — the spec's own named example of a substitution question

Find ∫exsin⁡x dx\int e^x\sin x\,dx∫exsinxdx — the case where parts never terminates, so you solve for the integral instead

Complete the chain — find ∫x(2x−1)4 dx\int x(2x-1)^4\,dx∫x(2x−1)4dx, using the substitution u=2x−1u=2x-1u=2x−1

Deep dive

Practice this for real

Find $\int \ln x\,dx$ — the special case where dv/dx = 1 is invisible until you need it

Find $\int x\sqrt{x-2}\,dx$ — the spec's own named example of a substitution question

Find $\int e^x\sin x\,dx$ — the case where parts never terminates, so you solve for the integral instead

Complete the chain — find $\int x(2x-1)^4\,dx$ , using the substitution $u=2x-1$