Integration (Pure 3)

• Divide by \(a\). Sine gets the minus. A square becomes a double angle. The top is the derivative of the bottom: a logarithm. Parts: \(u\) simplifies. Substitution: new limits. And whatever you did, differentiate the answer.
• Divide by a. The (x − a)−2 piece is not a logarithm. Parts: u simplifies. Substitution: new limits. n strips, n + 1 ordinates. Differentiate the answer.

• 3.5.1 2.5.1 extend the idea of ‘reverse differentiation’ to include the integration of eax+b, 1/(ax + b), sin(ax + b), cos(ax + b), sec²(ax + b) and 1/(x² + a²) — the 1/(x² + a²) part Paper 3 only
• 3.5.2 2.5.2 use trigonometrical relationships in carrying out integration
• 3.5.3 Paper 3 only integrate rational functions by means of decomposition into partial fractions
• 3.5.4 Paper 3 only recognise an integrand of the form k f′(x)/f(x), and integrate such functions
• 3.5.5 Paper 3 only recognise when an integrand can usefully be regarded as a product, and use integration by parts
• 3.5.6 Paper 3 only use a given substitution to simplify and evaluate either a definite or an indefinite integral
• 2.5.3 Paper 2 only understand and use the trapezium rule to estimate the value of a definite integral

Accuracy for this chapter. An integral that comes out as a logarithm, a multiple of \(\pi\), a surd or a power of \(e\) is left exact unless the question asks for a decimal: \(3\ln 3\), \(\tfrac{\pi}{8}\), \(\ln\tfrac{27}{4}\), \(2 - \tfrac{5}{e}\). Combine logarithms into one where the question asks for a single logarithm. Otherwise give 3 significant figures, with the unrounded value carried until the last line. For the trapezium rule, tabulate the ordinates to 4 decimal places, keep the sum unrounded, and round once at the end. Work inverse tangents and trigonometric ordinates in radians. Show the method line before every number: the integral with its limits substituted earns the method mark; a value from a calculator with no working earns nothing.

• “\(\displaystyle\int \dfrac{1}{2x + 1}\,dx = \ln|2x + 1| + c\).” Divide by \(a\), every time \(\tfrac{1}{2}\ln|2x + 1| + c\). The same \(\dfrac{1}{a}\) is in every \((ax + b)\) integral: \(\int e^{3x}\,dx = \tfrac{1}{3}e^{3x}\), \(\int \cos 4x\,dx = \tfrac{1}{4}\sin 4x\), \(\int \sec^2 2x\,dx = \tfrac{1}{2}\tan 2x\). Differentiating the wrong answer gives \(\dfrac{2}{2x + 1}\), twice the integrand, which is how you catch it.
• “\(\displaystyle\int \dfrac{3}{(x - 1)^2}\,dx = 3\ln(x - 1)^2 + c\).” The \((x - a)^{-2}\) piece is not a logarithm \(3(x - 1)^{-2}\) is a power, and the power rule gives \(\dfrac{3(x - 1)^{-1}}{-1} = -\dfrac{3}{x - 1} + c\). Only a first power of a linear factor in the denominator gives a logarithm. This is the repeated-factor term of every partial-fraction integral with a squared factor (section C, Paper 3 only).
• “Five ordinates from \(x = 0\) to \(x = 2\), so \(h = \dfrac{2}{5}\).” \(n\) strips means \(n + 1\) ordinates Five ordinates are the edges of four strips, so \(h = \dfrac{2 - 0}{4} = 0.5\). Count the gaps, not the fence posts. Get \(h\) wrong and every line after it is wrong (section G, Paper 2 only).
• “\(\displaystyle\int \sin 2x\,dx = \tfrac{1}{2}\cos 2x + c\).” Sine gets the minus \(-\tfrac{1}{2}\cos 2x + c\). Differentiation sends \(\cos \to -\sin\), so integration sends \(\sin \to -\cos\). Cosine integrates to \(+\sin\).
• “\(\displaystyle\int \sin^2 x\,dx = \tfrac{1}{3}\sin^3 x + c\).” Rewrite a square first The derivative of \(\tfrac{1}{3}\sin^3 x\) is \(\sin^2 x \cos x\), not \(\sin^2 x\). Use \(\sin^2 x = \tfrac{1}{2}(1 - \cos 2x)\): the integral is \(\tfrac{1}{2}x - \tfrac{1}{4}\sin 2x + c\).
• “With \(u = \sin x\): \(\displaystyle\int_0^{\pi/2} \sin^2 2x \cos x\,dx = \left[\tfrac{4}{3}u^3 - \tfrac{4}{5}u^5\right]_0^{\pi/2}\).” New variable, new limits The limits are values of \(x\); the integral is now in \(u\). Convert: \(x = 0 \Rightarrow u = 0\), \(x = \tfrac{\pi}{2} \Rightarrow u = 1\). The answer is \(\left[\tfrac{4}{3}u^3 - \tfrac{4}{5}u^5\right]_0^1 = \tfrac{8}{15}\) (section F, Paper 3 only).
• “\(\displaystyle\int_{-3}^{-1} \dfrac{1}{x}\,dx = [\ln x]_{-3}^{-1}\), which cannot be evaluated.” Keep the modulus MF19 prints \(\ln|x|\), and the modulus is part of the answer: \([\ln|x|]_{-3}^{-1} = \ln 1 - \ln 3 = -\ln 3\). The value is negative because the curve is below the axis there.
• “\(\displaystyle\int e^{x^2}\,dx = \dfrac{e^{x^2}}{2x} + c\).” The \(1/a\) is for a linear inside only Dividing by the derivative of the inside works only when that derivative is a constant. Differentiating \(\dfrac{e^{x^2}}{2x}\) by the quotient rule does not give \(e^{x^2}\). This integral has no answer in terms of standard functions; it can only be estimated, for example by the trapezium rule.
• Wrong: “\(\displaystyle\int \frac{1}{2x + 1}\,dx = \ln|2x + 1| + c\).” Repair Divide by the coefficient of \(x\): \(\tfrac{1}{2}\ln|2x + 1| + c\). Differentiating the wrong answer shows the extra factor 2.
• Wrong: “\(\displaystyle\int \sin 2x\,dx = \tfrac{1}{2}\cos 2x + c\).” Repair \(-\tfrac{1}{2}\cos 2x + c\): the integral of sine carries the minus sign.
• Paper 3 only Wrong: “\(\displaystyle\int \frac{1}{x^2 + 4}\,dx = \ln(x^2 + 4) + c\).” Repair There is no \(x\) in the numerator, so this is not a \(k\,\dfrac{f'}{f}\) pattern. It is the MF19 form with \(a = 2\): \(\tfrac{1}{2}\tan^{-1}\tfrac{x}{2} + c\).
• Paper 3 only Wrong: “\(\displaystyle\int \frac{1}{2 + 3x^2}\,dx = \frac{1}{\sqrt{2}}\tan^{-1}\frac{x}{\sqrt{2}} + c\).” Repair The coefficient of \(x^2\) must be 1 first: \(\tfrac{1}{3} \cdot \dfrac{1}{x^2 + \tfrac{2}{3}}\), \(a = \sqrt{\tfrac{2}{3}}\), giving \(\dfrac{1}{\sqrt{6}}\tan^{-1}\dfrac{\sqrt{6}\,x}{2} + c\).
• Wrong: “\(\displaystyle\int \sin^2 x\,dx = \tfrac{1}{3}\sin^3 x + c\).” Repair \(\dfrac{d}{dx}\left(\tfrac{1}{3}\sin^3 x\right) = \sin^2 x\cos x\), not \(\sin^2 x\). Use \(\sin^2 x = \tfrac{1}{2}(1 - \cos 2x)\): \(\tfrac{1}{2}x - \tfrac{1}{4}\sin 2x + c\).
• Wrong: “\(\cos^2 2x = \tfrac{1}{2}(1 + \cos 2x)\).” Repair The identity doubles the angle, whatever it is: \(\cos^2 2x = \tfrac{1}{2}(1 + \cos 4x)\), so \(\int \cos^2 2x\,dx = \tfrac{1}{2}x + \tfrac{1}{8}\sin 4x + c\).
• Paper 3 only Wrong: “\(\displaystyle\int \frac{x}{x^2 + 1}\,dx = \ln(x^2 + 1) + c\).” Repair \(f' = 2x\), so the numerator is \(\tfrac{1}{2}f'\) and the answer is \(\tfrac{1}{2}\ln(x^2 + 1) + c\).
• Paper 3 only Wrong: “\(\displaystyle\int \tan x\,dx = \ln|\cos x| + c\).” Repair The derivative of \(\cos x\) is \(-\sin x\), so the sign flips: \(-\ln|\cos x| + c\).
• Paper 3 only Wrong: “\(\displaystyle\int \frac{3}{(x - 1)^2}\,dx = 3\ln(x - 1)^2 + c\).” Repair \((x - 1)^{-2}\) integrates by the power rule to \(\dfrac{(x - 1)^{-1}}{-1}\): the answer is \(-\dfrac{3}{x - 1} + c\). Only a first power in the denominator gives a logarithm.
• Paper 3 only Wrong: choosing \(u = \sin 2x\), \(\dfrac{dv}{dx} = x\) for \(\displaystyle\int x\sin 2x\,dx\). Repair That makes the second integral \(\int \tfrac{1}{2}x^2 \cdot 2\cos 2x\,dx\), harder than the first. \(u\) is the factor that simplifies on differentiating: \(u = x\).
• Paper 3 only Wrong: “\(\displaystyle\int \ln x\,dx = \frac{1}{x} + c\).” Repair That is the derivative. Use parts with \(u = \ln x\) and \(\dfrac{dv}{dx} = 1\): \(x\ln x - x + c\).
• Paper 3 only Wrong: “\(\displaystyle\int_0^{\pi/2} \sin^2 2x\cos x\,dx = \left[\tfrac{4}{3}u^3 - \tfrac{4}{5}u^5\right]_0^{\pi/2}\).” Repair The limits must be converted: \(u = \sin 0 = 0\) and \(u = \sin\tfrac{\pi}{2} = 1\). Limits in \(x\) on an integral in \(u\) are meaningless.
• Paper 2 only Wrong: “Five ordinates, so \(h = \dfrac{b - a}{5}\).” Repair Five ordinates are four strips: \(h = \dfrac{b - a}{4}\).
• Paper 2 only Wrong: “\(\tfrac{1}{2}h\{y_0 + y_n + y_1 + \dots + y_{n-1}\}\)”, or leaving out the \(\tfrac{1}{2}\). Repair The interior ordinates are doubled and the whole bracket is halved: \(\tfrac{1}{2}h\{(y_0 + y_n) + 2(y_1 + \dots + y_{n-1})\}\).

• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.
• One check for the whole chapter. In chapter 8 you checked an integral by differentiating it. Here the check is the same, and it is the only one you need: whatever method produced the answer (a table, an identity, partial fractions, parts, a substitution), its derivative must be the integrand. Every worked integral in this chapter ends with that line.
• The wrong choice announces itself. For \(\int x\sin 2x\,dx\), taking \(u = \sin 2x\) and \(\dfrac{dv}{dx} = x\) gives \(v = \tfrac{1}{2}x^2\) and a new integral \(\int \tfrac{1}{2}x^2 \cdot 2\cos 2x\,dx\), a higher power of \(x\) than you started with. If the new integral is harder, swap \(u\) and \(\dfrac{dv}{dx}\).
• Interleave with the chapters that use this one. Chapter 16 (differential equations) separates variables and then needs every integral here: when you reach it, redo retrieval questions 11, 13 and 17. Chapter 30 (Paper 6) integrates probability density functions, and the syllabus states that the calculus of Paper 3 is assumed knowledge for Paper 6: when you reach it, redo worked examples 1 and 6. Recalling a method inside a new problem is worth more than another pass over this chapter on its own: later chapters use these methods without re-teaching them, and the syllabus says an individual examination question may involve ideas and methods from more than one section of the content for that paper, so nothing here is ever finished with.

• Cambridge International AS & A Level Mathematics 9709 has six components, and a candidate takes two of them for the AS Level and four for the A Level. This chapter is Pure Mathematics 3 content, examined in Paper 3 (and Paper 2 on the AS-only pure route). Paper 3 (Pure Mathematics 3) is compulsory for the A Level and is 30% of it. The Paper 2 (Pure Mathematics 2) outcomes this chapter also serves are marked in the syllabus map: Paper 2 is offered only in the AS-only Pure Mathematics route (Papers 1 and 2), where it is 40% of the AS Level, and that route cannot be carried forward to the A Level. Every paper is a written examination of compulsory structured questions, answered on the question paper, with MF19 (the list of formulae and statistical tables) supplied. Examinations are available in the June and November series, and in March in India.
• Across the whole qualification the assessment objectives are weighted AO1 55% (knowledge and understanding: concepts, terminology, notation and accurate manipulative technique) and AO2 45% (application and communication: choosing the procedure, combining techniques to solve problems, and presenting the work clearly and logically) at AS Level, and AO1 52%, AO2 48% at A Level. AS candidates are graded a–e; A Level candidates A*–E.
• A structured question can combine a method step (an identity, partial fractions, parts, a given substitution) with an exact value or a “show that” ending in a printed logarithm or multiple of π. When one part asks for partial fractions and the next says “hence”, the partial fractions are the intended route to the integral. For the trapezium rule, the syllabus includes deciding from a sketch whether the estimate is an over-estimate or an under-estimate.
• MF19 prints the integration table for a plain x, the by-parts formula and ∫ f′(x)/f(x) dx = ln|f(x)|, and the double-angle identities. It does not print the (ax + b) forms with their 1/a, the rearranged identities for sin²x and cos²x, the partial-fraction pieces, or the trapezium rule: those must be known. The MF19 card lists both sides.
• An answer such as 3 ln 3, π/8 or ln(27/4) is left exact unless a decimal is asked for; otherwise give 3 significant figures. The integral with its limits substituted is the method line and must be shown; the syllabus gives no marks for an unsupported answer from a calculator. Inverse tangents and trigonometric ordinates are in radians. Trapezium ordinates go to 4 decimal places, rounded once at the end.
• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.