Ch 5 · 2.5 / 3.5 Integration · A Level Mathematics 9709
Topic 5 of 9 in Pure Mathematics 2 and 3 — A Level Mathematics 9709 — 87 lessons.
Introduction
Every new derivative from section 3.4 gives a new integral, read backwards. On top of that, Paper 3 adds three techniques for integrals that do not look standard: partial fractions, integration by parts and substitution. Paper 2 instead adds a numerical method, the trapezium rule. Sections are marked (P2, P3), (P2) or (P3).
Key ideas — Standard integrals (P2, P3)
Note the modulus in ln|ax + b|: the log of a negative number does not exist.
🧑🏫 Teacher's explanation
Integration is essentially the 'reverse' of differentiation. If differentiation tells you the slope of a curve at any point, integration helps you find the area under that curve by undoing the derivative. Think of it like a movie being played in reverse: if you know how a car's position changes over time (the derivative), integration allows you to reconstruct the original path the car took.
📖 Definitions
🔍 Worked example — Easy
Find ∫ e^3x-5 dx
🔍 Worked example — Standard
Evaluate ∫ 1/x^2 + 9 dx
🔍 Worked example — Exam-style
Find ∫ cos(2x) dx
⚠️ Common errors
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🎯 Exam question
Key ideas — Using trigonometric identities (P2, P3)
sin^2 x and cos^2 x have no direct integral: use the double-angle formula first.
🧑🏫 Teacher's explanation
Integration is essentially 'reverse differentiation'. When you see a function like sin^2 x, you might notice that its derivative doesn't look like a simple power rule application. In fact, there is no direct 'reverse rule' for squared trigonometric functions. To solve this, we use a 'mathematical makeover'—we use trigonometric identities to rewrite the expression into a form that we actually know how to integrate.
📖 Definitions
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — The trapezium rule (P2)
With n strips of width h = b - a/n and ordinates y_0, y_1, dots, y_n:
🧑🏫 Teacher's explanation
When we integrate a function, we are finding the area under a curve. Finding the exact area is often difficult, so we use an approximation. Imagine you are trying to measure the area of a wavy lake. Instead of complex calculus, you could lay down several rectangular planks across the lake. However, rectangular planks leave gaps or hang over the edges. The trapezium rule is smarter: instead of flat-bottomed rectangles, we use trapeziums (four-sided shapes with two parallel sides) to better follow the slope of the curve.
📖 Definitions
🔍 Worked example — Exam-style
A function f(x) is concave down on the interval [a, b]. Using the trapezium rule with n strips, determine if the estimate is an over-estimate or an under-estimate, and justify your answer.
⚠️ Common errors
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🎯 Exam question
Key ideas — Integrals of the form kf′(x)/f(x) (P3)
∫ f'(x)/f(x) dx = lnlvert f(x)rvert + c
🧑🏫 Teacher's explanation
Think of integration as 'reverse engineering' a function. When you see a fraction where the top part looks like the derivative of the bottom part, you are looking at the 'footprint' of a natural logarithm. It is like seeing a puzzle piece that fits perfectly into a specific slot.
📖 Definitions
🔍 Worked example — Easy
Evaluate ∫ 3/2x+5 dx
🔍 Worked example — Standard
Find ∫ tan x dx
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — Partial fractions (P3)
Split the fraction (section 3.1), then integrate each part, usually as a log: ∫ 2/x - 1 + 3/x + 2 dx = 2ln|x - 1| + 3ln|x + 2| + c. A term C/(x-1)^2 integrates to -C/x - 1.
🧑🏫 Teacher's explanation
Imagine you are trying to swallow a large, solid block of food. It is difficult to digest in one piece. However, if you slice that block into smaller, manageable pieces, you can swallow them one by one. Integration of complex rational functions is exactly the same. Some fractions are 'indigestible' because they don't match our standard integration rules like ∫ 1/x dx = ln|x| + c.
📖 Definitions
🔍 Worked example — Easy
Find ∫ 5/x+2 dx
🔍 Worked example — Standard
Use partial fractions to find ∫ 3x+1/(x-1)(x+2) dx
🔍 Worked example — Exam-style
Evaluate ∫ 4/x^2-1 dx
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — Integration by parts (P3)
∫ udv/dx dx = uv - ∫ vdu/dx dx
🧑🏫 Teacher's explanation
Integration by parts is essentially the 'Product Rule' in reverse. When you differentiate a product of two functions, y = uv, you get dy/dx = udv/dx + vdu/dx. If we integrate both sides and rearrange, we get the integration by parts formula. It is a tool used when you are faced with an integrand that looks like a product of two different types of functions that don't easily simplify through substitution.
📖 Definitions
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — Substitution (P3)
The question gives the substitution. Steps: find du/dx and replace dx; write everything in terms of u; for a definite integral, change the limits to u values; integrate and evaluate (no need to return to x for a definite integral).
🧑🏫 Teacher's explanation
Integration by substitution is essentially the 'reverse' of the Chain Rule from differentiation. When you differentiate a composite function like f(g(x)), you multiply by the derivative of the inner function g'(x). Integration by substitution allows us to undo this process by identifying that 'inner' part and replacing it with a single variable, u.
📖 Definitions
🔍 Worked example — Easy
Evaluate ∫ 2x(x^2 + 3)^5 dx
🔍 Worked example — Exam-style
Use the substitution u = sin x to evaluate ∫_0^π/2 sin^2 x cos x dx
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Worked examples — Example 1: double angle and sec²
∫_0^π/4cos^2 x dx = [tfrac12x + tfrac14sin 2x ]_0^π/4 = π/8 + frac14. ∫_0^π/8sec^2 2x dx = [tfrac12tan 2x ]_0^π/8 = tfrac12.
🔍 Worked example — Easy
Evaluate ∫ sec^2(3x) dx
🔍 Worked example — Standard
Evaluate ∫_0^π/4 cos^2 x dx
🔍 Worked example — Exam-style
Evaluate ∫_0^π/8 sec^2(2x) dx
⚠️ Common errors
Worked examples — Example 2: trapezium rule (P2)
Estimate ∫_0^2√(1 + x^3) dx with 4 strips.
🔍 Worked example — Easy
Estimate ∫_0^1 x^2 dx using the trapezium rule with 2 strips.
🔍 Worked example — Standard
Estimate ∫_0^2 √(1+x^3) dx using 4 strips.
🔍 Worked example — Exam-style
Given f(x) = x^2 on [0, 2], use 4 strips to estimate ∫_0^2 f(x) dx and state if it is an over- or under-estimate.
⚠️ Common errors
Worked examples — Example 3: partial fractions (P3)
∫_2^3 5x + 1/(x - 1)(x + 2) dx = big[2ln(x - 1) + 3ln(x + 2)big]_2^3 = 2ln 2 + 3ln 5 - 3ln 4 = ln125/16.
🔍 Worked example — Easy
Find ∫ 3/x+1 dx
🔍 Worked example — Standard
Find ∫ 5x + 1/(x - 1)(x + 2) dx
⚠️ Common errors
Worked examples — Example 4: by parts (P3)
∫ xsin 2x dx: u = x, dv/dx = sin 2x, v = -tfrac12cos 2x. = -tfrac12xcos 2x + ∫tfrac12cos 2x dx = -tfrac12xcos 2x + tfrac14sin 2x + c.
🔍 Worked example — Easy
Evaluate ∫ x sin(2x) dx
🔍 Worked example — Standard
Evaluate ∫_1^e ln x dx
🔍 Worked example — Exam-style
Find ∫ x^2 e^-x dx
⚠️ Common errors
Worked examples — Example 5: a given substitution (P3)
Use u = sin x to find ∫_0^π/2sin^2 2xcos x dx.
🔍 Worked example — Easy
Evaluate ∫ sin(5x) dx using substitution.
🔍 Worked example — Standard
Use the substitution u = cos x to evaluate ∫_0^π/2 sin^3 x dx.
🔍 Worked example — Exam-style
Use the substitution u = sin x to find ∫_0^π/2 sin^2(2x) cos x dx.
⚠️ Common errors
Common mistakes
Practice
Answers: (1) -tfrac13e^-3x + 2ln|2x + 1| + c. (2) [tfrac12x - tfrac14sin 2x ]_0^π/2 = π/4. (3) [1/sqrt3tan^-1x/sqrt3 ]_0^1 = 1/sqrt3·π/6 = sqrt3 π/18. (4) [tfrac12xe^2x - tfrac14e^2x ]_0^1 = tfrac14e^2 + tfrac14. (5) ln|x^3 + 5| + c.