Library › Further Pure Mathematics 3 WFM03 › Integration

AS Level · Further Pure Mathematics 3 WFM03

Integration

Revise Integration for Further Pure Mathematics 3 WFM03 (AS Level) — revision notes and instant AI marking. Free to start.

📖 Revision notes · preview

FP3 · Chapter 4

Integration

Every integral in this chapter is really the same three moves in disguise: spot a pattern you already know, force the integrand into that shape, then execute — whether that's a clean substitution, a "reverse chain rule" guess, or building a reduction formula for when the power just won't quit.

📎 Builds on: Chapter 1 (Hyperbolic Functions) — you'll lean heavily on sinh, cosh, their derivatives, and the log forms of arsinh/arcosh/artanh. Chapter 3's derivatives of inverse functions also feed straight into the standard substitutions here.

Summary

Recognise a standard function — some integrals are just derivatives you already know, run backwards.
Using formulae to change the integrand — trig/hyperbolic identities (like double angle formulae) turn an unintegrable expression into one you can do term by term.
Reverse chain rule — spot the "function of a function times its derivative" pattern, guess the antiderivative shape, differentiate to fix the constant.
Standard substitutions — four go-to substitutions (trig or hyperbolic) for integrals containing $a² \pm b²x²$ , with or without a square root.
Integrating inverse functions and ln x — integration by parts with $dv/dx = 1$ , a trick that works because you can differentiate the inverse function even though you can't easily integrate it.
Reduction formulae — for integrals with a power n, express $Iₙ$ in terms of $Iₙ₋₁$ or $Iₙ₋₂$ instead of integrating from scratch every time.
Arc length — using Pythagoras on an infinitesimal triangle to find the length of a curve, in Cartesian or parametric form.
Area of a surface of revolution — spinning a curve around the x-axis and adding up the surface area of infinitely many thin cylindrical bands.

Standard Techniques

Before reaching for anything clever, always ask: "Have I seen this exact shape before, as a derivative?" This section builds up your toolkit in the order you should actually try it in an exam, from laziest to most involved.

1. Recognise a standard function

This is the "if it's staring you in the face, just write the answer down" category. If you can look at the integrand and immediately think "that's the derivative of something," you're done — no working required beyond stating the antiderivative and checking it by differentiating.

Recognise & reverse ∫ sec x tan x dx = sec x + c ∫ sech x tanh x dx = −sech x + c

These work because $d(sec x)/dx = sec x tan x$ and $d(sech x)/dx = -sech x tanh x$ — you memorised these derivatives in Chapter 3, and integration is just running the film backwards.

Worked Example

Find \int cosech x coth x dx. Recall d(cosech x)/dx = -cosech x coth x. So \int cosech x coth x dx = -cosech x + c

✏️ Practice Find ∫ sech²x dx.

2. Using formulae to change the integrand

Sometimes there's no standard derivative that matches — yet. But an identity (Pythagorean, double angle, Osborne's rule) can rewrite the integrand into a form that does match. This is the "disguise removal" technique.

🧠 Why this works You're not changing the value of the integral — you're rewriting the same function using an identity so that what you're integrating suddenly looks like something in your standard list.

Worked Example

Find \int tan²x dx. tan²x on its own isn't a standard integral. But 1 + tan²x = sec²x is. \Rightarrow tan²x = sec²x - 1 \Rightarrow \int tan²x dx = \int sec²x - 1 dx = tan x - x + c

Worked Example (hyperbolic version)

Find \int sinh²x dx. Use cosh 2x = 1 + 2sinh²x \Rightarrow sinh²x = ½(cosh 2x - 1) \int sinh²x dx = ½ \int cosh 2x - 1 dx = ½(½ sinh 2x - x) + c

💡 Osborne's rule reminder When converting a trig identity to its hyperbolic twin, flip the sign on any term that came from a product of two sines. That's exactly what happened above: cos 2x = 1 − 2sin²x → cosh 2x = 1 + 2sinh²x.

✏️ Practice Find ∫ cos²x dx.

3. Reverse chain rule

This is the technique for integrands that look like $f'(g(x)) \cdot g'(x)$ — a function of a function, multiplied by the derivative of the "inner" bit. The method: notice the pattern $uⁿ \cdot du/dx$ , guess that the answer involves $u n+1$ , differentiate your guess, and see what constant you need to divide by to make it match.

🔍 Spot the pattern Ask: "if I strip away one factor, is what's left the derivative of the rest?" If ∫ (stuff)ⁿ × (derivative of stuff) dx, try (stuff)ⁿ⁺¹ as your answer shape.

Worked Example

Find \int cos²x sin x dx. Looks like u² du/dx with u = cos x (since d(cos x)/dx = -sin x is "nearly" the sin x factor). Try u³ \Leftrightarrow cos³x. d(cos³x)/dx = 3cos²x(-sin x) = -3cos²x sin x So we need to divide by -3: \Rightarrow \int cos²x sin x dx = -⅓ cos³x + c

Worked Example

Find \int x² (2x³ - 7)⁴ dx. Looks like u⁴ du/dx with u = 2x³ - 7. Try u⁵ \Leftrightarrow (2x³ - 7)⁵. d[(2x³-7)⁵]/dx = 5(2x³-7)⁴ \times 6x² = 30x²(2x³-7)⁴ Divide by 30: \Rightarrow \int x² (2x³ - 7)⁴ dx = 1/30 (2x³ - 7)⁵ + c

Worked Example (hyperbolic flavour)

Find \int sech⁴x tanh x dx. Rewrite: \int sech³x (sech x tanh x) dx — looks like u³ du/dx with u = sech x. Try u⁴ \Leftrightarrow sech⁴x. d(sech⁴x)/dx = 4sech³x \times (-sech x tanh x) = -4sech³x sech x tanh x Divide by -4: \Rightarrow \int sech⁴x tanh x dx = -¼ sech⁴x + c

✏️ Practice Find ∫ sin³x cos x dx.

4. Standard substitutions

This is the big one — four "template" substitutions to memorise, each designed to knock out a square root or a quadratic denominator by turning it into a hyperbolic or trig identity (like $cosh²u - sinh²u = 1$ ). The choice depends on two things: is there a square root, and is it a sum or a difference?

Integral	Substitution	Why
∫ 1/(a² + b²x²) dx	bx = a tan u	No root present — use tan (avoids messy hyperbolic forms)
∫ 1/√(a² + b²x²) dx	bx = a sinh u	Root of a sum — sinh, since 1 + sinh²u = cosh²u
∫ 1/(a² − b²x²) dx	bx = a tanh u (or partial fractions)	No root, difference — tanh
∫ 1/√(b²x² − a²) dx	bx = a cosh u	Root of a difference, x² term first — cosh, since cosh²u − 1 = sinh²u

🗺️ The decision map No square root? Use tan (sum in denominator) or tanh (difference).
Square root present? Use sinh (sum under the root) or cosh (difference under the root, x² term positive and first).

Worked Example — completing the square first

Find \int 1/\sqrt(4x² - 8x - 5) dx. 4x² - 8x - 5 = 4(x² - 2x + 1) - 9 = 4(x - 1)² - 9 So the integral is \int 1/\sqrt[4(x-1)² - 9] dx — a "root of a difference," x² term first \to use cosh. Substitute 2(x - 1) = 3 cosh u \Rightarrow 2 dx = 3 sinh u du = \int [1/\sqrt(9(cosh²u - 1))] \times (3 sinh u / 2) du = \int [1/(3 sinh u)] \times (3 sinh u / 2) du = ½ \int du = ½u + c \Rightarrow ½ arcosh[(2x - 2)/3] + c

⚡ The "nice trick" shortcut If the quadratic under the root is already in the form (linear)² ± (number)², you don't need to complete the square from scratch — substitute u = the linear bit directly.

Worked Example — the shortcut in action

Find I = \int 1/\sqrt[(2x - 3)² + 25] dx. Let u = 2x - 3 \Rightarrow dx = ½ du I = \int 1/\sqrt(u² + 5²) \times ½ du = ½ arsinh(u/5) + c \Rightarrow I = ½ arsinh[(2x - 3)/5] + c

⚠️ Odd vs even powers rule For ∫ xⁿ/√(a² ± x²) dx: if n is odd, substitute u (or u²) = a² ± x² directly — much faster. If n is even, you need the full trig/hyperbolic substitution above.

✏️ Practice Find ∫ 1/√(9 + 4x²) dx.

5. Integrating inverse functions and ln x

You can't easily write down an antiderivative of arctan x or ln x by inspection — but you can differentiate them (Chapter 3 gave you all these derivatives). The trick: treat the whole function as "u" in integration by parts, and treat dv/dx = 1, so v = x. That converts an integral you can't do into one you can.

The trick ∫ f(x) dx = ∫ f(x) × 1 dx → by parts with u = f(x), dv/dx = 1, v = x

f(x)	f'(x)
arcsin x	1/√(1−x²)
arccos x	−1/√(1−x²)
arctan x	1/(1+x²)
arsinh x	1/√(1+x²)
arcosh x	1/√(x²−1)
artanh x	1/(1−x²)

Worked Example

Find \int arctan x dx. Let u = arctan x \Rightarrow du/dx = 1/(1+x²), and dv/dx = 1 \Rightarrow v = x \int arctan x dx = x arctan x - \int x \times 1/(1+x²) dx The remaining integral is a reverse chain rule: \int x/(1+x²) dx = ½ ln(1+x²) \Rightarrow \int arctan x dx = x arctan x - ½ ln(1 + x²) + c

Worked Example

Find \int arcosh x dx. Let u = arcosh x \Rightarrow du/dx = 1/\sqrt(x²-1), and dv/dx = 1 \Rightarrow v = x \int arcosh x dx = x arcosh x - \int x/\sqrt(x²-1) dx The remaining integral is reverse chain rule: \int x(x²-1)^(-½) dx = \sqrt(x²-1) \Rightarrow \int arcosh x dx = x arcosh x - \sqrt(x² - 1) + c

📌 Bonus: 1/(1−x²) two ways Partial fractions give ∫1/(1−x²) dx = ½ ln[(1+x)/(1−x)] + c, which is exactly artanh x in log form — a nice consistency check between the two methods.

✏️ Practice Find ∫ ln x dx (hint: use u = ln x, dv/dx = 1).

Reduction Formulae

Some integrals have a power n in them, and computing them from scratch for large n would be brutal. A reduction formula expresses $Iₙ$ in terms of a lower-power version — usually $Iₙ₋₁$ or $Iₙ₋₂$ — so you can "step down" to a base case you can integrate directly, then climb back up.

🪜 The general strategy Almost always: integrate by parts (sometimes twice), splitting the power so that differentiating reduces it and you can still integrate the rest. Occasionally, algebraic manipulation (like a Pythagorean identity or an angle-splitting trick) does the job instead.

Worked Example 1 — by parts once

Iₙ = \int xⁿ eˣ dx. Find a reduction formula, then find I₀ and I₄. (a) Reduction formula: u = xⁿ \Rightarrow du/dx = nx^(n-1); dv/dx = eˣ \Rightarrow v = eˣ Iₙ = xⁿeˣ - \int nx^(n-1)eˣ dx = xⁿeˣ - n Iₙ₋₁ (b) I₀ = \int eˣ dx = eˣ + c (c) Climb back up: I₄ = x⁴eˣ - 4I₃ = x⁴eˣ - 4(x³eˣ - 3I₂) = x⁴eˣ - 4x³eˣ + 12(x²eˣ - 2I₁) = x⁴eˣ - 4x³eˣ + 12x²eˣ - 24(xeˣ - I₀) \Rightarrow I₄ = x⁴eˣ - 4x³eˣ + 12x²eˣ - 24xeˣ + 24eˣ + c

Worked Example 2 — algebraic split, no parts needed

Find a reduction formula for Iₙ = \int tanⁿx dx. Iₙ = \int tan^(n-2)x tan²x dx = \int tan^(n-2)x (sec²x - 1) dx = \int tan^(n-2)x sec²x dx - \int tan^(n-2)x dx The first integral is reverse chain rule (u = tan x): gives (1/(n-1)) tan^(n-1)x. \Rightarrow Iₙ = (1/(n-1)) tan^(n-1)x - Iₙ₋₂

Worked Example 3 — definite integral with limits (Wallis-type)

Iₙ = \int₀^(π/2) sinⁿx dx. Find a reduction formula and use it to find I₆. Split sinⁿx = sin^(n-1)x sin x, take u = sin^(n-1)x, dv/dx = sin x \Rightarrow v = -cos x Iₙ = [-cos x sin^(n-1)x]₀^(π/2) + (n-1)\int₀^(π/2) cos²x sin^(n-2)x dx = 0 + (n-1)\int₀^(π/2)(1 - sin²x) sin^(n-2)x dx = (n-1)Iₙ₋₂ - (n-1)Iₙ \Rightarrow nIₙ = (n-1)Iₙ₋₂ \Rightarrow Iₙ = [(n-1)/n] Iₙ₋₂ Using it: I₆ = (5/6)I₄ = (5/6)(3/4)I₂ = (5/6)(3/4)(1/2)I₀ I₀ = \int₀^(π/2) 1 dx = π/2 \Rightarrow I₆ = 5π/32

💡 Recognise the "Wallis" pattern Whenever limits are 0 to π/2 and you're integrating a pure power of sin or cos, expect the boundary term to vanish (since sin 0 = 0 and cos(π/2) = 0), leaving a clean recursive relation.

Worked Example 4 — parts twice

Find a reduction formula for Iₙ = \int₀^(π/2) xⁿ cos x dx. First pass: u = xⁿ, dv/dx = cos x \Rightarrow v = sin x Iₙ = [xⁿ sin x]₀^(π/2) - n\int₀^(π/2) x^(n-1) sin x dx = (π/2)ⁿ - n\int₀^(π/2) x^(n-1) sin x dx Second pass on the remaining integral: u = x^(n-1), dv/dx = sin x \Rightarrow v = -cos x \int₀^(π/2) x^(n-1) sin x dx = [-x^(n-1)cos x]₀^(π/2) + (n-1)\int₀^(π/2) x^(n-2)cos x dx = 0 + (n-1)Iₙ₋₂ \Rightarrow Iₙ = (π/2)ⁿ - n(n-1) Iₙ₋₂

Worked Example 5 — a "fiddle" is needed

Find a reduction formula for Iₙ = \int₋₁⁰ xⁿ(1+x)² dx. u = xⁿ \Rightarrow du/dx = nx^(n-1); dv/dx = (1+x)² \Rightarrow v = ⅓(1+x)³ Iₙ = [xⁿ\cdot⅓(1+x)³]₋₁⁰ - \int₋₁⁰ nx^(n-1)\cdot⅓(1+x)³ dx = 0 - (n/3)\int₋₁⁰ x^(n-1)(1+x)³ dx Write (1+x)³ = (1+x)²(1+x) to split into two integrals in terms of Iₙ₋₁ and Iₙ: Iₙ = -(n/3)Iₙ₋₁ - (n/3)Iₙ \Rightarrow [(n+3)/3]Iₙ = -(n/3)Iₙ₋₁ \Rightarrow Iₙ = -[n/(n+3)] Iₙ₋₁

Worked Example 6 — a compound-angle trick (no parts at all)

Find a reduction formula for Iₙ = \int (sin nx / sin x) dx. Write sin nx = sin[(n-2)x + 2x] and expand using addition formulae, then use double angle formulae for cos 2x and sin 2x. After simplifying: Iₙ = \int [sin(n-2)x/sin x] dx + 2\int cos(n-1)x dx = Iₙ₋₂ + [2/(n-1)] sin(n-1)x \Rightarrow Iₙ = Iₙ₋₂ + [2/(n-1)] sin(n-1)x

✏️ Practice Iₙ = ∫ secⁿx dx. Show that (n−1)Iₙ = sec^(n−2)x tan x + (n−2)Iₙ₋₂. (Hint: split secⁿx = sec^(n−2)x sec²x, integrate by parts.)

Arc Length

Imagine zooming so far into a curve that a tiny piece of it looks like a straight line segment, forming the hypotenuse of a tiny right-angled triangle with legs $δx$ and $δy$ . Pythagoras gives $(δs)² \approx (δx)² + (δy)²$ . Divide through and let $δx \to 0$ , and you get the arc length formula.

Arc length — Cartesian form s = ∫ √[1 + (dy/dx)²] dx or s = ∫ √[(dx/dy)² + 1] dy

Arc length — parametric form s = ∫ √[(dx/dt)² + (dy/dt)²] dt = ∫ √(ẋ² + ẏ²) dt

🧠 One formula, three faces All three versions come from the same triangle — just choose whichever variable your curve is naturally given in terms of (x, y, or a parameter t).

Worked Example — Cartesian form

Find the length of y = (2/3)x^(3/2), from x = 3 to x = 8. dy/dx = \sqrtx s = \int₃⁸ \sqrt(1 + x) dx = [⅔(1+x)^(3/2)]₃⁸ = ⅔(9)^(3/2) - ⅔(4)^(3/2) = ⅔(27) - ⅔(8) = 18 - 16/3 \Rightarrow s = 12⅔

Worked Example — parametric form

Find the length of one arch of the cycloid x = a(t - sin t), y = a(1 - cos t). dx/dt = a(1 - cos t), dy/dt = a sin t ẋ² + ẏ² = a²(1 - 2cos t + cos²t + sin²t) = 2a²(1 - cos t) Use 1 - cos t = 2sin²(t/2): \sqrt(ẋ²+ẏ²) = a\sqrt[4sin²(t/2)] = 2a sin(t/2) (for 0 \leq t \leq 2π, sin(t/2) \geq 0) s = \int₀^(2π) 2a sin(t/2) dt = [-4a cos(t/2)]₀^(2π) = -4a - (-4a) \Rightarrow s = 8a

⚠️ Common slip Don't forget to square-root the sum of the squared derivatives — students sometimes forget the "+1" (Cartesian) or forget to add both squared derivatives together (parametric) before rooting.

✏️ Practice Find the arc length of the curve x = 3t², y = 2t³, from t = 0 to t = 2.

Area of a Surface of Revolution

Spin a curve around the x-axis and it sweeps out a 3D surface — think of a lampshade or a vase. To find its surface area, chop the curve into tiny arc pieces of length $δs$ , each sitting a distance $y$ from the axis. When rotated, each tiny piece traces out (approximately) a thin cylindrical band of circumference $2πy$ and width $δs$ , so its area is $\approx 2πy\cdotδs$ . Add them all up and take the limit — that's the integral.

Surface area of revolution about the x-axis A = ∫ 2πy (ds/dx) dx or A = ∫ 2πy (ds/dt) dt

where you plug in $ds/dx = \sqrt[1+(dy/dx)²]$ or $ds/dt = \sqrt[(dx/dt)²+(dy/dt)²]$ from the arc length section above — this formula is literally "2πy" glued onto the arc length integrand.

Worked Example — Cartesian, deriving the sphere formula

A sphere of radius r. Find the surface area between the planes x = a and x = b. x² + y² = r² \Rightarrow 2x + 2y(dy/dx) = 0 \Rightarrow dy/dx = -x/y ds/dx = \sqrt[1 + x²/y²] A = \intₐᵇ 2πy \sqrt(1 + x²/y²) dx = \intₐᵇ 2π\sqrt(y² + x²) dx = \intₐᵇ 2πr dx (since x²+y²=r²) \Rightarrow A = [2πrx]ₐᵇ = 2πr(b - a) Taking the whole sphere (a = -r to b = r) gives the famous 4πr² result.

🏺 Historical aside Archimedes proved that a sphere's surface area equals the curved surface area of the cylinder that exactly wraps around it — a neat sanity check for the formula above.

Worked Example — parametric

The parabola x = at², y = 2at, from t = 0 to t = 2, is rotated about the x-axis. Find the surface area. dx/dt = 2at, dy/dt = 2a ds/dt = \sqrt[(2at)² + (2a)²] = 2a\sqrt(t²+1) A = \int₀² 2π(2at) \times 2a\sqrt(t²+1) dt = 8πa² \int₀² t\sqrt(t²+1) dt Let u = t²+1: A = 8πa² \times [⅓(t²+1)^(3/2)]₀² = 8πa²[⅓(5^(3/2)) - ⅓] \Rightarrow A = (8πa²/3)(5\sqrt5 - 1)

⚠️ Common slip Always double-check which variable you're rotating about and whether y (the radius of each band) is expressed correctly in terms of that variable before you multiply by 2π — mixing up x and y here is the single most common error.

✏️ Practice The curve y = x³, for 0 ≤ x ≤ 1, is rotated about the x-axis. Set up (but don't fully evaluate numerically) the integral for the surface area formed.

What to Memorise

Standard substitutions

∫1/(a²+b²x²) dx	bx = a tan u
∫1/√(a²+b²x²) dx	bx = a sinh u
∫1/(a²−b²x²) dx	bx = a tanh u (or partial fractions)
∫1/√(b²x²−a²) dx	bx = a cosh u

Derivatives of inverse functions (needed for "parts with dv/dx=1")

arcsin x	1/√(1−x²)
arccos x	−1/√(1−x²)
arctan x	1/(1+x²)
arsinh x	1/√(1+x²)
arcosh x	1/√(x²−1)
artanh x	1/(1−x²)

Core formulae

Arc length (Cartesian): $s = \int\sqrt[1+(dy/dx)²] dx$
Arc length (parametric): $s = \int\sqrt(ẋ²+ẏ²) dt$
Surface of revolution: $A = \int2πy (ds/dx) dx$ or $\int2πy (ds/dt) dt$
Reduction formula strategy: integrate by parts to peel off one power at a time, relating Iₙ to Iₙ₋₁ or Iₙ₋₂
Odd power under a root → substitute u = a² ± x² directly; even power → use the full trig/hyperbolic substitution

Concepts Checklist

I can recognise when an integral is just a standard derivative reversed.

I can use trig/hyperbolic identities (double angle, Pythagorean, Osborne's rule) to rewrite an integrand into an integrable form.

I can spot the reverse chain rule pattern (uⁿ · du/dx) and find the correct constant by differentiating my guess.

I know all four standard substitutions and can choose the right one based on square root + sum/difference.

I can complete the square before substituting when the quadratic isn't already in a² ± b²x² form.

I can integrate inverse trig/hyperbolic functions and ln x using parts with dv/dx = 1.

I can derive a reduction formula using integration by parts (once or twice) or algebraic manipulation.

I can use a reduction formula recursively to evaluate a specific case (e.g. I₆ from I₀).

I can find arc length in both Cartesian and parametric form.

I can find the area of a surface of revolution in both Cartesian and parametric form.

Exam Tips

Always try the fastest method first. Before launching into a substitution, check if the integral is a standard result in disguise (Technique 1) or one identity away from being one (Technique 2). Reduction formula and substitution questions are usually flagged by the presence of a power n or a square root respectively.

Complete the square before choosing a substitution whenever the quadratic isn't already written as a² ± b²x². Marks are frequently lost by attempting the substitution on the un-simplified quadratic.

Check odd vs even powers before diving into a hyperbolic/trig substitution for ∫xⁿ/√(a²±x²) dx — an odd power almost always has a much faster route via u = a² ± x².

In reduction formula questions, always state which variable is "u" and which is "dv/dx" explicitly — examiners want to see the choice justified (usually: differentiate the power to reduce it, integrate the part you already know how to integrate).

Watch your limits when they appear. In definite reduction formula problems (like the Wallis-type ∫₀^(π/2) sinⁿx dx), the boundary term from integration by parts very often vanishes — but check this explicitly rather than assuming it, since not every reduction formula has that luxury.

For arc length and surface area, decide Cartesian vs parametric immediately based on how the curve is given — converting parametric to Cartesian first is usually far more work than just using the parametric formula directly.

Don't forget the "ds" is doing double duty. The surface-of-revolution formula A = ∫2πy ds is built directly from the arc length formula — if you've already found ds/dx or ds/dt for an arc length part of a question, reuse it rather than recomputing.

Simplify square roots using Pythagorean-style identities (like 1 − cos t = 2sin²(t/2), or cosh²u − 1 = sinh²u) before integrating — leaving an un-simplified root under the integral sign is the most common reason students get stuck partway through.

What's inside

📖 Revision notes ✦ AI flashcards ✓ Instant AI marking

Read the full Integration notes free

That's the preview — create a free account to read the rest, plus flashcards and practice questions with instant AI marking. No credit card.

Unlock the full notes free →

Integration

Integration

Summary

Standard Techniques

1. Recognise a standard function

2. Using formulae to change the integrand

3. Reverse chain rule

4. Standard substitutions

5. Integrating inverse functions and ln x

Reduction Formulae

Arc Length

Area of a Surface of Revolution

What to Memorise

Standard substitutions

Derivatives of inverse functions (needed for "parts with dv/dx=1")

Core formulae

Concepts Checklist

Exam Tips

More Further Pure Mathematics 3 topics