Further Integration

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Quick Overview

Integration by Substitution Simplify complex integrals by substituting u = f(x) and converting everything in terms of u

Harder Substitution When a substitution is given, apply the same process but expect trickier algebra

Integration by Parts For products of functions using ∫u dv/dx dx = uv - ∫v du/dx dx

Partial Fractions Split complex fractions into simpler pieces that integrate easily

Volumes of Revolution Find volumes when 2D regions rotate around an axis using V = π∫y²dx

Decision Making Use a systematic approach to choose which integration technique fits each problem

1. Integration by Substitution (Reverse Chain Rule)

When a function is too awkward to integrate directly, and you can't spot the reverse chain rule immediately, substitution transforms the integral into something manageable.

What is Integration by Substitution?

Substitution is useful when:

The reverse chain rule is hard to spot
You have a composite function (a function inside another function)
The substitution is "obvious" from context—you're expected to find it

The key insight: Look to substitute the "second" (inner) function rather than the "main" (outer) function.

Example: In ∫18x(x² + 2)⁵ dx, the main function is (...)⁵ and the second function is x² + 2

So u = x² + 2 would be a suitable substitution.

The Four-Step Method

STEP 1: Determine the substitution Ask yourself: What is the 'main' function? What is the 'second' function? Choose the second one as u.

STEP 2: Differentiate the substitution and rearrange Find du/dx, then treat it like a fraction. Rearrange to get dx in terms of du.
Example: If u = 5x³, then du/dx = 15x², so dx = du/(15x²)

STEP 3: Replace all parts of the integral Replace all x terms with u terms (including dx).
If it's a definite integral, change the limits from x values to u values.

STEP 4: Integrate and substitute back The result should be easy to integrate now. After integrating, substitute x back in.
Don't forget the constant c for indefinite integrals!

If u = f(x), then ∫g(f(x))·f'(x)dx = ∫g(u)du

Worked Example

Find: ∫₀¹ 15x² cos(5x³) dx

STEP 1: DETERMINE THE SUBSTITUTION
u = 5x³ (the inner function)

STEP 2: DIFFERENTIATE AND REARRANGE
du/dx = 15x²
du = 15x² dx

STEP 3: REPLACE ALL PARTS
When x = 0: u = 0
When x = 1: u = 5

∫₀¹ 15x² cos(5x³) dx = ∫₀⁵ cos(u) du

STEP 4: INTEGRATE AND EVALUATE
= [sin(u)]₀⁵
= sin(5) - sin(0)
= sin(5) ≈ -0.959 (3 s.f.)

Examiner Tip:

Practice Questions

What to Memorise

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Also in the full note

2. Harder Substitution
3. Integration by Parts
4. Integration using Partial Fractions
5. Volumes of Revolution
6. Adding & Subtracting Volumes of Revolution
7. Modelling with Volumes of Revolution
8. Integration Decision Making
Concepts Checklist

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Further Integration