Advanced Integration Techniques — AHL 5.9 Further integration techniques (substitution; integration by parts) (HL) - IB Mathematics Analysis and Approaches

SECTION 1

Historical Context & Motivation

Integration is fundamentally the reverse of differentiation, but that reversal is far from straightforward. While the chain rule and the product rule give us systematic recipes for differentiating composite and product functions, undoing those rules when integrating requires special techniques. Two of the most important methods — integration by substitution and integration by parts — were developed over centuries of mathematical innovation.

1665–1676 • Newton & Leibniz Develop Calculus

Isaac Newton and Gottfried Wilhelm Leibniz independently developed the foundations of calculus, establishing that differentiation and integration are inverse processes through the Fundamental Theorem of Calculus.

1680s • Leibniz Formalises Substitution

Leibniz's notation — writing integrals as ∫ f(x) dx — made it natural to treat dx as a manipulable quantity, paving the way for the substitution method (also called u-substitution).

1715 • Brook Taylor & Integration by Parts

Brook Taylor, expanding on the product rule of differentiation, published early forms of what we now call integration by parts. The technique reverses the product rule to convert difficult integrals into simpler ones.

1800s • Rigorous Foundations by Riemann & Others

Bernhard Riemann and Augustin-Louis Cauchy placed integration on rigorous footing, providing formal justifications for substitution and parts that remain central to modern analysis.

The central question these techniques answer is: How do we integrate functions that are products or compositions of simpler functions? Basic power, exponential, and trigonometric rules handle only the simplest integrands. For anything more complex — such as ∫ x·eˣ dx or ∫ 2x·cos(x²) dx — you need the tools covered in this lesson.

SECTION 2

Core Principles & Definitions

Both substitution and integration by parts rest on differentiation rules you already know, applied in reverse. Understanding why each method works will help you decide when to use it.

Reversing the Chain Rule

Integration by substitution undoes the chain rule. If a function looks like f(g(x))·g′(x), substituting u = g(x) collapses the integral into ∫ f(u) du, which is usually much simpler.

Reversing the Product Rule

Integration by parts undoes the product rule. It converts ∫ u dv into u·v − ∫ v du, trading one integral for another that (with the right choice) is easier to evaluate.

Choosing u and dv/du

In substitution, pick u = the 'inner function' whose derivative also appears. In parts, pick u as the function that simplifies when differentiated and dv as the part you can integrate.

Definite vs. Indefinite

Both techniques work with definite integrals. When using substitution on definite integrals, remember to change the limits of integration to match the new variable u.

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KEY TAKEAWAY

Think of integration by substitution like translating a sentence into a simpler language — you swap the variable, solve the easier integral, then translate back. Integration by parts is more like rearranging a messy room: you move one piece (the integral of the product) into two more manageable pieces (a boundary term minus a simpler integral). Neither creates new mathematics; both reorganise existing relationships to reveal the answer.

SECTION 3

Visual Explanation — Substitution in Action

The diagram below illustrates the idea behind u-substitution. On the left, you see the graph of f(x) = 2x·cos(x²), a product involving a composite function. By setting u = x², the right-hand graph shows the same area re-expressed as cos(u), which is straightforward to integrate.

Left: the original integrand 2x·cos(x²) with the area under the curve shaded in cyan. Right: after substituting u = x², the integrand simplifies to cos(u) (shaded violet). Both shaded regions represent the same accumulated value — the substitution merely repackages the area in an easier-to-compute form.

Notice how the complicated oscillating curve on the left becomes the familiar cosine wave on the right. The "extra" factor of 2x in the original integrand is exactly the derivative of x², which is why the substitution works perfectly. If that derivative were not present, the substitution would fail or require additional algebraic manipulation.

SECTION 4

Mathematical Framework

Integration by Substitution

SUBSTITUTION RULE

∫ f(g(x)) · g′(x) dx = ∫ f(u) du, where u = g(x)

Here u is the inner function g(x), and du = g′(x) dx. After integrating with respect to u, substitute back to express the result in terms of x.

The practical steps are: (1) identify a part of the integrand to call u; (2) compute du/dx and rearrange so that dx is expressed in terms of du; (3) rewrite the entire integral in terms of u; (4) integrate; (5) replace u with g(x).

DEFINITE INTEGRAL — SUBSTITUTION WITH CHANGED LIMITS

∫ₐᵇ f(g(x)) · g′(x) dx = ∫_{g(a)}^{g(b)} f(u) du

When evaluating a definite integral, convert the original limits a and b to the new variable: lower limit becomes g(a), upper limit becomes g(b). This removes the need to substitute back.

Integration by Parts

INTEGRATION BY PARTS FORMULA

∫ u (dv/dx) dx = u·v − ∫ v (du/dx) dx

Choose u = the function that becomes simpler when differentiated, and dv/dx = the function you can readily integrate. The goal is that ∫ v (du/dx) dx is easier than the original integral.

DEFINITE FORM — INTEGRATION BY PARTS

∫ₐᵇ u (dv/dx) dx = [u·v]ₐᵇ − ∫ₐᵇ v (du/dx) dx

The bracket notation [u·v]ₐᵇ means evaluate u·v at x = b and subtract its value at x = a. Then subtract the remaining integral evaluated between the same limits.

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LIATE Rule for Choosing u

A helpful mnemonic for integration by parts is LIATE: choose u in this priority order — Logarithmic → Inverse trig → Algebraic (polynomials) → Trigonometric → Exponential. The function highest on the list should be u, and the remainder is dv/dx.

SECTION 5

Deciding Which Technique to Use

One of the biggest challenges students face is recognising which technique to apply. The decision flowchart below provides a systematic approach. Start at the top and follow the arrows based on the structure of your integrand.

Follow this decision flowchart when facing an unfamiliar integral. Begin by checking whether the integrand is a composite function with its inner derivative present (substitution path, cyan). If not, check whether it is a product of two different function types (parts path, violet). In some IB problems, you may need to combine both techniques.

Integral Form	Technique	Why It Works
∫ 2x · eˣ² dx	Substitution	2x is the derivative of x², the inner function of eˣ²
∫ sin(3x + 1) dx	Substitution	Inner function 3x + 1 has constant derivative 3 (adjust with 1/3)
∫ x · eˣ dx	Parts	Product of algebraic (x) and exponential (eˣ); no chain rule link
∫ x² · ln x dx	Parts	Product of algebraic and logarithmic; LIATE says u = ln x
∫ ln x dx	Parts	Write as ∫ 1 · ln x dx; set u = ln x, dv/dx = 1

Common integral forms and the recommended technique

SECTION 6

Worked Examples

Example 1: Integration by Substitution

Evaluate ∫ 3x² · cos(x³) dx.

Substitution: ∫ 3x² · cos(x³) dx

1
Step 1 — Identify the inner function

The integrand contains cos(x³), a composite function. The inner function is x³, and its derivative 3x² appears as a factor. This is a strong signal for substitution.
2
Step 2 — Define u and find du

Let u = x³. Then du/dx = 3x², so du = 3x² dx.

u = x³, du = 3x² dx
3
Step 3 — Rewrite the integral in terms of u

Replacing 3x² dx with du and cos(x³) with cos(u), the integral becomes ∫ cos(u) du.

∫ cos(u) du
4
Step 4 — Integrate with respect to u

The integral of cos(u) is sin(u) + C.

sin(u) + C
5
Step 5 — Substitute back

Replace u with x³ to express the answer in terms of the original variable.

∫ 3x² · cos(x³) dx = sin(x³) + C

Example 2: Integration by Parts

Evaluate ∫ x · eˣ dx.

Integration by Parts: ∫ x · eˣ dx

1
Step 1 — Recognise the need for parts

The integrand is a product of an algebraic function (x) and an exponential (eˣ). There is no chain-rule relationship between them, so substitution will not simplify this. We use integration by parts.
2
Step 2 — Choose u and dv/dx using LIATE

Algebraic functions rank higher than exponential in LIATE. So let u = x (which simplifies to 1 when differentiated) and dv/dx = eˣ (which we can integrate easily).

u = x, du/dx = 1, dv/dx = eˣ, v = eˣ
3
Step 3 — Apply the formula

Using ∫ u (dv/dx) dx = u·v − ∫ v (du/dx) dx, we get:

∫ x · eˣ dx = x · eˣ − ∫ eˣ · 1 dx
4
Step 4 — Evaluate the remaining integral

The remaining integral ∫ eˣ dx is simply eˣ + C. Substituting back:

= x · eˣ − eˣ + C
5
Step 5 — Simplify

Factor out eˣ to present a clean final answer.

∫ x · eˣ dx = eˣ(x − 1) + C

SECTION 7

Strengths & Limitations of Each Technique

Feature	Substitution	Integration by Parts
Best for	Composite functions f(g(x)) where g′(x) appears as a factor	Products of functions from different families (e.g., polynomial × exponential)
Differentiation rule reversed	Chain rule	Product rule
Typical difficulty	Usually reduces the integral in one step	May need to be applied more than once (e.g., ∫ x² eˣ dx)
Common pitfall	Forgetting to change limits in definite integrals	Choosing u and dv/dx the wrong way round, making the new integral harder
Limitation	Fails if the derivative of the inner function is not present (even up to a constant)	Fails if neither choice of u produces a simpler integral

Comparison of substitution and integration by parts

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KEY TAKEAWAY

Think of these two techniques as tools in a toolkit. Substitution is like a wrench — it works best when the bolt (inner function) and the handle (its derivative) are already matched. Integration by parts is like a lever — it helps you pry apart a product by shifting the difficulty from one integral to another. Knowing which tool fits the problem is just as important as knowing how to use either one.

SECTION 8

Connections to Advanced Theory

Substitution and integration by parts are the building blocks for even more powerful techniques you will encounter in further study. In the IB HL course itself, these methods combine with trigonometric identities and partial fractions. At university level, they extend into multivariable calculus, differential equations, and theoretical physics.

This Lesson (AHL 5.9)	Where It Leads
u-substitution on single-variable integrals	Change of variables in double and triple integrals (multivariable calculus)
Integration by parts for ∫ u dv	Solving differential equations via integrating factors and reduction formulae
Choosing u = ln x, dv = rest	Integral representations of special functions (Gamma function, Beta function)
Repeated integration by parts	Tabular (DI) method and reduction formulas for ∫ sinⁿx dx, ∫ xⁿeˣ dx

How AHL 5.9 techniques extend into higher mathematics

In the IB exam, integration by parts may also appear in Paper 3 (HL) as part of proof-based or exploration-style questions. A strong grasp of these techniques is essential not only for the calculus option but also for probability distributions (e.g., finding the mean of a continuous random variable using ∫ x · f(x) dx, which often requires parts).

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain in your own words why integration by substitution works. Specifically, which differentiation rule does it reverse, and what role does the factor g′(x) play in the integrand?

Show answer

Integration by substitution reverses the chain rule of differentiation. When we differentiate a composite function f(g(x)), the chain rule tells us the result is f′(g(x)) · g′(x). So when we encounter an integrand of the form f(g(x)) · g′(x), we recognise it as the derivative of F(g(x)), where F is an antiderivative of f. The factor g′(x) is crucial because it accounts for the 'stretching' caused by the inner function — without it (or a constant multiple of it), the substitution cannot cleanly absorb all the x-dependent terms into du.

Question 2: BASIC CALCULATION

Evaluate ∫ 4x · (x² + 5)³ dx using an appropriate substitution.

Show answer

Let u = x² + 5, so du = 2x dx, meaning 4x dx = 2 du. The integral becomes ∫ 2u³ du = 2 · u⁴/4 = u⁴/2 + C. Substituting back: ∫ 4x · (x² + 5)³ dx = (x² + 5)⁴ / 2 + C. You can verify by differentiating the result using the chain rule: d/dx[(x² + 5)⁴ / 2] = 4(x² + 5)³ · 2x / 2 = 4x(x² + 5)³, which matches the original integrand.

Question 3: INTERMEDIATE

Evaluate ∫ x · cos(x) dx using integration by parts.

Show answer

Using LIATE, let u = x (algebraic, higher priority) and dv/dx = cos(x). Then du/dx = 1 and v = sin(x). Applying the formula: ∫ x · cos(x) dx = x · sin(x) − ∫ sin(x) · 1 dx = x · sin(x) − (−cos(x)) + C = x · sin(x) + cos(x) + C. To verify, differentiate: d/dx[x sin(x) + cos(x)] = sin(x) + x cos(x) − sin(x) = x cos(x), confirming the answer.

Question 4: APPLIED

A particle moves along a straight line with velocity v(t) = t · e⁻ᵗ m/s. Find the displacement of the particle between t = 0 and t = 3 seconds. Give your answer in exact form.

Show answer

Displacement = ∫₀³ t · e⁻ᵗ dt. Use integration by parts: let u = t, dv/dt = e⁻ᵗ. Then du/dt = 1, v = −e⁻ᵗ. Applying the formula: [t · (−e⁻ᵗ)]₀³ − ∫₀³ (−e⁻ᵗ)(1) dt = [−t · e⁻ᵗ]₀³ + ∫₀³ e⁻ᵗ dt = (−3e⁻³ − 0) + [−e⁻ᵗ]₀³ = −3e⁻³ + (−e⁻³ + 1) = 1 − 4e⁻³. This gives approximately 1 − 4(0.04979) ≈ 0.801 metres. The exact displacement is (1 − 4e⁻³) metres.

Question 5: CRITICAL THINKING

Evaluate ∫ eˣ · sin(x) dx. Hint: you will need to apply integration by parts twice, and the original integral will reappear. Solve for it algebraically.

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Let I = ∫ eˣ sin(x) dx. First application: u = sin(x), dv/dx = eˣ, so du/dx = cos(x), v = eˣ. Then I = eˣ sin(x) − ∫ eˣ cos(x) dx. Second application on ∫ eˣ cos(x) dx: u = cos(x), dv/dx = eˣ, so du/dx = −sin(x), v = eˣ. Then ∫ eˣ cos(x) dx = eˣ cos(x) − ∫ eˣ(−sin(x)) dx = eˣ cos(x) + I. Substituting back: I = eˣ sin(x) − [eˣ cos(x) + I] = eˣ sin(x) − eˣ cos(x) − I. Adding I to both sides: 2I = eˣ(sin(x) − cos(x)). Therefore I = eˣ(sin(x) − cos(x))/2 + C. This 'boomerang' technique — where the original integral reappears and you solve for it algebraically — is a classic IB HL exam pattern.

SUMMARY

Lesson Summary

Summary

This lesson covered two essential techniques in the IB HL calculus toolkit. Integration by substitution reverses the chain rule: you set u = g(x), replace dx with du/g′(x), and integrate a simpler function of u before substituting back. For definite integrals, you convert the limits to the u-variable, eliminating the need to convert back. The key requirement is that g′(x) (or a constant multiple of it) must appear as a factor in the integrand.

Integration by parts reverses the product rule using the formula ∫ u (dv/dx) dx = u·v − ∫ v (du/dx) dx. The LIATE mnemonic (Logarithmic → Inverse trig → Algebraic → Trigonometric → Exponential) guides your choice of u. In some problems — like ∫ eˣ sin(x) dx — you must apply parts twice and solve for the original integral algebraically, a pattern known as the boomerang technique. Mastering both methods and knowing when to deploy each one is a crucial skill for IB HL Paper 1, Paper 2, and Paper 3.