Integration by Parts

This lesson covers integration by parts — a technique for integrating the product of two functions — as required by the Edexcel A-Level Mathematics specification (9MA0). You need to know the formula, how to choose u and dv/dx, when to apply the method repeatedly, and the LIATE rule for guidance.

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The Formula

Integration by parts is derived from the product rule for differentiation. The formula is:

∫ u (dv/dx) dx = uv - ∫ v (du/dx) dx

Alternatively, using the shorthand notation:

∫ u dv = uv - ∫ v du

Derivation

The product rule states: d/dx(uv) = u(dv/dx) + v(du/dx)

Rearranging: u(dv/dx) = d/dx(uv) - v(du/dx)

Integrating both sides: ∫ u(dv/dx) dx = uv - ∫ v(du/dx) dx

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Choosing u and dv/dx

The key to integration by parts is choosing the right function for u and the right function for dv/dx. The goal is to make the resulting integral ∫ v(du/dx) dx simpler than the original.

The LIATE Rule

A useful guideline is the LIATE rule. Choose u to be the function that comes first in this list:

Priority	Type	Examples
1st	Logarithmic	ln x, log x
2nd	Inverse trigonometric	arcsin x, arctan x
3rd	Algebraic (polynomial)	x, x², 3x + 1
4th	Trigonometric	sin x, cos x
5th	Exponential	eˣ, 2ˣ

The remaining function becomes dv/dx.

Key Point: The LIATE rule is a guideline, not a strict rule. It works in the vast majority of cases at A-Level.

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Worked Examples

Example 1: ∫ x cos x dx

Choose u and dv/dx:

• u = x (algebraic — higher priority in LIATE)
• dv/dx = cos x

Find du/dx and v:

• du/dx = 1
• v = sin x

Apply 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

Example 2: ∫ x eˣ dx

Choose: u = x, dv/dx = eˣ

Find: du/dx = 1, v = eˣ

Apply: ∫ x eˣ dx = x eˣ - ∫ eˣ dx = x eˣ - eˣ + c = eˣ(x - 1) + c

Example 3: ∫ x² sin x dx (repeated application)

First application:

• u = x², dv/dx = sin x
• du/dx = 2x, v = -cos x

∫ x² sin x dx = x²(-cos x) - ∫ (-cos x)(2x) dx = -x² cos x + 2 ∫ x cos x dx

Second application (on ∫ x cos x dx):

• u = x, dv/dx = cos x
• du/dx = 1, v = sin x

∫ x cos x dx = x sin x - ∫ sin x dx = x sin x + cos x + c

Combining: ∫ x² sin x dx = -x² cos x + 2(x sin x + cos x) + c = -x² cos x + 2x sin x + 2 cos x + c

Example 4: ∫ ln x dx

This is a classic example — it looks like there is only one function, but we write it as 1 × ln x.

Choose: u = ln x, dv/dx = 1

Find: du/dx = 1/x, v = x

Apply: ∫ ln x dx = x ln x - ∫ x × (1/x) dx = x ln x - ∫ 1 dx = x ln x - x + c

Exam Tip: To integrate ln x, always use integration by parts with u = ln x and dv/dx = 1. This is a standard result worth memorising: ∫ ln x dx = x ln x - x + c.

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Integration by Parts with Definite Integrals

For definite integrals, apply the limits to the formula:

∫ from a to b of u(dv/dx) dx = [uv] from a to b - ∫ from a to b of v(du/dx) dx

Example: Evaluate ∫ from 0 to 1 of x e^(2x) dx

Choose: u = x, dv/dx = e^(2x) Find: du/dx = 1, v = (1/2)e^(2x)

∫ from 0 to 1 of x e^(2x) dx = [(1/2)x e^(2x)] from 0 to 1 - ∫ from 0 to 1 of (1/2)e^(2x) dx

= [(1/2)(1)e² - 0] - [(1/4)e^(2x)] from 0 to 1

= (1/2)e² - [(1/4)e² - (1/4)e⁰]

= (1/2)e² - (1/4)e² + 1/4

= (1/4)e² + 1/4

= (e² + 1)/4

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The "Cycling" Technique

Sometimes, applying integration by parts twice brings you back to the original integral. This happens with integrals like ∫ eˣ sin x dx and ∫ eˣ cos x dx.

Example: ∫ eˣ sin x dx

First application:

• u = sin x, dv/dx = eˣ
• du/dx = cos x, v = eˣ

∫ eˣ sin x dx = eˣ sin x - ∫ eˣ cos x dx ... (*)

Second application (on ∫ eˣ cos x dx):

• u = cos x, dv/dx = eˣ
• du/dx = -sin x, v = eˣ

∫ eˣ cos x dx = eˣ cos x - ∫ eˣ(-sin x) dx = eˣ cos x + ∫ eˣ sin x dx

Substitute back into (*): ∫ eˣ sin x dx = eˣ sin x - [eˣ cos x + ∫ eˣ sin x dx] ∫ eˣ sin x dx = eˣ sin x - eˣ cos x - ∫ eˣ sin x dx

Add ∫ eˣ sin x dx to both sides: 2 ∫ eˣ sin x dx = eˣ sin x - eˣ cos x

Divide by 2: ∫ eˣ sin x dx = (1/2)eˣ(sin x - cos x) + c

Exam Tip: When the original integral reappears after two applications, call it I and solve the resulting equation for I. This technique is tested regularly at A-Level.

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When to Use Integration by Parts vs Substitution

Use Integration by Parts	Use Substitution
Product of two different types of function (e.g., x × eˣ)	Composite function with inner derivative present (e.g., 2x × e^(x²))
∫ x sin x dx, ∫ x² eˣ dx	∫ sin(3x + 1) dx, ∫ x√(x² + 1) dx
∫ ln x dx, ∫ x ln x dx	∫ f'(x)/f(x) dx

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Common Mistakes

1. Wrong choice of u — If the integral becomes more complicated after applying parts, try swapping u and dv/dx.
2. Sign errors — Be very careful with the minus sign in the formula: uv minus ∫ v du/dx dx.
3. Forgetting to apply parts again — Sometimes one application is not enough (e.g., ∫ x² eˣ dx needs two applications).
4. Not recognising the cycling technique — If you get the original integral back, set up an equation and solve for it.

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Summary

• Integration by parts formula: ∫ u(dv/dx) dx = uv - ∫ v(du/dx) dx.
• Use the LIATE rule to choose u (logarithmic > inverse trig > algebraic > trig > exponential).
• For ∫ ln x dx, let u = ln x and dv/dx = 1.
• Apply the method repeatedly for higher powers of x (e.g., ∫ x² eˣ dx).
• Use the cycling technique when the original integral reappears (e.g., ∫ eˣ sin x dx).
• For definite integrals, apply limits to the [uv] term and to the remaining integral.

A-Level Deep Dive: Integration by Parts

Spec mapping

Edexcel 9MA0-02 specification section 10 — Integration, sub-strand 10.5: integration by parts is a Year 2 Pure technique used to integrate products of functions, derived by reversing the product rule. The formula $\int u \, dv = uv - \int v \, du$∫udv=uv−∫vdu — equivalently $\int u(x) \dfrac{dv}{dx}\,dx = u(x)v(x) - \int v(x)\dfrac{du}{dx}\,dx$∫u(x)dxdv​dx=u(x)v(x)−∫v(x)dxdu​dx — is listed in the Edexcel formula booklet, so candidates are not required to memorise it, but they are required to apply it fluently. By-parts is examined on 9MA0-02 (Pure Mathematics 2) and frequently surfaces in 9MA0-03 (Statistics & Mechanics) when computing expected values $E(X) = \int x f(x)\,dx$E(X)=∫xf(x)dx for continuous random variables and when integrating velocity expressions involving products. Synoptically the technique connects to section 7 (Differentiation, product rule), section 9 (Numerical methods, when an antiderivative cannot be found in closed form) and section 8 (Differential equations, where by-parts often features in solving non-separable forms).

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Worked example with full mark scheme

Question (8 marks): Find $\displaystyle\int x^2 e^x \, dx$∫x2exdx, simplifying your answer.

Solution with mark scheme:

Step 1 — choose $u$u and $dv$dv for the first application of by-parts.

Apply the LIATE rule: among Logarithmic, Inverse-trig, Algebraic, Trigonometric, Exponential factors, choose $u$u to be the type that appears earlier in the list. Here we have an algebraic factor $x^2$x2 and an exponential factor $e^x$ex. A beats E, so let:

$u = x^2, \quad \dfrac{dv}{dx} = e^x$u=x2,dxdv​=ex

Then $\dfrac{du}{dx} = 2x$dxdu​=2x and $v = e^x$v=ex.

M1 — correct identification of $u$u and $\dfrac{dv}{dx}$dxdv​ consistent with reducing the algebraic power. Choosing the reverse ($u = e^x$u=ex, $dv = x^2\,dx$dv=x2dx) does not reduce the difficulty of the integral and earns no marks for setup.

Step 2 — apply the by-parts formula.

$\int x^2 e^x \, dx = x^2 e^x - \int e^x \cdot 2x \, dx = x^2 e^x - 2\int x e^x \, dx$∫x2exdx=x2ex−∫ex⋅2xdx=x2ex−2∫xexdx

A1 — correct first application, with the new integral $\int x e^x \, dx$∫xexdx clearly written and the constant factor $2$2 extracted.

Step 3 — recognise that $\int x e^x \, dx$∫xexdx itself requires by-parts.

For the new integral, again apply LIATE: $u = x$u=x (algebraic), $\dfrac{dv}{dx} = e^x$dxdv​=ex (exponential), so $\dfrac{du}{dx} = 1$dxdu​=1 and $v = e^x$v=ex.

$\int x e^x \, dx = x e^x - \int e^x \cdot 1 \, dx = x e^x - e^x + C_1$∫xexdx=xex−∫ex⋅1dx=xex−ex+C1​

M1 — correct second application of by-parts to the residual integral. The fact that this question requires by-parts twice is the marker that distinguishes it from a 4-mark question.

A1 — correct evaluation of the inner antiderivative $x e^x - e^x$xex−ex.

Step 4 — substitute back and simplify.

$\int x^2 e^x \, dx = x^2 e^x - 2(x e^x - e^x) + C = x^2 e^x - 2x e^x + 2e^x + C$∫x2exdx=x2ex−2(xex−ex)+C=x2ex−2xex+2ex+C

M1 — correct substitution of the inner result, with brackets handled correctly so that the sign of $+2e^x$+2ex comes out positive (subtracting a negative).

A1 — correct expanded form.

Step 5 — factorise to present a tidy final answer.

$\int x^2 e^x \, dx = e^x(x^2 - 2x + 2) + C$∫x2exdx=ex(x2−2x+2)+C

A1 — final factorised form with constant of integration. Examiners reward the factorised presentation here as it makes verification by differentiation straightforward.

Total: 8 marks (M3 A5).

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Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks): Find $\displaystyle\int_0^{\pi/2} x \sin x \, dx$∫0π/2​xsinxdx, giving your answer in exact form.

Mark scheme decomposition by AO:

• M1 (AO1.1a) — correct LIATE choice: $u = x$u=x, $\dfrac{dv}{dx} = \sin x$dxdv​=sinx, so $\dfrac{du}{dx} = 1$dxdu​=1 and $v = -\cos x$v=−cosx. The minus sign on $v$v is the most common slip.
• M1 (AO1.1b) — correct application of the by-parts formula: $\int_0^{\pi/2} x \sin x \, dx = [-x \cos x]_0^{\pi/2} - \int_0^{\pi/2}(-\cos x)\,dx$∫0π/2​xsinxdx=[−xcosx]0π/2​−∫0π/2​(−cosx)dx.
• A1 (AO1.1b) — sign management on the residual integral: $-\int(-\cos x)\,dx = +\int \cos x \, dx = \sin x$−∫(−cosx)dx=+∫cosxdx=sinx.
• M1 (AO1.1b) — correct evaluation of the boundary term $[-x\cos x]_0^{\pi/2} = -\tfrac{\pi}{2}\cos(\tfrac{\pi}{2}) - (-0\cdot\cos 0) = 0$[−xcosx]0π/2​=−2π​cos(2π​)−(−0⋅cos0)=0.
• A1 (AO1.1b) — correct evaluation of $[\sin x]_0^{\pi/2} = 1 - 0 = 1$[sinx]0π/2​=1−0=1.
• A1 (AO2.5) — final exact answer $1$1, presented as an integer (not $1.000$1.000 or $1.00$1.00). The "exact form" instruction binds the candidate to non-decimal presentation.

Total: 6 marks split AO1 = 5, AO2 = 1. Edexcel typically reserves one AO2 mark on by-parts questions for the closing simplification or the recognition that a boundary term vanishes — the moves that demonstrate fluency rather than rote application.

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Synoptic links

Connects to:

1. Section 7 — Differentiation, product rule (reverse derivation): the by-parts formula is the integrated version of the product rule $\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$dxd​(uv)=udxdv​+vdxdu​. Integrating both sides and rearranging gives $\int u\,dv = uv - \int v\,du$∫udv=uv−∫vdu. Recognising the formula as "product rule run backwards" is the conceptual hook that A* candidates use to remember it without the formula booklet.

2. LIATE / strategic choice of $u$u: the LIATE heuristic (Logarithmic, Inverse-trig, Algebraic, Trigonometric, Exponential) is not a rule but a guide for which factor to differentiate. The deeper principle is "differentiate the factor whose derivative is simpler" — so $\ln x$lnx becomes $1/x$1/x, $x^n$xn reduces to $nx^{n-1}$nxn−1, while $e^x$ex and $\sin x$sinx stay essentially the same complexity under differentiation.

3. Reduction formulae (university extension): repeated by-parts on integrands like $\int x^n e^x\,dx$∫xnexdx produces a recurrence $I_n = x^n e^x - n I_{n-1}$In​=xnex−nIn−1​. This reduction formula perspective is implicit in the 8-mark example above and is studied formally in Year 1 university calculus.

4. Substitution combined with by-parts (section 10.4): integrals such as $\int x \ln(x^2 + 1)\,dx$∫xln(x2+1)dx require first a substitution ($w = x^2 + 1$w=x2+1) followed by by-parts on the resulting $\ln w$lnw integrand. Recognising which technique to apply first, and in which order, is a synoptic problem-solving skill rewarded with AO3 marks.

5. Section 8 — Differential equations: non-separable first-order ODEs solved by integrating factor lead to integrals like $\int x e^x \, dx$∫xexdx and $\int e^{-x}\sin x\,dx$∫e−xsinxdx that require by-parts. The "integrating-factor method" cannot be completed without fluent by-parts.

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Mark-scheme literacy

By-parts questions on 9MA0-02 split AO marks as follows: