Integration by Parts

Integrating products using the reverse of the product rule

SPEC 8.5 • INTEGRATION

Why Integration by Parts?

Some integrals cannot be solved using substitution or standard results. Consider these products:

Challenge 1

\int x e^{x} d x

Polynomial × Exponential

Challenge 2

\int x cos x d x

Polynomial × Trigonometric

Key Insight

Integration by parts is derived from the product rule for differentiation. It allows us to transform a difficult integral into a simpler one.

Deriving the Formula

Start with the product rule for differentiation:

Step 1: Product rule:

\frac{d}{d x} (uv) = u \frac{d v}{d x} + v \frac{d u}{d x}

Step 2: Rearrange:

u \frac{d v}{d x} = \frac{d}{d x} (uv) - v \frac{d u}{d x}

Step 3: Integrate both sides with respect to x:

Integration by Parts Formula

\int u \frac{d v}{d x} d x = uv - \int v \frac{d u}{d x} d x

Alternative Notation

\int u d v = uv - \int v d u

The Formula

Integration by Parts

\int u \frac{d v}{d x} d x = uv - \int v \frac{d u}{d x} d x

How to Apply

Identify the two parts of the product

Choose which part is u and which is dv/dx

Find du/dx by differentiating u

Find v by integrating dv/dx

Substitute into the formula

Simplify the remaining integral

Important

The goal is to make the new integral

\int v \frac{d u}{d x} d x

simpler than the original!

Choosing u and dv/dx

Use the LATE rule to decide which part should be u:

L

Logarithms — always choose as u (ln x differentiates to 1/x)

A

Algebraic — polynomials like x, x², x³ (differentiates to simpler)

T

Trigonometric — sin x, cos x (choose as dv/dx usually)

E

Exponentials — eˣ (choose as dv/dx — integrates to itself)

Rule of Thumb

Choose u as the function that becomes simpler when differentiated. Polynomials eventually differentiate to zero!

Worked Example01

Find $\int x e^{x} d x$

1

Identify: u = x (polynomial), dv/dx = eˣ (exponential)

2

Find derivatives/integrals: du/dx = 1, v = eˣ

3

Apply formula:

\int x e^{x} d x = x \cdot e^{x} - \int e^{x} \cdot 1 d x

4

Simplify:

= x e^{x} - e^{x} + C

Answer

x e^{x} - e^{x} + C

or

e^{x} (x - 1) + C

Worked Example02

Find $\int x cos x d x$

1

Identify: u = x (polynomial), dv/dx = cos x (trigonometric)

2

Find derivatives/integrals: du/dx = 1, v = sin x

3

Apply formula:

\int x cos x d x = x sin x - \int sin x \cdot 1 d x

4

Integrate:

= x sin x - (- cos x) + C

Answer

x sin x + cos x + C

Deriving ∫ln x dx

To integrate ln x, we use a clever trick: write it as 1 × ln x

Setup:

\int ln x d x = \int 1 \cdot ln x d x

Choose: u = ln x (differentiates to 1/x), dv/dx = 1 (integrates to x)

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

Apply:

= x ln x - \int x \cdot \frac{1}{x} d x = x ln x - \int 1 d x

Simplify:

= x ln x - x + C

Standard Result

\int ln x d x = x ln x - x + C

Memorise This!

This is a standard result that appears frequently in examinations.

Repeated Integration by Parts

Sometimes after applying IBP once, the new integral still requires integration by parts!

When to Apply IBP Again

If

\int v \frac{d u}{d x} d x

is still a product that cannot be integrated directly, apply IBP again.

Common Examples

\int x^{2} e^{x} d x

— requires IBP twice

\int x^{2} sin x d x

— requires IBP twice

\int x^{3} e^{x} d x

— requires IBP three times

Pattern

For

\int x^{n} e^{x} d x

or

\int x^{n} sin x d x

, you need to apply IBP n times (once for each power of x).

Worked Example03

Find $\int x^{2} sin x d x$

1

First IBP: u = x², dv/dx = sin x → du/dx = 2x, v = −cos x

2

= x^{2} (- cos x) - \int (- cos x) (2 x) d x = - x^{2} cos x + 2 \int x cos x d x

3

Second IBP: For

\int x cos x d x

: u = x, dv/dx = cos x

4

\int x cos x d x = x sin x + cos x

(from Example 02)

5

Substitute back:

= - x^{2} cos x + 2 (x sin x + cos x) + C

Answer

- x^{2} cos x + 2 x sin x + 2 cos x + C

Worked Example04

Find $\int (2 x + 1) e^{3 x} d x$

1

Identify: u = 2x + 1, dv/dx = e³ˣ

2

Find: du/dx = 2, v =

\frac{1}{3} e^{3 x}

3

Apply:

= (2 x + 1) \cdot \frac{1}{3} e^{3 x} - \int \frac{1}{3} e^{3 x} \cdot 2 d x

4

= \frac{( 2 x + 1 ) e ^{3 x}}{3} - \frac{2}{3} \int e^{3 x} d x

5

= \frac{( 2 x + 1 ) e ^{3 x}}{3} - \frac{2}{3} \cdot \frac{e ^{3 x}}{3} + C = \frac{( 2 x + 1 ) e ^{3 x}}{3} - \frac{2 e ^{3 x}}{9} + C

Answer

\frac{e ^{3 x} ( 6 x + 1 )}{9} + C

or

\frac{( 2 x + 1 ) e ^{3 x}}{3} - \frac{2 e ^{3 x}}{9} + C

Integration by Loop

Rare but Important

Sometimes when you apply IBP twice, you get the original integral back. This creates an equation you can solve algebraically!

When Does This Happen?

This occurs with products of exponentials and trigonometric functions, such as:

\int e^{a x} sin b x d x

\int e^{a x} cos b x d x

The Trick

Let I = the original integral. After applying IBP twice, you'll get an equation like:

I = [something] − kI

Then solve: (1 + k)I = [something] to find I.

Take Note

This technique is rare in A-Level exams but worth knowing. You must be consistent with your choice of u throughout both IBP applications!

Worked Example — Integration by Loop05

Find $\int e^{2 x} sin 4 x d x$

1

Let

I = \int e^{2 x} sin 4 x d x

. Choose u = sin 4x, dv/dx = e²ˣ

2

First IBP: du/dx = 4cos 4x, v = ½e²ˣ

I = \frac{1}{2} e^{2 x} sin 4 x - \int \frac{1}{2} e^{2 x} \cdot 4 cos 4 x d x = \frac{1}{2} e^{2 x} sin 4 x - 2 \int e^{2 x} cos 4 x d x

3

Second IBP: For

\int e^{2 x} cos 4 x d x

: u = cos 4x, dv/dx = e²ˣ

= \frac{1}{2} e^{2 x} cos 4 x - \int \frac{1}{2} e^{2 x} (- 4 sin 4 x) d x = \frac{1}{2} e^{2 x} cos 4 x + 2 \int e^{2 x} sin 4 x d x

4

Substitute back:

I = \frac{1}{2} e^{2 x} sin 4 x - 2 [\frac{1}{2} e^{2 x} cos 4 x + 2 I]

I = \frac{1}{2} e^{2 x} sin 4 x - e^{2 x} cos 4 x - 4 I

5

Solve for I:

5 I = \frac{1}{2} e^{2 x} sin 4 x - e^{2 x} cos 4 x

I = \frac{e ^{2 x} ( s i n 4 x - 2 c o s 4 x )}{10} + C

Answer

\frac{e ^{2 x} ( s i n 4 x - 2 c o s 4 x )}{10} + C

Worked Example06

Evaluate $\int_{0}^{π} x sin x d x$

1

Identify: u = x, dv/dx = sin x → du/dx = 1, v = −cos x

2

Apply:

= [- x cos x]_{0}^{π} - \int_{0}^{π} (- cos x) d x

3

= [- x cos x]_{0}^{π} + [sin x]_{0}^{π}

4

Evaluate:

= (- π cos π - 0) + (sin π - sin 0)

5

= (- π \cdot (- 1)) + (0 - 0) = π

Answer

π

Worked Example07

Find $\int x ln x d x$

1

By LATE rule: u = ln x (L before A), dv/dx = x

2

Find: du/dx = 1/x, v = x²/2

3

Apply:

= ln x \cdot \frac{x ^{2}}{2} - \int \frac{x ^{2}}{2} \cdot \frac{1}{x} d x

4

Simplify:

= \frac{x ^{2} l n x}{2} - \frac{1}{2} \int x d x

5

= \frac{x ^{2} l n x}{2} - \frac{1}{2} \cdot \frac{x ^{2}}{2} + C = \frac{x ^{2} l n x}{2} - \frac{x ^{2}}{4} + C

Answer

\frac{x ^{2}}{4} (2 ln x - 1) + C

or

\frac{x ^{2} l n x}{2} - \frac{x ^{2}}{4} + C

Practice

Question 1: Find

\int x e^{2 x} d x

Let u = x, dv/dx = e²ˣ → du/dx = 1, v = ½e²ˣ

\int x e^{2 x} d x = x \cdot \frac{1}{2} e^{2 x} - \int \frac{1}{2} e^{2 x} d x

= \frac{x e ^{2 x}}{2} - \frac{1}{2} \cdot \frac{e ^{2 x}}{2} + C

Answer

\frac{e ^{2 x}}{4} (2 x - 1) + C

or

\frac{x e ^{2 x}}{2} - \frac{e ^{2 x}}{4} + C

Practice

Question 2: Find

\int x sin 2 x d x

Let u = x, dv/dx = sin 2x → du/dx = 1, v = −½cos 2x

\int x sin 2 x d x = x \cdot (- \frac{1}{2} cos 2 x) - \int (- \frac{1}{2} cos 2 x) d x

= - \frac{x c o s 2 x}{2} + \frac{1}{2} \int cos 2 x d x

= - \frac{x c o s 2 x}{2} + \frac{1}{2} \cdot \frac{s i n 2 x}{2} + C

Answer

- \frac{x c o s 2 x}{2} + \frac{s i n 2 x}{4} + C

Practice

Question 3: Evaluate

\int_{1}^{e} x ln x d x

From Example 07:

\int x ln x d x = \frac{x ^{2} l n x}{2} - \frac{x ^{2}}{4} + C

Apply limits:

[\frac{x ^{2} l n x}{2} - \frac{x ^{2}}{4}]_{1}^{e}

At x = e:

\frac{e ^{2} l n e}{2} - \frac{e ^{2}}{4} = \frac{e ^{2}}{2} - \frac{e ^{2}}{4} = \frac{e ^{2}}{4}

At x = 1:

\frac{1 \cdot l n 1}{2} - \frac{1}{4} = 0 - \frac{1}{4} = - \frac{1}{4}

Answer

\frac{e ^{2}}{4} - (- \frac{1}{4}) = \frac{e ^{2} + 1}{4}

Summary

Key Formulas

FormulaApplication

\int u \frac{d v}{d x} d x = uv - \int v \frac{d u}{d x} d x

Integration by Parts formula

\int ln x d x = x ln x - x + C

Standard result (memorise)

\int x e^{x} d x = e^{x} (x - 1) + C

Standard result

\int x sin x d x = - x cos x + sin x + C

Standard result

\int x cos x d x = x sin x + cos x + C

Standard result

Exam Skills

Use LATE rule to choose u (Logarithms, Algebraic, Trig, Exponentials)

For

\int ln x d x

, write as 1 × ln x and use u = ln x

Apply IBP multiple times for higher powers (x², x³...)

For eˣ × trig: use integration by loop — solve algebraically for I