Integration by Substitution

Master the technique of changing variables to simplify complex integrals

SPEC 8.5 • INTEGRATION

What is Substitution?

Key Idea

Integration by substitution simplifies complex integrals by changing to a new variable. It transforms an integral we can't do directly into one we can.

The technique involves:

Introducing a new variable

u

to replace part of the expression

Finding

\frac{d u}{d x}

and rearranging to express

d x

in terms of

d u

Expressing ALL parts of the integral in terms of

u

Integrating with respect to

u

, then converting back to

x

Key Point

After substitution, there should be NO x's remaining in the integral — everything must be in terms of

u

and

d u

.

The Key Question

Ask Yourself

"What part of this integral is causing the difficulty?"

The answer to this question tells you what to substitute for. Look for:

Square roots: √(...)

The expression inside the root is usually causing problems

Brackets with powers: (...)ⁿ

The expression inside the brackets is the awkward part

√x in denominators

The √x itself is causing the difficulty

√(a² − x²) forms

This needs a trigonometric substitution

Golden Rule

Substitute for the most awkward part. If your substitution makes things more complicated, you've chosen wrong!

Pattern Recognition Guide

When You See This...Think This...Substitute This...

x^{n} \cdot x + a

"Root of linear is awkward"

u = x + a

, then

x = u - a

x^{n} \cdot a x + b

"Root of linear is awkward"

u = a x + b

, then

x = \frac{u - b}{a}

\frac{...}{1 + x}

or

\frac{x}{...}

"√x is in awkward position"

u = x

, so

x = u^{2}

\frac{1}{a ^{2} - x ^{2}}

"This is the arcsin form"

x = a sin θ

Key Insight

After substituting, you must express EVERY x in terms of u. This is why we need

x = u - a

or

x = u^{2}

— to replace any remaining x's!

The Substitution Method

Step-by-Step Process

Identify the awkward part — What makes this integral difficult?

Choose u — Usually the expression inside a root or bracket

Find du/dx — Differentiate u with respect to x

Express x in terms of u — Rearrange

u = ...

to get

x = ...

Substitute everything — Replace all x's and dx with u's and du

Integrate — With respect to u

Back-substitute — Replace u with the expression in x

Critical Check

Before integrating, verify that your integral contains only u and du — no x's should remain!

Type 1: Linear Expression in Root

Pattern

Integrals containing

x + a

or

a x + b

multiplied by powers of x

How to Spot It

A root containing

x + a

Powers of

x

multiplied by the root

Example:

\int x^{2} x + 1 d x

What to Do

Let

u = x + a

Rearrange:

x = u - a

Express powers of

x

using

u

Key Insight

When you let

u = x + a

, you get

x = u - a

. So

x^{2} = (u - a)^{2}

,

x^{3} = (u - a)^{3}

, etc. Expand these before integrating!

Worked Example01

Find $\int x^{2} x + 1 d x$

1

Identify: The awkward part is

x + 1

. Let

u = x + 1

2

Express x:

x = u - 1

, so

x^{2} = (u - 1)^{2} = u^{2} - 2 u + 1

3

Find dx:

\frac{d u}{d x} = 1

, so

d u = d x

4

Substitute:

\int (u^{2} - 2 u + 1) \cdot u^{1/2} d u = \int (u^{5/2} - 2 u^{3/2} + u^{1/2}) d u

5

Integrate:

= \frac{2 u ^{7/2}}{7} - \frac{4 u ^{5/2}}{5} + \frac{2 u ^{3/2}}{3} + C

6

Back-substitute:

= \frac{2 ( x + 1 ) ^{7/2}}{7} - \frac{4 ( x + 1 ) ^{5/2}}{5} + \frac{2 ( x + 1 ) ^{3/2}}{3} + C

Worked Example02

Find $\int x 2 x + 3 d x$

1

Identify: The awkward part is

2 x + 3

. Let

u = 2 x + 3

2

Express x:

2 x = u - 3

, so

x = \frac{u - 3}{2}

3

Find dx:

\frac{d u}{d x} = 2

, so

d x = \frac{d u}{2}

4

Substitute:

\int \frac{u - 3}{2} \cdot u^{1/2} \cdot \frac{d u}{2} = \frac{1}{4} \int (u^{3/2} - 3 u^{1/2}) d u

5

Integrate:

= \frac{1}{4} (\frac{2 u ^{5/2}}{5} - \frac{6 u ^{3/2}}{3}) + C = \frac{u ^{5/2}}{10} - \frac{u ^{3/2}}{2} + C

6

Back-substitute:

= \frac{( 2 x + 3 ) ^{5/2}}{10} - \frac{( 2 x + 3 ) ^{3/2}}{2} + C

Definite Integrals

Important Rule

For definite integrals, you MUST change the limits when you substitute!

When x = a (lower limit)

Calculate

u = g (a)

This becomes the new lower limit

When x = b (upper limit)

Calculate

u = g (b)

This becomes the new upper limit

Advantage

If you change the limits, you don't need to back-substitute at the end — just evaluate using the u-limits!

Worked Example03

Evaluate $\int_{0}^{3} x x + 1 d x$

1

Choose substitution: Let

u = x + 1

, so

x = u - 1

and

d u = d x

2

Change limits: When

x = 0

:

u = 1

. When

x = 3

:

u = 4

3

Substitute:

\int_{1}^{4} (u - 1) u^{1/2} d u = \int_{1}^{4} (u^{3/2} - u^{1/2}) d u

4

Integrate:

= [\frac{2 u ^{5/2}}{5} - \frac{2 u ^{3/2}}{3}]_{1}^{4}

5

Evaluate: At

u = 4

:

\frac{64}{5} - \frac{16}{3} = \frac{112}{15}

. At

u = 1

:

\frac{2}{5} - \frac{2}{3} = - \frac{4}{15}

6

Result:

\frac{112}{15} - (- \frac{4}{15}) = \frac{116}{15}

Type 2: Substituting for √x

When to Use

When

x

appears in an awkward position — especially in denominators or combined with other terms.

The Substitution

Let

u = x

Then:

x = u^{2}

and

d x = 2 u d u

Why This Works

This substitution eliminates the square root entirely. After substituting:

x

becomes

u

(no root!)

x

becomes

u^{2}

d x

becomes

2 u d u

The resulting integral is usually a rational function in

u

.

Worked Example04

Find $\int \frac{1}{1 + x} d x$

1

Identify:

x

is in the denominator. Let

u = x

2

Express x and dx:

x = u^{2}

and

d x = 2 u d u

3

Substitute:

\int \frac{1}{1 + u} \cdot 2 u d u = \int \frac{2 u}{1 + u} d u

4

Algebraic manipulation:

\frac{2 u}{1 + u} = \frac{2 ( u + 1 ) - 2}{1 + u} = 2 - \frac{2}{1 + u}

5

Integrate:

\int (2 - \frac{2}{1 + u}) d u = 2 u - 2 ln ∣1 + u ∣ + C

6

Back-substitute:

= 2 x - 2 ln (1 + x) + C

Worked Example05

Find $\int \frac{x}{1 + x} d x$

1

Identify:

x

and

x

both appear. Let

u = x

2

Express everything:

x = u^{2}

,

x = u

,

d x = 2 u d u

3

Substitute:

\int \frac{u}{1 + u ^{2}} \cdot 2 u d u = \int \frac{2 u ^{2}}{1 + u ^{2}} d u

4

Algebraic manipulation:

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

5

Integrate:

\int (2 - \frac{2}{1 + u ^{2}}) d u = 2 u - 2 tan^{- 1} (u) + C

6

Back-substitute:

= 2 x - 2 tan^{- 1} (x) + C

Type 3: Trigonometric Substitutions

When to Use

When you see expressions like

a^{2} - x^{2}

,

a^{2} + x^{2}

, or

x^{2} - a^{2}

ExpressionSubstitutionWhy It Works

a^{2} - x^{2}

x = a sin θ

a^{2} - a^{2} sin^{2} θ = a^{2} cos^{2} θ

a^{2} + x^{2}

x = a tan θ

a^{2} + a^{2} tan^{2} θ = a^{2} sec^{2} θ

x^{2} - a^{2}

x = a sec θ

a^{2} sec^{2} θ - a^{2} = a^{2} tan^{2} θ

Important

After integrating, convert back to

x

using inverse trig functions.

Worked Example06

Find $\int \frac{1}{9 - x ^{2}} d x$

1

Identify: This is

a^{2} - x^{2}

with

a = 3

. Let

x = 3 sin θ

2

Find dx:

d x = 3 cos θ d θ

3

Simplify the root:

9 - 9 sin^{2} θ = 3 cos^{2} θ = 3 cos θ

4

Substitute:

\int \frac{1}{3 c o s θ} \cdot 3 cos θ d θ = \int 1 d θ = θ + C

5

Back-substitute: Since

x = 3 sin θ

, we have

θ = sin^{- 1} (x /3)

Answer: $sin^{- 1} (\frac{x}{3}) + C$

Worked Example07

Evaluate $\int_{1}^{4} \frac{1}{x ( 1 + x ) ^{2}} d x$

1

Identify:

x

appears multiple times. Let

u = 1 + x

2

Find relationship:

\frac{d u}{d x} = \frac{1}{2 x}

, so

\frac{d x}{x} = 2 d u

3

Change limits: When

x = 1

:

u = 2

. When

x = 4

:

u = 3

4

Substitute:

\int_{2}^{3} \frac{2}{u ^{2}} d u

5

Integrate:

= 2 [- \frac{1}{u}]_{2}^{3} = - \frac{2}{3} + \frac{2}{2} = - \frac{2}{3} + 1 = \frac{1}{3}

Answer: $\frac{1}{3}$

Practice

Question 1: Find

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

The awkward part is $x - 2$ . Let $u = x - 2$

Then

x = u + 2

and

x^{2} = (u + 2)^{2} = u^{2} + 4 u + 4

, and

d u = d x

\int (u^{2} + 4 u + 4) u^{1/2} d u = \int (u^{5/2} + 4 u^{3/2} + 4 u^{1/2}) d u

= \frac{2 u ^{7/2}}{7} + \frac{8 u ^{5/2}}{5} + \frac{8 u ^{3/2}}{3} + C

Answer: $\frac{2 ( x - 2 ) ^{7/2}}{7} + \frac{8 ( x - 2 ) ^{5/2}}{5} + \frac{8 ( x - 2 ) ^{3/2}}{3} + C$

Practice

Question 2: Find

\int \frac{x}{1 + x} d x

Let $u = x$ , so

x = u^{2}

and

d x = 2 u d u

\int \frac{u ^{2}}{1 + u} \cdot 2 u d u = \int \frac{2 u ^{3}}{1 + u} d u

Polynomial division:

\frac{2 u ^{3}}{1 + u} = 2 u^{2} - 2 u + 2 - \frac{2}{1 + u}

\int (2 u^{2} - 2 u + 2 - \frac{2}{1 + u}) d u = \frac{2 u ^{3}}{3} - u^{2} + 2 u - 2 ln ∣1 + u ∣ + C

Answer: $\frac{2 x x}{3} - x + 2 x - 2 ln (1 + x) + C$

Practice

Question 3: Evaluate

\int_{0}^{5} \frac{x}{x + 4} d x

Let $u = x + 4$ , so

x = u - 4

and

d u = d x

Change limits: When

x = 0

:

u = 4

. When

x = 5

:

u = 9

\int_{4}^{9} \frac{u - 4}{u} d u = \int_{4}^{9} (u^{1/2} - 4 u^{- 1/2}) d u

= [\frac{2 u ^{3/2}}{3} - 8 u^{1/2}]_{4}^{9}

At

u = 9

:

18 - 24 = - 6

. At

u = 4

:

\frac{16}{3} - 16 = - \frac{32}{3}

= - 6 - (- \frac{32}{3}) = - 6 + \frac{32}{3} = \frac{14}{3}

Answer: $\frac{14}{3}$

Summary

Choosing Your Substitution

If You See...Substitute...

x + a

or

(x + a)^{n}

u = x + a

, then

x = u - a

a x + b

u = a x + b

, then

x = \frac{u - b}{a}

x

in denominator or mixed

u = x

, then

x = u^{2}

a^{2} - x^{2}

x = a sin θ

The Golden Rule

Substitute for the most awkward part. After substituting, ALL x's must be eliminated — only u and du should remain. If you can't eliminate all x's, try a different substitution!