a lecture on integration by substitution

a lecture on integration by substitution

The Method of Substitution (or Change of Variables)

The following is a general method to find

indefinite integrals

that look like the result of a

chain rule

.

•

When to use it:

We use the method of substitution for indefinite integrals which look like the result of
a chain rule. In particular, try to use this method when you see a

composition

of two

functions

.

•

How to use it:

In this method, we go from integrating with respect to

x

to integrating with respect to
a new variable,

u

, which makes the

integral

much easier.

(a)

Find inside the integral the composition of two functions and set

u

=

“the inner function”.

(b)

We also write

d

⁢

u

=

d

⁢

u

d

⁢

x

⁢

d

⁢

x

.

(c)

Substitute everything in the integral that depends on

x

in terms of

u

.

(d)

Integrate with respect to

u

.

(e)

Once we have the result of integration in terms of

u

(

+

C

), substitute back in terms of

x

.

The method is best explained through examples:

Example 0.1

.

We want to find

∫

e

2

⁢

x

⁢

𝑑

x

. The integrand is

e

2

⁢

x

, which is a composition of two functions.
The inner function is

2

⁢

x

so we set:

u

=

2

⁢

x

,

d

⁢

u

=

2

⁢

d

⁢

x

Thus,

x

=

u

/

2

,

d

⁢

x

=

d

⁢

u

/

2

Substitute into the integral:

∫

e

2

⁢

x

⁢

𝑑

x

=

∫

e

u

⁢

d

⁢

u

2

=

1

2

⁢

∫

e

u

⁢

𝑑

u

=

1

2

⁢

e

u

+

C

=

1

2

⁢

e

2

⁢

x

+

C

The following are typical examples where we use the subsitution method:

Example 0.2

.

∫

x

⁢

e

3

⁢

x

2

+

7

⁢

𝑑

x

The inner function is

u

=

3

⁢

x

2

+

7

and

d

⁢

u

=

6

⁢

x

⁢

d

⁢

x

. Thus

d

⁢

x

=

d

⁢

u

/

(

6

⁢

x

)

. Substitute:

∫

x

⁢

e

3

⁢

x

2

+

7

⁢

𝑑

x

=

∫

x

⁢

e

u

6

⁢

x

⁢

𝑑

u

=

∫

e

u

6

⁢

𝑑

u

=

e

u

6

+

C

=

e

3

⁢

x

2

+

7

6

+

C

.

Example 0.3

.

∫

sin

⁡

(

3

⁢

x

+

7

)

⁢

𝑑

x

The inner function is

u

=

3

⁢

x

+

7

and

d

⁢

u

=

3

⁢

d

⁢

x

. Therefore:

∫

sin

⁡

(

3

⁢

x

+

7

)

⁢

𝑑

x

=

∫

sin

⁡

(

u

)

3

⁢

𝑑

u

=

-

cos

⁡

(

u

)

3

+

C

=

-

cos

⁡

(

3

⁢

x

+

7

)

3

+

C

.

Example 0.4

.

∫

(

2

⁢

x

+

3

)

⁢

x

2

+

3

⁢

x

+

20

⁢

𝑑

x

Inner

u

=

x

2

+

3

⁢

x

+

20

and

d

⁢

u

=

(

2

⁢

x

+

3

)

⁢

d

⁢

x

. Thus:

∫

(

2

⁢

x

+

3

)

⁢

x

2

+

3

⁢

x

+

20

⁢

𝑑

x

=

∫

u

⁢

𝑑

u

=

∫

u

1

/

2

⁢

𝑑

u

=

2

⁢

u

3

/

2

3

+

C

=

2

⁢

(

x

2

+

3

⁢

x

+

20

)

3

/

2

3

+

C

.

Now another integral which is a little more difficult:

Example 0.5

.

∫

cos

⁡

(

ln

⁡

x

)

x

⁢

𝑑

x

The inner function here is

u

=

ln

⁡

x

and

d

⁢

u

=

1

x

⁢

d

⁢

x

.

∫

cos

⁡

(

ln

⁡

x

)

x

⁢

𝑑

x

=

∫

cos

⁡

(

u

)

⋅

1

x

⁢

𝑑

x

=

∫

cos

⁡

(

u

)

⁢

𝑑

u

=

sin

⁡

(

u

)

+

C

=

sin

⁡

(

ln

⁡

x

)

+

C

.

Example 0.6

.

∫

3

⁢

x

2

+

14

⁢

x

+

1

x

3

+

7

⁢

x

2

+

x

+

115

⁢

𝑑

x

This function is also a typical example of integration with substitution. Whenever there is a fraction, and the numerator
looks like the

derivative

of the denominator, we set

u

to be the denominator:

u

=

x

3

+

7

⁢

x

2

+

x

+

115

,

d

⁢

u

=

(

3

⁢

x

2

+

14

⁢

x

+

1

)

⁢

d

⁢

x

Thus:

∫

3

⁢

x

2

+

14

⁢

x

+

1

x

3

+

7

⁢

x

2

+

x

+

115

⁢

𝑑

x

=

∫

1

u

⁢

𝑑

u

=

ln

⁡

u

+

C

=

ln

⁡

(

x

3

+

7

⁢

x

2

+

x

+

115

)

+

C

.

Example 0.7

.

∫

7

1

+

3

⁢

x

⁢

𝑑

x

As in the example above, we set

u

=

1

+

3

⁢

x

,

d

⁢

u

=

3

⁢

d

⁢

x

:

∫

7

1

+

3

⁢

x

⁢

𝑑

x

=

∫

7

u

⁢

d

⁢

u

3

=

7

3

⁢

∫

1

u

⁢

𝑑

u

=

7

3

⁢

ln

⁡

u

+

C

=

7

3

⁢

ln

⁡

(

1

+

3

⁢

x

)

+

C

.

Example 0.8

.

∫

t

3

⁢

(

t

4

-

50

)

700

⁢

𝑑

t

Here the inner function is

u

=

t

4

-

50

and

d

⁢

u

=

4

⁢

t

3

⁢

d

⁢

t

. Thus

∫

t

3

⁢

(

t

4

-

50

)

700

⁢

𝑑

t

=

∫

u

700

4

⁢

𝑑

u

=

1

4

⁢

u

701

701

+

C

=

(

t

4

-

50

)

701

4

⋅

701

+

C

.

Some other examples (solve them!):

∫

e

x

⁢

sin

⁡

(

e

x

)

⁢

𝑑

x

,

∫

e

x

e

x

+

1

⁢

𝑑

x

,

∫

1

x

⁢

ln

⁡

x

⁢

𝑑

x

Title

a

lecture on integration by substitution

Canonical name

ALectureOnIntegrationBySubstitution

Date of creation

2013-03-22 15:38:29

Last modified on

2013-03-22 15:38:29

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

4

Author

alozano (2414)

Entry type

Feature

Classification

msc 26A36

Related topic

ALectureOnIntegrationByParts

Related topic

ALectureOnTrigonometricIntegralsAndTrigonometricSubstitution

Related topic

ALectureOnThePartialFractionDecompositionMethod

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