illustration of integration techniques

illustration of integration techniques

The following

integral

is an example that illustrates many

integration techniques

.

Problem.

Determine the

antiderivative

of

tan

⁡

x

.

.

We start with substitution (

http://planetmath.org/IntegrationBySubstitution

):

u

=

tan

⁡

x

u

2

=

tan

⁡

x

2

⁢

u

⁢

d

⁢

u

=

sec

2

⁡

x

⁢

d

⁢

x

Using the

Pythagorean identity

tan

2

⁡

x

+

1

=

sec

2

⁡

x

, we obtain:

2

⁢

u

⁢

d

⁢

u

=

(

tan

2

⁡

x

+

1

)

⁢

d

⁢

x

2

⁢

u

⁢

d

⁢

u

=

(

u

4

+

1

)

⁢

d

⁢

x

2

⁢

u

u

4

+

1

⁢

d

⁢

u

=

d

⁢

x

Thus,

∫

tan

⁡

x

⁢

𝑑

x

=

∫

u

⁢

2

⁢

u

u

4

+

1

⁢

𝑑

u

=

∫

2

⁢

u

2

(

u

2

-

u

⁢

2

+

1

)

⁢

(

u

2

+

u

⁢

2

+

1

)

⁢

𝑑

u

.

For this last integral, we use the method of

partial fractions

(

http://planetmath.org/ALectureOnThePartialFractionDecompositionMethod

):

2

⁢

u

2

(

u

2

-

u

⁢

2

+

1

)

⁢

(

u

2

+

u

⁢

2

+

1

)

=

A

+

B

⁢

u

u

2

-

u

⁢

2

+

1

+

C

+

D

⁢

u

u

2

+

u

⁢

2

+

1

2

⁢

u

2

=

(

A

+

B

⁢

u

)

⁢

(

u

2

+

u

⁢

2

+

1

)

+

(

C

+

D

⁢

u

)

⁢

(

u

2

-

u

⁢

2

+

1

)

=

(

B

+

D

)

⁢

u

3

+

(

A

+

C

+

(

B

-

D

)

⁢

2

)

⁢

u

2

+

(

B

+

D

+

(

A

-

C

)

⁢

2

)

⁢

u

+

A

+

C

From this, we obtain the following system of equations:

{

B

+

D

=

0

A

+

C

+

(

B

-

D

)

⁢

2

=

2

(

A

-

C

)

⁢

2

+

B

+

D

=

0

A

+

C

=

0

This can be into two smaller systems of equations:

{

A

+

C

=

0

A

⁢

2

-

C

⁢

2

=

0

{

B

+

D

=

0

B

⁢

2

-

D

⁢

2

=

2

It is clear that the first system yields

A

=

C

=

0

, and it can easily be verified that

B

=

1

2

and

D

=

-

1

2

. Therefore,

∫

tan

⁡

x

⁢

𝑑

x

=

1

2

⁢

∫

u

u

2

-

u

⁢

2

+

1

⁢

𝑑

u

-

1

2

⁢

∫

u

u

2

+

u

⁢

2

+

1

⁢

𝑑

u

=

1

2

⁢

∫

u

u

2

-

u

⁢

2

+

1

2

+

1

2

⁢

𝑑

u

-

1

2

⁢

∫

u

u

2

+

u

⁢

2

+

1

2

+

1

2

⁢

𝑑

u

=

1

2

⁢

∫

u

(

u

-

1

2

)

2

+

1

2

⁢

𝑑

u

-

1

2

⁢

∫

u

(

u

+

1

2

)

2

+

1

2

⁢

𝑑

u

.

Now we make the following substitutions:

v

=

u

-

1

2

w

=

u

+

1

2

d

⁢

v

=

d

⁢

u

d

⁢

w

=

d

⁢

u

Note that we have

v

+

1

2

=

u

=

w

-

1

2

. Therefore,

∫

tan

⁡

x

⁢

𝑑

x

=

1

2

⁢

∫

v

+

1

2

v

2

+

1

2

⁢

𝑑

v

-

1

2

⁢

∫

w

-

1

2

w

2

+

1

2

⁢

𝑑

w

=

1

2

⁢

∫

v

v

2

+

1

2

⁢

𝑑

v

-

1

2

⁢

∫

d

⁢

v

v

2

+

1

2

-

1

2

⁢

∫

w

w

2

+

1

2

⁢

𝑑

w

+

1

2

⁢

∫

d

⁢

w

w

2

+

1

2

.

For the first and third integrals in the last expression, note that the

numerator

is a of the

derivative

of the denominator. For these, we use the formula

∫

k

⁢

f

′

⁢

(

x

)

f

⁢

(

x

)

⁢

𝑑

x

=

k

⁢

ln

⁡

|

f

⁢

(

x

)

|

.

For the second and fourth integrals in the last expression, we use the formula

∫

d

⁢

x

x

2

+

a

2

=

1

a

⁢

arctan

⁡

(

x

a

)

with

a

=

1

2

. Hence,

∫

tan

⁡

x

⁢

𝑑

x

=

1

2

⁢

2

⁢

ln

⁡

(

v

2

+

1

2

)

+

1

2

⁢

arctan

⁡

(

v

⁢

2

)

-

1

2

⁢

2

⁢

ln

⁡

(

w

2

+

1

2

)

+

1

2

⁢

arctan

⁡

(

w

⁢

2

)

+

K

=

1

2

⁢

2

⁢

ln

⁡

(

v

2

+

1

2

w

2

+

1

2

)

+

1

2

⁢

(

arctan

⁡

(

v

⁢

2

)

+

arctan

⁡

(

w

⁢

2

)

)

+

K

=

1

2

⁢

2

⁢

ln

⁡

(

(

u

-

1

2

)

2

+

1

2

(

u

+

1

2

)

2

+

1

2

)

+

1

2

⁢

(

arctan

⁡

[

(

u

-

1

2

)

⁢

2

]

+

arctan

⁡

[

(

u

+

1

2

)

⁢

2

]

)

+

K

=

1

2

⁢

2

⁢

ln

⁡

(

u

2

-

u

⁢

2

+

1

u

2

+

u

⁢

2

+

1

)

+

1

2

⁢

[

arctan

⁡

(

u

⁢

2

-

1

)

+

arctan

⁡

(

u

⁢

2

+

1

)

]

+

K

=

1

2

⁢

2

⁢

ln

⁡

(

tan

⁡

x

-

2

⁢

tan

⁡

x

+

1

tan

⁡

x

+

2

⁢

tan

⁡

x

+

1

)

+

1

2

⁢

[

arctan

⁡

(

2

⁢

tan

⁡

x

-

1

)

+

arctan

⁡

(

2

⁢

tan

⁡

x

+

1

)

]

+

K

.

(We use

K

for the constant of integration to avoid confusion with

C

from the system of equations.)

Title

illustration of integration techniques

Canonical name

IllustrationOfIntegrationTechniques

Date of creation

2013-03-22 17:50:16

Last modified on

2013-03-22 17:50:16

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

13

Author

Wkbj79 (1863)

Entry type

Example

Classification

msc 26A36

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