Euler’s substitutions for integration

Euler’s substitutions for integration

In the integration task

∫

R

⁢

(

x

,

a

⁢

x

2

+

b

⁢

x

+

c

)

⁢

𝑑

x

,

where the

integrand

is a

rational function

of

x

and

a

⁢

x

2

+

b

⁢

x

+

c

, the integrand can be changed to a rational function of a new

variable

t

by using the following

substitutions of Euler

.

•

The first substitution of Euler.

If

a

>

0

,  we may write

a

⁢

x

2

+

b

⁢

x

+

c

=

±

x

⁢

a

+

t

.

(1)

When we take

a

with the minus sign, then

a

⁢

x

2

+

b

⁢

x

+

c

=

a

⁢

x

2

-

2

⁢

x

⁢

t

⁢

a

+

t

2

,

from which we get the

expression

x

=

t

2

-

c

b

+

2

⁢

t

⁢

a

;

thus also

d

⁢

x

is expressible rationally via

t

. We have

a

⁢

x

2

+

b

⁢

x

+

c

=

-

x

⁢

a

+

t

=

c

-

t

2

b

+

2

⁢

t

⁢

a

⁢

a

+

t

.

•

The second substitution of Euler.

If

c

>

0

,  we take

a

⁢

x

2

+

b

⁢

x

+

c

=

x

⁢

t

±

c

.

(2)

With the minus sign we obtain, similarly as above,

x

=

2

⁢

t

⁢

c

+

b

t

2

-

a

.

•

The third substitution of Euler.

If the

polynomial

a

⁢

x

2

+

b

⁢

x

+

c

has the real zeros

α

and

β

, we may chose

a

⁢

x

2

+

b

⁢

x

+

c

=

(

x

-

α

)

⁢

t

.

(3)

Now

a

⁢

x

2

+

b

⁢

x

+

c

=

a

⁢

(

x

-

α

)

⁢

(

x

-

β

)

=

(

x

-

α

)

2

⁢

t

2

,

whence

a

⁢

(

x

-

β

)

=

(

x

-

α

)

⁢

t

2

. This gives the expression

x

=

a

⁢

β

-

α

⁢

t

2

a

-

t

2

.

As in the preceding cases, we can express

d

⁢

x

and

a

⁢

x

2

+

b

⁢

x

+

c

rationally via

t

.

Examples.

1.

In the

integral

∫

d

⁢

x

x

2

+

c

we can use the first substitution:

x

2

+

c

=

-

x

+

t

;  then

x

2

+

c

=

x

2

-

2

⁢

x

⁢

t

+

t

2

and thus

x

=

t

2

-

c

2

⁢

t

,

d

⁢

x

=

t

2

+

c

2

⁢

t

2

⁢

d

⁢

t

,

x

2

+

c

=

-

t

2

-

c

2

⁢

t

+

t

=

t

2

+

c

2

⁢

t

.

Accordingly we obtain

∫

d

⁢

x

x

2

+

c

=

∫

t

2

+

c

2

⁢

t

2

⁢

d

⁢

t

t

2

+

c

2

⁢

t

=

∫

d

⁢

t

t

=

ln

⁡

|

t

|

+

C

=

ln

⁡

|

x

+

x

2

+

c

|

+

C

.

Especially the cases

c

=

±

1

give the formulas

∫

d

⁢

x

x

2

+

1

=

arsinh

x

+

C

,

∫

d

⁢

x

x

2

-

1

=

arcosh

x

+

C

(

x

>

1

)

.

2.

The integral

∫

c

2

-

x

2

x

⁢

𝑑

x

is needed in deriving the equation of the tractrix. We use for integrating the second substitution

c

2

-

x

2

=

x

⁢

t

-

c

;  then

c

2

-

x

2

=

x

2

⁢

t

2

-

2

⁢

c

⁢

x

⁢

t

+

c

2

, which implies

x

=

2

⁢

c

⁢

t

t

2

+

1

,

d

⁢

x

=

2

⁢

c

⁢

(

1

-

t

2

)

⁢

d

⁢

t

(

1

+

t

2

)

2

,

c

2

-

x

2

=

2

⁢

c

⁢

t

2

t

2

+

1

-

c

=

c

⁢

(

t

2

-

1

)

t

2

+

1

.

We then obtain

∫

c

2

-

x

2

x

⁢

𝑑

x

=

-

c

⁢

∫

(

1

-

t

2

)

2

t

⁢

(

1

+

t

2

)

2

⁢

𝑑

t

=

c

⁢

∫

(

4

⁢

t

(

1

+

t

2

)

2

-

1

t

)

⁢

𝑑

t

=

-

2

⁢

c

1

+

t

2

-

c

⁢

ln

⁡

|

t

|

+

C

1

.

The equation tying

x

and

t

gives

2

⁢

c

1

+

t

2

=

x

t

and

t

=

c

+

c

2

-

x

2

x

,  whence

∫

c

2

-

x

2

x

⁢

𝑑

x

=

-

x

2

c

+

c

2

-

x

2

-

c

⁢

ln

⁡

c

+

c

2

-

x

2

x

+

C

1

=

-

c

+

c

2

-

x

2

-

c

⁢

ln

⁡

c

+

c

2

-

x

2

x

+

C

1

,

i.e.

∫

c

2

-

x

2

x

⁢

𝑑

x

=

c

2

-

x

2

-

c

⁢

ln

⁡

c

+

c

2

-

x

2

x

+

C

.

3.

In the integral

∫

d

⁢

x

x

2

+

3

⁢

x

-

4

, the

radicand

is

(

x

+

4

)

⁢

(

x

-

1

)

. Using the third substitution of Euler, we take

x

2

+

4

⁢

x

-

3

=

(

x

+

4

)

⁢

t

. This simplifies to

x

-

1

=

(

x

+

4

)

⁢

t

2

. Then we get

x

=

1

+

4

⁢

t

2

1

-

t

2

,

d

⁢

x

=

10

⁢

t

(

1

-

t

2

)

2

⁢

d

⁢

t

,

x

2

+

3

⁢

x

-

4

=

(

1

+

4

⁢

t

⁢

r

1

-

t

2

+

4

)

⁢

t

=

5

⁢

t

1

-

t

2

.

And we obtain

∫

d

⁢

x

x

2

+

3

⁢

x

-

4

=

∫

10

⁢

t

⁢

(

1

-

t

2

)

(

1

-

t

2

)

2

⋅

5

⁢

t

⁢

𝑑

t

=

∫

2

1

-

t

2

⁢

𝑑

t

=

ln

⁡

|

1

+

t

1

-

t

|

+

C

=

ln

⁡

|

1

+

x

-

1

x

+

4

1

-

x

-

1

x

+

4

|

+

C

=

ln

⁡

|

x

+

4

+

x

-

1

x

+

4

-

x

-

1

|

+

C

.

References

1

N. Piskunov:

Diferentsiaal- ja integraalarvutus kõrgematele tehnilistele õppeasutustele

. Viies, täiendatud trükk.  Kirjastus “Valgus”, Tallinn (1965).

Title

Euler’s

substitutions for integration

Canonical

name

EulersSubstitutionsForIntegration

Date of creation

2013-03-22 17:19:43

Last modified on

2013-03-22 17:19:43

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

15

Author

pahio (2872)

Entry type

Topic

Classification

msc 26A36

Synonym

integration of expressions of square roots of quadratic polynomials

Related topic

IntegrationOfRationalFunctionOfSineAndCosine

Related topic

Tractrix

Related topic

Arsinh

Related topic

Arcosh

Defines

Euler’s substitutions

Defines

substitutions of Euler

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