Point 4. Integration by parts

Theorem 3. Let be two continuously differentiable functions. The next formula (formula of integration by parts) is true

( 12 )

■It’s known that the differential of a product of two functions equals

.

Integrating this equality we obtain

.■

According tо the formula (12) we represent the expression under the integral sign in the form of a product of two functions, namely u and dv. Then we differentiate the first function and integrate the second one.

Ex. 19.

.

Ex. 20.

Note 5. Integration by parts can be performed by necessity several times.

Ex. 21.

Note 6. Sometimes integration by parts leads to an equation in a required integral

Ex. 22.

We’ve got the equation in the sought integral I and hence

Ex. 23.

Note 7. There are no general rules to choose . But in some cases one can give certain advices.

In the cases of integrals

,

where is a polynomial, it’s well to put .

In the cases of integrals

it’s well to put .

Ex. 24.

.

Note 7. Evaluating indefinite integrals one often can combine both methods of changing a variable and integration by parts.

Ex. 25.

.

Ex. 26.

.

Ex. 27.

Lecture no.20. Classes of integrable functions

POINT 1. RATIONAL FRACTIONS (RATIONAL FUNCTIONS)

POINT 2. TRIGONOMETRIC FUNCTIONS

POINT 3. IRRATIONAL FUNCTIONS

Point 1. Rational functions (rational fractions)

Def. 1. A rational function (a rational fraction) is called a function which can be represented as a ratio of two polynomials

. ( 1 )

Def. 2. A rational function (a rational fraction) (1) is called proper one if and improper otherwise ( ).

Theorem 1 (extraction of integer part of an improper rational function). Every improper rational function can be represented as a sum of some polynomial (so-called integer part) and a proper rational fraction.

■Let . Dividing the numerator by the denominator we get

where polynomials are the quotient and the remainder respectively and

is a proper rational fraction.■

Ex. 1. Extract an integer part of an improper rational fraction

a) The first (theoretical) way. After division of by we get

b) The second way. Subtracting and adding 1 in the numerator we’ll have

There are partial [simplest, elementary] fractions of 1- 4 types.

1. ;

2. ;

3. ;

4. .

We’ve integrated the fractions 1, 3 in the Point 2 of the preceding lecture (ex. 14, 16, formulas (7), (10) and (11)). To integrate the fraction 2 we can put . Integration of the fraction 4 with the help of the substitution

leads to a linear combination of a simple integral

and the next one

.

As to evaluation of this latter see textbooks. For small values of k ( ) one can use the next substitution: .

Ex. 2.

.

Thus we can say that we able to integrate the partial fractions 1 – 4.

Theorem 2 (a partial decomposition of a proper rational function). Every pro-per rational function can be represented as a linear combination of partial fractions.

For example

Here are some unknown numbers (undetermined coefficients), which one can find by so-called method of undetermined coefficients.

Corollary. Every rational function can be integrated by virtue of the linear pro-perty of indefinite integral.

Rule of integration of a rational function. To integrate a rational function it’s necessary:

1. To extract an integer part of the function if it is improper one.

2. To factorize the denominator of obtained proper fraction into a product of polynomials of degree not higher than two.

3. To make a partial decomposition of the proper function.

4. To integrate all the terms of the obtained algebraic sum.

5. To write an answer.

Ex. 3.

.

Ex. 4. Evaluate the indefinite integral .

1 step (factorizing the denominator of the proper rational function).

.

2 step (a partial decomposition of the proper function).

Let’s assign two particular values to x in (*), namely .

3 step (integration of the given function by integration of its partial decomposition).

Ex. 5. Calculate the next indefinite integral

1 step (a partial decomposition of the proper rational function with factorized denominator).

,

. (**)

Assigning tree arbitrary values to x in (**), for example , we get a system of linear equations in ,

2 step (integration of all the terms of obtained partial decomposition of the integrand)

Answer.