A method for integrating an indefinite integral by parts is presented. Examples of integrals calculated by this method are given. Examples of solutions are analyzed.

Content

See also: Methods for calculating indefinite integrals
Table of indefinite integrals
Basic elementary functions and their properties

The formula for integration by parts is:
.

The method of integration by parts consists in applying this formula. In practical application, it is worth noting that u and v are functions of the integration variable. Let the integration variable be denoted as x (symbol after the differential sign d at the end of the integral notation). Then u and v are functions of x : u(x) and v(x) .
Then
, .
And the integration-by-parts formula takes the form:
.

That is, the integrand must consist of the product of two functions:
,
one of which we denote as u: g(x) \u003d u, and the integral must be calculated for the other (more precisely, the antiderivative must be found):
, then dv = f(x) dx .

In some cases f(x) = 1 . That is, in the integral
,
we can put g(x) = u, x = v .

Summary

So, in this method, the integration-by-parts formula should be remembered and applied in two forms:
;
.

Integrals calculated by integration by parts

Integrals containing logarithm and inverse trigonometric (hyperbolic) functions

Integrals containing the logarithm and inverse trigonometric or hyperbolic functions are often integrated by parts. In this case, the part that contains the logarithm or inverse trigonometric (hyperbolic) functions is denoted by u, the remaining part - by dv.

Here are examples of such integrals, which are calculated by the method of integration by parts:
, , , , , , .

Integrals containing the product of a polynomial and sin x, cos x, or e x

According to the formula for integrating parts, integrals of the form are found:
, , ,
where P(x) is a polynomial in x . In integration, the polynomial P(x) is denoted by u , and e ax dx , cos ax dx or sin ax dx- via dv.

Here are examples of such integrals:
, , .

Examples of calculating integrals by the method of integration by parts

Examples of integrals containing logarithm and inverse trigonometric functions

Example

Calculate integral:

Detailed Solution

Here the integrand contains the logarithm. Making substitutions
u= ln x,
dv=x 2dx.
Then
,
.

We calculate the remaining integral:
.
Then
.
At the end of the calculations, it is imperative to add the constant C, since the indefinite integral is the set of all antiderivatives. It could also be added in intermediate calculations, but this would only clutter up the calculations.

Shorter Solution

It is possible to present the solution in a shorter version. To do this, you do not need to do substitutions with u and v, but you can group the factors and apply the integration-by-parts formula in the second form.

.

Other examples

Examples of integrals containing the product of a polynomial and sin x, cos x or ex

Example

Calculate integral:
.

We introduce the exponent under the differential sign:
e - x dx = - e - x d(-x) = - d(e - x).

We integrate by parts.
.
We also use the integration by parts method.
.
.
.
Finally we have.

This method is based on the following formula: (*)

Let And are functions of x that have continuous derivatives and .

It is known that or ; or .

Integrals and , since by assumption the functions u and v are differentiable and hence continuous.

The formula (*) is called the formula of integration by parts.

The method based on its application is called the method of integration by parts.

It reduces the calculation to the calculation of another integral: .

The application of the method of integration by parts consists in the fact that under the integral expression of a given integral they try to represent in the form of a product, where and are some functions of x, and these functions are chosen so that was easier to calculate than the original integral. When to calculate previously found and .

(as “v” we take one of the original antiderivatives found from dv, therefore, in the future, when calculating “v”, we will omit the constant C in the notation).

Comment. When factoring under the integral expression, one must understand what and should contain.

Unfortunately, it is impossible to give general rules for factoring the integral expression into factors "u" and "dv". This can be taught by much and thoughtful practice.

With all this, it should be borne in mind that was simpler than the original integral.

Example 6.6.22.

Sometimes, to obtain the final result, the rule of integration by parts is applied successively several times.

The method of integration by parts is convenient to use, of course, not every time, and the ability to use it depends on experience.

When calculating integrals, it is important to correctly establish which integration method should be used (as in the previous example, trigonometric substitution leads to the goal faster).

Consider the most common integrals that are calculated by integration by parts.

1.Integrals of the form :

where is an integer (with respect to x) polynomial; a is a constant number.

If the product of a trigonometric or exponential function is an algebraic one under the integral sign, then the algebraic function is usually taken for "u".

Example 6.6.23.

Note that another breakdown into factors: does not lead to the goal.

Proved
.

We get a more complex integral.

2.Integrals of the form :

where is a polynomial.

If the integral sign is the product of the logarithm of a function or an inverse trigonometric function by an algebraic one, then the functions should be taken as "u".

Example 6.6.23.

3.Integrals of the form:

Here you can use any of the 2 possible breakdowns of the integral expression into factors: for "u" you can take both and .

Moreover, the calculation of such integrals using the method of integration by parts leads to the original integral, that is, an equation is obtained with respect to the desired integral.

Example 6.6.24 Calculate .

.

When integrating, it is often necessary to successively apply the substitution method and the method of integration by parts.

Example 6.6.25.

Integration of some functions containing a square trinomial

1)

.

and these are tabular integrals.

2) real number coefficients

in the numerator we select the derivative of the denominator.

a,b,c are real numbers

A) ; then we have:

b) . In this case, it makes sense to consider only when the discriminant trinomial positive:

Now we have:

Comment. In practice, they usually do not use ready-made results, but prefer to carry out similar calculations again every time.

Example.

4)

We transform the numerator so that the derivative of the square trinomial can be extracted from it:

Due to the fact that in practice there is no convenient general method for calculating indefinite integrals, along with particular methods of integration (see the previous lecture), we also have to consider methods for integrating some particular classes of functions, the integrals of which are often encountered in practice.

The most important class among them is the class of rational functions.

"Integration of Fractional-Rational Functions"

The integration of a proper rational fraction is based on the expansion of a rational fraction into a sum of elementary fractions.

Elementary (simple) fractions and their integration.

Definition. Fractions of the form: ; (1)

(2), where

(that is, the roots of the trinomial are complex), are called elementary.

Consider the integration of elementary fractions

2)

(where let ).

We calculate the integral

(*)

The last integral is calculated using a recursive formula.

Sometimes integration by parts allows you to get the relationship between an indefinite integral containing the degree of some function, and a similar integral, but with a smaller exponent of the same function. Such relations are called recursive formulas.

Denote by .

We have:

In the last integral we put:

That's why

where

Thus, we have come to a recursive formula: the repeated application of which ultimately leads to the "table" integral:

Then instead of "t" and "k" we substitute their values.

Example 6.6.26.

(according to the recurrence formula).=

.

A rational fraction is a function representable in the form ; where and are polynomials with real coefficients.

A rational fraction is called proper if the degree of the numerator is less than the degree of the denominator.

Any proper rational fraction can be represented as the sum of a finite number of elementary fractions.

The decomposition of a proper fraction into elementary ones is determined by the following theorem, which we consider without proof.

Theorem . If the fraction - correct and, (where the trinomial has no real roots), then the identity is true:

(I)

Note that each real root, for example a, of multiplicity “ ” of the polynomial in this expansion corresponds to the sum of elementary fractions of the form (1), and each pair of complex conjugate roots and (such that ) of multiplicity “ ” corresponds to the sum of elementary fractions of the form (2).

To carry out expansion (I), you need to learn how to determine the coefficients .

There are various ways to find them. We will consider the method of indeterminate coefficients and the method of partial values.

Integration by parts. Solution examples

Hello again. Today in the lesson we will learn how to integrate by parts. The method of integration by parts is one of the cornerstones of integral calculus. At the test, exam, the student is almost always offered to solve integrals of the following types: the simplest integral (see article) or an integral to change the variable (see article) or the integral just on method of integration by parts.

As always, on hand should be: Table of integrals And Derivative table. If you still do not have them, then please visit the storeroom of my site: Mathematical formulas and tables. I will not get tired of repeating - it is better to print everything. I will try to present all the material in a consistent, simple and accessible way; there are no particular difficulties in integrating by parts.

What problem does integration by parts solve? The method of integration by parts solves a very important problem, it allows you to integrate some functions that are not in the table, work functions, and in some cases - and private. As we remember, there is no convenient formula: . But there is this one: is the formula for integration by parts in person. I know, I know, you are the only one - with her we will work the whole lesson (it’s already easier).

And immediately the list in the studio. Integrals of the following types are taken by parts:

1) , , - logarithm, logarithm multiplied by some polynomial.

2) ,is an exponential function multiplied by some polynomial. This also includes integrals like - an exponential function multiplied by a polynomial, but in practice it is 97 percent, a pretty letter “e” flaunts under the integral. ... the article turns out to be something lyrical, oh yes ... spring has come.

3) , , are trigonometric functions multiplied by some polynomial.

4) , - inverse trigonometric functions (“arches”), “arches”, multiplied by some polynomial.

Also, some fractions are taken in parts, we will also consider the corresponding examples in detail.

Integrals of logarithms

Example 1

Classic. From time to time, this integral can be found in tables, but it is undesirable to use a ready-made answer, since the teacher has beriberi in the spring and he will scold a lot. Because the integral under consideration is by no means tabular - it is taken in parts. We decide:

We interrupt the solution for intermediate explanations.

We use the formula for integration by parts:

The formula is applied from left to right

We look at the left side:. Obviously, in our example (and in all the others that we will consider), something needs to be denoted by , and something by .

In integrals of the type under consideration, we always denote the logarithm.

Technically, the design of the solution is implemented as follows, we write in the column:

That is, for we denoted the logarithm, and for - the remaining part integrand.

Next step: find the differential:

The differential is almost the same as the derivative, we have already discussed how to find it in previous lessons.

Now we find the function . In order to find the function it is necessary to integrate right side lower equality :

Now we open our solution and construct the right side of the formula: .
By the way, here is an example of a final solution with a few notes:

The only moment in the product, I immediately rearranged and, since it is customary to write the multiplier before the logarithm.

As you can see, applying the integration-by-parts formula essentially reduced our solution to two simple integrals.

Please note that in some cases right after application of the formula, a simplification is necessarily carried out under the remaining integral - in the example under consideration, we reduced the integrand by "x".

Let's do a check. To do this, you need to take the derivative of the answer:

The original integrand has been obtained, which means that the integral has been solved correctly.

During the verification, we used the product differentiation rule: . And this is no coincidence.

Integration by parts formula and formula These are two mutually inverse rules.

Example 2

Find the indefinite integral.

The integrand is the product of the logarithm and the polynomial.
We decide.

I will once again describe in detail the procedure for applying the rule, in the future the examples will be made out more briefly, and if you have any difficulties in solving it yourself, you need to go back to the first two examples of the lesson.

As already mentioned, for it is necessary to designate the logarithm (the fact that it is in a degree does not matter). We denote the remaining part integrand.

We write in a column:

First we find the differential:

Here we use the rule of differentiation of a complex function . It is no coincidence that at the very first lesson of the topic Indefinite integral. Solution examples I focused on the fact that in order to master the integrals, you need to "get your hand" on the derivatives. Derivatives will have to face more than once.

Now we find the function , for this we integrate right side lower equality :

For integration, we applied the simplest tabular formula

Now you are ready to apply the formula . We open it with an "asterisk" and "design" the solution in accordance with the right side:

Under the integral, we again have a polynomial on the logarithm! Therefore, the solution is interrupted again and the rule of integration by parts is applied a second time. Do not forget that for in similar situations the logarithm is always denoted.

It would be nice if at this point you were able to find the simplest integrals and derivatives orally.

(1) Do not get confused in the signs! Very often a minus is lost here, also note that the minus applies to all bracket , and these brackets need to be opened correctly.

(2) Expand the brackets. We simplify the last integral.

(3) We take the last integral.

(4) “Combing” the answer.

The need to apply the rule of integration by parts twice (or even thrice) is not uncommon.

And now a couple of examples for an independent solution:

Example 3

Find the indefinite integral.

This example is solved by the change of variable method (or subsuming under the differential sign)! And why not - you can try to take it in parts, you get a funny thing.

Example 4

Find the indefinite integral.

But this integral is integrated by parts (the promised fraction).

These are examples for self-solving, solutions and answers at the end of the lesson.

It seems that in examples 3,4 the integrands are similar, but the solution methods are different! This is precisely the main difficulty in mastering integrals - if you choose the wrong method for solving the integral, then you can fiddle with it for hours, like with a real puzzle. Therefore, the more you solve various integrals, the better, the easier the test and the exam will be. In addition, in the second year there will be differential equations, and without experience in solving integrals and derivatives there is nothing to do there.

By logarithms, perhaps more than enough. For a snack, I can also remember that tech students call female breasts logarithms =). By the way, it is useful to know by heart the graphs of the main elementary functions: sine, cosine, arc tangent, exponent, polynomials of the third, fourth degree, etc. No, of course, a condom on a globe
I will not pull, but now you will remember a lot from the section Graphs and functions =).

Integrals of the exponent multiplied by the polynomial

General rule:

Example 5

Find the indefinite integral.

Using a familiar algorithm, we integrate by parts:

If you have any difficulties with the integral, then you should return to the article Variable change method in indefinite integral.

The only other thing to do is to "comb" the answer:

But if your calculation technique is not very good, then leave the most profitable option as an answer. or even

That is, the example is considered solved when the last integral is taken. It won’t be a mistake, it’s another matter that the teacher may ask to simplify the answer.

Example 6

Find the indefinite integral.

This is a do-it-yourself example. This integral is integrated twice by parts. Particular attention should be paid to the signs - it is easy to get confused in them, we also remember that - a complex function.

There is not much more to say about the exhibitor. I can only add that the exponential and the natural logarithm are mutually inverse functions, this is me on the topic of entertaining graphs of higher mathematics =) Stop-stop, don't worry, the lecturer is sober.

Integrals of trigonometric functions multiplied by a polynomial

General rule: always stands for the polynomial

Example 7

Find the indefinite integral.

Integrating by parts:

Hmmm... and nothing to comment on.

Example 8

Find the indefinite integral

This is an example for a do-it-yourself solution

Example 9

Find the indefinite integral

Another example with a fraction. As in the two previous examples, a polynomial is denoted by.

Integrating by parts:

If you have any difficulties or misunderstanding with finding the integral, then I recommend attending the lesson Integrals of trigonometric functions.

Example 10

Find the indefinite integral

This is a do-it-yourself example.

Hint: before using the integration by parts method, you should apply some trigonometric formula that turns the product of two trigonometric functions into one function. The formula can also be used in the course of applying the method of integration by parts, to whom it is more convenient.

That, perhaps, is all in this paragraph. For some reason, I recalled a line from the anthem of the Physics and Mathematics Department “And the sine graph wave after wave runs along the abscissa axis” ....

Integrals of inverse trigonometric functions.
Integrals of inverse trigonometric functions multiplied by a polynomial

General rule: always stands for the inverse trigonometric function.

I remind you that the inverse trigonometric functions include arcsine, arccosine, arctangent and arccotangent. For the sake of brevity, I will refer to them as "arches"

The method of integration by parts is used when it is necessary to simplify the existing indefinite integral or reduce it to a tabular value. Most often, it is used in the case of exponential, logarithmic, direct and inverse trigonometric formulas and their combinations in the integrand.

The basic formula needed to use this method looks like this:

∫ f (x) d x = ∫ u (x) d (v (x)) = u (x) v (x) - ∫ v (x) d (u (x))

It means that we first need to represent the expression under the integral as the product of the function u (x) and the differential of the function v (x) . After that, we calculate the value of the function v (x) by some method (the method of direct integration is most often used), and the resulting expressions are substituted into the indicated formula, reducing the original integral to the difference u (x) v (x) - ∫ v (x) d(u(x)) . The resulting integral can also be taken using any integration method.

Consider a problem in which you need to find the set of antiderivatives of the logarithm function.

Example 1

Compute the indefinite integral ∫ ln (x) d x .

Solution

We use the method of integration by parts. To do this, we take ln (x) as a function of u (x) , and the remainder of the integrand as d (v (x)) . As a result, we get that ln (x) d x = u (x) d (v (x)) , where u (x) = ln (x) , d (v (x)) = d x .

The differential of the function u(x) is d(u(x)) - u"(x) d x = d x x , and the function v(x) can be represented as v(x) = ∫ d(v(x)) = ∫ d x = x

Important: the constant C when calculating the function v (x) will be considered equal to 0 .

We substitute what we have obtained into the formula for integration by parts:

∫ ln (x) d x = u (x) v (x) - ∫ v (x) d (u (x)) = = ln (x) x - ∫ x d x x = ln (x) x - ∫ d x \u003d ln (x) x - x + C 1 \u003d \u003d x (ln (x) - 1) + C

where C \u003d - C 1

Answer:∫ ln (x) d x = x (ln (x) - 1) + C .

The most difficult thing in applying this method is the choice of which part of the original expression under the integral to take as u (x) and which - d (v (x)) .

Let's look at a few standard cases.

If we have integrals of the form ∫ P n (x) e a x d x , ∫ P n (x) sin (a x) d x or ∫ P n (x) cos (a x) d x , where a is a coefficient, and P n (x) is a polynomial of degree n , then P n (x) should be taken as the function u (x).

Example 2

Find the set of antiderivatives of the function f (x) = (x + 1) sin (2 x) .

Solution

We can take the indefinite integral ∫ (x + 1) sin (2 x) d x by parts. We take x + 1 as u (x) and sin (2 x) d x as d (v (x)) , that is, d (u (x)) = d (x + 1) = d x .

Using direct integration, we get:

v (x) = ∫ sin (2 x) d x = - 1 2 cos (2 x)

Substitute in the formula for integration by parts:

∫ (x + 1) sin (2 x) d x = u (x) v (x) - ∫ v (x) d (u (x)) = = (x + 1) - 1 2 cos (2 x ) - ∫ - 1 2 cos (2 x) d x = = - 1 2 (x + 1) cos (2 x) + 1 2 ∫ cos (2 x) d (x) = = - 1 2 (x + 1) cos (2 x) + 1 4 sin (2 x) + C

Answer:∫ (x + 1) sin (2 x) d x = - 1 2 (x + 1) cos (2 x) + 1 4 sin (2 x) + C .

Example 3

Compute the indefinite integral ∫ (x 2 + 2 x) e x d x .

Solution

We take a second order polynomial x 2 + 2 x as u (x) and d (v (x)) - e x d x .

∫ x 2 + 2 x e x d x = u (x) = x 2 + 2 x , d (v (x)) = e x d x d (u (x)) = (2 x + 2) d x , v (x) = ∫ e x d x = e x = = u (x) v (x) - ∫ v (x) d (u (x)) = (x 2 + 2 x) e x - ∫ (2 x + 2) e x d x

To what we have done, we must again apply the method of integration by parts:

∫ (2 x + 2) e x d x = (x 2 + 2 x) e x - ∫ 2 x + 2 e x d x = = u (x) = (2 x + 2) , d (v (x)) = e x d x d (u ( x)) = 2 d x , v (x) = ∫ e x d x = e x = = (x 2 + 2 x) e x - (2 x + 2) e x - ∫ v (x) d (u (x)) = = ( x 2 + 2 x) e x - (2 x + 2) e x - ∫ 2 e x d x = = (x 2 + 2 x - 2 x - 2) e x + 2 ∫ e x d x = (x 2 - 2) e x + 2 e x + C = x 2 e x + C

Answer:∫ (x 2 + 2 x) e x d x = x 2 e x + C .

Example 4

Calculate the integral ∫ x 3 cos 1 3 x d x .

Solution

According to the method of integration by parts, we take u (x) = x 3 and d (v (x)) = cos 1 3 x d x .

In this case, d (u (x)) = 3 x 2 d x and v (x) = ∫ cos 1 3 x d x = 3 sin 1 3 x .

Now we substitute the obtained expressions into the formula:

∫ x 3 cos 1 3 x d x = u (x) v (x) - ∫ v (x) d (u)) = = x 3 3 sin 1 3 x - ∫ 3 x 2 3 sin 1 3 x d x = = 3 x 3 sin 1 3 x - 9 ∫ x 2 sin 1 3 x d x

We have an indefinite integral, which again needs to be taken in parts:

∫ x 3 cos 1 3 x d x = 3 x 3 sin 1 3 x - 9 ∫ x 2 sin 1 3 x d x = = u (x) = x 2, d (v (x)) = sin 1 3 x d x d (u (x )) = 2 x d x , v (x) = ∫ sin 1 3 x d x = - 3 cos 1 3 x = = 3 x 3 sin 1 3 x - 9 - 3 x 2 cos 1 3 x - ∫ - 3 cos 1 3 x 2 x d x = = 3 x 3 sin 1 3 x + 27 x 2 cos 1 3 x - 54 ∫ x cos 1 3 x d x

We perform partial integration again:

∫ x 3 cos 1 3 x d x = 3 x 3 sin 1 3 x + 27 x 2 cos 1 3 x - 54 ∫ x cos 1 3 x d x = = u (x) = x, d (v (x)) = cos 1 3 x d x d (u (x)) = d x , v (x) = ∫ cos 1 3 x - ∫ 3 sin 1 3 x d x = = 3 x 3 - 162 x sin 1 3 x + 27 x 2 cos 1 3 x + 162 ∫ sin 1 3 x d x = = (3 x 3 - 162 x) sin 1 3 x + 27 x 2 cos 1 3 x - 486 cos 1 3 x + C = = (3 x 3 - 162 x) sin 1 3 x + (27 x 2 - 486) cos 1 3 x + C

Answer:∫ x 3 cos 1 3 x d x = (3 x 3 - 162 x) sin 1 3 x + (27 x 2 - 486) cos 1 3 x + C .

If we have integrals of the form ∫ P n (x) ln (a x) d x , ∫ P n (x) a r c sin (a x) d x , ∫ P n (x) a r c cos (a x) d x , ∫ P n (x) a r c t g (a x) d x , ∫ P n (x) a r c c t g (a x) d x

then we should take as u (x) the functions a r c t g (a x) , a r c c t g (x) , ln (a x) , a r c sin (a x) , a r cos (a x) .

Example 5

Calculate the set of antiderivatives of the function (x + 1) ln (2 x) .

Solution

We take ln (2 x) as u (x) and (x + 1) d x as d (v (x)) . We get:

d (u (x)) = (ln (2 x)) " d x = 1 2 x (2 x) " d x = d x x v (x) = ∫ (x + 1) d x = x 2 2 + x

Substitute these expressions into the formula:

∫ (x + 1) ln (2 x) d x = u (x) v (x) - ∫ v (x) d (u (x)) = = x 2 2 + x ln 2 x - ∫ x 2 2 + x d x x = = x 2 2 + x ln (2 x) - ∫ x 2 + 1 d x = x 2 2 + x ln 2 x - 1 2 ∫ x d x - ∫ d x = = x 2 2 + x ln (2 x) - x 2 4 - x + C

Answer:∫ (x + 1) ln (2 x) d x = x 2 2 + x ln (2 x) - x 2 4 - x + C .

Example 6

Compute the indefinite integral ∫ x · a r c sin (2 x) d x .

Solution

We decide which part to take for u (x) and which part for d (v (x)) . According to the rule above, as the first function, you need to take a r c sin (2 x) , and d (v (x)) = x d x . We get:

d (u (x)) = (a r c sin (2 x) "d x = 2 x" d x 1 - (2 x) 2 = 2 d x 1 - (2 x) 2 , v (x) = ∫ x d x = x 2 2

Substitute the values ​​in the formula:

∫ x a r c sin (2 x) d x = u (x) v (x) - ∫ v (x) d (u (x)) = = x 2 2 a r c sin (2 x) - ∫ x 2 2 - 2 d x 1 - (2 x) 2 = x 2 2 a r c sin (2 x) - ∫ x 2 d x 1 - 4 x 2

As a result, we arrived at the following equality:

∫ x a r c sin (2 x) d x = x 2 2 a r c sin (2 x) - ∫ x 2 d x 1 - 4 x 2

Now we calculate the resulting integral ∫ x 2 d x 1 - 4 x 2:

∫ x 2 d x 1 - 4 x 2 = ∫ x 2 d x 4 1 4 - x 2 = 1 2 ∫ x 2 d x 1 4 - x 2 = - 1 2 ∫ - x 2 d x 1 4 - x 2 = = - 1 2 ∫ 1 4 - x 2 - 1 4 1 4 - x 2 d x = - 1 2 1 4 - x 2 d x + 1 8 ∫ d x 1 4 - x 2 = = - 1 2 ∫ 1 4 - x 2 d x + 1 8 a r c sin (2x)

Here you can apply the method of integration by parts and get:

∫ x 2 d x 1 - 4 x 2 = - 1 2 ∫ 1 4 - x 2 d x + 1 8 a r c sin (2 x) = = u (x) = 1 4 - x 2 , d (v (x)) = d x d (u (x)) = 1 4 - x 2 "d x 2 1 4 - x 2 = - x d x 1 4 - x 2 , v (x) = ∫ d x = x = = - 1 2 u (x) v ( x) - ∫ v (x) d (u (x)) + 1 8 a r c sin (2 x) = = - 1 2 x 1 4 - x 2 - ∫ - x 2 d x 1 4 - x 2 + 1 8 a r c sin (2 x) = = - 1 2 x 1 4 - x 2 - 1 2 ∫ x 2 d x 1 4 - x 2 + 1 8 a r c sin (2 x) = = - 1 2 x 1 4 - x 2 - ∫ x 2 d x 1 - 4 x 2 + 1 8 a r c sin (2 x)

Now our equality looks like this:

∫ x 2 d x 1 - 4 x 2 = - 1 2 x 1 4 - x 2 - ∫ x 2 d x 1 - 4 x 2 + 1 8 a r c sin (2 x)

We see that the integral on the right is similar to that on the left. We transfer it to another part and get:

2 ∫ x 2 d x 1 - 4 x 2 = - 1 2 x 1 4 - x 2 + 1 8 a r c sin (2 x) + C 1 ⇒ x 2 d x 1 - 4 x 2 = - 1 4 x 1 4 - x 2 + 1 16 a r c sin (2 x) + C 2 x 2 d x 1 - 4 x 2 = - 1 8 x 1 4 - x 2 + 1 16 a r c sin (2 x) + C 2

where C 2 = C 1 2

Let's go back to the original variables:

∫ x a r c sin (2 x) d x = x 2 2 a r c sin (2 x) - ∫ x 2 d x 1 - 4 x 2 = = x 2 2 a r c sin (2 x) - - 1 8 x 1 - 4 x 2 + 1 16 a r c sin (2 x) + C 2 = = 1 2 x 2 - 1 8 a r c sin (2 x) + 1 8 x 1 - 4 x 2 + C

where C \u003d - C 2

Answer:∫ x a r c sin (2 x) d x = 1 2 x 2 - 1 8 a r c sin (2 x) + 1 8 x 1 - 4 x 2 + C .

If we have an integral of the form ∫ e a x sin (b x) d x or ∫ e a x cos (b x) d x in the problem, then any function can be chosen as u (x).

Example 7

Calculate the indefinite integral ∫ e x · sin (2 x) d x .

Solution

∫ e x sin (2 x) d x = u (x) = sin (2 x) , d (v (x)) = e x d x d (u (x)) = 2 cos (2 x) d x , v (x) = ∫ e x d x = e x = = u (x) v (x) - ∫ v (x) d (u (x)) = sin (2 x) e x - ∫ e x 2 cos 2 x d x = = sin (2 x) e x - 2 ∫ e x cos (2 x) d x = u (x) = cos (2 x) , d (v (x)) = e x d x d (u (x)) = - 2 sin (2 x) d x , v (x) = ∫ e x d x = e x = = sin (2 x) e x - 2 cos (2 x) e x - ∫ (e x (- 2 sin (2 x) d x)) = = sin (2 x) e x = 2 cos (2 x ) e x - 4 ∫ e x sin (2 x) d x

As a result, we will get:

∫ e x sin (2 x) d x = sin (2 x) e x - 2 cos (2 x) e x - 4 ∫ e x sin (2 x) d x

We see the same integrals on the left and on the right, which means we can bring similar terms:

5 ∫ e x sin (2 x) d x = sin (2 x) e x - 2 cos (2 x) e x ⇒ ∫ e x sin (2 x) d x = 1 5 sin (2 x) e x - 2 5 cos (2 x) e x + C

Answer: ∫ e x sin (2 x) d x = 1 5 sin (2 x) e x - 2 5 cos (2 x) e x + C

This way of solving is standard, and on the right, an integral is often obtained, which is identical to the original one.

We examined the most typical tasks in which you can determine exactly which part of the expression to take for d (v (x)) and which for u (x) . In other cases, this has to be determined independently.

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Indefinite integral

1Antiderivative and indefinite integral 1

2The simplest properties of the indefinite integral. 3

Table of basic integrals 3

2.1Additional table of integrals 4

3Change of variable in indefinite integral 5

3.1Method of integrating functions of the form and (a≠ 0). 6

4Integration by parts in the indefinite integral 7

4.1Method of integrating functions of the form. 7

4.2 Method of integrating functions of the form: 8

5Integrating rational fractions 8

5.1Method of integration of the simplest fractions of the 4th type. eleven

6Integrating irrational expressions 12

6.1Integration of trigonometric expressions 14

1. Antiderivative and indefinite integral

Solve the differential equation

on the interval , i.e. find a function such that . Since , equation (1) can be rewritten in differentials:

Any solution to such an equation is called an antiderivative function. So the function is called antiderivative function on the interval if for all . Cases and/or are not excluded. It is clear that if antiderivative, then also antiderivative. Our task is to find all solutions of equation (1). The function of two variables is called the general solution of equation (1) or, in other words, indefinite integral functions if, when substituting for any number, we obtain a particular solution of equation (1) and any particular solution of equation (1) is obtained in this way.

The indefinite integral is denoted by . The function is called integrand, the differential is called integrand, and is the sign of the integral (stretched Latin letter S, the first letter of the word Sum is the sum). The question arises about the existence of an antiderivative and an indefinite integral. In the section "Definite integral", § Newton-Leibniz formula, it will be proved that the antiderivative of a continuous function always exists.

Lemma.Let it be identical for all . Then is a constant on this interval.

Proof. Let us denote for any point . Let's take an arbitrary point and apply Lagrange's theorem to the difference: for some point . Hence the lemma is proved.□

The theorem on antiderivatives. Two antiderivatives of the same function defined on an interval differ by a constant.

Proof. Let and be antiderivative functions. Then from where, by the lemma -- constant. Hence, . □

Consequence. If is the antiderivative of the function, then .

Note that if we take not an interval as the ODZ function, but, for example, such a disconnected set as the union of two intervals , That any function of the form

has a zero derivative, and thus the lemma and the antiderivative theorem cease to be valid in this case.

1. The simplest properties of the indefinite integral.

1. The integral of the sum is equal to the sum of the integrals:

2. The constant can be taken out of the integral sign:

3. The derivative of the integral is equal to the integrand.

4. The differential from the integral is equal to the integrand.

5. (Linear change of variables) If , That (Here ).

Table of basic integrals

In particular,

For the exceptional case we have:

1. Additional table of integrals

1. Change of variable in the indefinite integral

Let us extend the definition of the indefinite integral to a more general case: we assume by definition . Thus, for example

Theorem. Let be a differentiable function. Then

Proof. Let . Then

which was to be proved.□

In the particular case when we obtain a linear change of variables (see property 5, §1). The application of formula (1) "from left to right" will mean a change of variable. The application of formula (1) in the opposite direction, "from right to left" is called entering under the differential sign.

Examples. A.

1. We select the derivative of the square trinomial in the numerator:

3. To calculate the first integral in (2), we use the entry under the sign of the differential:

To calculate the second integral, we select a full square in a square trinomial and reduce it to a tabular one by a linear change of variables.

Integrals of the form

Examples

1. Integration by parts in the indefinite integral

Theorem. For differentiable functions and we have the relation

Proof. Integrating the left and right sides of the formula , we get:

Since by definition and , formula (1) follows.□

Example.

To integrate such functions, we put the polynomial under the differential sign and apply the integration-by-parts formula. The procedure is repeated k times.

Example.

1. Integration of rational fractions

Rational fraction is called a function of the form , where are polynomials. If , then a rational fraction is called correct. Otherwise, it is called wrong.

The following rational fractions are called the simplest

(type 2)

(type 3)

(4 type) ,

Theorem 1. Any fraction can be decomposed into the sum of a polynomial and a proper rational fraction.

Proof. Let be an improper rational fraction. Divide the numerator by the denominator with a remainder: Here are polynomials, and Then

The fraction is correct due to the inequality. □

Theorem 2. Any proper rational fraction can be decomposed into a sum of simplest ones.

Decomposition algorithm.

a) We expand the denominator of a proper fraction into a product of irreducible polynomials (linear and quadratic with negative discriminant):

Here and -- multiplicities of the corresponding roots.

b) We decompose the fraction into the sum of the simplest ones with indefinite coefficients according to the following principles:

We do this for every linear factor and for every quadratic factor.

c) The resulting expansion is multiplied by a common denominator, and the indefinite coefficients are found from the condition that the left and right parts are identical. Working with a combination of two methods

??? – substantiation of the algorithm

Examples. A. Decompose in the sum of the simplest

Hence it follows that . Substituting into this ratio, we immediately find . So

B. Expand the rational fraction in the sum of the simplest. The expansion of this fraction with indefinite coefficients has the form

Multiplying by a common denominator, we get the ratio

Substituting here , we find where . Substituting we find . Equating the coefficients at , we obtain the system

From here and . Adding the equalities of the last system, we obtain and . Then And

Hence,

/**/ Task. Generalize the result of example A and prove the equality

1. Method of integration of the simplest fractions of the 4th type.

a) Separating the derivative of the denominator in the numerator, we expand the integral to the sum of two integrals.

b) The first of the resulting integrals, after being entered under the sign of the differential, will become tabular.

c) In the second denominator, select the full square and reduce the calculation to an integral of the form . We apply the following recursive procedure to this integral

We apply the integration-by-parts formula to the last integral:

So, if we designate , That

This is a recursive formula for calculating integrals given the initial value .

Example

1. Integration of irrational expressions

Integrals of the form , where m/n,...,r/s are rational numbers with a common denominator k, are reduced to the integral of a rational function by the change

Then the essence of rational expressions, therefore, after substitution, we get the integral of the rational fraction:

Calculating this integral (see par. 4) and making the reverse substitution , we get the answer.

Similarly, integrals of the form

where ad-bc≠ 0, and k has the same meaning as above, are reduced to integrals of a rational fraction by replacing

Examples. A. Calculate the integral

B. Calculate the integral

A simpler method of integration (but requiring a guess) for the same function is this:

1. Integration of trigonometric expressions

Integrals of the form are reduced to integrals of a rational function by the universal change

so we get the integral of the rational expression

In special cases  R(sin x) cos x dx,  R(cos x) sin x dx and R(sin 2 x, cos 2 x, tg x, ctg x) dx, it is better to use substitutions, respectively.