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Integration by substitution of variable solution examples. Integration by variable replacement. Indefinite integral of a power function

We turn to the consideration of the general case - the method of changing variables in an indefinite integral.

Example 5

As an example, take the integral that we considered at the very beginning of the lesson. As we already said, for solving the integral we liked the table formula ,

and I would like to reduce the whole matter to her.

The idea behind the replacement method is to replace a complex expression (or some function) with one letter.

In this case, it begs:

The second most popular replacement letter is the letter z... In principle, you can use other letters, but we will still stick to tradition.

But when replacing, we still have dx! Probably many have guessed that if a transition is made to a new variable t, then in the new integral everything should be expressed through the letter t, and the differential dxthere is no place at all. It follows a logical conclusion that dx need to turn into some expression that depends only ont.

The action is as follows. After we have found a replacement, in this example - this, we need to find the differential dt.

Now, according to the rules of proportion, we express dx:

In this way:

And this is already the most that neither is the tabular integral

(the table of integrals is naturally also valid for the variable t).

In conclusion, it remains to carry out the reverse replacement. Remember that.

The final design of the considered example should look something like this:

Let's replace:, then

The icon has no mathematical meaning, it means that we have interrupted the solution for intermediate explanations.

When writing an example in a notebook, it is better to superscript the reverse replacement with a simple pencil.

Attention! In the following examples, finding the differential of a new variable will not be described in detail.

Remember the first solution:

What is the difference? There is no fundamental difference. They are actually the same thing.

But, from the point of view of the design of the task, the method of bringing the function under the differential sign is much shorter.

The question arises. If the first way is shorter, then why use the replace method? The point is that for a number of integrals it is not so easy to "fit" the function under the sign of the differential.

Example 6

Find the indefinite integral.

Let's replace:

;

As you can see, as a result of the replacement, the original integral has become much simpler - reduced to an ordinary power function. This is the purpose of the replacement - to simplify the integral.

Lazy advanced people can easily solve this integral by putting the function under the differential sign:

Another thing is that such a solution is not obvious for all students. In addition, already in this example, the use of the method of bringing a function under the differential sign significantly increases the risk of confusion in the solution.

Example 7

Find the indefinite integral

Check.

Example 8

Find the indefinite integral.

Decision: We make a replacement:.

It remains to find out what will become xdx? From time to time, when solving integrals, the following trick occurs: xwe express from the same replacement:

Example 9

Find the indefinite integral.

This is an example for independent decision... The answer is at the end of the lesson.

Example 10

Find the indefinite integral.

Surely some have noticed that there is no variable replacement rule in the lookup table. This was done deliberately. The rule would cause confusion in explanation and understanding, since it does not appear explicitly in the examples above.

Now is the time to talk about the basic premise of using the variable replacement method: the integrand must contain some function and its derivative ... For example how : .

Ffunctions may not be in the work, but in a different combination.

In this regard, when finding integrals, one often has to look into the table of derivatives.

In this Example 10, note that the degree of the numerator is one less than the degree of the denominator. In the table of derivatives, we find the formula that just lowers the degree by one. So, if we denote by tdenominator, then chances are good that the numerator xdx turns into something good:

Replacement: .

By the way, here it is not so difficult to bring the function under the differential sign:

It should be noted that for fractions like, such a trick will no longer work (more precisely, it will be necessary to apply not only the replacement technique).

You can learn how to integrate some fractions in the lesson Integration of compound fractions... Here are a couple of typical examples for an independent solution to the same method.

Example 11

Find the indefinite integral

Example 12

Find the indefinite integral

Solutions at the end of the lesson.

Example 13

Find the indefinite integral

We look at the table of derivatives and find our arccosine: , since we have the arccosine and something similar to its derivative in the integrand.

General rule:

Per tdenote the function itself(and not its derivative).

In this case: . It remains to find out what the rest of the integrand will turn into

In this example, finding d t we will write in detail, since it is a complex function:

Or, in short:

According to the rule of proportion, we express the remainder we need: .

In this way:

Example 14

Find the indefinite integral.

An example for an independent solution. The answer is very close.

Attentive readers will have noticed that we have looked at few examples with trigonometric functions. And this is not accidental, since under and integrals of trigonometric functions separate lessons are allocated 7.1.5, 7.1.6, 7.1.7. Moreover, some useful guidelines for changing a variable are given below, which is especially important for dummies who do not always and do not immediately understand what kind of replacement needs to be carried out in a particular integral. Also, some types of substitutions can be found in article 7.2.

More experienced students can familiarize themselves with a typical replacement in integrals with irrational functions

Example 12: Solution:

Let's replace:

Example 14: Solution:

Let's replace:

Integration by substitution of a variable (substitution method) is one of the most common methods for finding integrals.

The purpose of introducing a new variable is to simplify integration. The best option is to replace the variable and obtain a tabular integral with respect to the new variable. How to determine which replacement to make? Skills come with experience. The more examples are solved, the faster the next ones are solved. At the initial stage, we use the following reasoning:

I.e. if under the integral sign we see the product of some function f (x) and its derivative f '(x), then this function f (x) must be taken as a new variable t, since the differential dt \u003d f' (x) dx is already ...

Let's see how the variable replacement method works with specific examples.

Calculate integrals by substitution with a variable:

Here 1 / (1 + x²) is the derivative of the function arctan x. Therefore, we take arctan x as the new variable t. Next, let's use:

After finding the integral of t, we perform the reverse replacement:

If we take sine as t, then there must be its derivative, cosine (up to sign). But there is no cosine in the integrand. But if we take the exponent as t, everything turns out:

To get the desired differential dt, we change the sign in the numerator and in front of the integral:

(Here (ln (cosx)) '-. )

We turn to the consideration of the general case - the method of changing variables in an indefinite integral.

Example 5

As an example, I took the integral that we considered at the very beginning of the lesson. As we have already said, for solving the integral we liked the tabular formula, and I would like to reduce the whole matter to it.

The idea behind the replacement method is to replace a complex expression (or some function) with one letter.
In this case, it begs:
The second most popular replacement letter is the letter.
In principle, you can use other letters, but we will still stick to tradition.

So:
But when replacing, we still have! Probably, many have guessed that if a transition to a new variable is carried out, then in the new integral everything should be expressed through a letter, and there is no place for the differential there.
It follows a logical conclusion that you need turn into some expression that depends only on.

The action is as follows. After we have found a replacement, in this example, we need to find the differential. With differentials, I think everyone has already established friendship.

Since then

After the showdown with the differential, I recommend rewriting the final result as short as possible:
Now, according to the rules of proportion, we express what we need:

Eventually:
In this way:

And this is already the most tabular integral ( integral table, of course, is also valid for a variable).

In conclusion, it remains to carry out the reverse replacement. Remember that.

Done.

The final design of the considered example should look something like this:

“

Let's replace:

“

The icon has no mathematical meaning, it means that we have interrupted the solution for intermediate explanations.

When writing an example in a notebook, it is better to superscript the reverse replacement with a simple pencil.

Attention! In the following examples, finding the differential will not be described in detail.

Now is the time to remember the first solution:

What is the difference? There is no fundamental difference. They are actually the same thing. But from the point of view of the design of the task, the method of bringing the function under the differential sign is much shorter.

Example 6

Find the indefinite integral.

Let's make a replacement: (it's hard to think of another replacement here)

Lazy advanced people can easily solve this integral by putting the function under the differential sign:

Example 7

Find the indefinite integral. Check.

Example 8

Find the indefinite integral.

Replacement:
It remains to find out what will become

Ok, we expressed it, but what to do with the “X” remaining in the numerator ?!
From time to time in the course of solving integrals the following trick occurs: we express from the same substitution!

Example 9

Find the indefinite integral.

This is an example for a do-it-yourself solution. The answer is at the end of the lesson.

Example 10

Find the indefinite integral.

Surely some have noticed that there is no variable replacement rule in my lookup table. This was done deliberately. The rule would cause confusion in explanation and understanding, since it does not appear explicitly in the examples above.

Now is the time to talk about the basic premise of using the variable replacement method: the integrand must contain some function and its derivative : (functions may not be in the work)

In this regard, when finding integrals, one often has to look into the table of derivatives.

In this example, we notice that the degree of the numerator is one less than the degree of the denominator. In the table of derivatives, we find the formula that just lowers the degree by one. So, if you designate for the denominator, then the chances are great that the numerator will turn into something good.

Replacement:

By the way, here it is not so difficult to bring the function under the differential sign:

It should be noted that for fractions like, such a trick will no longer work (more precisely, it will be necessary to apply not only the replacement technique). You can learn how to integrate some fractions in the lesson Integration of some fractions.

Here are a couple more typical examples for an independent solution from the same opera:

Example 11

Find the indefinite integral.

Example 12

Find the indefinite integral.

Solutions at the end of the lesson.

Example 13

Find the indefinite integral.

We look at the table of derivatives and find our arccosine: ... In our integrand we have the inverse cosine and something similar to its derivative.

General rule:
Per denote the function itself (and not its derivative).

In this case: . It remains to find out what the rest of the integrand will turn into.

In this example, I will describe the finding in detail since it is a complex function.

Or shorter:
According to the rule of proportion, we express the remainder we need:

In this way:

Here it is no longer so easy to bring the function under the differential sign.

Example 14

Find the indefinite integral.

An example for an independent solution. The answer is very close.

Astute readers will have noticed that I have looked at few examples with trigonometric functions. And this is no coincidence, since under integrals of trigonometric functions a separate lesson is given. Moreover, this lesson provides some useful guidelines for changing a variable, which is especially important for dummies who do not always and do not immediately understand what kind of replacement needs to be done in a particular integral. Also, some types of replacements can be found in the article Definite integral. Solution examples.

More experienced students can familiarize themselves with a typical replacement in integrals with irrational functions... Replacing when integrating roots is specific, and its execution technique differs from the one we discussed in this lesson.

Wish you luck!

Example 3:Decision :

Example 4:Decision :

Example 7:Decision :

Example 9:Decision :

Replacement:

Example 11:Decision :

Let's replace:

Example 12:Decision :

Let's replace:

Example 14:Decision :

Let's replace:

Integration by parts. Solution examples

Hello again. Today in the lesson we will learn how to integrate in parts. Integration by parts is one of the cornerstones of integral calculus. On the test, the exam, the student is almost always asked to solve integrals of the following types: the simplest integral (see articleIndefinite integral. Solution examples ) or the variable change integral (see articleVariable change method in indefinite integral ) or the integral is just on method of integration by parts.

As always, you should have at hand: Integral table and Derivatives table... If you still do not have them, then please visit the storeroom of my website: Mathematical formulas and tables... I will not tire of repeating - it is better to print everything. I will try to present all the material consistently, simply and easily, there are no special difficulties in integration by parts.

What problem does the method of integration by parts solve? The method of integration by parts solves a very important problem, it allows you to integrate some functions that are missing in the table, composition functions, and in some cases - and the quotient. As we recall, there is no convenient formula: ... But there is this: - the formula of integration by parts in person. I know, I know, you are the only one - we will work with her for the whole lesson (it's already easier).

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

Find the indefinite integral.

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 spring vitamin deficiency and he will swear strongly. Because the integral under consideration is by no means tabular - it is taken piece by piece. We decide:

We interrupt the solution for intermediate explanations.

We use the formula for integration by parts:

2. Variable substitution (substitution method)

The essence of the substitution method is that as a result of introducing a new variable, the given complicated the integral is reduced to a tabular or such, the calculation method of which is known.

Let it be required to calculate the integral. There are two substitution rules:

General rule for selection of functions
does not exist, but there are several types of integrands for which there are recommendations for selecting a function
.

Variable substitution can be applied multiple times until the result is obtained.

Example 1. Find integrals:

and)
; b)
; at)
;

d)
; e)
; e)
.

Decision.

a) Among the tabular integrals, there are no radicals of various degrees, so "I want to get rid of", first of all, from
and
... This will require replacing x such an expression from which both roots would be easily extracted:

b) A typical example when there is a desire to "get rid" of the exponential function
... But in this case, it is more convenient to take the entire expression in the denominator of the fraction as a new variable:

;

c) Noting that the numerator contains the product
, which is part of the differential of the radical expression, replace the whole expression with a new variable:

;

d) Here, as in case a), one wants to get rid of the radical. But since, unlike point a), there is only one root, then we will replace it with a new variable:

e) Here, the choice of replacement is facilitated by two circumstances: on the one hand, an intuitive desire to get rid of logarithms, on the other hand, the presence of the expression , which is the differential of the function
... But, as in the previous examples, it is better to include constants accompanying the logarithm in the replacement:

f) Here, as in the previous example, the intuitive desire to get rid of the cumbersome exponent in the integrand is consistent with the well-known fact:
(formula 8 of table 3). Therefore, we have:

Change of variables for some classes of functions

Let's consider some classes of functions for which certain substitutions can be recommended.

Table 4.Rational functions

Integral form	Integration method
1.1.
1.2.
1.3.	Selecting a full square:
1.4.	Recurrent formula

Transcendental functions:

1.5.
- substitution t = e x ;

1.6.
- substitution t \u003d log a x.

Example 2. Find integrals of rational functions:

and)
; b)
;

at)
; e)
.

Decision.

a) There is no need to calculate this integral using a change of variables, here it is easier to use the summation under the differential sign:

b) Similarly, we use the summation under the differential sign:

;

c) Before us is an integral of type 1.3 of Table 4, we will use the appropriate recommendations:

e) Similar to the previous example:

Example 3. Find integrals

and)
; b)
.

Decision.

b) The integrand contains the logarithm, so we will use recommendation 1.6. Only in this case it is more convenient to replace not just a function
, and the whole expression is:

Table 6. Trigonometric functions (R

Integral form	Integration method
3.1.	Universal substitution , , ,
3.1.1. , if	Substitution
3.1.2. , if	Substitution .
3.1.3. . , if (i.e. there are only even powers of the functions )	Substitution
3.2.	If - odd, then see 3.1.1; if - odd, then see 3.1.2; if - even, then see 3.1.3; if - even, then use the degree reduction formulas ,
3.3. , ,	Use formulas

Example 4. Find integrals:

and)
; b)
; at)
; e)
.

Decision.

a) Here we integrate the trigonometric function. Let's apply the universal substitution (table 6, 3.1):

b) Here we also apply generic substitution:

Note that the change of variables in the considered integral had to be applied twice.

c) Calculate in the same way:

e) Consider two methods for calculating this integral.

As you can see, we got different antiderivative functions. This does not mean that one of the techniques used is giving the wrong result. The fact is that using the well-known trigonometric identities connecting the tangent of a half angle with the trigonometric functions of the total angle, we have

Thus, the found antiderivatives coincide with each other.

Example 5. Find integrals:

and)
; b)
; at)
; d)
.

Decision.

a) In this integral, one can also apply the universal substitution
, but since the cosine included in the integrand is in an even degree, it is more rational to use the recommendations of paragraph 3.1.3 of Table 6:

b) First, let us bring all trigonometric functions included in the integrand to one argument:

In the resulting integral, a universal substitution can be applied, but we note that the integrand does not change sign when the signs of the sine and cosine change:

Therefore, the function has the properties specified in clause 3.1.3 of Table 6, so the most convenient substitution is
... We have:

c) If the sign of the cosine is changed in the given integrand, then the whole function will change sign:

Hence, the integrand has the property described in Section 3.1.2. Therefore, it is rational to use the substitution
... But first, as in the previous example, we transform the integrand:

d) If the sign of the sine is changed in a given integrand, then the whole function will change sign, which means that we have the case described in paragraph 3.1.1 of Table 6, therefore, the new variable must be designated the function
... But since neither the presence of the function is observed in the integrand
, nor its differential, we first transform:

Example 6. Find integrals:

and)
; b)
;

at)
d)
.

Decision.

a) This integral refers to integrals of the form 3.2 of Table 6. Since the sine is of an odd power, then, according to the recommendations, it is convenient to replace the function
... But first, we transform the integrand:

b) This integral is of the same type as the previous one, but here the functions
and
have even degrees, so you need to apply the degree reduction formulas:
,
... We get:

c) We transform the function:

d) According to recommendations 3.1.3 of Table 6, in this integral it is convenient to make the replacement
... We get:

Table 5.Irrational functions (R Is a rational function of its arguments)

Integral form	Integration method
	Substitution where k – common denominator of fractions …, .
	Substitution where k - common denominator of fractions …,
2.3.	Substitution, , where k - common denominator of exponent fractions …,
2.4.	Substitution .
2.5.	Substitution ,
2.6.	Substitution , .
2.7.	Substitution , .
2.8. (differential bin), is integrated only in three cases: and) r - integer (substitution x = t k where k - common denominator of fractions t and p); b) - whole (replacement = t k where k - fraction denominator r); at) - whole (replacement = t k where k - fraction denominator r).

Example 7. Find integrals:

and)
; b)
; at)
.

Decision.

a) This integral can be attributed to integrals of the form 2.1, therefore, we perform the appropriate substitution. Recall that the meaning of the substitution in this case is to get rid of irrationality. And this means that the radical expression should be replaced with such a power of the new variable, from which all the roots under the integral would be extracted. In our case it is obvious :

The integral is an incorrect rational fraction. The integration of such fractions presupposes, first of all, the selection of the whole part. Therefore, we divide the numerator by the denominator:

Then we get
, from here

Calculate the given integral by direct integration

does not always work out. One of the most effective techniques

is a method of substitution or replacement of the variable of integration.

The essence of this method lies in the fact that by introducing a new variable of integration, it is possible to reduce the given integral to

new integral, which is taken by direct integration.

Consider this method:

Let be a continuous function

to find: (1)

Let's change the variable of integration:

where φ (t) is a monotone function that has a continuous derivative

and there is a complex function f (φ (t)).

Applying to F (x) \u003d F (φ (t)) the formula for differentiating the complex

functions, we get:

﴾F (φ (t))﴿ ′ \u003d F ′ (x) ∙ φ ′ (t)

But F ′ (x) \u003d f (x) \u003d f (φ (t)), therefore

﴾F (φ (t))﴿ ′ \u003d f (φ (t)) ∙ φ ′ (t) (3)

Thus, the function F (φ (t)) is the antiderivative for the function

f (φ (t)) ∙ φ ′ (t), therefore:

∫ f (φ (t)) ∙ φ ′ (t) dt \u003d F (φ (t)) + C (4)

Taking into account that F (φ (t)﴿ \u003d F (x), from (1) and (4) follows the replacement formula

variable in the indefinite integral:

∫ f (x) dx \u003d ∫ f (φ (t)) φ ′ (t) dt (5)

Formally, formula (5) is obtained by replacing x with φ (t) and dx with φ ′ (t) dt

The result obtained after integration by formula (5) follows

go back to variable x. This is always possible, since according to preference

the function x \u003d φ (t) is monotone.

A good choice of substitution is usually a well-known work.

nosti. To overcome them, it is necessary to master the technique of differential

citation and good knowledge of table integrals.

But it is still possible to set the series general rules and some tricks

integration.

Integration rules by substitution:

1. Determine to which tabular integral the given integral is reduced (after transforming the integrand, if necessary).

2. Determine which part of the integrand needs to be replaced

a new variable, and write this replacement.

3. Find the differentials of both parts of the record and express the differentials

old variable (or an expression containing this dif-

renial) through the differential of the new variable.

4. Make a substitution under the integral.

5. Find the resulting integral.

6. As a result, go to the old variable.

Examples of solving integrals using the substitution method:

1. Find: ∫ x² (3 + 2x) dx

Decision:

make the substitution 3 + 2x \u003d t

Let's find the differential of both parts of the substitution:

6x dx \u003d dt, whence

Consequently:

∫ x (3 + 2x) dx \u003d ∫ t ∙ dt \u003d ∫ t dt \u003d ∙ + C \u003d t + C

Replacing t with its expression from the substitution, we get:

∫ x (3 + 2x) dx \u003d (3 + 2x) + С

Decision:

\u003d \u003d ∫ e \u003d e + C \u003d e + C

Decision:

The concept of a definite integral.

The difference in values \u200b\u200bfor any antiderivative function when the argument changes from to is called a definite integral of this function in the range from a to b and is denoted:

a and b are called the lower and upper limits of integration.

To calculate a definite integral you need:

1. Find the corresponding indefinite integral

2. Substitute in the resulting expression instead of x, first the upper limit of integration at, and then the lower one - a.

3. Subtract the second from the first substitution result.

Briefly, this rule is written in the form of formulas as follows:

This formula is called the Newton-Leibniz formula.

Basic properties of a definite integral:

1., where K \u003d const

3. If, then

4. If the function is non-negative on the interval, where, then

When replacing in a certain integral the old variable of integration with a new one, it is necessary to replace the old limits of integration with new ones. These new limits are determined by the selected substitution.

Application of a definite integral.

Area of \u200b\u200ba curved trapezoid bounded by a curve, an abscissa axis and two straight lines andcalculated by the formula:

The volume of a body formed by rotation around the abscissa axis of a curvilinear trapezoid bounded by a curve that does not change its sign to, an abscissa axis and two straight lines andcalculated by the formula:

A number of physical problems can also be solved with the help of a definite integral.

For instance:

If the speed of a rectilinearly moving body is a known function of time t, then the path S traversed by this body from time t \u003d t 1 to time t \u003d t 2 is determined by the formula:

If the variable force is a known function of the path S (it is assumed that the direction of the force does not change), then the work A performed by this force on the path from to is determined by the formula:

Examples:

1. Calculate the area of \u200b\u200bthe figure bounded by the lines:

y \u003d; y \u003d (x-2) 2; 0x.

Decision:

a) Let's build graphs of functions: y \u003d; y \u003d (x-2) 2

b) Define the figure whose area you want to calculate.

c) Determine the limits of integration by solving the equation: \u003d (x-2) 2; x \u003d 1;

d) We calculate the area of \u200b\u200ba given figure:

S \u003d dx + 2 dx \u003d 1 unit 2

2. Calculate the area of \u200b\u200bthe figure bounded by the lines:

Y \u003d x 2; x \u003d y 2.

Decision:

x 2 \u003d; x 4 \u003d x;

x (x 3 - 1) \u003d 0

x 1 \u003d 0; x 2 \u003d 1

S \u003d - x 2) dx \u003d (x 3 \\ 2 -) │ 0 1 \u003d unit 2

3. Calculate the volume of the body obtained by rotating the figure around the 0x axis, bounded by the lines: y \u003d; x \u003d 1.

Decision:

V \u003d π dx \u003d π) 2 dx \u003d π \u003d π │ \u003d π / 2 units. 3

Home test mathematics
Job options.

Option number 1

y \u003d (x + 1) 2; y \u003d 1 - x; 0x

Option number 2