Method of changing a variable in an indefinite integral. Examples of solutions. Variable replacement method. Detailed theory with examples How to make substitutions in integrals

2. Variable replacement (substitution method)

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

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

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

Substitution of variables can be applied several times until the result is obtained.

Example 1. Find the integrals:

A)
; b)
; V)
;

G)
;
d)
.

;

e)
Solution.
a) Among the table integrals there are no radicals of various degrees, so “I want to get rid”, first of all, of And. To do this you will need to replace

X
such an expression from which both roots could 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) Noticing that the numerator contains the product

, which is part of the differential of the radical expression, replace this entire expression with a new variable: d) Here, as in case a), I want to get rid of the radical. But since, unlike point a), there is only one root, we will replace it with a new variable:
e) Here two circumstances contribute to the choice of replacement: on the one hand, the intuitive desire to get rid of logarithms, on the other hand, the presence of the expression

, which is the differential of the function
. But just as in the previous examples, it is better to include the 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:

Replacing variables for some function classesLet's look at some classes of functions for which certain substitutions may be recommended.

Selecting a complete square:

1.5.
Recurrence formula Transcendental functions: = – substitution t ;

1.6.
e Transcendental functions: x – substitution t.

=log a

A)
Example 2.
;

Find integrals of rational functions:
;
.

;

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

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

;

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

e) Similar to the previous example:

Example 3. Find integrals

A)
Example 2.
.

;

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

.

Table 6. Trigonometric functions (R

Example 4. Find the integrals:

A)
Example 2.
; V)
;
.

;

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

.

b) Here we also apply a universal substitution:

.

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

c) We calculate similarly:

e) Let's consider two methods for calculating this integral.

1)

.

As you can see, we have obtained different primitive functions. This does not mean that one of the techniques used gives 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 a full angle, we have

Thus, the found antiderivatives coincide with each other.

Example 5. Find the integrals:

A)
; b)
;
V)
.

;

;
G)

a) In this integral we can also apply the universal substitution

, but since the cosine included in the integrand is to an even power, it is more rational to use the recommendations of paragraph 3.1.3 of Table 6:

b) First, let’s reduce all the trigonometric functions included in the integrand to one argument:
In the resulting integral, we can apply a universal substitution, but we note that the integrand does not change sign when the signs of the sine and cosine change:

Consequently, the function has the properties specified in paragraph 3.1.3 of Table 6, so the most convenient substitution will be

.

. We have:
c) If in a given integrand the sign of the cosine is changed, then the entire function changes sign:

d) If in a given integrand the sign of the sine is changed, then the entire function will change sign, which means we have the case described in paragraph 3.1.1 of Table 6, therefore the new variable must be designated as a function
. But since in the integrand there is no presence of the function
, nor its differential, we first transform:

Example 6. Find the integrals:

A)
Example 2.
;

Find integrals of rational functions:
G)
.

;

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

.

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

=

c) 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– rational function of its arguments)

Example 7. Find the integrals:

A)
;
;
.

;

b) :

a) This integral can be classified as integrals of type 2.1, so let’s make the appropriate substitution. Let us recall that the point of replacement in this case is to get rid of irrationality. And this means that the radical expression should be replaced by such a power of a new variable from which all the roots under the integral would be extracted. In our case it is obvious

Under the integral we get an improper rational fraction. Integrating such fractions involves, first of all, isolating the whole part. So let's divide the numerator by the denominator:
Then we get

, from here

The method is based on the following formula: ò f(x)dx = ò f(j(t)) j`(t) dt, where x = j(t) is a function differentiable on the interval under consideration.

Proof. Let's find the derivatives with respect to the variable t from the left and right sides of the formula.

Note that on the left side there is a complex function whose intermediate argument is x = j(t). Therefore, to differentiate it with respect to t, we first differentiate the integral with respect to x, and then take the derivative of the intermediate argument with respect to t.

(ò f(x)dx)` t = (ò f(x)dx)` x *x` t = f(x) j`(t)

Derivative from the right side:

Since these derivatives are equal, by corollary to Lagrange’s theorem, the left and right sides of the formula being proved differ by a certain constant. Since the indefinite integrals themselves are defined up to an indefinite constant term, this constant can be omitted from the final notation. Proven.

A successful change of variable allows you to simplify the original integral, and in the simplest cases, reduce it to a tabular one. In the application of this method, a distinction is made between linear and nonlinear substitution methods.

a) Let us consider the method of linear substitution using an example.

Example 1.. Let t = 1 – 2x, then

dx = d(½ - ½ t) = - ½ dt

It should be noted that the new variable does not need to be written out explicitly. In such cases, they talk about transforming a function under the differential sign or about introducing constants and variables under the differential sign, i.e. O implicit variable replacement.

Example 2. For example, let's find òcos(3x + 2)dx. According to the properties of the differential
dx = (1/3)d(3x) = (1/3)d(3x + 2), then òcos(3x + 2)dx = ò(1/3)cos(3x + 2)d(3x +
+ 2) = (1/3)òcos(3x + 2)d(3x + 2) = (1/3)sin(3x + 2) + C.

In both examples considered, linear substitution t = kx + b (k ¹ 0) was used to find the integrals.

In the general case, the following theorem is valid.

Linear substitution theorem. Let F(x) be some antiderivative of the function f(x). Then òf(kx + b)dx = (1/k)F(kx + b) + C, where k and b are some constants, k ¹ 0.

Proof.

By definition of the integral, òf(kx + b)d(kx + b) = F(kx + b) + C. Ho
d(kx + b)= (kx + b)`dx = kdx. Let us take the constant factor k outside the integral sign: kòf(kx + b)dx = F(kx + b) + C. Now we can divide the left and right sides of the equality by k and obtain the statement to be proved up to the designation of the constant term.

This theorem states that if in the definition of the integral ò f(x)dx = F(x) + C instead of the argument x we ​​substitute the expression (kx + b), this will lead to the appearance of an additional factor 1/k in front of the antiderivative.

Using the proven theorem, we solve the following examples.

Example 3.

Let's find it. Here kx + b = 3 – x, i.e. k = -1, b = 3. Then

Example 4.

Let's find it. Here kx + b = 4x + 3, i.e. k = 4, b = 3. Then

Example 5.

Let's find it. Here kx + b = -2x + 7, i.e. k = -2, b = 7. Then

.

Example 6. Let's find it. Here kx + b = 2x + 0, i.e. k = 2, b = 0.

.

Let us compare the result obtained with example 8, which was solved by the decomposition method. Solving the same problem using a different method, we got the answer . Let's compare the results: . Thus, these expressions differ from each other by a constant term, i.e. The answers received do not contradict each other.

Example 7. We'll find . Let's select a perfect square in the denominator.

In some cases, changing a variable does not reduce the integral directly to a tabular one, but can simplify the solution, making it possible to use the expansion method at a subsequent step.

Example 8. For example, let's find . We replace t = x + 2, then dt = d(x + 2) = dx. Then

where C = C 1 – 6 (when substituting the expression (x + 2) instead of t, instead of the first two terms we get ½x 2 -2x – 6).

Example 9. Let's find it. Let t = 2x + 1, then dt = 2dx; dx = ½ dt; x = (t – 1)/2.

Let's substitute the expression (2x + 1) for t, open the brackets and give similar ones.

Note that in the process of transformations we moved to another constant term, because the group of constant terms could be omitted during the transformation process.

b) Let us consider the method of nonlinear substitution using an example.

Example 1.. Let t = - x 2 . Next, one could express x in terms of t, then find an expression for dx and implement a change of variable in the desired integral. But in this case it’s easier to do things differently. Let's find dt = d(-x 2) = -2xdx. Note that the expression xdx is a factor of the integrand of the desired integral. Let us express it from the resulting equality xdx = - ½ dt. Then

= ò (- ½)e t dt = (- ½)ò e t dt = (- ½)e t + C = (- ½) + C

Let's look at a few more examples.

Example 2. Let's find it. Let t = 1 - x 2 . Then

Example 3. Let's find it. Let t = . Then

Example 4. In the case of nonlinear substitution, it is also convenient to use implicit variable substitution.

For example, let's find . Let's write xdx =
= (-1/4)d(3 - 2x 2) (implicitly replaced by the variable t = 3 - 2x 2). Then

Example 5. We'll find . Here we also introduce a variable under the differential sign: (implicit replacement t = 3 + 5x 3). Then

Example 6. Let's find it. Because the ,

Example 7. Let's find it. Since then

Let's look at a few examples in which it becomes necessary to combine various substitutions.

Example 8. We'll find . Let
t = 2x + 1, then x = (t – 1)/2; dx = ½ dt.

Example 9. We'll find . Let
t = x - 2, then x = t + 2; dx = dt.

Integration by change of 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 a variable and obtain a tabular integral with respect to the new variable. How to determine what replacement needs to be made? 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:

That is. 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=f '(x)dx already exists .

Let's look at how the variable replacement method works using specific examples.

Calculate integrals using the variable replacement method:

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

After we have found the integral of t, we perform the reverse substitution:

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

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

(Here (ln(cosx))’ - . )

Replacing a polynomial or. Here is a polynomial of degree, for example, the expression is a polynomial of degree.

Let's say we have an example:

Let's use the variable replacement method. What do you think should be taken for? Right, .

The equation becomes:

We perform a reverse change of variables:

Let's solve the first equation:

Let's decide second the equation:

… What does this mean? Right! That there are no solutions.

Thus, we received two answers - ; .

Do you understand how to use the variable replacement method for a polynomial? Practice doing this yourself:

Decided? Now let's check the main points with you.

You need to take it.

We get the expression:

Solving a quadratic equation, we find that it has two roots: and.

The solution to the first quadratic equation is the numbers and

Solving the second quadratic equation - numbers and.

Answer: ; ; ;

Let's sum it up

The variable replacement method has the main types of variable replacements in equations and inequalities:

1. Power substitution, when we take as some unknown, raised to a power.

2. Replacement of a polynomial, when we take for an entire expression containing an unknown.

3. Fractional-rational replacement, when we take any relation containing an unknown variable.

Important adviсe when introducing a new variable:

1. Replacement of variables must be done immediately, at the first opportunity.

2. The equation for a new variable must be solved to the end and only then returned to the old unknown.

3. When returning to the original unknown (and indeed throughout the entire solution), do not forget to check the roots for ODZ.

A new variable is introduced in a similar way, both in equations and in inequalities.

Let's look at 3 problems

Answers to 3 problems

1. Let, then the expression takes the form.

Since, it can be both positive and negative.

Answer:

2. Let, then the expression takes the form.

there is no solution because...

Answer:

3. By grouping we get:

Let then the expression take the form
.

Answer:

REPLACEMENT OF VARIABLES. AVERAGE LEVEL.

Replacing variables- this is the introduction of a new unknown, with respect to which the equation or inequality has a simpler form.

I will list the main types of replacements.

Power substitution

Power substitution.

For example, using a substitution, a biquadratic equation is reduced to a quadratic one: .

In inequalities everything is similar.

For example, we make a replacement in the inequality and get a quadratic inequality: .

Example (decide for yourself):

Solution:

This is a fractional-rational equation (repeat), but solving it using the usual method (reduction to a common denominator) is inconvenient, since we will obtain an equation of degree, so a change of variables is used.

Everything will become much easier after replacing: . Then:

Now let's do it reverse replacement:

Answer: ; .

Replacing a polynomial

Replacing a polynomial or.

Here is a polynomial of degree, i.e. expression of the form

(for example, the expression is a polynomial of degree, that is).

The most commonly used substitution for the quadratic trinomial is: or.

Example:

Solve the equation.

Solution:

And again, substitution of variables is used.

Then the equation will take the form:

The roots of this quadratic equation are: and.

We have two cases. Let's make a reverse substitution for each of them:

This means that this equation has no roots.

The roots of this equation are: i.

Answer. .

Fractional-rational substitution

Fractional-rational replacement.

and are polynomials of degrees and, respectively.

For example, when solving reciprocal equations, that is, equations of the form

replacement is usually used.

Now I'll show you how it works.

It is easy to check what is not the root of this equation: after all, if we substitute it into the equation, we get what contradicts the condition.

Let's divide the equation into:

Let's regroup:

Now we make a replacement: .

The beauty of it is that when squaring the double product of the terms, x is reduced:

It follows that.

Let's return to our equation:

Now it is enough to solve the quadratic equation and make the reverse substitution.

Example:

Solve the equation: .

Solution:

When equality does not hold, therefore. Let's divide the equation into:

The equation will take the form:

Its roots:

Let's make a reverse replacement:

Let's solve the resulting equations:

Answer: ; .

Another example:

Solve the inequality.

Solution:

By direct substitution we are convinced that it is not included in the solution of this inequality. Divide the numerator and denominator of each fraction by:

Now the replacement of the variable is obvious: .

Then the inequality will take the form:

We use the interval method to find y:

in front of everyone, because

in front of everyone, because

So the inequality is equivalent to the following:

in front of everyone, because...

This means that the inequality is equivalent to the following: .

So, inequality turns out to be equivalent to aggregate:

Answer: .

Replacing variables- one of the most important methods for solving equations and inequalities.

Finally, I’ll give you a couple of important tips:

REPLACEMENT OF VARIABLES. SUMMARY AND BASIC FORMULAS.

Replacing variables- a method for solving complex equations and inequalities, which allows you to simplify the original expression and bring it to a standard form.

Types of variable replacement:

1. Power substitution: is taken to be some unknown, raised to a power - .
2. Fractional-rational replacement: is taken to be any relation containing an unknown variable - , where and are polynomials of degrees n and m, respectively.
3. Replacing a polynomial: the whole expression containing the unknown is taken as - or, where is a polynomial of degree.

After solving a simplified equation/inequality, it is necessary to make a reverse substitution.

Type of lesson: learning new material.

Educational tasks:

• teach students to use the method of integration by substitution;
• continue to develop skills in the use of integration of functions;
• continue to develop an interest in mathematics through problem solving;
• cultivate a conscious attitude towards the learning process, instill a sense of responsibility for the quality of knowledge, exercise self-control over the process of solving and designing exercises;
• remind that only the conscious use of algorithms for calculating the indefinite integral will allow students to qualitatively master the topic being studied.

Providing classes:

• table of basic integration formulas;
• task cards for test work.

The student must know: algorithm for calculating the indefinite integral using the substitution method.

The student must be able to: apply the acquired knowledge to the calculation of indefinite integrals.

Motivation of cognitive activity of students.

The teacher reports that in addition to the direct integration method, there are other methods for calculating indefinite integrals, one of which is the substitution method. This is the most common method of integrating a complex function, consisting of transforming the integral by moving to another integration variable.

Progress of the lesson

I. Organizing time.

II. Checking homework.

Frontal survey:

III. Repetition of students' basic knowledge.

1) Repeat the table of basic integration formulas.

2) Repeat what the direct integration method is.

Direct integration is a method of integration in which a given integral is reduced to one or more table integrals by means of identical transformations of the integrand and the application of the properties of the indefinite integral.

IV. Learning new material.

It is not always possible to calculate a given integral by direct integration, and sometimes this is associated with great difficulties. In these cases, other techniques are used. One of the most effective techniques is the method of substitution or replacement of the integration variable. The essence of this method is that by introducing a new integration variable it is possible to reduce a given integral to a new integral, which is relatively easy to take directly. If after changing the variable the integral becomes simpler, then the purpose of the substitution has been achieved. Integration by substitution method is based on the formula

Let's consider this method.

Calculation algorithmindefinite integral by substitution method:

1. Determine which table integral this integral is reduced to (after first transforming the integrand, if necessary).
2. Determine which part of the integrand to replace with a new variable, and write down this replacement.
3. Find the differentials of both parts of the record and express the differential of the old variable (or an expression containing this differential) in terms of the differential of the new variable.
4. Make a substitution under the integral.
5. Find the resulting integral.
6. As a result, a reverse replacement is made, i.e. go to the old variable. It is useful to check the result by differentiation.

Let's look at examples.

Examples. Find the integrals:

1) )4

Let's introduce the substitution:

Differentiating this equality, we have:

V. Application of knowledge when solving typical examples.

VI. Independent application of knowledge, skills and abilities.

Option 1

Find the integrals:

Option 2

Find the integrals:

VII. Summing up the lesson.

VIII. Homework:

G.N. Yakovlev, part 1, §13.2, paragraph 2, No. 13.13 (1,4,5), 13.15 (1,2,3)