Integration by introducing a new variable. Integration by substitution method

In this lesson we will get acquainted with one of the most important and most common techniques that is used when solving indefinite integrals - the change of variable method. To successfully master the material you need basic knowledge and integration skills. If there's a feeling of empty full kettle in integral calculus, then you should first familiarize yourself with the material, where I explained in an accessible form what an integral is and analyzed in detail basic examples for beginners.

Technically, the method of changing a variable in an indefinite integral is implemented in two ways:

– Subsuming the function under the differential sign;
– Actually replacing the variable.

Essentially, these are the same thing, but the design of the solution looks different.

Let's start with a simpler case.

Subsuming a function under the differential sign

In class Indefinite integral. Examples of solutions we learned how to open the differential, I remind you of the example I gave:

That is, revealing a differential is formally almost the same as finding a derivative.

Example 1

Perform check.

We look at the table of integrals and find a similar formula: . But the problem is that under the sine we have not just the letter “X”, but a complex expression. What to do?

We bring the function under the differential sign:

By opening the differential, it is easy to check that:

In fact and is a recording of the same thing.

But, nevertheless, the question remained, how did we come to the idea that at the first step we need to write our integral exactly like this: ? Why is this and not otherwise?

Formula (and all other table formulas) are valid and applicable NOT ONLY for the variable, but also for any complex expression ONLY AS A FUNCTION ARGUMENT( – in our example) AND THE EXPRESSION UNDER THE DIFFERENTIAL SIGN WERE THE SAME .

Therefore, the mental reasoning when solving should be something like this: “I need to solve the integral. I looked in the table and found a similar formula . But I have a complex argument and I can’t immediately use the formula. However, if I manage to get it under the differential sign, then everything will be fine. If I write it down, then. But in the original integral there is no factor-three, therefore, in order for the integrand function not to change, I need to multiply it by ". In the course of approximately such mental reasoning, the following entry is born:

Now you can use the tabular formula :

Ready

The only difference is that we do not have the letter “X”, but a complex expression.

Let's check. Open the table of derivatives and differentiate the answer:

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

Please note that during the verification we used the differentiation rule complex function . In essence, subsuming the function under the differential sign and - these are two mutually inverse rules.

Example 2

Let's analyze the integrand function. Here we have a fraction, and the denominator is a linear function (with “X” to the first power). We look at the table of integrals and find the most similar thing: .

We bring the function under the differential sign:

Those who find it difficult to immediately figure out which fraction to multiply by can quickly reveal the differential in a draft: . Yeah, it turns out that this means that in order for nothing to change, I need to multiply the integral by .
Next we use the tabular formula :

Examination:

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

Example 3

Find indefinite integral. Perform check.

Example 4

Find the indefinite integral. Perform check.

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

With some experience in solving integrals, similar examples will seem light and crack like nuts:

At the end of this paragraph I would like to dwell on the “free” case when in linear function the variable is included with a unit coefficient, for example:

Strictly speaking, the solution should look like this:

As you can see, subsuming the function under the differential sign was “painless”, without any multiplication. Therefore, in practice, such a long solution is often neglected and immediately written down that . But be prepared, if necessary, to explain to the teacher how you solved it! Because there is actually no integral in the table.

Variable change method in indefinite integral

Let's move on to consider the general case - the method of changing variables in the indefinite integral.

Example 5

Find the indefinite integral.

As an example, I took the integral that we looked at at the very beginning of the lesson. As we have already said, to solve the integral we liked the tabular 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 a single letter.
IN in this case this suggests itself:
The second most popular replacement letter is the letter .
In principle, you can use other letters, but we will still adhere to traditions.

So:
But when we replace it, we are left with ! Probably, many guessed that if a transition is made to a new variable, then in the new integral everything should be expressed through the letter, and there is no place for a differential there at all.
The logical conclusion is that you need turn into some expression that depends only on.

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

Since then

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

As a result:
Thus:

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

Finally, all that remains is to carry out the reverse replacement. Let us remember that.

Ready.

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

“

Let's replace:

“

The icon does not have any mathematical meaning; it means that we have interrupted the solution for intermediate explanations.

When preparing an example in a notebook, it is better to mark the reverse substitution with a simple pencil.

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

And now it’s time to remember the first solution:

A question arises. If the first method is shorter, then why use the replacement method? The fact is that for a number of integrals it is not so easy to “fit” the function to the sign of the differential.

Example 6

Find the indefinite integral.

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

As you can see, as a result of the replacement, the original integral was significantly simplified - reduced to the usual power function. This is the purpose of the replacement - to simplify the integral.

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

Another thing is that such a solution is obviously not for all students. In addition, already in this example, the use of the method of subsuming a function under the differential sign significantly increases the risk of getting confused in a decision.

Example 7

Find the indefinite integral. Perform check.

Example 8

Find the indefinite integral.

Replacement:
It remains to be seen what it will turn into

Okay, we have expressed it, but what to do with the “X” remaining in the numerator?!
From time to time, when solving integrals, we encounter the following trick: we will express from the same replacement !

Example 9

Find the indefinite integral.

This is an example for you to solve on your own. The answer is at the end of the lesson.

Example 10

Find the indefinite integral.

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

Now it's time to talk about the basic premise of using the variable substitution method: the integrand must contain some function and its derivative:(functions may not be in the product)

In this regard, when finding integrals, you often have to look at the table of derivatives.

In the example under consideration, we notice that the degree of the numerator is one less than the degree of the denominator. In the table of derivatives we find a formula that just reduces the degree by one. And that means that if you designate it as the denominator, then the chances are high that the numerator will turn into something good.

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If the function x=φ(t) has a continuous derivative, then in a given indefinite integral ∫f(x)dx you can always go to a new variable t using the formula

∫f(x)dx=∫f(φ(t))φ"(t)dt

Then find the integral from the right side and return to the original variable. In this case, the integral on the right side of this equality may turn out to be simpler than the integral on the left side of this equality, or even tabular. This method of finding the integral is called the change of variable method.

Example 7. ∫x√x-5dx

To get rid of the root, we set √x-5=t. Hence x=t 2 +5 and therefore dx=2tdt. Making the substitution, we consistently have:

∫x√x-5dx=∫(t 2 +5) 2tdt=∫(2t 4 +10t 2)dt=2∫t 4 dt+10∫t 2 dt=

III. Method of integration by parts

The integration by parts method is based on the following formula:

∫udv=uv-∫vdu

where u(x),v(x) are continuously differentiable functions. The formula is called the integration by parts formula. This formula shows that the integral ∫udv leads to the integral ∫vdu, which may turn out to be simpler than the original one, or even tabular.

Example 12. Find the indefinite integral ∫xe -2x dx

Let's use the method of integration by parts. Let's put u=x, dv=e -2x dx. Then du=dx, v=∫xe -2x dx=-e -2x +C Therefore, according to the formula we have: ∫xe -2x dx=x(-e -2x)-∫- -2 dx=-e -2x -e -2x +C

23 . Rational fraction is a fraction whose numerator and denominator are polynomials.

Rational fractions. The simplest rational fractions and their integration

Any rational function can be represented as a rational fraction, that is, as a ratio of two polynomials:

If the degree of the numerator is lower than the degree of the denominator, then the fraction is called correct, otherwise the fraction is called wrong.

If the fraction is improper, then by dividing the numerator by the denominator (according to the rule for dividing polynomials), you can represent this fraction as the sum of a polynomial and some proper fraction: , Where M(x)- a polynomial, but a proper fraction.

Example: Let us be given an improper rational fraction.

Then , since when dividing by a corner we get a remainder (4x-6).

Since the integration of polynomials does not present any fundamental difficulties, the main difficulty in integrating rational fractions lies in the integration of proper rational fractions.

There are several types of rational fractions:

II. Type: (k-positive integer ³2).

IY. View: (k-integer³2).

Let's consider integrals of the simplest rational fractions.

I. .

II. =A .

24 .Integrating rational fractions

Let the integrand be a rational fraction where and are polynomials (polynomials) of degrees k And n respectively. Without loss of generality, we can assume that k < n, since otherwise you can always present the numerator in the form P(x) = Q(x)R(x) + S(x) where R(x) and S(x) are polynomials, usually called, as in the case real numbers, quotient and remainder, and the degree of the polynomial S(x) is less n. Then

, (1.1)

and we can calculate the integral of the polynomial R(x). Let us show by example how to obtain expansion (1.1). Let P(x) = x 7 + 3x 6 + 3x 5 – 3x 3 + 4x 2 + x -2, Q(x) = x + 3x 2 + x-2. Let's divide the polynomial P(x) by the polynomial Q(x) in the same way as we divide real numbers (we obtain the solution using a column division calculator). Thus, we received the whole part of the fraction (the quotient from the division of the polynomial P by the polynomial Q) R(x) = x 4 + 2x 2 – 4x + 7 and the remainder S(x) = 9x 2 – 14x +12 from this division. According to the basic theorem of algebra, any polynomial can be decomposed into simple factors, that is, represented in the form , where are the roots of the polynomial Q(x) repeated as many times as their multiplicity. Let the polynomial Q(x) have n different roots. Then a proper rational fraction can be represented as , where are numbers to be determined. If is a root of multiplicity α, then in the decomposition into simple fractions there corresponds α terms . If x j is the complex root of the multiplicity of a polynomial with real coefficients, then the complex conjugate number is also the root of the multiplicity α of this polynomial. In order not to deal with complex numbers when integrating rational fractions, the terms in the expansion of a proper rational fraction corresponding to pairs of complex conjugate roots are combined and written as one term of the form , if - roots of multiplicity one. If are roots of multiplicity , then they correspond to terms and the corresponding expansion has the form

Thus, the integration of proper rational fractions has been reduced to the integration of the simplest fractions, of which the tabular ones can be found using the recurrence formula, which is obtained by integration by parts. Integrals, in the case where the denominator has complex roots (discriminant), are reduced, using the selection of a complete square, to integrals, by replacement. One way to find coefficients in the expansion of a proper rational fraction is as follows. The right side of the resulting expansion with undetermined coefficients is reduced to a common denominator. Since the denominators of the right and left sides are equal, the numerators, which are polynomials, must also be equal. Equating the coefficients at the same degrees (since polynomials are equal if the coefficients at the same degrees are equal), we obtain a system of linear equations for determining these coefficients.

25. Integration of irrational functions - The general principle of integrating irrational expressions is to replace the variable, which allows you to get rid of the roots in the integrand. For some function classes this goal is achieved using standard substitutions.

Integrals of the form .

Integrals of the form are calculated by replacing or .

26 . Integration of irrational functions - The general principle of integrating irrational expressions is to replace the variable, which allows you to get rid of the roots in the integrand. For some function classes this goal is achieved using standard substitutions.

Integrals of the form , where is the rational function of its arguments, are calculated by replacing .

Integrals of the form are calculated by replacing or .

Integrals of the form are calculated by replacing or . Integrals of the form are calculated by replacing or .

A ways to reduce integrals to tabular ones We have listed for you:

variable replacement method;

method of integration by parts;

Direct integration method

methods of representing indefinite integrals through tabular ones for integrals of rational fractions;

methods for representing indefinite integrals through table integrals for integrals of irrational expressions;

ways to express indefinite integrals through tabular ones for integrals of trigonometric functions.

Indefinite integral of a power function

Indefinite integral of the exponential function

But the indefinite integral of the logarithm is not a tabular integral; instead, the formula is tabular:

Indefinite integrals of trigonometric functions: Integrals of sine, cosine and tangent

Indefinite integrals with inverse trigonometric functions

Reduction to tabular form or direct integration method. Using identical transformations of the integrand, the integral is reduced to an integral to which the basic rules of integration are applicable and it is possible to use a table of basic integrals.

Example

Exercise. Find the integral

Solution. Let's use the properties of the integral and reduce this integral to tabular form.

Answer.

Technically variable replacement method in the indefinite integral is implemented in two ways:

– Subsuming a function under the differential sign. – Actually changing the variable.

Subsuming a function under the differential sign

Example 2

Find the indefinite integral. Perform check.

We bring the function under the differential sign:

Examination: The original integrand function has been obtained, which means that the integral has been found correctly.

Variable change method in indefinite integral

Example 5

Find the indefinite integral.

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

So: But when we replace it, we are left with ! Probably, many guessed that if a transition is made to a new variable, then in the new integral everything should be expressed through the letter, and there is no place for a differential there at all. The logical conclusion is that you need turn into some expression that depends only on.

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

Since then

After disassembling the differential, I recommend rewriting the final result as briefly as possible: Now, according to the rules of proportion, we express the one we need:

As a result: Thus: And this is already the most table integral (the table of integrals is, of course, also valid for the variable ).

Finally, all that remains is to carry out the reverse replacement. Let us remember that.

Ready.

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

“

Let's replace:

“

The icon does not have any mathematical meaning; it means that we have interrupted the solution for intermediate explanations.

When preparing an example in a notebook, it is better to mark the reverse substitution with a simple pencil.

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

And now it’s time to remember the first solution:

What's the difference? There is no fundamental difference. It's actually the same thing. But from the point of view of designing the task, the method of subsuming a function under the differential sign is much shorter. A question arises. If the first method is shorter, then why use the replacement method? The fact is that for a number of integrals it is not so easy to “fit” the function to the sign of the differential.

Integration by parts. Examples of solutions

Integrals of logarithms

Example 1

Find the indefinite integral.

Classic. From time to time this integral can be found in tables, but it is not advisable to use a ready-made answer, since the teacher has spring vitamin deficiency and will swear heavily. 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 integration by parts formula:

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 designated as , and something as .

In integrals of the type under consideration foralways denoted by logarithm.

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

That is, we denoted the logarithm by, and by - the rest integrand expression.

Next stage: find the differential:

A differential is almost the same as a derivative; we have already discussed how to find it in previous lessons.

Now we find the function. In order to find the function you need to integrate right side lower equality:

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

Changing a variable in an indefinite integral. Formula for converting differentials. Examples of integration. Examples of linear substitutions.

Variable Replacement Method

Variable changes can be used to evaluate simple integrals and, in some cases, to simplify the calculation of more complex ones.

The variable replacement method is that we move from the original integration variable, let it be x, to another variable, which we denote as t. In this case, we believe that the variables x and t are related by some relation x = x (t), or t = t (x). For example, x = ln t, x = sint, t = 2 x + 1, etc. Our task is to select such a relationship between x and t that the original integral is either reduced to a tabular one or becomes simpler.

Basic variable replacement formula

Let's consider the expression that stands under the integral sign. It consists of the product of the integrand, which we denote as f (x) and differential dx: . Let us move to a new variable t by choosing some relation x = x (t). Then we must express the function f (x) and the differential dx through the variable t.

To express the integrand function f (x) through the variable t, you just need to substitute the selected relation x = x instead of the variable x (t).

The differential conversion is done like this:
.
That is, the differential dx is equal to the product of the derivative of x with respect to t and the differential dt.

Then
.

In practice, the most common case is in which we perform a replacement by choosing a new variable as a function of the old one: t = t (x). If we guessed that the integrand function can be represented as
,
where t′ (x) is the derivative of t with respect to x, then
.

So, the basic formula for replacing a variable can be presented in two forms.
(1) ,
where x is a function of t.
(2) ,
where t is a function of x.

Important Note

In tables of integrals, the integration variable is most often denoted as x. However, it is worth considering that the integration variable can be denoted by any letter. Moreover, any expression can be used as an integration variable.

As an example, consider the table integral
.

Here x can be replaced by any other variable or function of a variable. Here are examples of possible options:
;
;
.

In the last example, you need to take into account that when moving to the integration variable x, the differential is transformed as follows:
.
Then
.

This example captures the essence of integration by substitution. That is, we must guess that
.
After which the integral is reduced to a tabular one.
.

You can evaluate this integral using a change of variable using the formula (2) . Let's put t = x 2+x. Then
;
;

.

Examples of integration by change of variable

1) Let's calculate the integral
.
We notice that (sin x)′ = cos x. Then

.
Here we have used the substitution t = sin x.

2) Let's calculate the integral
.
We notice that . Then

.
Here we performed the integration by changing the variable t = arctan x.

3) Let's integrate
.
We notice that . Then

. Here, during integration, the variable t = x is replaced 2 + 1 .

Linear substitutions

Perhaps the most common are linear substitutions. This is a replacement for a variable of the form
t = ax + b,
where a and b are constants. With such a replacement, the differentials are related by the relation
.

Examples of integration by linear substitutions

A) Calculate integral
.
Solution.
.

B) Find the integral
.
Solution.
Let's use the properties of the exponential function.
.
ln 2- this is a constant. We calculate the integral.

.

C) Calculate integral
.
Solution.
Let us reduce the quadratic polynomial in the denominator of the fraction to the sum of squares.
.
We calculate the integral.

.

D) Find the integral
.
Solution.
Let's transform the polynomial under the root.

.
We integrate using the variable replacement method.

.
Previously we received the formula
.
From here
.
Substituting this expression, we get the final answer.