Solve the integral using the change of variable method. Variable change method in indefinite integral

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

Leading to tabular view 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 V indefinite integral 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.

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.

Variable change method in 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 it suggests itself: The second most popular letter for replacement 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 it is necessary 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 sorting out 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:

Eventually: Thus: And this is already the most tabular 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.

Now it’s time to remember the first solution:

What is 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. The 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

Let's look at 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 remaining part 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 from the original integration variable, let it be x, let's move on 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

, and so on. 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. , or t = t Basic variable replacement formula In this case, we believe that the variables x and t are related by some relation x = x Let's consider the expression that stands under the integral sign. It consists of the product of the integrand, which we denote as f , or t = t and differential dx: .

Let us move to a new variable t by choosing some relation x = x , or t = t. In this case, we believe that the variables x and t are related by some relation x = x.

Then we must express the function f
.
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 , or t = t.
,
If we guessed that the integrand function can be represented as , or t = t where t′
.

is the derivative of t with respect to x, then
(1) ,
So, the basic variable replacement formula can be presented in two forms.
(2) ,
where x is a function of t.

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 (2) After which the integral is reduced to a tabular one. You can evaluate this integral using a change of variable using the formula. Let's put t = x
;
;

.

2+x

1) .
.
Then Examples of integration by change of variable. Let's put t = x

.
Let's calculate the integral We notice that.

2) .
.
(sin x)′ = cos x

.
Here we used the substitution t = sin x.

3) We notice that .
.
(sin x)′ = cos x

Then 2 + 1 .

Here we performed the integration by changing the variable t =

arctan x
Let's integrate
.
.

Here, during integration, the variable t = x is replaced

Linear substitutions Perhaps the most common are linear substitutions. This is a replacement for a variable of the form
.
Solution.
.

t = ax + b, where a and b are constants. With such a replacement, the differentials are related by the relation
.
Solution.
Examples of integration by linear substitutions
.
A) Calculate integral

.

B) Perhaps the most common are linear substitutions. This is a replacement for a variable of the form
.
Solution.
Find the integral
.
Let's use the properties of the exponential function.

.

ln 2 where a and b are constants. With such a replacement, the differentials are related by the relation
.
Solution.
- this is a constant. We calculate the integral.

.
C)

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

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. , 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 this case it begs:
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 it is necessary 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:

Eventually:
Thus:

And this is already the most tabular 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.

Now it’s time to remember the first solution:

The 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 an ordinary 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 independent decision. 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 the formula, which 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.

Replacement:

By the way, it’s not so difficult to subsume the function under the differential sign:

It should be noted that for fractions like , this trick will no longer work (more precisely, it will be necessary to apply not only the replacement technique). You can learn to integrate some fractions in class. Integrating Some Fractions.

Here's 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 arc cosine: . In our integrand we have the arc cosine and something similar to its derivative.

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

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

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

Or in short:
Using the rule of proportion, we express the remainder we need:

Thus:

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

Example 14

Find the indefinite integral.

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

Attentive readers will have noticed that I have considered few examples with trigonometric functions. And this is no coincidence, since a separate lesson is devoted to integrals of trigonometric functions. Moreover, this lesson provides some useful guidelines for replacing a variable, which is especially important for dummies, who do not always and do not immediately understand what kind of replacement needs to be made in a particular integral. You can also see some types of substitutions in the article Definite Integral. Examples of solutions.

More experienced students can familiarize themselves with typical substitutions in integrals with irrational functions. Substitution when integrating roots is specific, and its technique is different from the one we discussed in this lesson.

I wish you success!

Solutions and answers:

Example 3: Solution:

Example 4: Solution:

Example 7: Solution:

Example 9: Solution:

Replacement:

Example 11: Solution:

Let's replace:

(see article Variable change method in indefinite integral ) or the integral is just on integration by parts method.

As always, you should have on hand: Table of integrals And Derivatives table. If you still don’t have them, then please visit the storage room of my website: Mathematical formulas and tables. I won’t tire of repeating – it’s better to print everything out. I will try to present all the material consistently, simply and clearly; there are no particular difficulties in integrating the 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 not in the table, work

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

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

Some fractions are also taken in parts; we will also consider the corresponding examples in detail.

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Direct integration

Basic integration formulas

1. C – constant	1*.
2. , n ≠ –1
3. +C
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.

Calculation of integrals by direct use of the table of simple integrals and the basic properties of indefinite integrals is called direct integration.

Example 1.

Example 2.

Example 3.

This is the most common integration method complex function, which consists of transforming the integral by moving to another integration variable.

If it is difficult to reduce the integral to a tabular one using elementary transformations, then in this case the substitution method is used. The essence of this method is that by introducing a new variable it is possible to reduce this integral to a new integral, which is relatively easy to take directly.

To integrate by substitution method, use the solution scheme:

2) find the differential from both replacement parts;

3) express the entire integrand through a new variable (after which a table integral should be obtained);

4) find the resulting table integral;

5) perform reverse replacement.

Find the integrals:

Example 1 . Substitution:cosx=t,-sinxdx=dt,

Solution:

Example 2.∫e -x3 x 2 dx Substitution:-x 3 =t, -3x 2 dx=dt, Solution:∫e -x3 x 2 dx=∫e t (-1/3)dt=-1/3e t +C=-1/3e -x3 +C

Example 3.Substitution: 1+sinx=t , cosxdx=dt ,

Solution: .

SECTION 1.5. Definite integral, methods of its calculation.

clause 1 Concept definite integral

Task. Find the increment of a function that is antiderivative of a function f(x), when passing the argument x from the value a to value b.

Solution. Let us assume that integration has found: ∫ (x)dx = F(x)+C.

Then F(x)+C 1, Where C 1- any given number, will be one of the antiderivative functions for this function f(x). Let's find its increment when the argument moves from the value a to value b. We get:

x=b - x=a =F(b) +C 1 - F(a) -C 1 =F(b)-F(a)

As we see, in the expression for the increment of the antiderivative function F(x)+C 1 no constant value C 1. And since under C 1 any given number was implied, the result obtained leads to the following conclusion: on argument transition x from the value x=a to value x=b all functions F(x)+C, antiderivatives for a given function f(x), have the same increment equal to F(b)-F(a).

This increment is usually called the definite integral and denoted by the symbol: and reads: integral of A before b from the function f(x) over dх or, in short, the integral of A before b from f(x)dx.

Number A called lower limit integration, number b - top; segment a ≤ x ≤ b – segment of integration. It is assumed that the integrand function f(x) continuous for all values x, satisfying the conditions: a x b

Definition. Increment of antiderivative functions F(x)+C on argument transition x from the value x=a to value x=b, equal to the difference F(b)-F(a), is called a definite integral and is denoted by the symbol: so that if ∫ (x)dx = F(x)+C, then = F(b)-F(a) - given the equality is called the Newton-Leibniz formula.

item 2 Basic properties of the definite integral

All properties are formulated in the proposition that the functions under consideration are integrable in the corresponding intervals.

item 3 Direct calculation of the definite integral

To calculate the definite integral, when you can find the corresponding indefinite integral, use the Newton–Leibniz formula

those. a definite integral is equal to the difference between the values of any antiderivative function at the upper and lower limits of integration.

This formula shows the procedure for calculating a definite integral:

1) find the indefinite integral of this function;

2) into the resulting antiderivative, substitute first the upper and then the lower limit of the integral instead of the argument;

3) subtract the result of substituting the lower limit from the result of substituting the upper limit.

Example 1: Calculate the integral:

Example 2: Calculate the integral:

p.4 Calculation of a definite integral by substitution method

The calculation of the definite integral by the substitution method is as follows:

1) replace part of the integrand with a new variable;

2) find new limits of the definite integral;

3) find the differential from both replacement parts;

4) express the entire integrand through a new variable (after which a table integral should be obtained); 5) calculate the resulting definite integral.

Example 1: Calculate the integral:

Substitution: 1+cosx=t,-sinxdx=dt,

SECTION 1.6. Geometric meaning of a definite integral.

Area of a curved trapezoid:

It is known that a definite integral on a segment is the area of a curvilinear trapezoid bounded by the graph of the function f(x).

The area of a figure bounded by certain lines can be found using certain integrals if the equations of these lines are known.

Let on the segment [a; b] a continuous function is given y = ƒ(x) ≥ 0. Let us find the area of this trapezoid.

Area of the figure bounded by axis 0 x, two vertical straight lines x = a, x = b and the graph of the function y = ƒ(x) (figure), determined by the formula:

This is the geometric meaning of the definite integral.

Example 1: Calculate the area of the figure bounded by the lines: y=x2.+2, y=0, x= -2, x=1.

Solution: Let's draw the drawing (note that the equation y=0 defines the Ox axis).

Answer: S = 9 units 2

Example 2: Calculate the area of the figure bounded by the lines: y= - e x, x=1 and coordinate axes.

Solution: Let's make a drawing.
If a curved trapezoid completely located under the Ox axis, then its area can be found using the formula:

In this case:

Attention! If you are asked to find the area of a figure using a definite integral, then the area is always positive! That is why the minus appears in the formula just discussed.

SECTION 1.7. Application of the definite integral

p.1 Calculation of the volume of a body of revolution

If a curved trapezoid is adjacent to the Ox axis, and straight lines y=a, y=b and the graph of the function y= F(x) (Fig. 1), then the volume of the body of rotation is determined by a formula containing an integral.

The volume of the body of revolution is equal to:

Example:

Find the volume of the body limited by the surface of rotation of the line around the Ox axis at 0≤ x ≤4.

Solution: V

units 3. Answer: unit 3.

SECTION 3.1. Ordinary differential equations

item 1 The concept of a differential equation

Definition. Differential equation is an equation containing a function of a set of variables and their derivatives.

General form such an equation =0, where F is a known function of its arguments, specified in a fixed area; x - independent variable (variable by which it is differentiated); y - dependent variable (the one from which derivatives are taken and the one that needs to be determined); - derivative of the dependent variable y with respect to the independent variable x.

item 2 Basic concepts of differential equation

In order of a differential equation is called the order of the highest derivative included in it.

For example:

A second order equation is a first order equation.

Any function that connects variables and turns a differential equation into a true equality is called decision differential equation.

General solution of a first-order differential equation is a function of and an arbitrary constant C that turns this equation into an identity in .

The general solution, written in the implicit form =0, is called general integral.

Private decision equation =0 is the solution obtained from the general solution for fixed value- fixed number.

The problem of finding a particular solution to a differential equation of nth order (n= 1,2,3,...), satisfying initial conditions type

called Cauchy problem.

clause 3 Differential equations first order with separable variables

A first order differential equation is called a separable equation if it can be represented in the form can be rewritten in the form . If . Let's integrate: .

To solve an equation of this type you need:

1. Separate variables;

2. Integrating the separated variable equation, find common decision given equation;

3. Find a particular solution that satisfies the initial conditions (if they are given).

Example 1. Solve the equation. Find a particular solution that satisfies the condition y=4 at x=-2.

Solution: This is a separated variable equation. Integrating, we find the general solution to the equation: . To obtain a simpler general solution, we represent the constant term on the right side in the form C/2. We have or is a general solution. Substituting the values y=4 and x=-2 into the general solution, we get 16=4+C, from which C=12.

So, a particular solution of the equation satisfying this condition, has the form

Example 2. Find a particular solution to the equation if .

Solution: , , , , , common decision.

We substitute the values of x and y into the particular solution: , , private solution.

Example 3. Find the general solution to the equation . Solution: ,, , - common decision.

item 4 Differential equations of order higher than the first

An equation of the form or is solved by double integration: , , whence . Integrating this function, we get new feature from f(x), which we denote by F(x). Thus, ; . Let's integrate again: or y=Ф(x). We obtained a general solution to the equation containing two arbitrary constants and .