Class equipment: lecture notes.

Evaluation Criteria

Work order

Exercise 1.

Read Lecture 9

Task 2.

Lecture 9

indefinite integral from this function:

10 .

( dx)" = d ( dx)=f(x) dx

20. The indefinite integral of the differential of a function is equal to this function plus an arbitrary constant:

30. A constant factor can be taken out of the sign of the indefinite integral.

40. The indefinite integral of the algebraic sum of functions is equal to the algebraic sum of the indefinite integrals of the terms of the functions:

50. If a is a constant, then the formula is valid

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"Integration technique Direct integration"

Practical work№ 7

Topic: Integration technique. Direct integration

Goals:

study formulas and rules for calculating the indefinite integral

learn to solve examples for direct integration

Class equipment: lecture notes.

Evaluation Criteria

grade "5" is set for the correct execution of all tasks of the work

mark "4" is given for the completion of task 1 and the correct solution of any ten examples from task 2.

mark "3" is given for the completion of task 1 and the correct solution of any seven examples from task 2.

Work order

Exercise 1.

Read Lecture 9

Using the lectures, answer the questions and write down the answers in your notebook:

1. What properties of the indefinite integral do you know?

2. Write out in the main integration formulas

3. What cases are possible with direct integration?

Task 2.

Solve examples for self-solving

Lecture 9

Topic “Indefinite integral. Direct Integration»

The function F(x) is called antiderivative for the function f(x) if F "(x) = f(x).

Any continuous function f(x) has an infinite set of antiderivatives that differ from each other by a constant term.

The general expression F(x) + C of the totality of all antiderivatives for the function f(x) is called indefinite integral from this function:

dx \u003d F (x) + C, if d (F (x) + C) \u003d dx

Basic properties of the indefinite integral

1 0 .The derivative of the indefinite integral is equal to the integrand and the differential from it is equal to the integrand:

( dx)" = d ( dx)=f(x) dx

2 0 . The indefinite integral of the differential of a function is equal to that function plus an arbitrary constant:

3 0 . The constant factor can be taken out of the sign of the indefinite integral.

4 0 .The indefinite integral of the algebraic sum of functions is equal to the algebraic sum of the indefinite integrals of the terms of the functions:

+dx

5 0 . If a is a constant, then the formula is valid

Basic integration formulas (tabular integrals)

4.

5.

7.

9.=-ctgx+C

12. = arcsin + C

When applying formulas (3), (10). (11) the sign of the absolute value is written only in cases where the expression under the sign of the logarithm can have a negative value.

Each of the formulas is easy to check. As a result of differentiation of the right side, an integrand is obtained.

Direct integration.

Direct integration is based on the direct use of the table of integrals. Here are the following cases:

1) this integral is found directly by the corresponding tabular integral;

2) after applying properties 3 0 and 4 0 this integral is reduced to one or more table integrals;

3) after elementary identical transformations over the integrand and applying properties 3 0 and 4 0, this integral is reduced to one or more table integrals.

Examples.

Based on property 3 0, the constant factor 5 is taken out of the integral sign and, using formula 1, we obtain

Decision. Using property 3 0 and formula 2, we obtain

6

Decision. Using properties 3 0 and 4 0 and formulas 1 and 2, we have

X + 3) = 4 + 12 = 4 - 4 + 12x + C = + 12x + C

The integration constant C is equal to the algebraic sum of three integration constants, since each integral has its own arbitrary constant (C 1 - C 2 + C 3 \u003d C)

Decision. Squaring and integrating each term, we have

Using the trigonometric formula 1 + ctg 2 x =

= = - ctgx - x + C

Decision. Subtracting and adding the number 9 in the numerator of the integrand, we get

= = + = - =

X + 9 + C = - x +

Examples for self-solving

Compute integrals using direct integration:

Control of students' knowledge:

check practical work;

Requirements for the design of practical work:

The task must be completed in a notebook for practical work.

Submit work after class

The method of direct integration is based on the transformation of the integrand, the application of the properties of the indefinite integral, and the reduction of the integrand to a tabular form.

For example:

Examination

Examination

2. Substitution method (variable substitution)

This method is based on the introduction of a new variable. Let's make a substitution in the integral:

;

Therefore, we get:

For example:

1)

Examination:

2)

Examination(based on property #2 of the indefinite integral):

Integration by parts

Let be u and v are differentiable functions. Let us reveal the differential of the product of these functions:

,

where

We integrate the resulting expression:

For example:

Examination(based on property #1 of the indefinite integral):

2)

We decide

Examination(based on property #1 of the indefinite integral):

PRACTICAL PART

Tasks for home solution

Find the integral:

a) ; e) ;

in) ; h)

G) ; and)

e) ; to)

a) ; e) ;

in) ; h) ;

e) ; to) .

a) ; in) ; e)

b) ; G) ; e)

Tasks for solving in practical classes:

I. Method of direct integration

a) ; g);

b) ; h) ;

in) ; and)

G) ; to)

e) ; m)

II. Substitution method (variable substitution)

G) ; to) ;

e) ; l);

III. Method of integration by parts

THEME #4

DEFINITE INTEGRAL

In mathematical calculations, it is often required to find the increment of the antiderivative function when its argument changes within given limits. Such a problem has to be solved when calculating the areas and volumes of various figures, when determining the average value of a function, when calculating the work of a variable force. These problems can be solved by calculating the corresponding definite integrals.

Purpose of the lesson:

1. Learn to calculate a definite integral using the Newton-Leibniz formula.

2. Be able to apply the concept of a definite integral to solve applied problems.

THEORETICAL PART

THE CONCEPT OF A DEFINITE INTEGRAL AND ITS GEOMETRIC MEANING

Consider the problem of finding the area of ​​a curvilinear trapezoid.

Let some function be given y=f(x), whose graph is shown in the figure.

Fig 1. The geometric meaning of a definite integral.

on axle 0x select points “a" and "in" and restore the perpendiculars from them to the intersection with the curve. Figure bounded by curve, perpendiculars and axis 0x called a curvilinear trapezoid. Let's break the interval into a number of small segments. Let's choose an arbitrary segment. We complete the curvilinear trapezoid corresponding to this segment to a rectangle. The area of ​​such a rectangle is defined as:

Then the area of ​​all completed rectangles in the interval will be equal to:

;

If each of the segments is small enough and tends to zero, then the total area of ​​the rectangles will tend to the area of ​​the curvilinear trapezoid:

;

So, the problem of calculating the area of ​​a curvilinear trapezoid is reduced to determining the limit of the sum.

The integral sum is the sum of the products of the increment of the argument by the value of the function f(x) taken at some point of the interval within which the argument changes. Mathematically, the problem of finding the limit of the integral sum, if the increment of the independent variable tends to zero, leads to the concept of a definite integral.

Function f(x ) some interval from x=a before x=v is integrable if there exists a number to which the integral sum tends as Dх®0 . In this case, the number J called definite integral functions f(x) in the interval:

;

where ] a, in[ is the area of ​​integration,

a is the lower limit of integration,

in is the upper limit of integration.

Thus, from the point of view of geometry, the definite integral is the area of ​​\u200b\u200bthe figure bounded by the graph of the function in a certain interval] a, in [ and the x-axis.

In this topic, we will talk in detail about the properties of the indefinite integral and about finding the integrals themselves using the mentioned properties. We will also work with the table of indefinite integrals. The material presented here is a continuation of the topic "Indefinite integral. Beginning". To be honest, integrals are rarely found in tests that can be taken using typical tables and (or) simple properties. These properties can be compared with the alphabet, the knowledge and understanding of which is necessary to understand the mechanism for solving integrals in other topics. Often, integration using tables of integrals and properties of the indefinite integral is called direct integration.

What I'm leading to: the functions change, but the formula for finding the derivative remains unchanged, unlike the integral, for which two methods have already been listed.

Let's go further. To find the derivative $y=x^(-\frac(1)(2))\cdot(1+x^(\frac(1)(4)))^\frac(1)(3)$ all the same applies the formula $(u\cdot v)"=u"\cdot v+u\cdot v"$, into which you have to substitute $u=x^(-\frac(1)(2))$, $v=( 1+x^(\frac(1)(4)))^\frac(1)(3)$. But to find the integral $\int x^(-\frac(1)(2))\cdot( 1+x^(\frac(1)(4)))^\frac(1)(3) dx$ requires a new method - Chebyshev substitutions.

And finally: to find the derivative of the function $y=\sin x\cdot\frac(1)(x)$, the formula $(u\cdot v)"=u"\cdot v+u\cdot v"$ is again applicable, into which we substitute $\sin x$ and $\frac(1)(x)$ instead of $u$ and $v$, respectively, while $\int \sin x\cdot\frac(1)(x) dx$ is not is taken in. More precisely, it is not expressed in terms of a finite number of elementary functions.

To summarize: where one formula was needed to find the derivative, four formulas were required for the integral (and this is not the limit), and in the latter case, the integral refused to be found at all. We changed the function - a new integration method was needed. From here we have multi-page tables in reference books. The absence of a general method (suitable for solving "manually") leads to an abundance of particular methods that are applicable only for integrating their own, extremely limited class of functions (in further topics, we will deal with these methods in detail). Although I cannot fail to note the presence of the Risch algorithm (I advise you to read the description on Wikipedia), it is only suitable for programmatic processing of indefinite integrals.

Question #3

But if there are so many of these properties, how can I learn to take integrals? With derivatives it was easier!

So far, there is only one way for a person: to solve as many examples as possible using various integration methods, so that when a new indefinite integral appears, you can choose a solution method for it, based on your experience. I understand that the answer is not very encouraging, but there is no other way.

Properties of the indefinite integral

Property #1

The derivative of the indefinite integral is equal to the integrand, i.e. $\left(\int f(x) dx\right)"=f(x)$.

This property is quite natural, because the integral and the derivative are mutually inverse operations. For example, $\left(\int \sin 3x dx\right)"=\sin 3x$, $\left(\int \left(3x^2+\frac(4)(\arccos x)\right) dx\ right)"=3x^2+\frac(4)(\arccos x)$ and so on.

Property #2

The indefinite integral of the differential of some function is equal to this function, i.e. $\int \mathrm d F(x) =F(x)+C$.

Usually, this property is perceived somewhat difficult, since it seems that there is "nothing" under the integral. To avoid this, you can write the specified property as follows: $\int 1\mathrm d F(x) =F(x)+C$. An example of using this property: $\int \mathrm d(3x^2+e^x+4)=3x^2+e^x+4+C$ or, if you like, in this form: $\int 1\; \mathrm d(3x^2+e^x+4) =3x^2+e^x+4+C$.

Property #3

The constant factor can be taken out of the integral sign, i.e. $\int a\cdot f(x) dx=a\cdot\int f(x) dx$ (we assume that $a\neq 0$).

The property is quite simple and, perhaps, does not require comments. Examples: $\int 3x^5 dx=3\cdot \int x^5 dx$, $\int (2x+4e^(7x)) dx=2\cdot\int(x+2e^(7x))dx $, $\int kx^2dx=k\cdot\int x^2dx$ ($k\neq 0$).

Property #4

The integral of the sum (difference) of two functions is equal to the sum (difference) of the integrals of these functions:

$$\int(f_1(x)\pm f_2(x))dx=\int f_1(x)dx\pm\int f_2(x)dx$$

Examples: $\int(\cos x+x^2)dx=\int \cos xdx+\int x^2 dx$, $\int(e^x - \sin x)dx=\int e^xdx -\ int \sin x dx$.

In standard tests, properties No. 3 and No. 4 are usually used, so we will dwell on them in more detail.

Example #3

Find $\int 3 e^x dx$.

We use property No. 3 and take out the constant, i.e. the number $3$, for the integral sign: $\int 3 e^x dx=3\cdot\int e^x dx$. Now let's open the table of integrals and substituting $u=x$ into formula No. 4 we get: $\int e^x dx=e^x+C$. This implies that $\int 3 e^x dx=3\cdot\int e^x dx=3e^x+C$. I suppose that the reader will immediately have a question, so I will formulate this question separately:

Question #4

If $\int e^x dx=e^x+C$ then $\int 3 e^x dx=3\cdot\int e^x dx=3\cdot\left(e^x+C\right) =3e^x+3C$! Why was it written just $3e^x+C$ instead of $3e^x+3C$?

The question is perfectly reasonable. The point is that an integral constant (i.e., the same number $C$) can be represented as any expression: the main thing is that this expression "run through" the entire set of real numbers, i.e. changed from $-\infty$ to $+\infty$. For example, if $-\infty≤ C ≤ +\infty$, then $-\infty≤ \frac(C)(3) ≤ +\infty$, so the constant $C$ can be represented as $\frac(C)( 3)$. We can write that $\int e^x dx=e^x+\frac(C)(3)$ and then $\int 3 e^x dx=3\cdot\int e^x dx=3\cdot\left (e^x+\frac(C)(3)\right)=3e^x+C$. As you can see, there is no contradiction here, but care must be taken when changing the form of the integral constant. For example, if you represent the constant $C$ as $C^2$, it would be an error. The point is that $C^2 ≥ 0$, i.e. $C^2$ does not change from $-\infty$ to $+\infty$, does not "run through" all real numbers. Similarly, it would be an error to represent a constant as $\sin C$, because $-1≤ \sin C ≤ 1$, i.e. $\sin C$ does not "run through" all values ​​of the real axis. In the future, we will not specifically discuss this issue, but we will simply write the constant $C$ for each indefinite integral.

Example #4

Find $\int\left(4\sin x-\frac(17)(x^2+9)-8x^3 \right)dx$.

We use property number 4:

$$\int\left(4\sin x-\frac(17)(x^2+9)-8x^3 \right) dx=\int 4\sin x dx-\int\frac(17)(x ^2+9)dx-\int8x^3dx$$

Now we take the constants (numbers) out of the signs of the integrals:

$$\int 4\sin x dx-\int\frac(17)(x^2+9)dx-\int8x^3dx=4\int \sin x dx-17\int\frac(dx)(x^ 2+9)-8\int x^3dx$$

Next, we work with each integral obtained separately. The first integral, i.e. $\int \sin x dx$, is easy to find in the table of integrals under No. 5. Substituting $u=x$ into formula #5, we get: $\int \sin x dx=-\cos x+C$.

To find the second integral $\int\frac(dx)(x^2+9)$, you need to apply formula No. 11 from the table of integrals. Substituting $u=x$ and $a=3$ into it, we get: $\int\frac(dx)(x^2+9)=\frac(1)(3)\cdot \arctg\frac(x)( 3)+C$.

And, finally, to find $\int x^3dx$, we use formula No. 1 from the table, substituting $u=x$ and $\alpha=3$ into it: $\int x^3dx=\frac(x^(3 +1))(3+1)+C=\frac(x^4)(4)+C$.

All integrals included in the expression $4\int \sin x dx-17\int\frac(dx)(x^2+9)-8\int x^3dx$ are found. It remains only to substitute them:

$$4\int \sin x dx-17\int\frac(dx)(x^2+9)-8\int x^3dx=4\cdot(-\cos x)-17\cdot\frac(1) (3)\cdot\arctg\frac(x)(3)-8\cdot\frac(x^4)(4)+C=\\ =-4\cdot\cos x-\frac(17)(3 )\cdot\arctg\frac(x)(3)-2\cdot x^4+C.$$

Problem solved, answer is: $\int\left(4\sin x-\frac(17)(x^2+9)-8x^3 \right)dx=-4\cdot\cos x-\frac(17 )(3)\cdot\arctg\frac(x)(3)-2\cdot x^4+C$. Let me add one little note to this problem:

Just a small note

Perhaps no one will need this insert, but still I will mention that $\frac(1)(x^2+9)\cdot dx=\frac(dx)(x^2+9)$. Those. $\int\frac(17)(x^2+9)dx=17\cdot\int\frac(1)(x^2+9)dx=17\cdot\int\frac(dx)(x^2 +9)$.

Let's look at an example in which we use formula No. 1 from the table of integrals for intercalating irrationalities (roots, in other words).

Example #5

Find $\int\left(5\cdot\sqrt(x^4)-\frac(14)(\sqrt(x^6))\right)dx$.

To begin with, we will do the same actions as in example No. 3, namely: we decompose the integral into two and take the constants out of the signs of the integrals:

$$\int\left(5\cdot\sqrt(x^4)-\frac(14)(\sqrt(x^6)) \right)dx=\int\left(5\cdot\sqrt(x^ 4) \right)dx-\int\frac(14)(\sqrt(x^6)) dx=\\ =5\cdot\int\sqrt(x^4) dx-14\cdot\int\frac( dx)(\sqrt(x^6)) $$

Since $\sqrt(x^4)=x^(\frac(4)(7))$, then $\int\sqrt(x^4) dx=\int x^(\frac(4)(7 ))dx$. To find this integral, we apply formula No. 1, substituting $u=x$ and $\alpha=\frac(4)(7)$ into it: $\int x^(\frac(4)(7))dx=\ frac(x^(\frac(4)(7)+1))(\frac(4)(7)+1)+C=\frac(x^(\frac(11)(7)))(\ frac(11)(7))+C=\frac(7\cdot\sqrt(x^(11)))(11)+C$. You can optionally represent $\sqrt(x^(11))$ as $x\cdot\sqrt(x^(4))$, but this is not required.

Let us now turn to the second integral, i.e. $\int\frac(dx)(\sqrt(x^6))$. Since $\frac(1)(\sqrt(x^6))=\frac(1)(x^(\frac(6)(11)))=x^(-\frac(6)(11) )$, then the considered integral can be represented in the following form: $\int\frac(dx)(\sqrt(x^6))=\int x^(-\frac(6)(11))dx$. To find the resulting integral, we apply formula No. 1 from the table of integrals, substituting $u=x$ and $\alpha=-\frac(6)(11)$ into it: $\int x^(-\frac(6)(11 ))dx=\frac(x^(-\frac(6)(11)+1))(-\frac(6)(11)+1)+C=\frac(x^(\frac(5) (11)))(\frac(5)(11))+C=\frac(11\cdot\sqrt(x^(5)))(5)+C$.

Substituting the results obtained, we get the answer:

$$5\cdot\int\sqrt(x^4) dx-14\cdot\int\frac(dx)(\sqrt(x^6))= 5\cdot\frac(7\cdot\sqrt(x^( 11)))(11)-14\cdot\frac(11\cdot\sqrt(x^(5)))(5)+C= \frac(35\cdot\sqrt(x^(11)))( 11)-\frac(154\cdot\sqrt(x^(5)))(5)+C. $$

Answer: $\int\left(5\cdot\sqrt(x^4)-\frac(14)(\sqrt(x^6))\right)dx=\frac(35\cdot\sqrt(x^(11 )))(11)-\frac(154\cdot\sqrt(x^(5)))(5)+C$.

And finally, let's take the integral that falls under formula No. 9 of the table of integrals. Example 6, which we will now turn to, could be solved in another way, but this will be discussed in subsequent topics. For now, we will remain within the framework of the application of the table.

Example #6

Find $\int\frac(12)(\sqrt(15-7x^2))dx$.

To begin with, let's do the same operation as before: taking the constant (number $12$) out of the integral sign:

$$ \int\frac(12)(\sqrt(15-7x^2))dx=12\cdot\int\frac(1)(\sqrt(15-7x^2))dx=12\cdot\int \frac(dx)(\sqrt(15-7x^2)) $$

The resulting integral $\int\frac(dx)(\sqrt(15-7x^2))$ is already close to the tabular integral $\int\frac(du)(\sqrt(a^2-u^2))$ (formula No. 9 of the table of integrals). The difference of our integral is that before $x^2$ under the root there is a coefficient $7$, which the table integral does not allow. Therefore, you need to get rid of this seven by moving it beyond the root sign:

$$ 12\cdot\int\frac(dx)(\sqrt(15-7x^2))=12\cdot\int\frac(dx)(\sqrt(7\cdot\left(\frac(15)( 7)-x^2\right)))= 12\cdot\int\frac(dx)(\sqrt(7)\cdot\sqrt(\frac(15)(7)-x^2))=\frac (12)(\sqrt(7))\cdot\int\frac(dx)(\sqrt(\frac(15)(7)-x^2)) $$

If we compare the table integral $\int\frac(du)(\sqrt(a^2-u^2))$ and $\int\frac(dx)(\sqrt(\frac(15)(7)-x^ 2)) $ it becomes clear that they have the same structure. Only in the integral $\int\frac(dx)(\sqrt(\frac(15)(7)-x^2))$ instead of $u$ is $x$, and instead of $a^2$ is $\frac (15)(7)$. Well, if $a^2=\frac(15)(7)$ then $a=\sqrt(\frac(15)(7))$. Substituting $u=x$ and $a=\sqrt(\frac(15)(7))$ into the formula $\int\frac(du)(\sqrt(a^2-u^2))=\arcsin\ frac(u)(a)+C$, we get the following result:

$$ \frac(12)(\sqrt(7))\cdot\int\frac(dx)(\sqrt(\frac(15)(7)-x^2))= \frac(12)(\sqrt (7))\cdot\arcsin\frac(x)(\sqrt(\frac(15)(7)))+C $$

If we take into account that $\sqrt(\frac(15)(7))=\frac(\sqrt(15))(\sqrt(7))$, then the result can be rewritten without "three-story" fractions:

$$ \frac(12)(\sqrt(7))\cdot\arcsin\frac(x)(\sqrt(\frac(15)(7)))+C=\frac(12)(\sqrt(7 ))\cdot\arcsin\frac(x)(\frac(\sqrt(15))(\sqrt(7)))+C= \frac(12)(\sqrt(7))\cdot\arcsin\frac (\sqrt(7)\;x)(\sqrt(15))+C $$

Problem solved, answer received.

Answer: $\int\frac(12)(\sqrt(15-7x^2))dx=\frac(12)(\sqrt(7))\cdot\arcsin\frac(\sqrt(7)\;x) (\sqrt(15))+C$.

Example #7

Find $\int\tg^2xdx$.

There are methods for integrating trigonometric functions. However, in this case, you can get by with the knowledge of simple trigonometric formulas. Since $\tg x=\frac(\sin x)(\cos x)$, then $\left(\tg x\right)^2=\left(\frac(\sin x)(\cos x) \right)^2=\frac(\sin^2x)(\cos^2x)$. Given $\sin^2x=1-\cos^2x$, we get:

$$ \frac(\sin^2x)(\cos^2x)=\frac(1-\cos^2x)(\cos^2x)=\frac(1)(\cos^2x)-\frac(\ cos^2x)(\cos^2x)=\frac(1)(\cos^2x)-1 $$

Thus $\int\tg^2xdx=\int\left(\frac(1)(\cos^2x)-1\right)dx$. Expanding the resulting integral into the sum of integrals and applying tabular formulas, we will have:

$$ \int\left(\frac(1)(\cos^2x)-1\right)dx=\int\frac(dx)(\cos^2x)-\int 1dx=\tg x-x+C . $$

Answer: $\int\tg^2xdx=\tg x-x+C$.

Since now we will only talk about the indefinite integral, we will omit the term “indefinite” for brevity.

In order to learn how to calculate integrals (or, as they say, integrate functions), you must first learn the table of integrals:

Table 1. Table of integrals

2. ( ),u>0. 2a. (α=0); 2b. (α=1); 2c. (α= ). 3. 3a. 4. 5. 5a) 6a. 7. 7a.

8. 9. 10. 10a. 11. 11a. 12. 13. 13a.

In addition, you will need the ability to calculate the derivative of a given function, which means you need to remember the rules of differentiation and the table of derivatives of the main elementary functions:

Table 2. Table of derivatives and differentiation rules:

6.a .

(sin and) = cos and  and (cos u) = – sin and and

And we also need the ability to find the differential of a function. Recall that the differential of the function
find by formula
, i.e. the differential of a function is equal to the product of the derivative of this function and the differential of its argument. It is useful to keep in mind the following known relations:

Table 3. Table of differentials

1. (b= Const) 2. ( ) 3. 4. 5. (b= Const) 6. 7. 8. 9.

10. 11. 12. 14. 15. 16. 17.

Moreover, you can use these formulas, both reading them from left to right, and from right to left.

Let us consider successively three basic methods for calculating the integral. The first one is called direct integration method. It is based on the use of the properties of the indefinite integral and includes two main techniques: expansion of an integral into an algebraic sum simpler and bringing under the sign of the differential, and these methods can be used both independently and in combination.

BUT) Consider algebraic sum decomposition- this technique involves the use of identical transformations of the integrand and the linearity properties of the indefinite integral:
and.

Example 1 Find integrals:

a)
; b)
;

in)
G)

e)
.

Decision.

a)We transform the integrand by dividing term by term the numerator by the denominator:

Here the property of degrees is used:
.

b) First, we transform the numerator of the fraction, then we divide the numerator by the denominator term by term:

The property of degrees is also used here:
.

Here is the property used:
,
.

.

Formulas 2 and 5 of Table 1 are used here.

Example 2 Find integrals:

a)
; b)
;

in)
G)

e)
.

Decision.

a)We transform the integrand using the trigonometric identity:

.

Here, the term-by-term division of the numerator by the denominator and formulas 8 and 9 of Table 1 are again used.

b) We similarly transform using the identity
:

.

c) First, we divide the numerator by the denominator term by term and take the constants out of the integral sign, then we use the trigonometric identity
:

d) Apply the formula for lowering the degree:

,

e) Using trigonometric identities, we transform:

B) Consider the integration technique, which is called p subtracting under the sign of the differential. This technique is based on the invariance property of the indefinite integral:

if
, then for any differentiable function and=and(X) takes place:
.

This property allows you to significantly expand the table of simplest integrals, since, by virtue of this property, the formulas in Table 1 are valid not only for the independent variable and, but also in the case when and is a differentiable function of some other variable.

For example,
, but also
, and
, and
.

Or
and
, and
.

The essence of the method is to extract the differential of a certain function in a given integrand so that this distinguished differential, together with the rest of the expression, forms a tabular formula for this function. If necessary, constants can be added appropriately for such a transformation. For example:

(in the last example it is written ln(3 + x 2) instead of ln|3 + x 2 | , since the expression 3 + x 2 is always positive).

Example 3 Find integrals:

a)
; b)
; in)
;

G)
; e)
; e)
;

g)
; h)
.

Decision.

a).

Here, formulas 2a, 5a and 7a of Table 1 are used, the last two of which are obtained just by substituting under the differential sign:

Integrate View Functions
occurs very often in the calculation of integrals of more complex functions. In order not to repeat the steps described above each time, we recommend that you remember the corresponding formulas given in Table 1.

.

Formula 3 of Table 1 is used here.

c) Similarly, taking into account that , we transform:

.

Formula 2 in Table 1 is used here.

G)

.

e) ;

e)

.

g);

h)

.

Example 4 Find integrals:

a)
b)

in)
.

Decision.

a) Let's transform:

Formula 3 of Table 1 is also used here.

b) Use the reduction formula
:

Formulas 2a and 7a of Table 1 are used here.

Here, along with formulas 2 and 8 of Table 1, the formulas of Table 3 are also used:
,
.

Example 5 Find integrals:

a)
; b)

in)
; G)
.

Decision.

a) The work
can be supplemented (see formulas 4 and 5 of Table 3) to the differential of the function
, where a and b- any constants,
. Indeed, where
.

Then we have:

.

b) Using formula 6 of table 3, we have
, as well as
, which means that the presence in the integrand of the product
means a hint: under the differential sign, you need to add an expression
. Therefore, we get

c) As in paragraph b), the product
can be supplemented to the differential of the function
. Then we get:

.

d) First, we use the linearity properties of the integral:

Example 6 Find integrals:

a)
; b)
;

in)
; G)
.

Decision.

a)Given that
(formula 9 of table 3), we transform:

b) Using formula 12 of table 3, we get

c) Taking into account formula 11 of Table 3, we transform

d) Using formula 16 of table 3, we get:

.

Example 7 Find integrals:

a)
; b)
;

in)
; G)
.

Decision.

a)All the integrals presented in this example have a common feature: the integrand contains a square trinomial. Therefore, the method of calculating these integrals will be based on the same transformation - the selection of the full square in this square trinomial.

.

b)

.

in)

G)

The method of summing under the sign of the differential is an oral implementation of a more general method of calculating the integral, called the substitution method or the change of variable. Indeed, each time, choosing the appropriate formula of Table 1 to the function obtained as a result of bringing it under the differential sign, we mentally replaced with the letter and function under the differential sign. Therefore, if integration by subsuming under the sign of the differential does not work out very well, you can directly make a change of variable. More on this in the next paragraph.