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, test papers rarely contain integrals that can be taken using typical tables and/or simple properties. These properties can be compared to the alphabet, 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 getting at: the functions change, but the formula for finding the derivative remains unchanged, unlike the integral, for which we already had to list two methods.
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 same formula $(u\cdot v)"=u"\cdot v+u\cdot v"$, into which you will have to substitute $u=x^(-\frac(1)(2))$, $v=( 1+x^(\frac(1)(4)))^\frac(1)(3)$. But to find the integral $\int x^(-\frac(1)(2))\cdot( 1+x^(\frac(1)(4)))^\frac(1)(3) dx$ will require the use of 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 instead of $u$ and $v$ we substitute $\sin x$ and $\frac(1)(x)$, respectively. But $\int \sin x\cdot\frac(1)(x) dx$ is not is taken, or more precisely, is not expressed in terms of a finite number of elementary functions.
Let's summarize: where one formula was needed to find the derivative, four were required for the integral (and this is not the limit), and in the latter case the integral refused to be found at all. The function was changed - a new integration method was needed. This is where we have multi-page tables in reference books. The lack of a general method (suitable for solving “manually”) leads to an abundance of private 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 help but note the presence of the Risch algorithm (I advise you to read the description on Wikipedia), it is only suitable for program processing of indefinite integrals.
Question #3
But if there are so many of these properties, how can I learn to take integrals? It was easier with derivatives!
For a person, there is only one way so far: 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 reassuring, but there is no other way.
Properties of the indefinite integral
Property No. 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, since the integral and 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 No. 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 as somewhat difficult, since it seems that there is “nothing” under the integral. To avoid this, you can write the indicated 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 No. 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 No. 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 No. 3
Find $\int 3 e^x dx$.
Let's use property No. 3 and take out the constant, i.e. 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: $\int e^x dx=e^x+C$. It follows that $\int 3 e^x dx=3\cdot\int e^x dx=3e^x+C$. I assume 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 did they just write $3e^x+C$ instead of $3e^x+3C$?
The question is completely reasonable. The point is that the integral constant (i.e. that same number $C$) can be represented in the form of any expression: the main thing is that this expression “runs through” the entire set of real numbers, i.e. varied from $-\infty$ to $+\infty$. For example, if $-\infty≤ C ≤ +\infty$, then $-\infty≤ \frac(C)(3) ≤ +\infty$, so the constant $C$ can be represented in the form $\frac(C)( 3)$. 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 you need to be careful when changing the form of the integral constant. For example, representing the constant $C$ as $C^2$ would be an error. The point is that $C^2 ≥ 0$, i.e. $C^2$ does not change from $-\infty$ to $+\infty$ and does not “run through” all real numbers. Likewise, it would be a mistake 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 what follows, we will not discuss this issue in detail, but will simply write the constant $C$ for each indefinite integral.
Example No. 4
Find $\int\left(4\sin x-\frac(17)(x^2+9)-8x^3 \right)dx$.
Let's use property No. 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 let’s take the constants (numbers) outside the integral signs:
$$\int 4\sin x dx-\int\frac(17)(x^2+9)dx-\int8x^3dx=4\int \sin x dx-17\int\frac(dx)(x^ 2+9)-8\int x^3dx$$
Next, we will work with each obtained integral separately. The first integral, i.e. $\int \sin x dx$, can be easily found in the table of integrals under No. 5. Substituting $u=x$ into formula No. 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 $4\int \sin x dx-17\int\frac(dx)(x^2+9)-8\int x^3dx$ have been found. All that remains is to substitute them:
$$4\int \sin x dx-17\int\frac(dx)(x^2+9)-8\int x^3dx=4\cdot(-\cos x)-17\cdot\frac(1) (3)\cdot\arctg\frac(x)(3)-8\cdot\frac(x^4)(4)+C=\\ =-4\cdot\cos x-\frac(17)(3 )\cdot\arctg\frac(x)(3)-2\cdot x^4+C.$$
The problem is solved, the answer is: $\int\left(4\sin x-\frac(17)(x^2+9)-8x^3 \right)dx=-4\cdot\cos x-\frac(17 )(3)\cdot\arctg\frac(x)(3)-2\cdot x^4+C$. I will add one small note to this problem:
Just a small note
Perhaps no one will need this insert, but I’ll still 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 to interpose irrationalities (roots, in other words).
Example No. 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 will decompose the integral into two and move the constants beyond 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: $\int x^(\frac(4)(7))dx=\ frac(x^(\frac(4)(7)+1))(\frac(4)(7)+1)+C=\frac(x^(\frac(11)(7)))(\ frac(11)(7))+C=\frac(7\cdot\sqrt(x^(11)))(11)+C$. If you wish, you can represent $\sqrt(x^(11))$ as $x\cdot\sqrt(x^(4))$, but this is not necessary.
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 integral under consideration 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 No. 6, which we will now move on to, could be solved in another way, but this will be discussed in subsequent topics. For now, we will remain within the framework of using the table.
Example No. 6
Find $\int\frac(12)(\sqrt(15-7x^2))dx$.
First, let's do the same operation as before: moving the constant (the number $12$) outside 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 one $\int\frac(du)(\sqrt(a^2-u^2))$ (formula No. 9 table of integrals). The difference in our integral is that before $x^2$ under the root there is a coefficient $7$, which the table integral does not allow. Therefore, we 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 $u$ there is $x$, and instead of $a^2$ there 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 the “three-story” fractions:
$$ \frac(12)(\sqrt(7))\cdot\arcsin\frac(x)(\sqrt(\frac(15)(7)))+C=\frac(12)(\sqrt(7 ))\cdot\arcsin\frac(x)(\frac(\sqrt(15))(\sqrt(7)))+C= \frac(12)(\sqrt(7))\cdot\arcsin\frac (\sqrt(7)\;x)(\sqrt(15))+C $$
The problem is solved, the answer is received.
Answer: $\int\frac(12)(\sqrt(15-7x^2))dx=\frac(12)(\sqrt(7))\cdot\arcsin\frac(\sqrt(7)\;x) (\sqrt(15))+C$.
Example No. 7
Find $\int\tg^2xdx$.
There are methods for integrating trigonometric functions. However, in this case, you can get by with knowledge of simple trigonometric formulas. Since $\tg x=\frac(\sin x)(\cos x)$, then $\left(\tg x\right)^2=\left(\frac(\sin x)(\cos x) \right)^2=\frac(\sin^2x)(\cos^2x)$. Considering $\sin^2x=1-\cos^2x$, 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: $\int\tg^2xdx=\tg x-x+C$.
Lesson equipment: lecture notes.
Evaluation criteria
Work order
Exercise 1.
Read lecture No. 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. The 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 indefinite integrals of the terms of the functions:
50. If a is a constant, then the formula is valid
View document contents
“Technique of Integration Direct Integration”
Practical work№ 7
Topic: Integration technique. Direct integration
Goals:
study formulas and rules for calculating the indefinite integral
learn to solve examples using direct integration
Lesson equipment: lecture notes.
Evaluation criteria
A grade of “5” is given for the correct completion of all work tasks
A grade of “4” is given for completing task 1 and correctly solving any ten examples from task 2.
A grade of “3” is given for completing task 1 and correctly solving any seven examples from task 2.
Work order
Exercise 1.
Read lecture No. 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 into the basic integration formulas
3. What cases are possible with direct integration?
Task 2.
Solve examples for independent solution
Lecture 9.
Topic: “Indefinite integral. Direct integration"
A function F(x) is called an antiderivative of a function f(x) if F "(x) = f(x).
Any continuous function f(x) has an infinite number of antiderivatives, which differ from each other by a constant term.
The general expression F(x) +C of the set of all antiderivatives for the function f(x) is called indefinite integral from this function:
dx = F(x) +С, if d(F(x) +С) = dx
Basic properties of the indefinite integral
1 0 .The derivative of the indefinite integral is equal to the integrand and its differential 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 this 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 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 absolute value sign is written only in cases where the expression under the logarithm sign can have a negative value.
Each of the formulas is easy to check. As a result of differentiating the right side, an integrand is obtained.
Direct integration.
Direct integration is based on the direct use of the table of integrals. The following cases may arise here:
1) this integral can be found directly from the corresponding table integral;
2) this integral, after applying the properties 3 0 and 4 0, is reduced to one or more tabular integrals;
3) this integral, after elementary identity transformations over the integrand and application of the properties 3 0 and 4 0, is reduced to one or more tabular integrals.
Examples.
Based on property 3 0, the constant factor 5 is taken out of the integral sign and, using formula 1, we obtain
Solution. Using property 3 0 and formula 2, we get
6 
Solution. 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 = C)
Solution. Squaring and integrating each term, we have
Using the trigonometric formula 1 + cot 2 x =
= = - ctgx – x + C
Solution. Subtracting and adding the number 9 to the numerator of the integrand, we get
= = + = - =
X + 9 + C = - x +
Examples for self-solution
Evaluate the integrals using direct integration:
Monitoring students' knowledge:
check practical work;
Requirements for completing practical work:
The task must be completed in a notebook for practical work
Submit work after class
Since now we will only talk about the indefinite integral, for the sake of brevity we will omit the term “indefinite”.
In order to learn how to calculate integrals (or, as they say, integrate functions), you first need to learn the table of integrals:
Table 1. Table of integrals
|
2.
2a. 2b. 2c. 3.
3a. 4.
5.
5a) 6a. 7.
7a. |
8.
9.
10.
10a. 11.
11a. 12.
13.
13a. |
(
),u>0.
(α=0);
(α=1);
(α= 
).
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 basic elementary functions:
Table 2. Table of derivatives and differentiation rules:
 6.a .
|
(sin And) = cos And And (cos u) = – sin 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 relationships:
Table 3. Differential table
|
1.
2.
3.
4.
5.
6.
7.
8.
9.
|
10.
11.
12.
14.
15.
16.
17.
|
(b=
Const)
(
)
(b=
Const)
Moreover, these formulas can be used either by reading them from left to right or from right to left.
Let us consider sequentially the three main methods of calculating the integral. The first of them is called by 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 subscribing to the differential sign, and these techniques can be used both independently and in combination.
A) Let's consider algebraic sum expansion– this technique involves the use of identical transformations of the integrand and the linearity properties of the indefinite integral: 
And.
Example 1. Find the integrals:
A)
;
b)
;
V)
G)
d) 
.
Solution.
A)Let's transform the integrand by dividing the numerator term by term:
The property of powers is used here: 
.
b) First, we transform the numerator of the fraction, then we divide the numerator term by term by the denominator:
The property of degrees is also used here: 
.
The property used here is: 
,
.
.
Formulas 2 and 5 of Table 1 are used here.
Example 2. Find the integrals:
A)
;
b)
;
V)
G)
d) 
.
Solution.
A)Let's transform the integrand using the trigonometric identity:
.
Here we again use term-by-term division of the numerator by the denominator and formulas 8 and 9 of Table 1.
b) We transform similarly, using the identity 
:
.
c) First, divide the numerator term by term by the denominator and take the constants out of the integral sign, then use the trigonometric identity 
:
d) Apply the formula for reducing the degree:
,
e) Using trigonometric identities, we transform:
B) Let's consider the integration technique, which is called n by placing it under the differential sign. This technique is based on the invariance property of the indefinite integral:
If 
, then for any differentiable function And=And(X) occurs: 
.
This property allows us to significantly expand the table of simple integrals, since due to 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 isolate the differential of a certain function in a given integrand so that this isolated differential, together with the rest of the expression, forms a tabular formula for this function. If necessary, during such a conversion, constants can be added accordingly. For example:
(in the last example written ln(3 + x 2) instead of ln|3 + x 2 | , since the expression is 3 + x 2 is always positive).
Example 3.
Find the integrals:
A)
;
b)
; V)
;
G)
; d)
; e)
;
and)
; h)
.
Solution.
A).
Formulas 2a, 5a and 7a of Table 1 are used here, the last two of which are obtained precisely by subsuming the differential sign:
Integrate view functions 
occurs very often within the framework of calculating 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 2c in Table 1 is used here.
G)
.
d) ;
e)
.
and) ;
h)
.
Example 4. Find the integrals:
A)
b)
V) 
.
Solution.
a) Let's transform:
Formula 3 of Table 1 is also used here.
b) We use the formula for reducing the degree 
:
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 the integrals:
A) 
;
b) 
V) 
; G) 
.
Solution.
a) 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, from where 
.
Then we have:
.
b) Using formula 6 of table 3, we have 
, and 
, which means the presence in the integrand of the product 
means a hint: under the differential sign you need to enter the expression 
. Therefore we get
c) Same as in point b), the product 
can be extended to differential functions 
. Then we get:
.
d) First we use the linearity properties of the integral:
Example 6. Find the integrals:
A)
;
b)
;
V)
; G)
.
Solution.
A)Considering 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 obtain:
.
Example 7. Find the integrals:
A)
;
b)
;
V)
;
G)
.
Solution.
A)All integrals presented in this example have a common feature: The integrand contains a quadratic trinomial. Therefore, the method of calculating these integrals will be based on the same transformation - isolating the complete square in this quadratic trinomial.
.
b)
.
V)
G)
The method of substituting a differential sign is an oral implementation of a more general method of calculating an integral, called the substitution method or change of variable. Indeed, each time, selecting a suitable formula in Table 1 for the one obtained as a result of subsuming the function differential sign, we mentally replaced the letter And function introduced under the differential sign. Therefore, if integration by subsuming the differential sign does not work out very well, you can directly change the variable. More details about this in the next paragraph.
The direct integration method is based on transforming the integrand function, applying the properties of the indefinite integral and reducing the integrand expression to tabular form.
For example:
Examination
Examination
2. Substitution method (variable replacement)
This method is based on introducing a new variable. Let's make a substitution in the integral:
;
Therefore, we get:
For example:
1)
Examination:
2)
Examination(based on property No. 2 of the indefinite integral):
Integrated piece by piece
Let u And v - differentiable functions. Let us reveal the differential of the product of these functions:
,
where 
Let's integrate the resulting expression:
For example:
Examination(based on property No. 1 of the indefinite integral):
2)
Let's decide 
Examination(based on property No. 1 of the indefinite integral):
PRACTICAL PART
Problems to solve at home
Find the integral:
A) 
; e) 
;
V) 
; h) 
G) 
; And) 
d) ; To) 
A) ; e) 
;
V) ; h) 
;
d) ; To) 
.
A) ; V) ; d)
b) ; G) ; e)
Problems to be solved during practical classes:
I. Direct integration method
A) 
; and) ;
b) 
; h) ;
V) 
; And)
G) 
; To)
e) 
; m) 
II. Substitution method (variable replacement)
G) ; To) 
;
d) 
; l) ;
III. Method of integration by parts
TOPIC No. 4
DEFINITE INTEGRAL
In mathematical calculations, it is often necessary to find the increment of an antiderivative function when its argument changes within specified limits. This 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 computing 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 DETERMINED INTEGRAL AND ITS GEOMETRICAL MEANING
Consider the problem of finding the area of a curvilinear trapezoid.
Let some function be given y=f(x), the graph of which is shown in the figure.
Figure 1. Geometric meaning of a definite integral.
On axis 0x select points “a" And "V" and restore perpendiculars from them until they intersect with the curve. A figure bounded by a curve, perpendiculars and an axis 0x called a curved trapezoid. Let's divide the interval into a number of small segments. Let's choose an arbitrary segment. Let's build a curved trapezoid corresponding to this segment to a rectangle. The area of such a rectangle is determined 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 curved trapezoid:
;
So, the problem of calculating the area of a curvilinear trapezoid comes down to determining the limit of the sum.
The integral sum is the sum of the products of the increment of the argument and the value of the function f(x) , taken at some point in the interval within the boundaries of 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 ) in some interval from x=a before x=b integrable if there is 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, c[ – area of integration,
A–lower limit of integration,
V–upper limit of integration.
Thus, from the point of view of geometry, a definite integral is the area of a figure limited by the graph of a function in a certain interval] a, c [ and x-axis.