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0:19

Повторение прошлого семинара

1:25

Замена переменной в неопределенном интеграле

6:04

Зачем мы пишем dx при интегрировании?

8:08

Задача №1674

12:46

Задача №1680

16:34

Задача №1697

18:47

Задача №1766

22:35

Задача №1776

26:21

Задача №1778: тригонометрические замены

32:02

Немного о замене корня в интеграле

33:20

Задача №1784: тригонометрические замены

40:58

Задача №1786: гиперболические замены

48:28

Интегрирование по частям

51:39

Задача №1791

53:52

Интегрирование по частям: замечание

55:28

Задача №1796

59:52

Задача №1828

1:10:00

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косухин

о.н

математический

анализ

часть

семинары

-.неопределенный

интеграл:замена

переменной

00:00:04

[music]

00:00:19

so let’s give a brief summary of the

00:00:21

previous episodes for what we discussed with you

00:00:23

we discussed with you the main table of

00:00:25

this Demidovich table, it’s called

00:00:28

the table of simple integrals, although it must be

00:00:30

said that not all of them are simple, that

00:00:32

is, there are also quite complex integrals,

00:00:34

such as yes here are the inverse

00:00:37

hyperbolic functions avia sine and

00:00:40

cosine

00:00:41

area tangent yes, this is what I will give you

00:00:44

for yes they are, in principle, quite

00:00:46

informative

00:00:48

and we discussed some of the

00:00:52

simplest methods, this is what

00:00:53

Dimitrovich calls the expansion method,

00:00:55

this is essentially working with linear

00:00:57

combinations and the simplest replacement

00:01:00

we discussed with you this is when a linear

00:01:02

function is substituted inside

00:01:04

and then it means it’s calculated

00:01:07

so let’s now move on to a

00:01:11

discussion of the two main methods of how

00:01:14

integrals are reduced more complex Kant the

00:01:17

game is simpler for you, that is, in

00:01:19

fact, there are two main techniques and we

00:01:21

just let's take it off, let's work on this, so the

00:01:24

first main technique is the replacement of a

00:01:26

variable, which means and what does the

00:01:30

corresponding formula look like? the

00:01:45

derivative it is equal to on the

00:01:49

interval on ours and of is small, as

00:01:52

then we know that instead of x we can

00:01:56

substitute another function inside,

00:01:58

for example, x is equal to f y t here x is equal to f y

00:02:02

t and the

00:02:04

derivative of a complex function f from the same

00:02:07

is expressed by the formula

00:02:10

f prime from t multiply

00:02:14

by the same prime from t

00:02:16

well, firstly, we know that the big

00:02:19

prime can be replaced simply naively and

00:02:22

then we get that we know the

00:02:26

antiderivative for functions like this f from the

00:02:30

same from t multiply

00:02:32

by the same prime from t dt

00:02:36

like this it turns out that this is an

00:02:39

antiderivative, it is equal to a large from

00:02:42

the belly,

00:02:43

well, then all the antiderivatives still

00:02:46

differ from it by a constant, I’ll start with

00:02:48

this one, if you know how to count the

00:02:50

antiderivative of a function f of x

00:02:53

dx, then you know how to count this kind of

00:02:58

antiderivative, too, and the scheme of honor is the

00:03:00

following first you count here is this

00:03:03

integral and then at the end you

00:03:07

can substitute X instead of living you, that is, they write

00:03:10

this equality to that one

00:03:13

indefinite integral is equal to another

00:03:15

indefinite integral and how is

00:03:17

this equality understood equality is understood as

00:03:19

follows: what if you write

00:03:21

here the antiderivative in big attacks and

00:03:24

then instead of x, substitute x y then you will

00:03:27

get the correct answer, the

00:03:30

following formula for replacing a

00:03:32

variable will begin to appear, that is, this is some kind of bridge along

00:03:36

which you can travel between

00:03:37

different integrals, that is, the integral of

00:03:39

one variable x, in this sense, it

00:03:42

coincides with the integral of another

00:03:44

variable t

00:03:45

like this

00:03:47

well, that means that here after all, well,

00:03:51

here we still

00:03:54

need to check,

00:03:56

yes, but behind the scenes there remains, as

00:04:00

always, the interval over which we carry out

00:04:02

the integration, that is, of course, we

00:04:04

mean that we have some kind of

00:04:06

interval and .b. so, in particular, we

00:04:09

need to demand that it

00:04:12

take values, so it must

00:04:15

take values

00:04:20

value on this interval

00:04:22

from well from this interval, then if it is

00:04:25

correct to say so, not on but from this

00:04:27

interval, that is, we cannot crawl

00:04:30

out

00:04:31

because we have this is the function of

00:04:34

big attacks, we don’t know what happens to it beyond

00:04:37

this interval, but this is outside this interval,

00:04:40

so x equals victims

00:04:44

must also take values from this

00:04:45

interval, we will remember this when we

00:04:47

talk about the definite integral,

00:04:49

that is, in our now the problem, as

00:04:51

a rule, this limitation will not come up,

00:04:53

so this means that this is a special case of this

00:04:57

theorem, we actually discussed it with you

00:04:59

last time, but if it lived, then this is a

00:05:01

linear function, then you and I know

00:05:04

that there is such a formula, that is, we

00:05:07

had it

00:05:08

f from and t + b multiplied

00:05:12

by dt well, we also had equal f from from and

00:05:17

plus b last time we discussed this

00:05:19

formula and even some problems when

00:05:22

I left you with a homework on this formula,

00:05:25

this formula is more general means

00:05:27

traveling through it you can do it in both

00:05:28

directions,

00:05:29

or you can say, but if you see a

00:05:33

problem like a rebus like this, that is,

00:05:35

there is a problem like puzzles for you, yes,

00:05:37

this is an expression and you need to see in it

00:05:40

the function of victims and the prime from t then you

00:05:42

can collapse a more complex expression

00:05:44

and to get a simpler expression,

00:05:47

this is one type of problem, another type of

00:05:50

problem can also be called a substitution,

00:05:52

this is when you start with a simpler

00:05:54

form like this, and here you can

00:05:57

insert then any function

00:06:00

that you like, so this means that a

00:06:04

question may arise and

00:06:07

why do we

00:06:09

use this very dx in the notation of the integral? Why do

00:06:12

n’t we just write, for example, the

00:06:13

integral f of x, why write this

00:06:16

dx at the end, but just one of the explanations,

00:06:20

yes, you can give several explanations for

00:06:22

this fact, one of the explanations is

00:06:24

that but it will concern a definite

00:06:26

integral that we will have a

00:06:28

delta x multiplier there, but we will get to that,

00:06:32

but another explanation can be given

00:06:34

now that if you write it like this, the

00:06:38

variable change formula will

00:06:40

look unnatural, that is,

00:06:42

look at this entry for the integral f from

00:06:44

x and here is the integral from here and yes if we

00:06:48

just replace them on the stomach or we

00:06:51

will get it wrong and equality, that

00:06:53

is, we cannot write that this is equal to

00:06:54

this or we need to

00:06:57

add an artificial factor from somewhere here and

00:06:59

also a prime from t and where it comes from is

00:07:01

not clear, but if we have

00:07:04

dx written here, then

00:07:07

this equality looks quite

00:07:09

natural, let me try to

00:07:11

explain to you why,

00:07:13

because in fact, if we write

00:07:16

the differential of a function, then how do we

00:07:19

know the differential of a function, it’s the same

00:07:22

prime from t multiply by dt and we just

00:07:25

get the correct formula, that is,

00:07:27

it turns out that if we assign this

00:07:29

dx under the integral sign, then when we replace the

00:07:32

variable, we get it quite

00:07:35

naturally and the equality then and the replacement of the

00:07:37

variable will simply consist in

00:07:39

taking x and replacing it with a function

00:07:42

stomach and so that means our formula

00:07:45

integral on f from x dx

00:07:49

is equal to integral f here you are

00:07:53

knives stroke from t dt

00:07:57

here is one of the two main forms of how

00:08:00

integrals are calculated let's see

00:08:03

how it works for example let's

00:08:05

take you

00:08:08

1674

00:08:09

problem 1674 in this problem we are asked to

00:08:14

calculate the integral

00:08:16

x dx

00:08:19

divided by the root of 1 minus x square,

00:08:21

this entry is the same as

00:08:24

this x divided by the root of 1 minus x

00:08:27

square dx, it’s just convenient to perceive

00:08:30

this dx as a multiplier by which everything

00:08:33

is multiplied, but when it is a multiplier it

00:08:35

can be written as a multiplier of a fraction, or it can be

00:08:38

written as a multiplier in the numerator of a fraction,

00:08:41

that is, it’s basically the same thing,

00:08:44

so they ask us to calculate this

00:08:47

integral,

00:08:48

one of the methods that helps one of the

00:08:52

methods that helps in solving problems

00:08:55

if you don’t like something in the integral,

00:08:57

you can choose this as a new

00:08:59

change, let's say that now we

00:09:04

'll try to make a change of

00:09:07

variable here so that that is, see,

00:09:10

let's change the letter to t, I

00:09:14

told you last time that we can

00:09:16

change the name of the letter to, let's say this for

00:09:18

now we’ll just change the letter there

00:09:22

so that we have something like this and

00:09:25

now we’ll try to see

00:09:28

the formula, that is, to see this

00:09:31

expression, that is, it’s like a rebus problem,

00:09:34

I suggest that as at&t

00:09:38

as at&t

00:09:40

we take

00:09:42

this function one minus t square

00:09:45

to give we will take one minus t square

00:09:48

which then will be the same prime from t

00:09:53

this is minus 2 t and we practically have

00:09:58

everything we need, but

00:10:01

instead of the multiplier minus 2 t which we

00:10:03

want to have here, we have the

00:10:06

multiplier written simply t let's add minus 2

00:10:08

artificially; we'll multiply what's

00:10:10

inside the integral by minus 2, which

00:10:12

means outside we'll add a factor of minus

00:10:14

one-half by the property of linearity, yes, we

00:10:17

can do that,

00:10:20

that is, while we're doing this

00:10:22

transformation, and now if we look at

00:10:25

this integral in this form then we

00:10:27

can see that what we have

00:10:30

written here let's rewrite minus one

00:10:33

second integral

00:10:36

one divided by the root is taken from

00:10:39

just our function lives and

00:10:41

multiplies this is all the knives stroke from t dt and

00:10:44

according to the formula that is written above we

00:10:47

understand that if we move to a new

00:10:49

variable but let's not confuse the

00:10:52

letter x with the previous one, for example, let's take

00:10:54

the variable y, we can choose

00:10:56

the name for the new variables ourselves, for example, y,

00:10:59

let us have the same otter, we get

00:11:01

that this is minus one-second

00:11:04

integral of one divided by the root

00:11:06

of y to y,

00:11:09

that is, we a more complex expression

00:11:11

has been turned into a simpler one, which means and now

00:11:14

this integral is already tabular, which means

00:11:17

that this is how it turns out, this is y to the power of

00:11:20

minus 1 2

00:11:23

our minus 1 2 we had a multiplier before

00:11:26

the integral, here we increase the power

00:11:28

by one and it becomes y to the power of 1

00:11:31

2 and we divide by this very one second, that

00:11:33

is, a factor of 2 appears, and that means

00:11:37

plus c can be written as a plus here right in

00:11:39

parentheses, but of course you can write it

00:11:41

outside, that is, it turns out minus y to the

00:11:43

power of 1 2 + c c is an arbitrary

00:11:46

constant, so I’m using it by one second

00:11:48

minus one second I can not multiply, but

00:11:51

at the end, in order to get an answer to the same, I was

00:11:53

interested in the answer in the original

00:11:55

variable given here you called it here you

00:11:57

called it x let's go back, that is,

00:12:00

instead of the y we will substitute this here,

00:12:02

well, let's write from x, that is, this

00:12:05

will be minus the root of 1 minus x square

00:12:08

plus c, so we got the answer, that is, we

00:12:12

started with this integral,

00:12:14

then changing the variable turned it

00:12:17

into this integral,

00:12:19

then we calculated this integral and in the

00:12:23

end we returned to the old variables, this is

00:12:25

how the calculation of most

00:12:28

integrals works only usually you

00:12:30

will need to make more than one replacement, but sometimes there may be

00:12:33

5 replacements before you have to do more in

00:12:36

order to calculate and

00:12:38

calculate all sorts of transformations,

00:12:40

but in general, the

00:12:42

main diagram for this task is very

00:12:44

well shown 1680 for example, the problem

00:12:49

in this problem we are asked to calculate

00:12:52

integral

00:12:53

dx

00:12:55

divided by 1 plus x root of x

00:13:01

and, moreover, Demidovich gives us

00:13:04

a hint that dx divided by the root of x

00:13:07

and that is, two differentials of the function

00:13:12

root of x, that

00:13:14

is, he gives us a hint such that we

00:13:17

need to go to a new function to a new

00:13:20

variable y is equal to the root of x so

00:13:24

let's take y is equal to the root of x

00:13:27

then what is before y here is the

00:13:31

differential of this function well as

00:13:34

Demidovich wrote for us yes it is 1 2 dx divided

00:13:39

by the root of x

00:13:40

then we can then make a replacement with you

00:13:44

here let's look at y equals

00:13:46

root language means and what is equal to y

00:13:48

square means everything you see here

00:13:53

dx yes we can replace through to y that

00:13:56

is we can even see yes this is

00:13:58

the expression dx divided by the root of x it

00:14:00

turns out 2

00:14:03

dy-2 d y well 1 plus x is what it is

00:14:09

one plus y square that's how we

00:14:11

made the replacement, that is,

00:14:14

in order to make the replacement we need to

00:14:16

know how to turn x and y here is

00:14:19

the formula and how to turn dx to to y

00:14:23

or in this case we will immediately replace

00:14:25

replace divide the expression dx by the root of the

00:14:27

language, we immediately replace it with 2d games and

00:14:29

so this integral turns into

00:14:32

this, which means and so we replaced it, we got

00:14:35

almost a tabular integral, that is,

00:14:37

two where you carry it and we see that

00:14:39

the arctangent remains, that is, two

00:14:41

arctangents

00:14:42

y + c

00:14:45

the problem will be solved when we substitute

00:14:48

x for the game, that is, when we

00:14:50

return to the old coordinates, it

00:14:53

turns out that these are two arctangents

00:14:55

of the root of x + c, the

00:14:59

whole problem is solved

00:15:03

like this, but if you have questions, if you have

00:15:06

questions, ask, that is, so far our

00:15:09

problem was such that the replacement was hidden in them, it

00:15:11

was necessary to see it, that is, it’s

00:15:14

something like an Olympiad problem for this, you

00:15:16

need to see it, but we are in

00:15:19

the process of solving problems, we will learn how to

00:15:23

make the correct replacements ourselves, that is,

00:15:25

we will learn to determine by the type of integrals

00:15:27

what replacement needs to be made in that or in

00:15:30

another case, this is what the

00:15:32

art of integration actually consists of, that is, we can

00:15:35

say that integration is

00:15:36

somewhat reminiscent of a

00:15:38

chess game before or a

00:15:41

sketch of some kind, the data you need

00:15:43

means to checkmate there in three moves,

00:15:45

respectively, you also need to understand

00:15:47

these three moves you need to make what kind of

00:15:49

replacements to turn the more

00:15:52

complex integral into an integral that you can

00:15:54

calculate and in general this is what we will do with you,

00:15:56

that is, we will study

00:15:58

some schemes by which these

00:16:00

integrals can be simplified so well, it means that we are

00:16:04

essentially traveling from

00:16:06

right to left. that is, in this formula

00:16:09

we tried to see what is written

00:16:12

on the right and turn it into what is

00:16:14

written on the left, but more often, in fact, when

00:16:16

solving problems, the reverse process occurs,

00:16:19

that is, we are given such an integral, for

00:16:21

some reason we substitute here the function w

00:16:24

and It seems that we get an integral that looks

00:16:26

more complex, but in practice it turns out

00:16:28

that it is this integral that we can

00:16:30

calculate, let’s now look at an example like this, let

00:16:33

’s solve this one, in

00:16:35

any case, it will be useful to us 1697 is a

00:16:39

good function up to a known tangent, that’s

00:16:43

what it’s like to integrate, it turns

00:16:46

out also using the replacement formula

00:16:49

variable, that is, let us first

00:16:50

remember that the tangent is the ratio of

00:16:53

sine to cosine,

00:17:01

so let’s remember the relationship between

00:17:03

functions, that is, the derivative of cosine

00:17:06

is just the same and minus sine,

00:17:11

we see that we have here the factor

00:17:13

sine x to y dix and in addition is

00:17:18

present also the cosine function itself

00:17:20

is a clear hint that we can

00:17:22

see here an expression of this type,

00:17:24

and so we can say the following:

00:17:28

if y is equal to cosine x, then it

00:17:32

turns out that d y it will be equal to minus

00:17:36

sine x dx and we have a very similar

00:17:40

expression not all that is enough is a

00:17:42

minus sign, but if we put this minus sign,

00:17:44

then we will see that it is yes y

00:17:51

but y

00:17:54

means our rebus has been resolved and you and I

00:17:58

have obtained a simpler integral, which means it

00:18:00

turns out minus the logarithm modulus y +

00:18:05

c and now we remember what y was equal to

00:18:08

y it was equal to cosine, which means the answer is

00:18:10

this minus the logarithm modulo cosine

00:18:14

x plus c this turned out to be the integral

00:18:19

of the tangent, that is, unexpectedly yes, it seems like a

00:18:21

trigonometric function and in the answer it

00:18:23

turned out that its antiderivative is

00:18:26

logarithm, let's take this

00:18:29

problem 1766

00:18:32

1766

00:18:49

so it means in this problem we are asked to

00:18:51

calculate the integral

00:18:53

now I would like to figure out what

00:18:56

degree is written there, which means in my opinion there is a

00:18:59

second degree here x square you are

00:19:01

written the

00:19:02

cubic root of 1 minus x dx but

00:19:07

let's now calculate such an integral

00:19:09

means here we will travel in the

00:19:12

opposite direction, that is, we will substitute

00:19:16

here, instead of X, there is some expression so

00:19:18

that,

00:19:20

although perhaps let’s now

00:19:22

figure out what replacement we should make,

00:19:24

then there is a rule that

00:19:27

often helps if you don’t

00:19:30

like something in the integral, select it

00:19:32

as a new variable, for example, but

00:19:34

I don’t like the cube root of 1

00:19:36

minus x so let me say that

00:19:39

let it be that I have a new variable y,

00:19:42

that is, y is equal to the cube root of 1

00:19:45

minus x then I can say what

00:19:48

x is equal to, that is, on the contrary, it means that it

00:19:51

turns out that y cubed is equal to 1 minus x,

00:19:54

which means x is equal to 1 minus y cubed well, I

00:19:58

also need what is equal to dx let's

00:20:01

calculate dx what is equal to minus 3y squared

00:20:05

to y that's when I got these two

00:20:08

equalities

00:20:09

I can do a change of variable and

00:20:12

so x turns into 1 minus y uber

00:20:17

but I had x squared that means I also

00:20:20

shouldn’t forget to put

00:20:22

a square here, then I have this very

00:20:25

root because of which I chose this

00:20:27

change of variable, that is, I just have

00:20:28

a factor y and

00:20:31

finally dx and when I write dx I have to

00:20:35

replace it with minus 3y squared

00:20:38

before y here is

00:20:40

ours, I made a change of variables in

00:20:43

this situation, I had freedom of

00:20:45

choice, that is, in previous tasks I had

00:20:48

practically no freedom, it was necessary,

00:20:50

like in a rebus, to see what the author intended

00:20:52

in this problem and to implement his idea,

00:20:55

here I already have it

00:20:57

some complete freedom, I can make

00:21:00

any substitutions I want, but not all substitutions are

00:21:02

equally useful, so I made a

00:21:05

substitution so that this root disappeared from me

00:21:07

by choosing this root as a

00:21:09

new variable and my idea

00:21:11

was realized, that is, it turned out that in the

00:21:13

new integral there were no roots

00:21:16

left, of course the degrees have become quite

00:21:19

high, but this doesn’t scare us, so

00:21:22

let’s take minus 3 outside and under the

00:21:25

integral sign we will open the parentheses,

00:21:27

otherwise here is the square of the difference, that is,

00:21:30

one minus 2y cubed plus y 6 multiplied

00:21:34

all this by playing in cubes, and the

00:21:36

degrees are of course high but nevertheless,

00:21:41

this integral is easy to read, so I

00:21:43

open the brackets, now I integrate

00:21:47

each term separately -3

00:21:50

I am left with a factor, which means y 4

00:21:53

divided by 4 minus 2y in 7 divided by 7

00:21:57

plus y in 10 divided by 10 and in the end

00:22:02

we also write plus c, that is, in general, this is

00:22:05

already the answer, but in order for the problem to be

00:22:07

finally solved, we need to

00:22:09

substitute x instead of this game, we physically wo

00:22:13

n’t do it directly, we’ll just

00:22:15

draw these arrows, and of

00:22:18

course I think that you can handle this

00:22:20

task, instead of y, write this

00:22:22

very root, that is, to the answer is not very

00:22:24

beautiful, but it is exactly the one that

00:22:27

should be, that is, when in mathematics

00:22:29

you have to choose between beauty and

00:22:31

truth, a mathematician must choose the truth

00:22:34

problem 1776 means here we have an integral from

00:22:39

above dx and from below the root of 1 plus e to the

00:22:44

power of x but here in this

00:22:48

integral I just don’t like this

00:22:50

root of 1 plus e to the power of x so

00:22:52

let me say let this be a new

00:22:55

change on y equal to the root of 1 plus e to the

00:22:58

power of x then what will it be is equal to x, well,

00:23:02

we solve this equation

00:23:03

y square is equal to 1 + e to the power of x e to the

00:23:07

power of x is equal to y squared minus 1, which means

00:23:11

x is equal to the logarithm of y squared minus 1,

00:23:15

of course you should understand that in

00:23:19

fact this implies a

00:23:22

transformation of not only the integral but also

00:23:24

some intervals, that is, in

00:23:26

particular, this integral was

00:23:28

definitely everywhere for us, that is, x for us

00:23:31

could take any values, but at the same time,

00:23:34

when we made a change of variable,

00:23:35

of course, we get that y is

00:23:38

greater than one because there are no

00:23:42

other values like this the root

00:23:43

cannot take all such values

00:23:45

greater than one, so in fact,

00:23:48

this replacement that we make, we

00:23:49

mean that y now

00:23:51

changes in the interval from one to

00:23:54

plus infinity, and in this interval

00:23:55

this replacement will be correct for us

00:23:58

so that to make a replacement, it’s

00:24:01

not enough for us to get what x is equal to, we need to

00:24:04

know what dx is equal to, so let’s

00:24:06

calculate, so let’s take the derivative, it

00:24:09

turns out 2y, divide by the fact that under the sign

00:24:12

of the logarithm on y, the square minus 1 to y, that

00:24:15

is, this is how the differential is replaced,

00:24:18

which means, according to the formula for replacing variables, we

00:24:21

see that this is

00:24:23

2y d y divided by y square minus one

00:24:28

this is what was from the differential but

00:24:31

we also have y itself too,

00:24:34

you see this game and of course this

00:24:37

y can be shortened for us y here is not equal to

00:24:40

zero yes it is greater than one so you can

00:24:42

shorten ours, we got the

00:24:45

integral

00:24:47

that we calculated last time,

00:24:50

that is, a tabular integral, which means we

00:24:53

get that this is a high logarithm,

00:24:55

well, there was one second, it is multiplied by

00:24:58

2, the logarithm of y minus 1

00:25:01

divided by y + 1 remains plus c

00:25:05

again, in order for the problem to be solved,

00:25:07

you also need to add that y here is a

00:25:10

designation for such a root, that is, you do

00:25:14

n’t have to physically substitute it. I

00:25:16

won’t demand from you when solving a problem on

00:25:18

a test, but it’s important for me that when you

00:25:20

get to the answer, you then sign

00:25:24

as new variables are connected with the original

00:25:27

variables in this case, we made

00:25:29

only one replacement and it turns out that

00:25:32

the connection is like this, that is, we

00:25:34

solved this problem like this, so so far

00:25:36

we have discussed with you one method before the

00:25:38

method is if something don’t

00:25:41

like it, select this as a new

00:25:42

variable, but of course there are other

00:25:46

ways, so make substitutions,

00:25:49

let’s now understand how to get rid of the

00:25:52

simplest roots, that is, a root is

00:25:55

quite bad from a computational point of view,

00:25:58

so there are some ways to

00:26:01

get rid of roots, so let’s

00:26:04

look at an example

00:26:06

problems

00:26:08

1778

00:26:19

1778

00:26:22

in this problem we are asked to calculate the

00:26:24

integral from above dx and from below one minus x

00:26:30

square to the power of three second, so this

00:26:35

means that we essentially have here the

00:26:38

root one minus x square appears,

00:26:41

but it is then raised again, together we

00:26:44

don’t like this expression, but here we go in this

00:26:47

case, if you want to make a change

00:26:49

of variable, that is, say that well,

00:26:51

let’s designate the root as a new

00:26:53

variable as we did last time,

00:26:55

if you make a transformation to express

00:26:58

and what x is equal to through y, then you will get

00:27:01

that you did not get rid of the root,

00:27:07

it will turn out that x you have equal the root of 1

00:27:10

minus y squared and when you calculate

00:27:12

the differential the root may

00:27:15

remain, so this replacement

00:27:17

will not be very good, but good, what kind of

00:27:20

replacement will be good then

00:27:23

trigonometric functions come to our aid,

00:27:24

so we see that in general this

00:27:28

expression is defined if our x

00:27:30

is within the range between minus

00:27:32

one one, in this case they

00:27:34

cannot even be equal modulo units,

00:27:37

this is a strict inequality, and under such

00:27:39

conditions we can say that our x is

00:27:42

equal, for example, to the sine of

00:27:45

some variable t, that is with

00:27:49

such it's, we will always find that

00:27:52

its sine is equal to fakro sex and what

00:27:54

will be the advantage of such a replacement, that is,

00:27:56

why it is convenient to take such a replacement, the

00:27:58

fact is that one minus x squared

00:28:01

will be 1 minus, which means sine squared

00:28:04

you and this just let us have a square, you can

00:28:09

extract the root from this expression well, that is, this

00:28:12

formulation of x equals sine m1 allows

00:28:16

us to get rid of this root, that is, the

00:28:19

root of the cosine square, and generally

00:28:23

speaking this will be the modulus of the cash register but the flock,

00:28:26

but since we have the right to choose the

00:28:28

variable ourselves t then we can say that for

00:28:30

example we take t which changes from

00:28:33

minus and in half to pi in half

00:28:36

and of course our sine will

00:28:39

run through all the values from minus one

00:28:41

to one but let us have a

00:28:42

positive one so we can just

00:28:45

lower the module with you and so on So here’s

00:28:49

a hint, why is it convenient to

00:28:52

use a trigonometric function here? It

00:28:53

means that with such a

00:28:55

replacement, the

00:28:56

root of 1 minus x square

00:28:59

simply turns into another trigonometric

00:29:01

function, let’s see what this

00:29:03

turns into, that is, what will it

00:29:06

lead us to night if x is equal to sine t- then

00:29:09

yes x is equal to the cosine of t dt and let's

00:29:14

make a replacement, that is, we do it in the

00:29:16

hope that by getting rid of the root we

00:29:19

will get a more pleasant integral, that is,

00:29:22

this is just some idea, that is, we

00:29:24

hope for this that it will turn out this way, let's

00:29:26

see how this is implemented, which means

00:29:29

dx is our cosine tdt,

00:29:35

which means how you and I figured out the root of

00:29:38

1 minus x square is just cosine t and

00:29:41

here it was raised into a cube, which means it

00:29:44

will be cosine cubed d

00:29:47

well, and the result is an integral

00:29:49

tabular unit per cosine square and

00:29:52

this is tangent

00:29:55

tangent t + c

00:29:59

and the problem has already been practically solved, that

00:30:03

is, we only need, yes, we

00:30:06

only need to express the tangent t through

00:30:09

sine, well, of course, we could use the

00:30:11

arcsine function to write that we have

00:30:14

ted arcsine x it would be true, we would

00:30:17

substitute it here but the expression here is the

00:30:21

tangent taken from the arcsine,

00:30:25

it’s not very beautiful, so let’s

00:30:27

now simplify this expression and this is

00:30:29

the correct one, but let’s simplify, remember

00:30:32

the tangent is 100 divided on the belt,

00:30:35

but 100, but we already know that sine

00:30:37

is just x and cosine is the root from

00:30:42

1 minus x square means this

00:30:44

tangent from the arcsine x from is actually

00:30:46

just x divided by the root of 1 minus x

00:30:49

square let’s write another plus c, the

00:30:51

problem is solved, that is, we managed to

00:30:55

get rid of the root to

00:30:57

[music]

00:31:00

why was the module omitted here about this,

00:31:03

yes, look, it means that we had

00:31:06

such an integral, from the form of the integral itself,

00:31:09

we understand that we are talking not just about

00:31:11

some x’s, but specifically about the x’s between

00:31:14

minus one one, otherwise the

00:31:15

integrand does not make

00:31:17

sense,

00:31:18

so we’ll make the replacement x equals

00:31:21

sine ted we have the right to choose t ourselves,

00:31:25

if from what interval it will be, and so

00:31:27

I propose to consider you from the interval

00:31:29

from minus and in half to pi in half on this

00:31:32

interval sine you run through all the

00:31:34

values from minus one to one, but

00:31:37

at the same time 17 has more than zero for

00:31:41

so when we got with you that 1

00:31:44

minus x square stage sine square d

00:31:46

we extract the root we get the modulus cosine

00:31:49

t but we have already agreed with you

00:31:51

which one we take and it turns out that the modulus

00:31:54

cosine tf of this situation is just

00:31:56

cosine 3

00:32:00

to you can choose yourself, that is, the

00:32:03

advantages of substitutions are that

00:32:05

you choose, you are the masters of

00:32:08

your own happiness, so to speak, what replacement

00:32:11

you make is the result you will get, that

00:32:13

is, some art already appears here,

00:32:16

that is, you need to know which

00:32:18

replacement will allow you, for example, to get rid

00:32:21

of the root before or something else to do and here

00:32:23

we are just learning

00:32:25

here, the replacement x

00:32:28

equals cosine 3 would also be suitable, that is, if you see the

00:32:30

root of 1 minus x square, then this is an

00:32:35

indicator that the

00:32:37

replacement x equals sine can help you x equals x

00:32:41

sine t.r. because these replacements kill

00:32:44

puri and

00:32:45

this is how a drop of nicotine kills

00:32:48

a horse to the axis of the bridge and the cosine you kill is

00:32:52

such a root,

00:32:54

so let's immediately strengthen this

00:32:57

replacement a little, that is, we will discuss with you if

00:32:59

you see an expression like this root

00:33:02

of a square minus x square,

00:33:05

that is, some

00:33:06

positive parameter appears, and

00:33:08

then the replacement x is equal to and sine

00:33:12

is or x is equal to and cosine is suitable

00:33:17

because they kill this

00:33:20

root, which means and let’s get acquainted with another

00:33:23

similar replacement that allows you to

00:33:25

get rid of the root of the square

00:33:28

trinomial which can take on both

00:33:30

positive and negative

00:33:31

values,

00:33:32

let's figure this out using the example of problem 1784

00:33:38

1784 in this problem we are asked to calculate

00:33:42

an integral of this type from above simply

00:33:46

dx and from below the root of this

00:33:50

product x minus a multiplied by b

00:33:53

minus x

00:33:55

in this problem we are given a hint they

00:34:00

tell us, let’s make a substitution like

00:34:01

this x minus a equals b minus a

00:34:07

multiplied by sine squared you,

00:34:12

let’s try to analyze

00:34:14

why they suddenly give us such a

00:34:18

hint, that is, why such a substitution

00:34:20

will be good, we see that

00:34:24

we really have one factor x

00:34:26

minus a and it will be easy to substitute

00:34:28

minus a instead of x, here’s the expression, but we also

00:34:30

have another factor minus x, so that’s what

00:34:33

b minus x is, let’s understand b

00:34:36

minus x and that is, a plus b from which

00:34:39

we already let’s be more precise

00:34:43

we have b minus and b minus a

00:34:46

from which we subtract

00:34:49

x minus and that is, well, you can easily

00:34:51

check that minus a minus x minus and

00:34:54

this would be just minus x, but if you

00:34:56

make such a substitution, it turns out

00:35:00

that this factor

00:35:02

also has a very nice view

00:35:07

according to the basic trigonometric

00:35:09

identity, well, if only with a minus you can

00:35:11

take out 1 minus sine squared, this

00:35:13

cosine squared, and that is, this is b minus a

00:35:15

on the pass nose square t, that is, look, it

00:35:19

turns out that if we substitute

00:35:22

x instead of minus a, we will have b minus a

00:35:25

on sine squared you

00:35:28

mean b minus x is b minus a times

00:35:33

cosine squared x squared etc. you and

00:35:37

I believe that we have less than that, that

00:35:43

is, it turns out just where

00:35:47

the meaning of the expression that is written means

00:35:49

x minus and multiplied by b minus x,

00:35:52

we get a square trinomial

00:35:54

whose roots a and b

00:35:56

between these roots, both factors are

00:35:58

positive, then here is the sign + and when

00:36:02

passing through these points, the

00:36:04

sign of the square trinomial changes to negative,

00:36:06

that is, in fact, when we

00:36:08

talk about this integral, it is meant

00:36:10

that we are working on the interval from a to b, and

00:36:13

on this interval from a to b

00:36:16

we can - firstly, make this replacement before the

00:36:19

indicated replacement and secondly, this replacement

00:36:22

will turn out to be very productive because

00:36:24

we can extract the root from this

00:36:26

product, let’s, let’s

00:36:28

understand, but with the root what’s happening is already

00:36:30

clear, yes, the root will

00:36:32

simply turn into b minus a multiplied for sine

00:36:35

theta cosine t

00:36:37

that is, this is what

00:36:40

the root will turn into, that is, agree that the expression is

00:36:42

much more pleasant, more understandable,

00:36:45

but now let us understand why

00:36:49

we can really make such a replacement,

00:36:51

if we have you run

00:36:53

through the values, let us choose

00:36:55

which values we want t to

00:36:57

run through us, for example, it runs through the

00:36:59

value from 0 to

00:37:03

pi in half from 0 to pi in half then it

00:37:08

turns out that sine squared t will run through

00:37:10

all possible values from zero to

00:37:12

one

00:37:14

from zero to one

00:37:16

means multiplied by b minus a but

00:37:19

will run through all values from 0 to b minus a

00:37:22

and then x which is equal to a + b minus a on the

00:37:27

sine square t it will

00:37:29

run through all sorts of values from a to the

00:37:32

teeth, just like this we would have and it will be b

00:37:35

minus a + which means that this is really the

00:37:38

replacement that Demidovich offers us

00:37:40

very good, firstly, so to speak, we

00:37:43

can run through the entire interval we need

00:37:45

with this and replacement, and secondly, we

00:37:47

will get rid of the root, of course this is not an

00:37:50

obvious replacement, that is, the

00:37:53

instruction is given for a reason, here it

00:37:55

was quite difficult to guess it, that is, you

00:37:57

need to have it some experience in solving problems,

00:37:59

but now we are gaining exactly this experience

00:38:01

with you, that is, we, using the example of

00:38:03

our predecessors who

00:38:05

trained themselves, so to speak, got their

00:38:08

teeth into solving such problems, so they

00:38:11

came to the conclusion that if you

00:38:12

see a quadratic trinomial on this

00:38:14

type, you know, yes, a popular sign is

00:38:17

that you see the root of a square

00:38:19

trinomial, which means it is not

00:38:22

positive everywhere, so you can make a

00:38:24

replacement like this x minus and equal to

00:38:26

minus sine square, and so we

00:38:29

have seen the benefit of this replacement, let's

00:38:32

get down to implementation, it means how to

00:38:34

replace we understand the root and we also need to

00:38:37

calculate dx, that is, what is dx equal to in

00:38:40

this case dx will be equal to us, well, we

00:38:43

need to take the differential from this

00:38:45

function, which means we will have b minus a

00:38:49

sine square and if we differentiate there

00:38:51

will be 2 axes by 100 multiplied by cosine

00:38:55

then Well, naturally dt, so if we

00:38:59

make this replacement, here we have dx and replaces a

00:39:01

with b minus a with two sons flock cosine

00:39:06

tdt and

00:39:08

the root, as we saw, is replaced with b

00:39:11

minus a

00:39:12

sine because but the flock, that is, a very similar

00:39:15

expression and is reduced in essence, everything

00:39:18

remains the integral 2 dt,

00:39:21

well, it’s clear that it will be 2 t + c, but

00:39:25

instead of you, now you need to substitute

00:39:27

the expression, but we can’t

00:39:30

write anything simpler here, which means you, we

00:39:33

get this is the arcsine

00:39:36

means the arcsine

00:39:39

of the root

00:39:41

of x minus a divided by b minus a that

00:39:45

is, if we express from this formula

00:39:47

you in terms of x then we will get this

00:39:51

answer to know the answer in this problem is 2

00:39:54

arcsine of the

00:39:56

root x minus a divided by b minus a and

00:39:59

+ c

00:40:02

that is, and so about trigonometric

00:40:06

substitutions and how to get rid of roots,

00:40:07

remember the following means if

00:40:12

we factor a square trinomial of this type into factors, then the

00:40:14

indicated substitution like this with a square helps

00:40:16

if we have a root of this type

00:40:20

and it is written square minus there x is a square,

00:40:22

as I told you, x is equal to and

00:40:25

sine is you or x is equal to cosine you, that

00:40:30

is, I think that

00:40:32

we have dealt with getting rid of these roots,

00:40:34

let’s deal with other roots,

00:40:37

which means when the root is taken from a

00:40:39

square trinomial of a different type here

00:40:43

hyperbolic substitutions will help, these

00:40:46

hyperbolic substitutions just like

00:40:48

sine cosine helped us deal with

00:40:50

these, also now we will see

00:40:53

how you can do hyperbolic substitutions,

00:40:56

so let's analyze problem 1786,

00:41:00

so the problem is the

00:41:03

integral of the root of a square plus x

00:41:07

square dx

00:41:10

here we have that the same problem, that

00:41:13

is, we also want to get rid of the root and

00:41:15

we want to make such a replacement, that is,

00:41:18

substitute the state for X so that the

00:41:20

root can then be easily extracted from this expression

00:41:22

and so that the differential does

00:41:24

not look too complicated from this function, which means

00:41:27

here hyperbolic functions come to our aid,

00:41:28

let’s I

00:41:30

’ll remind you again, it means sine

00:41:33

hyperbolic you are e to the power of t

00:41:36

minus e to the power of minus 3 in half axis

00:41:39

brains pir painfully t this is e to the power of t

00:41:41

plus e to the power of minus t in half the

00:41:45

advantage of these functions is

00:41:47

that it is precisely the derivative of this function

00:41:49

is equal to this one and vice versa, but in addition

00:41:52

there are some relationships, we are

00:41:55

now interested in the main

00:41:56

hyperbolic identity, that is, if

00:41:58

you take the hyperbolic cosine

00:42:00

squared, subtract the hyperbolic sine squared from it,

00:42:02

then you

00:42:04

will get one, just like with the main

00:42:07

trigonometric identity, so there

00:42:09

was a sign plus here is a minus sign, so that

00:42:12

means we want to play on the fact that this

00:42:15

hyperbolic identity will work and

00:42:18

therefore I propose to make a replacement such that

00:42:20

x is equal to a hyperbolic sine

00:42:24

t-then what will a square

00:42:28

plus x square turn into? this will be a square plus a

00:42:32

square of a hyperbolic sine

00:42:35

squared t and we see with you that one

00:42:38

plus the sine of the hyperbolic squared

00:42:39

is the hyperbolic cosine squared

00:42:41

that is, this is the squared cosine of the

00:42:43

hyperbolic squared you mean the

00:42:46

root then is a square plus x square

00:42:50

will be equal to the root of a squared

00:42:55

cosine hyperbolic squared t well

00:42:58

and this is the hyperbolic cosine, in

00:43:01

this case it’s not even important to clarify

00:43:02

which ones we take the

00:43:04

hyperbolic cosine, it is always positive

00:43:07

with the differential of this function,

00:43:09

everything will be fine too, the differential x dx

00:43:11

will be equal to and the hyperbolic cosine

00:43:14

tdt means what our

00:43:17

integral root turns into and the

00:43:19

hyperbolic cosine t differential is also

00:43:22

and the hyperbolic cosine tdt

00:43:27

but well, that means our problem has already turned out to be

00:43:33

simpler and the square can be taken out the

00:43:36

cosine of the hyperbolic square

00:43:37

t-t-t-here remains, in principle, to

00:43:40

solve this problem you can simply

00:43:42

remember what the hyperbolic cosine is equal to,

00:43:44

that is write what it

00:43:46

is, but this 1 2 if taken out will be 1

00:43:49

4 e to the power of t plus e to the power of minus t

00:43:52

squared, well, of course, you can

00:43:55

open the brackets and then, in essence, it

00:43:58

will be a table integral, but

00:44:01

maybe we will also need some

00:44:02

relationships that there is for

00:44:04

hyperbolic functions, for example,

00:44:07

just like for the

00:44:09

cosine of the usual axis, we have

00:44:11

hyperbolic 2t and that is, two

00:44:15

hyperbolic cosine squared t minus 1,

00:44:18

you can easily check this relationship,

00:44:20

then it turns out that you can lower

00:44:23

the degree here, that is, write that this is the

00:44:26

hyperbolic cosine of 2 t plus one in

00:44:29

half and this integral will also

00:44:33

practically become tabular, so let’s

00:44:36

now understand what will happen, which

00:44:38

means we have a square in half

00:44:40

cosine hyperbolic 2 t plus one dt

00:44:44

we need to calculate this integral, which means we

00:44:47

leave the square in half,

00:44:49

so here we will have a sine

00:44:51

hyperbolic and 2t is the integral of

00:44:54

this, but you still need to divide by 2, remember,

00:44:57

last time we discussed with you about

00:45:00

linear substitution, it means there is one by and a

00:45:02

multiplier appeared, but in this

00:45:04

case it’s two, it’s not the one that

00:45:07

here means 1 2 and plus Well, the integral of

00:45:11

this money is those to from one it’s just

00:45:15

t and + c, here we get the answer, but in

00:45:21

order to solve the problem finally we

00:45:22

need to return to the previous variables,

00:45:26

that is, we then need to write what

00:45:28

t is equal to, but we can

00:45:30

use the inverse functions we

00:45:32

talked about yes what is an aria sine, that

00:45:35

is, it turns out that this is a logarithm, the long

00:45:39

logarithm to is taken from us,

00:45:43

which

00:45:45

means we need x divided by

00:45:50

plus the root of, which means x divided by

00:45:55

squared plus 1, in

00:45:57

my opinion, if I

00:45:59

remembered something correctly we have the inverse function, that

00:46:02

is, you can substitute it here, well, sine

00:46:05

2 t is this sine up to which it

00:46:07

occurs,

00:46:09

it is equal to 2 sine hyperbolic cosine

00:46:13

hyperbolic t well, you and I

00:46:16

also know what it is 2 multiplied by x

00:46:20

divided by

00:46:21

hyperbolic cosine you here we

00:46:24

have a formula, yes, this is one on the root

00:46:28

of a square plus x square, that is, we

00:46:32

know how to transform

00:46:34

new variables here is the function that

00:46:37

turned out into old ones, and that means we

00:46:39

will get the answer with you, that is, this is what

00:46:41

this root is equal to maybe before

00:46:43

maybe yes I can sometimes it’s here we

00:46:46

can yes maybe there’s a minus I

00:46:49

showed it last time but now I’ve already

00:46:50

forgotten where it’s plus you’re where it’s minus

00:46:52

3 of the sine minus we get it

00:46:57

but it’s good to know if it’s a sine, well that

00:46:59

means there’s a minus here too so

00:47:03

here well, that means we figured out

00:47:08

these replacements, but similarly, there is a

00:47:12

replacement for the quadratic trinomial of

00:47:14

x plus x plus b, but in general,

00:47:18

Demidovich tells us problem 1790

00:47:20

to make a similar replacement with the usual sine,

00:47:22

I think that you will already figure this out at home,

00:47:25

but the only one before the comment

00:47:28

which I will give this if you see a

00:47:30

root like this root and for the square plus x

00:47:32

square then the substitution a hyperbolic sine helps

00:47:34

and if you see a root

00:47:37

like this x square minus x square

00:47:40

then the substitution x equals and the

00:47:43

hyperbolic cosine will help you then that is

00:47:46

these are the simplest trigonometric

00:47:48

and hyperbolic substitutions that

00:47:49

allow you to get rid of such roots,

00:47:52

so well, I think that we’ll probably finish with this

00:47:56

block of problems and

00:47:59

let’s now move on to discussing the

00:48:02

2nd method of the main one, how

00:48:06

integrals are calculated, that is, substitutions of variables, we

00:48:08

got to know each other a little, but we still

00:48:11

we will return to it more than once, we

00:48:13

will discuss with you how different types of

00:48:15

functions are integrated, this is just the

00:48:18

beginning; such a basis is training, that is,

00:48:20

how to get rid of such roots, but

00:48:23

let’s now get acquainted with the second

00:48:25

method, transforming integrals as

00:48:27

the main one, which means the second main method, it

00:48:30

stems from another formula for

00:48:32

derivatives, here is a change of variables, this is the

00:48:35

derivative of a complex function,

00:48:36

and

00:48:37

now we will introduce you to

00:48:39

integration by parts

00:48:48

by parts,

00:48:52

this method is based on the

00:48:55

derivative products carried out, that is, if

00:48:57

we have two functions at&t and v.t. and

00:49:02

we want to calculate their derivative, then it

00:49:05

is calculated like this: y prime from t

00:49:07

multiplied by v2 + from t multiplied by you

00:49:12

prime from t

00:49:14

means what conclusion do we come to by looking

00:49:17

at this formula? Well, if suddenly you were

00:49:19

asked to calculate the integral of y prime from

00:49:22

t dt

00:49:23

plus a you are a prime from t dt then you know

00:49:29

the answer and that is, y from t multiplied by v2 + an

00:49:33

undefined constant,

00:49:36

but we also have the property of linearity

00:49:39

of the integral, that is, if we have an

00:49:41

integral not just from the sum, but

00:49:43

suppose it exists from one

00:49:45

-that term, that is, you

00:49:47

know that there is such an integral from

00:49:50

one term, then that means how it is

00:49:53

connected with the second term, that is, as

00:49:55

soon as ga ima wa itself exists from one term,

00:49:56

then immediately this second

00:49:58

term will also exist

00:49:59

because the integral from their sum is there

00:50:01

certainly is, here it is written, it means how

00:50:04

then they are related to each other, you can

00:50:07

break this integral then into two parts

00:50:09

1 and 2,

00:50:17

we know the sum, and

00:50:21

then we take one of the integrals and

00:50:23

move it to the other part and we get

00:50:26

the equality,

00:50:27

let me write it somewhere so that

00:50:30

it hangs here and remains here

00:50:32

on the board, it means the integral from u prime from t

00:50:36

in from t dt

00:50:38

it is equal to y from and in a t

00:50:44

minus the integral from and in prime from t dt

00:50:50

this is your formula for integration by

00:50:53

parts a question may arise,

00:50:56

but c we had water here, where did it

00:51:00

go, why am I not writing it here, and

00:51:02

it turns out that the integral is really what

00:51:04

you and I said, that this is the

00:51:06

totality of all antiderivatives, that

00:51:08

is, this c, it ran through all

00:51:09

possible values, so

00:51:11

the equality is written here two sets, now

00:51:14

if we were here we could not go down

00:51:16

if we closed c here we would write

00:51:18

that the sum of two sets is some

00:51:21

one element, just one representative,

00:51:22

this would be an incorrect equality, but

00:51:26

here when we moved the integral to

00:51:28

another part we get that an

00:51:30

indefinite constant is hidden

00:51:32

here and it is hidden here, and

00:51:35

indeed watts, the equality of such a

00:51:36

set is already

00:51:38

1791, which

00:51:40

means in this problem we are asked

00:51:43

to integrate the logarithm,

00:51:47

that is, they are asked to calculate what the antiderivative will be,

00:51:50

well, here are the antiderivatives

00:51:52

this function has,

00:51:54

how to approach this it is not

00:51:56

clear to the problem, that is, if it were not for the method

00:51:58

of integration by parts, then it is completely

00:52:00

unclear what would be done in this case,

00:52:02

but here we start, let's see

00:52:06

what the advantages of this method are, you see in

00:52:09

this formula y was a prime was a prime and you are

00:52:13

without a prime and after the transformation the stroke

00:52:16

migrates to in and that is, the essence of this

00:52:19

method is that we

00:52:21

transfer the derivative from one

00:52:22

factor to another, here we

00:52:25

don’t have 2 factors, but in fact, if

00:52:27

you look closely, you can see it, it’s

00:52:29

just a unit and

00:52:31

that means what’s the problem with the logarithm that

00:52:35

the logarithm it’s not included in the table

00:52:37

of integrals, we don’t know how to calculate the

00:52:39

integral from it, but if the logarithm

00:52:42

were differentiated then there

00:52:44

wouldn’t be any problems, so we say,

00:52:47

let’s say, we’ll have a unit, we’ll have

00:52:49

a prime, and we’ll have a logarithm then

00:52:53

after the transformation we will take this

00:52:56

stroke and throw Ivanov and the logarithm will

00:52:59

disappear, which means that to implement

00:53:02

our plan we just need to understand what

00:53:04

we can take as y, well, it’s clear y

00:53:06

from x can be taken x that is, the derivative

00:53:09

exo is just equal to one then the formula

00:53:12

integration by parts tells us

00:53:14

the following: this is x multiplied by

00:53:17

the logarithm of x, that is, this is multiplied by you,

00:53:20

then comes minus, then the integral now

00:53:24

we already have without a prime, that is, y is now

00:53:27

just x, but by the logarithm and now

00:53:30

we add a prime and this prime

00:53:32

turns the logarithm into one on x,

00:53:35

which means that there remains a much simpler

00:53:38

integral than it was, that is, there remains an

00:53:40

integral simply from one to x, which

00:53:43

we of course can calculate and

00:53:48

now we got the answer, so let’s

00:53:52

now have such a popular sign, that is,

00:53:55

when you need to use integration by

00:53:57

parts, but if you see an inverse

00:54:00

function of some kind, that is, it could

00:54:02

be a logarithm,

00:54:03

it could be an arc sine

00:54:08

arc cosine

00:54:10

there arc tangent area sines area tangent

00:54:14

that is, if you see inverse functions

00:54:16

to logarithmic well these

00:54:19

exponential functions and including

00:54:22

hyperbolic electrician functions on

00:54:24

metric ones then this is a clear hint that it is

00:54:26

advisable to use integration

00:54:28

by parts,

00:54:30

that is, if you take the derivative

00:54:32

of such a function, it will change its type,

00:54:35

that is, it will no longer be an inverse

00:54:37

logarithmic or an inverse trigan

00:54:39

such as an algorithmic one or a

00:54:40

trigonometric inverse

00:54:42

trigonometric another sign is

00:54:44

that that you have two

00:54:47

dissimilar factors, that is, for example,

00:54:50

you can multiply x by something,

00:54:52

for example, x can be multiplied by an exponent, or

00:54:55

x can be multiplied by a sine, or distinguish

00:54:59

x, not necessarily to the first power, but

00:55:01

maybe to the second power, it can be

00:55:03

multiplied by a cosine, for example, here

00:55:07

these are clear hints that you need

00:55:08

to integrate by parts, but

00:55:11

the task will also be assigned; this can also help when

00:55:14

calculating the integral, for example, from such

00:55:16

factors e to the power of x multiplied by

00:55:18

sine x, let's now look

00:55:21

at these examples and see how this

00:55:24

formula works like this for example, let's

00:55:27

take you on a problem like this 1796

00:55:41

means the integral

00:55:43

of x, in my opinion, there is e squared to the power of

00:55:48

minus 2 x dx, which means we see that we

00:55:52

really have the product of two

00:55:54

diverse factors, that is, there is a

00:55:57

power function, there is an exponential

00:55:59

function, which means a hint that we need to

00:56:02

use integration by parts and

00:56:04

now we need to understand which one to use,

00:56:06

that is, we want to get rid of one of

00:56:09

these factors

00:56:10

so that we don’t have completely

00:56:12

different factors, but we wo

00:56:14

n’t be able to get rid of the exponential, no matter how you

00:56:16

integrate it, it

00:56:18

remains undifferentiated, but x it is

00:56:22

quite possible to destroy a square, to do this it is

00:56:24

enough to differentiate it twice,

00:56:25

which means we will

00:56:28

transfer the derivative of the exponential to the

00:56:30

x square and so we say that we have a

00:56:34

prime from x and this is in our attacks,

00:56:38

but well, what function do we choose, well, of

00:56:43

course we need some

00:56:44

take the antiderivative, that is, e to the power of minus 2x, but

00:56:46

if you just take it, you won’t get

00:56:49

the derivative of e to the power of minus 2, which

00:56:51

means you still need to divide them in half and

00:56:53

take the minus sign, so

00:56:56

we take this function, which means then, according to the formula

00:56:59

integration by parts,

00:57:00

we must write the product y by you

00:57:03

minus x squared in half e to the power of

00:57:06

minus 2x this is multiplied by you then according to the

00:57:10

formula we must write the sign, but we

00:57:13

see that we also have a minus sign

00:57:15

here too, which means let me immediately replace

00:57:17

this with a plus sign, that is, minus from the

00:57:19

formula and minus from here, here I am learning,

00:57:21

taking into account, which means I’m taking from here one more

00:57:25

second you on the neck for the sign of the integral, and

00:57:28

inside I have what remains of attacks and I

00:57:31

need to multiply by v the stroke from x, that is,

00:57:34

I will also have y that’s what from what remains of it is that

00:57:37

which has not yet been taken out, this is

00:57:39

e to the power of minus 2x, but it still appears in the

00:57:42

prime, that is, 2 x dx,

00:57:45

well, here you can reduce the two, that is, like

00:57:49

this, at first glance it may

00:57:51

seem that we have not solved the

00:57:54

problem because - as before, we

00:57:56

still have this bad factor up

00:57:58

instead of x, the square has only now become x, but

00:58:01

we are one step closer to solving our

00:58:03

problem because if we repeat our

00:58:06

scheme again, then we will get rid of x;

00:58:09

finally, this means we again

00:58:11

declare that this is three photex and this is you

00:58:14

from x and again we apply the

00:58:16

integration formula by parts, which means that

00:58:19

in the end we will have minus x squared in

00:58:21

half e to the power of minus 2x as it was

00:58:24

and is, which means then we write a

00:58:26

multiplied by in that is, we have a new

00:58:29

minus but now we have x

00:58:32

in half

00:58:33

means e to the power of minus

00:58:37

2x then again, according to the formula, we must

00:58:40

write a minus sign to the formula and the minus sign

00:58:43

10 becomes a sign + sign + again I

00:58:47

take out one second from under the sign of the

00:58:50

integral and in the integral I am left with

00:58:53

what expression but here e to the power of minus

00:58:55

2x as it was and still remains

00:58:58

the derivative of X, that is, just

00:59:00

one means this is what happened if

00:59:04

we integrate again by parts, this is

00:59:06

how we did it, we have already obtained an integral

00:59:08

in which there is no X left, this is just a

00:59:11

table integral, that is, we can

00:59:14

write that this is e to the power of minus

00:59:17

2x divided by 4 with a minus sign, well,

00:59:20

1 2 came from here, not a single second, that’s

00:59:23

when we calculated this integral than so,

00:59:25

and that’s it, that’s the answer, that is, we end

00:59:29

up with it, well, we can

00:59:31

simplify it a little to take out that is,

00:59:33

minus x squared in half minus x in half

00:59:37

means it becomes that there is minus 1 4 e to the

00:59:41

power of minus 2 x plus c, this is the

00:59:45

answer we got, that is,

00:59:49

we needed to integrate this problem twice

00:59:50

by parts, let's also

00:59:53

analyze one

00:59:55

rather tricky problem before in in which the

01:00:00

method of integration by parts also

01:00:02

works, but in a more unusual way,

01:00:06

so the problem is like this

01:00:09

1828

01:00:12

1828

01:00:15

in this problem they ask you to calculate an integral of

01:00:19

this type e to the power of x

01:00:22

by the cosine bx

01:00:27

means dx

01:00:29

let's the main thing is that we require that a

01:00:32

and b do not equal zero, so if they k

01:00:35

are equal to zero, then this will be a

01:00:37

simpler integral and there is no need to calculate it like this,

01:00:38

as we are doing now, so starting with

01:00:41

such an integral,

01:00:44

we see that here is a heterogeneous function e

01:00:47

to the power of x and cosine bx, but unlike the

01:00:51

previous problem, if you

01:00:53

differentiate the cosine, then you get the

01:00:56

sine to differentiate and the sine

01:00:58

turns out to be a cosine and it seems like we are

01:01:00

reaching a dead end, that is, it seems like we can’t get rid of the x

01:01:02

square,

01:01:04

but another effect is observed,

01:01:06

let’s now do

01:01:09

the integration by parts, which means it

01:01:12

doesn’t matter in which direction to move or

01:01:14

what to declare and what to declare in let's

01:01:17

say so let it be y stroke and

01:01:20

it will be in attacks

01:01:22

what then will this

01:01:24

formula for integration by parts look like in

01:01:26

this case y from x this is what is this

01:01:28

unit divided by e to the power of x then

01:01:32

we get it like this one on e to the

01:01:35

power x cosine would be x minus well, now

01:01:40

we get the derivative we need to

01:01:42

transfer with this function anapu

01:01:44

sine bx derivative of cosine

01:01:49

is minus b si nos bx that is, we have

01:01:54

this minus, let's take it into account right away, that is,

01:01:57

put here plus plus would divide by

01:02:00

separation by it would be water

01:02:03

integral e to the power x sine bx dx that

01:02:10

is, nothing seems to

01:02:11

have changed much, let's repeat our

01:02:14

action, that is, we again declare that this

01:02:16

is y prime from x and this is you attacks

01:02:20

then we can repeat the transformation

01:02:28

this I’ll open the parentheses here, which means

01:02:31

we repeat the transformation of one by e to the

01:02:34

power of x, now sine bx will be

01:02:38

minus, but in the same way, yes, if

01:02:41

we differentiate and

01:02:43

another factor appears, b

01:02:45

division by we also have e to the

01:02:48

power of x, cosine would be x dx, well and it seems

01:02:54

like and it seems like, well, we’ve come to the same place where we

01:02:57

started, that is, we can decide that

01:02:59

this is a dead end for us, but let’s think about

01:03:02

what’s happening, that is, and so let’s

01:03:04

denote this integral by and let’s

01:03:08

denote this intagram by and that we

01:03:10

got and is equal to

01:03:13

one over e to the power of x cosine bx

01:03:19

plus b divided by the square of e to the power of x

01:03:24

sine b x

01:03:26

minus b square divided by the square and

01:03:32

that is, we got some

01:03:35

relations, that is, we can say that

01:03:37

some equation for a set of functions

01:03:40

for a set of these antiderivatives and

01:03:42

that means then we can take it and

01:03:47

move it to another part, that is,

01:03:50

what we get is if we take this

01:03:52

integral multiplied by b square

01:03:53

divided by square, move it to another

01:03:55

part, we will have 1 plus b square

01:03:58

divided by the square and

01:04:00

equal to what is written here let’s

01:04:04

use the repetition sign so as not to

01:04:06

rewrite, that is, we need to

01:04:08

repeat exactly what is written here, but an

01:04:10

undefined

01:04:12

constant c appears here, that is, let’s

01:04:15

try to explain what’s going on here, we have a

01:04:17

certain relation of the cat that

01:04:19

tells us that if we take some then the

01:04:21

antiderivative is from this class from

01:04:23

this representative, then it can be

01:04:26

represented in the following form, what is

01:04:28

written here plus there minus b square

01:04:31

divided by a square and some other

01:04:34

possible representative of this class of

01:04:36

indefinite integrals, so here there

01:04:38

was a relation to factors, we need a

01:04:40

set but when we moved it to another

01:04:43

part, we need to add the constant

01:04:45

c here because no one told us that

01:04:47

this equality holds for identical

01:04:49

representatives of this

01:04:51

set, that is, this one, and therefore the

01:04:54

constant c appears, and in the end we

01:04:56

get the answer to what then

01:04:59

the integral is equal and equal must be multiplied by the

01:05:02

square and divided by the square plus b

01:05:05

square and the left and right sides, then

01:05:08

just this factor will cancel,

01:05:10

which means that we will then end up with and

01:05:13

equal to a

01:05:15

divided by a square plus b square

01:05:18

let’s put it here yes, this is here here is

01:05:22

one by if you multiply by this, then it

01:05:25

will be a square simply divided by a

01:05:27

square b square cosine bx

01:05:31

plus here what would happen if they were multiplied by

01:05:34

divided by a square plus b square

01:05:39

sine bx

01:05:42

all this is multiplied by e to the power of x and c

01:05:46

is added here

01:05:48

is the formula, that is, in this problem we

01:05:52

managed to get to the answer, but at the same time it’s

01:05:55

cunning, that is, we essentially got such

01:05:58

an equation for the function we were looking for and

01:06:01

having solved this equation we found it

01:06:04

well, so look, we

01:06:08

must remember that integration by

01:06:10

parts and that is, a relationship for a

01:06:11

set of solutions, that is, here is

01:06:14

some kind of representative,

01:06:16

that is, some kind of antiderivative for this

01:06:19

function, it is equal to y from pvp minus

01:06:22

some kind of antiderivative from here from here on

01:06:26

them and this fact let us understand what

01:06:28

happened and so we had a set of

01:06:31

antiderivatives, so we are looking for a

01:06:34

set of antiderivatives, we got

01:06:36

the relation that this is a certain

01:06:38

function plus b divided by multiply by a

01:06:41

set of other antiderivatives already for

01:06:43

this function,

01:06:44

here we again integrated by

01:06:47

parts and as a result we got the following

01:06:49

relation that these are the antiderivatives

01:06:52

which we are looking for are these and they are related to the

01:06:55

same antiderivatives that we are looking for,

01:06:57

but how should we understand this equality that

01:07:01

some specific antiderivative of

01:07:04

this type is a specific function

01:07:07

from here minus b square divided by the

01:07:09

square but another antiderivative is possible,

01:07:12

that is, in fact The

01:07:14

following equality is written here: if you take

01:07:16

some antiderivative function, it is equal to what is

01:07:18

written here, asterisk,

01:07:21

let me not rewrite it, but

01:07:22

much more this function, asterisk minus

01:07:25

b square divided by square, but who

01:07:28

told you that exactly the same antiderivative is

01:07:29

needed take no one said this

01:07:31

means it is f + c, that is, some

01:07:34

indefinite constant, that is, we

01:07:36

received some relations of

01:07:37

this type and from it we understood what

01:07:40

the functions look like,

01:07:42

that is, it turns out that we need to remember that this is

01:07:44

this set and for

01:07:47

different representatives it can

01:07:48

this equality is fulfilled,

01:07:50

which means we got this

01:07:53

answer, that is, what is this integral equal to?

01:07:56

I’ll set you a similar problem at home,

01:07:58

but you need to calculate the sine bx so that you can

01:08:01

check for yourself how this circuit works, and it

01:08:04

turns out that this formula for

01:08:07

integration by parts is is given

01:08:09

in relation to some antiderivatives you

01:08:11

that some antiderivative from here and

01:08:13

the facts it is equal to a from t multiplied by p

01:08:17

minus some antiderivative from here but of

01:08:21

course this antiderivative can be

01:08:23

shifted by a constant, that is, you need to

01:08:25

understand this so that if the Vater is some kind - then the

01:08:27

antiderivative of what is written with a

01:08:31

fea t is some antiderivative of what is

01:08:33

written here on the right, then you may also

01:08:36

need some additional

01:08:38

term that will balance them, that is, the

01:08:41

left-right side, that is, this is how

01:08:43

this integration by parts formula works,

01:08:45

remember we have this constant c it

01:08:47

was when these integrals were on one

01:08:50

side, we had this constant c, it was

01:08:52

just hidden here in the notation, but

01:08:54

when we solved this problem, this constant

01:08:57

c left again, that is, it

01:08:59

appeared again, you know, as they say, God from the

01:09:03

machine means this among the ancient Greeks when

01:09:07

some situation was unsolvable,

01:09:09

Andromeda’s exit at the end of the scene,

01:09:11

God appeared who had already

01:09:14

said to everyone, you’re right, you’re wrong,

01:09:16

well done, just like that,

01:09:18

this constant c appears here unexpectedly, and as a

01:09:21

result, it means we got an answer with

01:09:24

this very constant c

01:09:26

yes please, well, let me move on to

01:09:31

the formulation of the homework now, that

01:09:33

is, on Wednesday we managed to do

01:09:36

quite a lot today, that is, we discussed two

01:09:37

main methods, that is,

01:09:39

what we will actually do, that is,

01:09:41

we will not have anything fundamentally new

01:09:43

in this topic anymore we will just

01:09:46

learn new substitutions with you and

01:09:49

also train before integrating by

01:09:52

parts, of course we will still have some

01:09:54

auxiliary ones in the process, but in general

01:09:56

this is the main thing that happens with integrals,

01:09:59

so that means in an hour I will write to you and the

01:10:02

problem numbers

01:10:05

for

01:10:10

so here let's use some roots and how

01:10:13

we got rid of some roots

01:10:16

we still had a guessing game before we had to

01:10:19

guess a replacement game like this means 1690 here

01:10:23

let's use this replacement problem

01:10:30

1695

01:10:40

means seventeen zero

01:10:44

how much is 03 let me give you

01:10:46

a hint go to the half corner then

01:10:49

there is sine x imagine as 2 sine x

01:10:53

in half cosine x in half this is

01:10:55

a hint that you then made the

01:10:59

correct replacement

01:11:02

so that is the replacement will need to be selected

01:11:06

one of these functions the hint is

01:11:10

try this at least more precisely more

01:11:15

precisely even not like that let’s

01:11:17

write it like this yes, let me tell you right away

01:11:19

then make a replacement like this d is

01:11:22

equal to tan x in half here is a new

01:11:25

function let it be like this

01:11:28

so good well here are 3 problems for this diagram I

01:11:31

think will be enough now let’s

01:11:36

do something else

01:11:41

so that

01:11:43

means substitutions where you need to choose a

01:11:46

new variable so that the integral in

01:11:50

where y beautifully means

01:11:52

let's do a problem like this, for example, I

01:11:56

think figure out what needs to be selected

01:11:59

as a function

01:12:01

so then on trigonometric ones, that is, for

01:12:05

example, like this up to

01:12:08

1780

01:12:11

for example 1785

01:12:15

times two three 4 5 6 but let's give a couple more

01:12:19

problems means 1790 and

01:12:23

that's how I I promised you that integration

01:12:25

by parts is given, we will work on it for you again,

01:12:28

let’s do integration by parts, this is

01:12:30

the problem

01:12:32

1829,

01:12:34

so I think that probably these eight

01:12:37

problems are not enough today, there is

01:12:40

some content there, next time we will continue to

01:12:42

discuss methods of integration with you

01:12:49

[music]

Description:

Неопределенный интеграл: замена переменной и интегрирование по частям 00:00:19 Повторение прошлого семинара 00:01:25 Замена переменной в неопределенном интеграле 00:06:04 Зачем мы пишем dx при интегрировании? 00:08:08 Задача №1674 00:12:46 Задача №1680 00:16:34 Задача №1697 00:18:47 Задача №1766 00:22:35 Задача №1776 00:26:21 Задача №1778: тригонометрические замены 00:32:02 Немного о замене корня в интеграле 00:33:20 Задача №1784: тригонометрические замены 00:40:58 Задача №1786: гиперболические замены 00:48:28 Интегрирование по частям 00:51:39 Задача №1791 00:53:52 Интегрирование по частям: замечание 00:55:28 Задача №1796 00:59:52 Задача №1828 01:10:00 Домашнее задание Ссылка на плейлист: https://www.youtube.com/playlist?list=PLcsjsqLLSfNBjTbeeiawjQlL7ok4ZWGdL

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