Integration by parts. Examples of solutions

Hello again. Today in the lesson we will learn how to integrate by parts. The method of integration by parts is one of the cornerstones of integral calculus. During tests or exams, students are almost always asked to solve the following types of integrals: the simplest integral (see article) or an integral by replacing a variable (see article) or the integral is just on integration by parts method.

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

What problem does the method of integration by parts solve? The method of integration by parts solves a very important problem; it allows you to integrate some functions that are not in the table, work functions, and in some cases – even quotients. As we remember, there is no convenient formula: . But there is this one: – formula for integration by parts in person. I know, I know, you’re the only one - we’ll work with her throughout the lesson (it’s easier now).

And immediately the list to the studio. The integrals of the following types are taken by parts:

1) , , – logarithm, logarithm multiplied by some polynomial.

2) ,is an exponential function multiplied by some polynomial. This also includes integrals like - an exponential function multiplied by a polynomial, but in practice this is 97 percent, under the integral there is a nice letter “e”. ... the article turns out to be somewhat lyrical, oh yes ... spring has come.

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

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

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

Integrals of logarithms

Example 1

Classic. From time to time this integral can be found in tables, but it is not advisable to use a ready-made answer, since the teacher has spring vitamin deficiency and will swear heavily. Because the integral under consideration is by no means tabular - it is taken in parts. We decide:

We interrupt the solution for intermediate explanations.

We use the integration by parts formula:

The formula is applied from left to right

We look at the left side: . Obviously, in our example (and in all the others that we will consider), something needs to be designated as , and something as .

In integrals of the type under consideration, the logarithm is always denoted.

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

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

Next stage: find the differential:

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

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

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

The only point in the work is that I immediately swapped and , since it is customary to write the factor before the logarithm.

As you can see, applying the integration by parts formula essentially reduced our solution to two simple integrals.

Please note that in some cases immediately after application of the formula, a simplification is necessarily carried out under the remaining integral - in the example under consideration, we reduced the integrand to “x”.

Let's check. To do this, you need to take the derivative of the answer:

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

During the test, we used the product differentiation rule: . And this is no coincidence.

Formula for integration by parts and formula – these are two mutually inverse rules.

Example 2

Find the indefinite integral.

The integrand is the product of a logarithm and a polynomial.
Let's decide.

I will once again describe in detail the procedure for applying the rule; in the future, examples will be presented more briefly, and if you have difficulties in solving it on your own, you need to go back to the first two examples of the lesson.

As already mentioned, it is necessary to denote the logarithm (the fact that it is a power does not matter). We denote by the rest integrand expression.

We write in the column:

First we find the differential:

Here we use the rule for differentiating a complex function . It is no coincidence that at the very first lesson of the topic Indefinite integral. Examples of solutions I focused on the fact that in order to master integrals, it is necessary to “get your hands on” derivatives. You will have to deal with derivatives more than once.

Now we find the function, for this we integrate right side lower equality:

For integration we used the simplest tabular formula

Now everything is ready to apply the formula . Open with an asterisk and “construct” the solution in accordance with the right side:

Under the integral we again have a polynomial for the logarithm! Therefore, the solution is again interrupted and the rule of integration by parts is applied a second time. Do not forget that in similar situations the logarithm is always denoted.

It would be good if by now you knew how to find the simplest integrals and derivatives orally.

(1) Don't get confused about the signs! Very often the minus is lost here, also note that the minus refers to to all bracket , and these brackets need to be expanded correctly.

(2) Open the brackets. We simplify the last integral.

(3) We take the last integral.

(4) “Combing” the answer.

The need to apply the rule of integration by parts twice (or even three times) does not arise very rarely.

And now a couple of examples for your own solution:

Example 3

Find the indefinite integral.

This example is solved by changing the variable (or substituting it under the differential sign)! Why not - you can try taking it in parts, it will turn out to be a funny thing.

Example 4

Find the indefinite integral.

But this integral is integrated by parts (the promised fraction).

These are examples for you to solve on your own, solutions and answers at the end of the lesson.

It seems that in examples 3 and 4 the integrands are similar, but the solution methods are different! This is the main difficulty in mastering integrals - if you choose the wrong method for solving an integral, then you can tinker with it for hours, like with a real puzzle. Therefore, the more you solve various integrals, the better, the easier the test and exam will be. In addition, in the second year there will be differential equations, and without experience in solving integrals and derivatives there is nothing to do there.

In terms of logarithms, this is probably more than enough. As an aside, I can also remember that engineering students use logarithms to call female breasts =). By the way, it is useful to know by heart the graphs of the main elementary functions: sine, cosine, arctangent, exponent, polynomials of the third, fourth degree, etc. No, of course, a condom on the globe
I won’t stretch it, but now you will remember a lot from the section Charts and functions =).

Integrals of an exponential multiplied by a polynomial

General rule:

Example 5

Find the indefinite integral.

Using a familiar algorithm, we integrate by parts:

If you have difficulties with the integral, then you should return to the article Variable change method in indefinite integral.

The only other thing you can do is tweak the answer:

But if your calculation technique is not very good, then the most profitable option is to leave it as an answer or even

That is, the example is considered solved when the last integral is taken. It won’t be a mistake; it’s another matter that the teacher may ask you to simplify the answer.

Example 6

Find the indefinite integral.

This is an example for you to solve on your own. This integral is integrated twice by parts. Particular attention should be paid to the signs - it’s easy to get confused in them, we also remember that this is a complex function.

There is nothing more to say about the exhibitor. I can only add that the exponential and the natural logarithm are mutually inverse functions, this is me on the topic of entertaining graphs of higher mathematics =) Stop, stop, don’t worry, the lecturer is sober.

Integrals of trigonometric functions multiplied by a polynomial

General rule: for always denotes a polynomial

Example 7

Find the indefinite integral.

Let's integrate by parts:

Hmmm...and there is nothing to comment on.

Example 8

Find the indefinite integral

This is an example for you to solve yourself

Example 9

Find the indefinite integral

Another example with a fraction. As in the two previous examples, for denotes a polynomial.

Let's integrate by parts:

If you have any difficulties or misunderstandings with finding the integral, I recommend attending the lesson Integrals of trigonometric functions.

Example 10

Find the indefinite integral

This is an example for you to solve on your own.

Hint: Before using the integration by parts method, you should apply some trigonometric formula that turns the product of two trigonometric functions into one function. The formula can also be used when applying the method of integration by parts, whichever is more convenient for you.

That's probably all in this paragraph. For some reason I remembered a line from the physics and mathematics hymn “And the sine graph runs wave after wave along the abscissa axis”….

Integrals of inverse trigonometric functions.
Integrals of inverse trigonometric functions multiplied by a polynomial

General rule: always denotes the inverse trigonometric function.

Let me remind you that the inverse trigonometric functions include arcsine, arccosine, arctangent and arccotangent. For the sake of brevity of the record I will call them "arches"

Examples of solutions of integrals by parts, the integrand of which contains the logarithm, arcsine, arctangent, as well as the logarithm to the integer power and the logarithm of the polynomial, are considered in detail.

Content

See also: Method of integration by parts
Table of indefinite integrals
Methods for calculating indefinite integrals
Basic elementary functions and their properties

Formula for integration by parts

Below, when solving examples, the integration by parts formula is used:
;
.

Examples of integrals containing logarithms and inverse trigonometric functions

Here are examples of integrals that are integrated by parts:
, , , , , , .

When integrating, that part of the integrand that contains the logarithm or inverse trigonometric functions is denoted by u, the rest by dv.

Below are examples with detailed solutions of these integrals.

Simple example with logarithm

Let's calculate the integral containing the product of a polynomial and a logarithm:

Here the integrand contains a logarithm. Making substitutions
u = ln x, dv = x 2 dx . Then
,
.

Let's integrate by parts.
.

.
Then
.
At the end of the calculations, add the constant C.

Example of a logarithm to the power of 2

Let's consider an example in which the integrand includes a logarithm to an integer power. Such integrals can also be integrated by parts.

Making substitutions
u = (ln x) 2, dv = x dx . Then
,
.

We also calculate the remaining integral by parts:
.
Let's substitute
.

An example in which the logarithm argument is a polynomial

Integrals can be calculated by parts, the integrand of which includes a logarithm whose argument is a polynomial, rational or irrational function. As an example, let's calculate an integral with a logarithm whose argument is a polynomial.
.

Making substitutions
u = ln( x 2 - 1), dv = x dx .
Then
,
.

We calculate the remaining integral:
.
We do not write the modulus sign here ln | x 2 - 1|, since the integrand is defined at x 2 - 1 > 0 . Let's substitute
.

Arcsine example

Let's consider an example of an integral whose integrand includes the arcsine.
.

Making substitutions
u = arcsin x,
.
Then
,
.

Next, we note that the integrand is defined for |x|< 1 . Let us expand the sign of the modulus under the logarithm, taking into account that 1 - x > 0 And 1 + x > 0.

Arc tangent example

Let's solve the example with arctangent:
.

Let's integrate by parts.
.
Let's select the whole part of the fraction:
x 8 = x 8 + x 6 - x 6 - x 4 + x 4 + x 2 - x 2 - 1 + 1 = (x 2 + 1)(x 6 - x 4 + x 2 - 1) + 1;
.
Let's integrate:
.
Finally we have.

Complex integrals

This article concludes the topic of indefinite integrals, and includes integrals that I find quite complex. The lesson was created at the repeated requests of visitors who expressed their wish that more difficult examples be analyzed on the site.

It is assumed that the reader of this text is well prepared and knows how to apply basic integration techniques. Dummies and people who are not very confident in integrals should refer to the very first lesson - Indefinite integral. Examples of solutions, where you can master the topic almost from scratch. More experienced students can become familiar with techniques and methods of integration that have not yet been encountered in my articles.

What integrals will be considered?

First we will consider integrals with roots, for the solution of which we successively use variable replacement And integration by parts. That is, in one example two techniques are combined at once. And even more.

Then we will get acquainted with interesting and original method of reducing the integral to itself. Quite a few integrals are solved this way.

The third issue of the program will be integrals from complex fractions, which flew past the cash desk in previous articles.

Fourthly, additional integrals from trigonometric functions will be analyzed. In particular, there are methods that avoid time-consuming universal trigonometric substitution.

(2) In the integrand function, we divide the numerator by the denominator term by term.

(3) We use the linearity property of the indefinite integral. In the last integral immediately put the function under the differential sign.

(4) We take the remaining integrals. Note that in a logarithm you can use parentheses rather than a modulus, since .

(5) We carry out a reverse replacement, expressing “te” from the direct replacement:

Masochistic students can differentiate the answer and get the original integrand, as I just did. No, no, I did the check in the right sense =)

As you can see, during the solution we had to use even more than two solution methods, so to deal with such integrals you need confident integration skills and quite a bit of experience.

In practice, of course, the square root is more common; here are three examples for solving it yourself:

Example 2

Find the indefinite integral

Example 3

Find the indefinite integral

Example 4

Find the indefinite integral

These examples are of the same type, so the complete solution at the end of the article will only be for Example 2; Examples 3-4 have the same answers. Which replacement to use at the beginning of decisions, I think, is obvious. Why did I choose examples of the same type? Often found in their role. More often, perhaps, just something like .

But not always, when under the arctangent, sine, cosine, exponential and other functions there is a root of a linear function, you have to use several methods at once. In a number of cases, it is possible to “get off easy,” that is, immediately after the replacement, a simple integral is obtained, which can be easily taken. The easiest of the tasks proposed above is Example 4, in which, after replacement, a relatively simple integral is obtained.

By reducing the integral to itself

A witty and beautiful method. Let's take a look at the classics of the genre:

Example 5

Find the indefinite integral

Under the root is a quadratic binomial, and trying to integrate this example can give the teapot a headache for hours. Such an integral is taken in parts and reduced to itself. In principle, it’s not difficult. If you know how.

Let us denote the integral under consideration by a Latin letter and begin the solution:

Let's integrate by parts:

(1) Prepare the integrand function for term-by-term division.

(2) We divide the integrand function term by term. It may not be clear to everyone, but I’ll describe it in more detail:

(3) We use the linearity property of the indefinite integral.

(4) Take the last integral (“long” logarithm).

Now let's look at the very beginning of the solution:

And to the end:

What happened? As a result of our manipulations, the integral was reduced to itself!

Let's equate the beginning and the end:

Move to the left side with a change of sign:

And we move the two to the right side. As a result:

The constant, strictly speaking, should have been added earlier, but I added it at the end. I strongly recommend reading what the rigor is here:

Note: More strictly, the final stage of the solution looks like this:

Thus:

The constant can be redesignated by . Why can it be redesignated? Because he still accepts it any values, and in this sense there is no difference between constants and.
As a result:

A similar trick with constant renotation is widely used in differential equations. And there I will be strict. And here I allow such freedom only in order not to confuse you with unnecessary things and to focus attention precisely on the integration method itself.

Example 6

Find the indefinite integral

Another typical integral for independent solution. Full solution and answer at the end of the lesson. There will be a difference with the answer in the previous example!

If under the square root there is a square trinomial, then the solution in any case comes down to two analyzed examples.

For example, consider the integral . All you need to do is first select a complete square:
.
Next, a linear replacement is carried out, which does “without any consequences”:
, resulting in the integral . Something familiar, right?

Or this example, with a quadratic binomial:
Select a complete square:
And, after linear replacement, we obtain the integral, which is also solved using the algorithm already discussed.

Let's look at two more typical examples of how to reduce an integral to itself:
– integral of the exponential multiplied by sine;
– integral of the exponential multiplied by the cosine.

In the listed integrals by parts you will have to integrate twice:

Example 7

Find the indefinite integral

The integrand is the exponential multiplied by the sine.

We integrate by parts twice and reduce the integral to itself:

As a result of double integration by parts, the integral was reduced to itself. We equate the beginning and end of the solution:

We move it to the left side with a change of sign and express our integral:

Ready. At the same time, it is advisable to comb the right side, i.e. take the exponent out of brackets, and in brackets place the sine and cosine in a “beautiful” order.

Now let's go back to the beginning of the example, or more precisely, to integration by parts:

We designated the exponent as. The question arises: is it the exponent that should always be denoted by ? Not necessarily. In fact, in the considered integral fundamentally doesn't matter, what do we mean by , we could have gone the other way:

Why is this possible? Because the exponential turns into itself (both during differentiation and integration), sine and cosine mutually turn into each other (again, both during differentiation and integration).

That is, we can also denote a trigonometric function. But, in the example considered, this is less rational, since fractions will appear. If you wish, you can try to solve this example using the second method; the answers must match.

Example 8

Find the indefinite integral

This is an example for you to solve on your own. Before you decide, think about what is more profitable in this case to denote by , an exponential function or a trigonometric function? Full solution and answer at the end of the lesson.

And, of course, do not forget that most of the answers in this lesson are quite easy to check by differentiation!

The examples considered were not the most complex. In practice, integrals are more common where the constant is both in the exponent and in the argument of the trigonometric function, for example: . Many people will get confused in such an integral, and I often get confused myself. The fact is that there is a high probability of fractions appearing in the solution, and it is very easy to lose something through carelessness. In addition, there is a high probability of an error in the signs; note that the exponent has a minus sign, and this introduces additional difficulty.

At the final stage, the result is often something like this:

Even at the end of the solution, you should be extremely careful and correctly understand the fractions:

Integrating Complex Fractions

We are slowly approaching the equator of the lesson and begin to consider integrals of fractions. Again, not all of them are super complex, it’s just that for one reason or another the examples were a little “off topic” in other articles.

Continuing the theme of roots

Example 9

Find the indefinite integral

In the denominator under the root there is a quadratic trinomial plus an “appendage” in the form of an “X” outside the root. An integral of this type can be solved using a standard substitution.

We decide:

The replacement here is simple:

Let's look at life after replacement:

(1) After substitution, we reduce the terms under the root to a common denominator.
(2) We take it out from under the root.
(3) The numerator and denominator are reduced by . At the same time, under the root, I rearranged the terms in a convenient order. With some experience, steps (1), (2) can be skipped by performing the commented actions orally.
(4) The resulting integral, as you remember from the lesson Integrating Some Fractions, is being decided complete square extraction method. Select a complete square.
(5) By integration we obtain an ordinary “long” logarithm.
(6) We carry out the reverse replacement. If initially , then back: .
(7) The final action is aimed at straightening the result: under the root we again bring the terms to a common denominator and take them out from under the root.

Example 10

Find the indefinite integral

This is an example for you to solve on your own. Here a constant is added to the lone “X”, and the replacement is almost the same:

The only thing you need to do additionally is to express the “x” from the replacement being carried out:

Full solution and answer at the end of the lesson.

Sometimes in such an integral there may be a quadratic binomial under the root, this does not change the method of solution, it will be even simpler. Feel the difference:

Example 11

Find the indefinite integral

Example 12

Find the indefinite integral

Brief solutions and answers at the end of the lesson. It should be noted that Example 11 is exactly binomial integral, the solution method of which was discussed in class Integrals of irrational functions.

Integral of an indecomposable polynomial of the 2nd degree to the power

(polynomial in denominator)

A more rare type of integral, but nevertheless encountered in practical examples.

Example 13

Find the indefinite integral

But let’s return to the example with lucky number 13 (honestly, I didn’t guess correctly). This integral is also one of those that can be quite frustrating if you don’t know how to solve.

The solution starts with an artificial transformation:

I think everyone already understands how to divide the numerator by the denominator term by term.

The resulting integral is taken in parts:

For an integral of the form ( – natural number) we derive recurrent reduction formula:
, Where – integral of a degree lower.

Let us verify the validity of this formula for the solved integral.
In this case: , , we use the formula:

As you can see, the answers are the same.

Example 14

Find the indefinite integral

This is an example for you to solve on your own. The sample solution uses the above formula twice in succession.

If under the degree is indivisible square trinomial, then the solution is reduced to a binomial by isolating the perfect square, for example:

What if there is an additional polynomial in the numerator? In this case, the method of indefinite coefficients is used, and the integrand is expanded into a sum of fractions. But in my practice there is such an example never met, so I missed this case in the article Integrals of fractional-rational functions, I'll skip it now. If you still encounter such an integral, look at the textbook - everything is simple there. I don’t think it’s advisable to include material (even simple ones), the probability of encountering which tends to zero.

Integrating complex trigonometric functions

The adjective “complex” for most examples is again largely conditional. Let's start with tangents and cotangents in high powers. From the point of view of the solving methods used, tangent and cotangent are almost the same thing, so I will talk more about tangent, implying that the demonstrated method for solving the integral is valid for cotangent too.

In the above lesson we looked at universal trigonometric substitution for solving a certain type of integrals of trigonometric functions. The disadvantage of universal trigonometric substitution is that its use often results in cumbersome integrals with difficult calculations. And in some cases, universal trigonometric substitution can be avoided!

Let's consider another canonical example, the integral of one divided by sine:

Example 17

Find the indefinite integral

Here you can use universal trigonometric substitution and get the answer, but there is a more rational way. I will provide the complete solution with comments for each step:

(1) We use the trigonometric formula for the sine of a double angle.
(2) We carry out an artificial transformation: Divide in the denominator and multiply by .
(3) Using the well-known formula in the denominator, we transform the fraction into a tangent.
(4) We bring the function under the differential sign.
(5) Take the integral.

A couple of simple examples for you to solve on your own:

Example 18

Find the indefinite integral

Note: The very first step should be to use the reduction formula and carefully carry out actions similar to the previous example.

Example 19

Find the indefinite integral

Well, this is a very simple example.

Complete solutions and answers at the end of the lesson.

I think now no one will have problems with integrals:
etc.

What is the idea of ​​the method? The idea is to use transformations and trigonometric formulas to organize only tangents and the tangent derivative into the integrand. That is, we are talking about replacing: . In Examples 17-19 we actually used this replacement, but the integrals were so simple that we got by with an equivalent action - subsuming the function under the differential sign.

Similar reasoning, as I already mentioned, can be carried out for the cotangent.

There is also a formal prerequisite for applying the above replacement:

The sum of the powers of cosine and sine is a negative integer EVEN number, For example:

for the integral – a negative integer EVEN number.

! Note : if the integrand contains ONLY a sine or ONLY a cosine, then the integral is also taken for a negative odd degree (the simplest cases are in Examples No. 17, 18).

Let's look at a couple of more meaningful tasks based on this rule:

Example 20

Find the indefinite integral

The sum of the powers of sine and cosine: 2 – 6 = –4 is a negative integer EVEN number, which means that the integral can be reduced to tangents and its derivative:

(1) Let's transform the denominator.
(2) Using the well-known formula, we obtain .
(3) Let's transform the denominator.
(4) We use the formula .
(5) We bring the function under the differential sign.
(6) We carry out replacement. More experienced students may not carry out the replacement, but it is still better to replace the tangent with one letter - there is less risk of getting confused.

Example 21

Find the indefinite integral

This is an example for you to solve on your own.

Hang in there, the championship rounds are about to begin =)

Often the integrand contains a “hodgepodge”:

Example 22

Find the indefinite integral

This integral initially contains a tangent, which immediately leads to an already familiar thought:

I will leave the artificial transformation at the very beginning and the remaining steps without comment, since everything has already been discussed above.

A couple of creative examples for your own solution:

Example 23

Find the indefinite integral

Example 24

Find the indefinite integral

Yes, in them, of course, you can lower the powers of sine and cosine, and use the universal trigonometric substitution, but the solution will be much more efficient and shorter if it is carried out through tangents. Full solution and answers at the end of the lesson

Table of antiderivatives ("integrals"). Table of integrals. Tabular indefinite integrals. (The simplest integrals and integrals with a parameter). Formulas for integration by parts. Newton-Leibniz formula.

Table of antiderivatives ("integrals"). Tabular indefinite integrals. (The simplest integrals and integrals with a parameter).
Integral of a power function.	Integral of a power function.
An integral that reduces to the integral of a power function if x is driven under the differential sign.
	Integral of an exponential, where a is a constant number.
Integral of a complex exponential function.	Integral of an exponential function.
An integral equal to the natural logarithm.	Integral: "Long logarithm".
	Integral: "Long logarithm".
Integral: "High logarithm".	An integral, where x in the numerator is placed under the differential sign (the constant under the sign can be either added or subtracted), is ultimately similar to an integral equal to the natural logarithm.
Integral: "High logarithm".
Cosine integral.	Sine integral.
Integral equal to tangent.	Integral equal to cotangent.
Integral equal to both arcsine and arccosine
An integral equal to both arcsine and arccosine.	An integral equal to both arctangent and arccotangent.
Integral equal to cosecant.	Integral equal to secant.
Integral equal to arcsecant.	Integral equal to arccosecant.
Integral equal to arcsecant.	Integral equal to arcsecant.
Integral equal to the hyperbolic sine.	Integral equal to hyperbolic cosine.
Integral equal to the hyperbolic sine, where sinhx is the hyperbolic sine in the English version.	Integral equal to the hyperbolic cosine, where sinhx is the hyperbolic sine in the English version.
Integral equal to the hyperbolic tangent.	Integral equal to the hyperbolic cotangent.
Integral equal to the hyperbolic secant.	Integral equal to the hyperbolic cosecant.

Formulas for integration by parts. Integration rules.

Formulas for integration by parts. Newton-Leibniz formula. Rules of integration.
Integrating a product (function) by a constant:
Integrating the sum of functions:
indefinite integrals:
Formula for integration by parts definite integrals:
Newton-Leibniz formula definite integrals:	Where F(a),F(b) are the values ​​of the antiderivatives at points b and a, respectively.

Table of derivatives. Tabular derivatives. Derivative of the product. Derivative of the quotient. Derivative of a complex function.

If x is an independent variable, then:

Table of derivatives. Tabular derivatives."table derivative" - ​​yes, unfortunately, this is exactly how they are searched for on the Internet
	Derivative of a power function
	Derivative of the exponent
Derivative of a complex exponential function	Derivative of exponential function
Derivative of a logarithmic function	Derivative of the natural logarithm
	Derivative of the natural logarithm of a function
Derivative of sine	Derivative of cosine
Derivative of cosecant	Derivative of a secant
Derivative of arcsine	Derivative of arc cosine
Derivative of arcsine	Derivative of arc cosine
Tangent derivative	Derivative of cotangent
Derivative of arctangent	Derivative of arc cotangent
Derivative of arctangent	Derivative of arc cotangent
Derivative of arcsecant	Derivative of arccosecant
Derivative of arcsecant	Derivative of arccosecant
Derivative of the hyperbolic sine Derivative of the hyperbolic sine in the English version	Derivative of hyperbolic cosine Derivative of hyperbolic cosine in English version
Derivative of hyperbolic tangent	Derivative of hyperbolic cotangent
Derivative of the hyperbolic secant	Derivative of the hyperbolic cosecant

Rules of differentiation. Derivative of the product. Derivative of the quotient. Derivative of a complex function.
Derivative of a product (function) by a constant:
Derivative of sum (functions):
Derivative of the product (functions):
Derivative of the quotient (of functions):
Derivative of a complex function:

Properties of logarithms. Basic formulas for logarithms. Decimal (lg) and natural logarithms (ln).

Basic logarithmic identity
Let's show how any function of the form a b can be made exponential. Since a function of the form e x is called exponential, then
Any function of the form a b can be represented as a power of ten

Natural logarithm ln (logarithm to base e = 2.718281828459045...) ln(e)=1; ln(1)=0

Taylor series. Taylor series expansion of a function.

It turns out that the majority practically encountered mathematical functions can be represented with any accuracy in the vicinity of a certain point in the form of power series containing powers of a variable in increasing order. For example, in the vicinity of the point x=1:

When using series called Taylor's rows mixed functions containing, say, algebraic, trigonometric and exponential functions can be expressed as purely algebraic functions. Using series, you can often quickly perform differentiation and integration.

The Taylor series in the neighborhood of point a has the form:

1) , where f(x) is a function that has derivatives of all orders at x=a. R n - the remainder term in the Taylor series is determined by the expression

2)

The k-th coefficient (at x k) of the series is determined by the formula

3) A special case of the Taylor series is the Maclaurin (=McLaren) series (the expansion occurs around the point a=0)

at a=0

members of the series are determined by the formula

Conditions for using Taylor series.

1. In order for the function f(x) to be expanded into a Taylor series on the interval (-R;R), it is necessary and sufficient that the remainder term in the Taylor (Maclaurin (=McLaren)) formula for this function tends to zero as k →∞ on the specified interval (-R;R).

2. It is necessary that there are derivatives for a given function at the point in the vicinity of which we are going to construct the Taylor series.

Properties of Taylor series.

If f is an analytic function, then its Taylor series at any point a in the domain of definition of f converges to f in some neighborhood of a.

There are infinitely differentiable functions whose Taylor series converges, but at the same time differs from the function in any neighborhood of a. For example:

Taylor series are used in approximation (approximation is a scientific method that consists of replacing some objects with others, in one sense or another close to the original ones, but simpler) of a function by polynomials. In particular, linearization ((from linearis - linear), one of the methods of approximate representation of closed nonlinear systems, in which the study of a nonlinear system is replaced by the analysis of a linear system, in some sense equivalent to the original one.) equations occurs by expanding into a Taylor series and cutting off all terms above first order.

Thus, almost any function can be represented as a polynomial with a given accuracy.

Examples of some common expansions of power functions in Maclaurin series (=McLaren, Taylor in the vicinity of point 0) and Taylor in the vicinity of point 1. The first terms of expansions of the main functions in Taylor and McLaren series.

Examples of some common expansions of power functions in Maclaurin series (=McLaren, Taylor in the vicinity of point 0)

Examples of some common Taylor series expansions in the vicinity of point 1

Antiderivative and integral

1. Antiderivative. The function F(x) is called antiderivative for the function f (x) on the interval X if for any x from X the equality F"(x)=f(x) holds.

T.7.13 (If F(x) is an antiderivative for a function f(x) on the interval X, then the function f(x) has infinitely many antiderivatives, and all these antiderivatives have the form F (x) + C, where C is an arbitrary constant (the main property of the antiderivative).

2. Table of antiderivatives. Considering that finding an antiderivative is the inverse operation of differentiation, and starting from the table of derivatives, we obtain the following table of antiderivatives (for simplicity, the table shows one antiderivative F(x), and not the general form of antiderivatives F(x) + C:

	Antiderivative		Antiderivative

Antiderivative and logarithmic function

Logarithmic function, the inverse of the exponential function. L. f. denoted by

its value y, corresponding to the value of the argument x, is called the natural logarithm of the number x. By definition, relation (1) is equivalent

(e is a Neper number). Since ey > 0 for any real y, then the L.f. is defined only for x > 0. In a more general sense, the L. f. call the function

antiderivative power integral logarithm

where a > 0 (a? 1) is an arbitrary base of logarithms. However, in mathematical analysis the InX function is of particular importance; the logaX function is reduced to it using the formula:

where M = 1/In a. L. f. - one of the main elementary functions; its graph (Fig. 1) is called logarithmics. Basic properties of L. f. follow from the corresponding properties of the exponential function and logarithms; for example, L. f. satisfies the functional equation

For - 1< х, 1 справедливо разложение Л. ф. в степенной ряд:

Many integrals are expressed in terms of linear functions; For example

L. f. occurs constantly in mathematical analysis and its applications.

L. f. was well known to mathematicians of the 17th century. For the first time, the dependence between variable quantities, expressed by L. f., was considered by J. Napier (1614). He represented the relationship between numbers and their logarithms using two points moving along parallel lines (Fig. 2). One of them (Y) moves uniformly, starting from C, and the other (X), starting from A, moves with a speed proportional to its distance to B. If we put SU = y, XB = x, then, according to this definition,

dx/dy = - kx, from where.

L. f. on the complex plane is a multi-valued (infinite-valued) function defined for all values ​​of the argument z? 0 is denoted by Lnz. The single-valued branch of this function, defined as

Inz = In?z?+ i arg z,

where arg z is the argument of the complex number z, which is called the main value of the linear function. We have

Lnz = lnz + 2kpi, k = 0, ±1, ±2, ...

All meanings of L. f. for negative: real z are complex numbers. The first satisfactory theory of L. f. in the complex plane was given by L. Euler (1749), who proceeded from the definition