In practice, it is often necessary to calculate integrals of transcendental functions that contain trigonometric functions. As part of this material, we will describe the main types of integrand functions and show what methods can be used to integrate them.

Integrating sine, cosine, tangent and cotangent

Let's start with methods for integrating the basic trigonometric functions - sin, cos, t g, c t g. Using the table of antiderivatives, we immediately write that ∫ sin x d x = - cos x + C, and ∫ cos x d x = sin x + C.

To calculate the indefinite integrals of the functions t g and c t g, you can use the differential sign:

∫ t g x d x = ∫ sin x cos x d x = d (cos x) = - sin x d x = = - ∫ d (cos x) cos x = - ln cos x + C ∫ c t g x d x = ∫ cos x sin x d x = d (sin x) = cos x d x = = ∫ d (sin x) sin x = ln sin x + C

How did we get the formulas ∫ d x sin x = ln 1 - cos x sin x + C and ∫ d x cos x = ln 1 + sin x cos x + C, taken from the table of antiderivatives? Let us explain only one case, since the second will be clear by analogy.

Using the substitution method, we write:

∫ d x sin x = sin x = t ⇒ x = a r c sin y ⇒ d x = d t 1 - t 2 = d t t 1 - t 2

Here we need to integrate the irrational function. We use the same substitution method:

∫ d t t 1 - t 2 = 1 - t 2 = z 2 ⇒ t = 1 - z 2 ⇒ d t = - z d z 1 - z 2 = = ∫ - z d z z 1 - z 2 1 - z 2 = ∫ d z z 2 - 1 = ∫ d z (z - 1) (z +) = = 1 2 ∫ d z z - 1 - 1 2 ∫ d z z + 1 = 1 2 ln z - 1 - 1 2 z + 1 + C = = 1 2 ln z - 1 z + 1 + C = ln z - 1 z + 1 + C

Now we make the reverse substitution z = 1 - t 2 and t = sin x:

∫ d x sin x = ∫ d t t 1 - t 2 = ln z - 1 z + 1 + C = = ln 1 - t 2 - 1 1 - t 2 + 1 + C = ln 1 - sin 2 x - 1 1 - sin 2 x + 1 + C = = ln cos x - 1 cos x + 1 + C = ln (cos x - 1) 2 sin 2 x + C = = ln cos x - 1 sin x + C

We will separately analyze cases with integrals that contain powers of trigonometric functions, such as ∫ sin n x d x, ∫ cos n x d x, ∫ d x sin n x, ∫ d x cos n x.

You can read about how to calculate them correctly in the article on integration using recurrence formulas. If you know how these formulas are derived, you can easily take integrals like ∫ sin n x · cos m x d x with natural m and n.

If we have a combination of trigonometric functions with polynomials or exponential functions, then they will have to be integrated by parts. We recommend reading an article devoted to methods for finding integrals ∫ P n (x) · sin (a x) d x , ∫ P n (x) · cos (a x) d x , ∫ e a · x · sin (a x) d x , ∫ e a · x · cos (a x) d x .

The most difficult problems are those in which the integrand includes trigonometric functions with different arguments. To do this, you need to use basic trigonometry formulas, so it is advisable to memorize them or keep a note of them at hand.

Example 1

Find the set of antiderivatives of the function y = sin (4 x) + 2 cos 2 (2 x) sin x · cos (3 x) + 2 cos 2 x 2 - 1 · sin (3 x) .

Solution

Let's use the formulas for reducing the degree and write that cos 2 x 2 = 1 + cos x 2, and cos 2 2 x = 1 + cos 4 x 2. Means,

y = sin (4 x) + 2 cos 2 (2 x) sin x cos (3 x) + 2 cos 2 x 2 - 1 sin (3 x) = sin (4 x) + 2 1 + cos 4 x 2 sin x cos (3 x) + 2 1 + cos x 2 - 1 sin (3 x) = = sin (4 x) + cos (4 x) + 1 sin x cos (3 x) + cos x sin (3 x)

In the denominator we have the formula for the sine of the sum. Then you can write it like this:

y = sin (4 x) + cos (4 x) + 1 sin x cos (3 x) + cos x sin (3 x) = sin (4 x) + cos (4 x) + 1 sin (4 x ) = = 1 + cos (4 x) sin (4 x)

We got the sum of 3 integrals.

∫ sin (4 x) + cos (4 x) + 1 sin x · cos (3 x) + cos x · sin (3 x) d x = = ∫ d x + cos (4 x) d x sin (4 x) + ∫ d x sin (4 x) = = x + 1 4 ln ∫ d (sin (4 x)) sin (4 x) + 1 4 ln cos (4 x) - 1 sin (4 x) = = 1 4 ln sin ( 4 x) + 1 4 ln cos (4 x) - 1 sin (4 x) + C = x + 1 4 ln cos 4 x - 1 + C

In some cases, trigonometric functions under the integral can be reduced to fractional rational expressions using the standard substitution method. First, let's take formulas that express sin, cos and t g through the tangent of the half argument:

sin x = 2 t g x 2 1 + t g 2 x 2 , sin x = 1 - t g 2 x 2 1 + t g 2 x 2 , t g x = 2 t g x 2 1 - t g 2 x 2

We will also need to express the differential d x in terms of the tangent of the half angle:

Since d t g x 2 = t g x 2 "d x = d x 2 cos 2 x 2, then

d x = 2 cos 2 x 2 d t g x 2 = 2 d t g x 2 1 cos 2 x 2 = 2 d t g x 2 cos 2 x 2 + sin 2 x 2 cos 2 x 2 = 2 d t g x 2 1 + t g 2 x 2

Thus, sin x = 2 z 1 + z 2, cos x 1 - z 2 1 + z 2, t g x 2 z 1 - z 2, d x = 2 d z 1 + z 2 at z = t g x 2.

Example 2

Find the indefinite integral ∫ d x 2 sin x + cos x + 2 .

Solution

We use the method of standard trigonometric substitution.

2 sin x + cos x + 2 = 2 2 z 1 + z 2 + 1 - z 2 1 + z 2 = z 2 + 4 z + 3 1 + z 2 ⇒ d x 2 sin x + cos x + 2 = 2 d z 1 + z 2 z 2 + 4 z + 3 1 + z 2 = 2 d z z 2 + 4 z + 3

We obtain that ∫ d x 2 sin x + cos x + 2 = 2 d z z 2 + 4 z + 3 .

Now we can expand the integrand into simple fractions and obtain the sum of two integrals:

∫ d x 2 sin x + cos x + 2 = 2 ∫ 2 d z z 2 + 4 z + 3 = 2 ∫ 1 2 1 z + 1 - 1 z + 3 d z = = ∫ d z z + 1 - ∫ C z + 3 = ln z + 1 - ln z + 3 + C = ln z + 1 z + 3 + C

∫ d x 2 sin x + cos x + 2 = ln z + 1 z + 3 + C = ln t g x 2 + 1 t g x 2 + 3 + C

Answer: ∫ d x 2 sin x + cos x + 2 = ln t g x 2 + 1 t g x 2 + 3 + C

It is important to note that those formulas that express functions through the tangent of a half argument are not identities, therefore, the resulting expression ln t g x 2 + 1 t g x 2 + 3 + C is the set of antiderivatives of the function y = 1 2 sin x + cos x + 2 only on the domain of definition.

To solve other types of problems, you can use basic integration methods.

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Examples of solutions of integrals by parts are considered in detail, the integrand of which is the product of a polynomial by an exponential (e to the x power) or by a sine (sin x) or a cosine (cos x).

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

When solving examples in this section, the integration by parts formula is used:
;
.

Examples of integrals containing the product of a polynomial and sin x, cos x or e x

Here are examples of such integrals:
, , .

To integrate such integrals, the polynomial is denoted by u, and the remaining part by v dx. Next, apply the integration by parts formula.

Below is a detailed solution to these examples.

Examples of solving integrals

Example with exponent, e to the power of x

Determine the integral:
.

Let us introduce the exponent under the differential sign:
e - x dx = - e - x d(-x) = - d(e - x).

Let's integrate by parts.

Here
.
We also integrate the remaining integral by parts.
.
.
.
Finally we have:
.

An example of defining an integral with sine

Calculate the integral:
.

Let's introduce sine under the differential sign:

Let's integrate by parts.

here u = x 2 , v = cos(2 x+3), du = ( x 2 )′ dx

We also integrate the remaining integral by parts. To do this, introduce the cosine under the differential sign.


here u = x, v = sin(2 x+3), du = dx

Finally we have:

Example of the product of a polynomial and cosine

Calculate the integral:
.

Let's introduce the cosine under the differential sign:

Let's integrate by parts.

here u = x 2 + 3 x + 5, v = sin 2 x, du = ( x 2 + 3 x + 5 )′ dx

Integrals of trigonometric functions.
Examples of solutions

In this lesson we will look at integrals of trigonometric functions, that is, the filling of the integrals will be sines, cosines, tangents and cotangents in various combinations. All examples will be analyzed in detail, accessible and understandable even for a teapot.

To successfully study integrals of trigonometric functions, you must have a good understanding of the simplest integrals, as well as master some integration techniques. You can get acquainted with these materials in lectures Indefinite integral. Examples of solutions And .

And now we need: Table of integrals, Derivatives table And Directory of trigonometric formulas. All teaching aids can be found on the page Mathematical formulas and tables. I recommend printing everything out. I especially focus on trigonometric formulas, they should be in front of your eyes– without this, work efficiency will noticeably decrease.

But first, about what integrals are in this article No. There are no integrals of the form , - cosine, sine, multiplied by some polynomial (less often something with a tangent or cotangent). Such integrals are integrated by parts, and to learn the method, visit the lesson Integration by parts. Examples of solutions. Also here there are no integrals with “arches” - arctangent, arcsine, etc., they are also most often integrated by parts.

When finding integrals of trigonometric functions, a number of methods are used:

(4) We use the tabular formula , the only difference is that instead of “X” we have a complex expression.

Example 2

Example 3

Find the indefinite integral.

A classic of the genre for those who are drowning in the competition. As you probably noticed, in the table of integrals there is no integral of tangent and cotangent, but, nevertheless, such integrals can be found.

(1) We use the trigonometric formula

(2) We bring the function under the differential sign.

(3) We use the table integral .

Example 4

Find the indefinite integral.

This is an example for an independent solution, the full solution and answer are at the end of the lesson.

Example 5

Find the indefinite integral.

Our degrees will gradually increase =).
First the solution:

(1) We use the formula

(2) We use the main trigonometric identity , from which it follows that .

(3) Divide the numerator by the denominator term by term.

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

(5) We integrate using the table.

Example 6

Find the indefinite integral.

This is an example for an independent solution, the full solution and answer are at the end of the lesson.

There are also integrals of tangents and cotangents, which are in higher powers. The integral of the tangent cubed is discussed in the lesson How to calculate the area of ​​a flat figure? Integrals of tangent (cotangent) to the fourth and fifth powers can be obtained on the page Complex integrals.

Reducing the degree of the integrand

This technique works when the integrand functions are stuffed with sines and cosines in even degrees. To reduce the degree, use trigonometric formulas , and , and the last formula is often used in the opposite direction: .

Example 7

Find the indefinite integral.

Solution:

In principle, there is nothing new here, except that we applied the formula (lowering the degree of the integrand). Please note that I have shortened the solution. As you gain experience, the integral of can be found orally; this saves time and is quite acceptable when finishing assignments. In this case, it is advisable not to describe the rule , first we verbally take the integral of 1, then of .

Example 8

Find the indefinite integral.

This is an example for an independent solution, the full solution and answer are at the end of the lesson.

This is the promised degree increase:

Example 9

Find the indefinite integral.

First the solution, then the comments:

(1) Prepare the integrand to apply the formula .

(2) We actually apply the formula.

(3) We square the denominator and take the constant out of the integral sign. It could have been done a little differently, but, in my opinion, it was more convenient.

(4) We use the formula

(5) In the third term we again reduce the degree, but using the formula .

(6) We present similar terms (here I divided term by term and did the addition).

(7) Actually, we take the integral, the linearity rule and the method of subsuming a function under the differential sign is performed orally.

(8) Combing the answer.

! In an indefinite integral, the answer can often be written in several ways

In the example just considered, the final answer could have been written differently - opening the brackets and even doing this before integrating the expression, that is, the following ending to the example is quite acceptable:

It’s quite possible that this option is even more convenient, I just explained it the way I’m used to solving it myself). Here is another typical example for an independent solution:

Example 10

Find the indefinite integral.

This example can be solved in two ways, and you may succeed two completely different answers(more precisely, they will look completely different, but from a mathematical point of view they will be equivalent). Most likely, you will not see the most rational method and will suffer with opening brackets and using other trigonometric formulas. The most effective solution is given at the end of the lesson.

To summarize the paragraph, we conclude: any integral of the form , where and – even numbers, is solved by the method of reducing the degree of the integrand.
In practice, I came across integrals with 8 and 10 degrees, and I had to solve their terrible mess by lowering the degree several times, resulting in long, long answers.

Variable Replacement Method

As mentioned in the article Variable change method in indefinite integral, the main prerequisite for using the replacement method is the fact that in the integrand there is a certain function and its derivative:
(functions are not necessarily in the product)

Example 11

Find the indefinite integral.

We look at the table of derivatives and notice the formulas, , that is, in our integrand there is a function and its derivative. However, we see that during differentiation, cosine and sine mutually transform into each other, and the question arises: how to perform a change of variable and what do we mean by sine or cosine?! The question can be solved by scientific poking: if we perform the replacement incorrectly, then nothing good will come of it.

A general guideline: in similar cases, you need to designate the function that is in the denominator.

We interrupt the solution and make a replacement

Everything is fine in the denominator, everything depends only on , now it remains to find out what it will turn into.
To do this, we find the differential:

Or, in short:
From the resulting equality, using the rule of proportion, we express the expression we need:

So:

Now our entire integrand depends only on and we can continue solving

Ready. Let me remind you that the purpose of the replacement is to simplify the integrand; in this case, everything came down to integrating the power function according to the table.

It is no coincidence that I described this example in such detail; this was done for the purpose of repetition and reinforcement of the lesson materials Variable change method in indefinite integral.

And now two examples for your own solution:

Example 12

Find the indefinite integral.

Example 13

Find the indefinite integral.

Complete solutions and answers at the end of the lesson.

Example 14

Find the indefinite integral.

Here again, in the integrand, there are sine and cosine (a function with a derivative), but in a product, and a dilemma arises - what do we mean by sine or cosine?

You can try to carry out a replacement using the scientific method, and if nothing works, then designate it as another function, but there is:

General guideline: you need to designate the function that, figuratively speaking, is in an “uncomfortable position”.

We see that in this example, the student cosine “suffers” from the degree, and the sine sits freely, on its own.

Therefore, let's make a replacement:

If anyone still has difficulties with the algorithm for replacing a variable and finding the differential, then you should return to the lesson Variable change method in indefinite integral.

Example 15

Find the indefinite integral.

Let's analyze the integrand, what should be denoted by ?
Let's remember our guidelines:
1) The function is most likely in the denominator;
2) The function is in an “inconvenient position”.

By the way, these guidelines are valid not only for trigonometric functions.

The sine fits both criteria (especially the second), so a replacement suggests itself. In principle, the replacement can already be carried out, but first it would be nice to figure out what to do with? First, we “pinch off” one cosine:

We reserve for our “future” differential

And we express it through sine using the basic trigonometric identity:

Now here's the replacement:

General rule: If in the integrand one of the trigonometric functions (sine or cosine) is in odd degree, then you need to “bite off” one function from the odd degree, and designate another function behind it. We are talking only about integrals where there are cosines and sines.

In the example considered, we had a cosine at an odd power, so we plucked one cosine from the power, and designated it as a sine.

Example 16

Find the indefinite integral.

Degrees are taking off =).
This is an example for you to solve on your own. Full solution and answer at the end of the lesson.

Universal trigonometric substitution

Universal trigonometric substitution is a common case of the variable replacement method. You can try to use it when you “don’t know what to do.” But in fact there are some guidelines for its application. Typical integrals where the universal trigonometric substitution needs to be applied are the following integrals: , , , etc.

Example 17

Find the indefinite integral.

The universal trigonometric substitution in this case is implemented in the following way. Let's replace: . I don’t use the letter , but the letter , this is not some kind of rule, it’s just that, again, I’m used to solving things this way.

Here it is more convenient to find the differential; for this, from equality, I express:
I attach an arctangent to both parts:

Arctangent and tangent cancel each other out:

Thus:

In practice, you don’t have to describe it in such detail, but simply use the finished result:

! The expression is valid only if under the sines and cosines we simply have “X’s”, for the integral (which we'll talk about later) everything will be a little different!

When replacing, sines and cosines turn into the following fractions:
, , these equalities are based on well-known trigonometric formulas: ,

So, the final design could look like this:

Let's carry out a universal trigonometric substitution:

There will also be problems for you to solve on your own, to which you can see the answers.

The integrand can be converted from the product of trigonometric functions to the sum

Let us consider integrals in which the integrand is the product of sines and cosines of the first degree of x multiplied by different factors, that is, integrals of the form

Using well-known trigonometric formulas

(2)
(3)
(4)
one can transform each of the products in integrals of the form (31) into an algebraic sum and integrate according to the formulas

(5)

(6)

Example 1. Find

Solution. According to formula (2) at

Example 2. Find integral of a trigonometric function

Solution. According to formula (3) at

Example 3. Find integral of a trigonometric function

Solution. According to formula (4) at we obtain the following transformation of the integrand:

Applying formula (6), we obtain

Integral of the product of powers of sine and cosine of the same argument

Let us now consider integrals of functions that are the product of powers of sine and cosine of the same argument, i.e.

(7)

In special cases, one of the indicators ( m or n) may be zero.

When integrating such functions, it is used that an even power of cosine can be expressed through sine, and the differential of sine is equal to cos x dx(or even power of sine can be expressed in terms of cosine, and the differential of cosine is equal to - sin x dx ) .

Two cases should be distinguished: 1) at least one of the indicators m And n odd; 2) both indicators are even.

Let the first case take place, namely the indicator n = 2k+ 1 - odd. Then, given that

The integrand is presented in such a way that one part of it is a function of only the sine, and the other is the differential of the sine. Now using variable replacement t= sin x the solution reduces to integrating the polynomial with respect to t. If only the degree m is odd, then they do the same, isolating the factor sin x, expressing the rest of the integrand in terms of cos x and believing t=cos x. This technique can also be used when integrating the quotient powers of sine and cosine , When at least one of the indicators is odd . The whole point is that the quotient of the powers of sine and cosine is a special case of their product : When a trigonometric function is in the denominator of an integrand, its degree is negative. But there are also cases of partial trigonometric functions, when their powers are only even. About them - in the next paragraph.

If both indicators m And n– even, then, using trigonometric formulas

reduce the exponents of sine and cosine, after which an integral of the same type as above is obtained. Therefore, integration should be continued according to the same scheme. If one of the even exponents is negative, that is, the quotient of even powers of sine and cosine is considered, then this scheme is not suitable . Then a change of variable is used depending on how the integrand can be transformed. Such a case will be considered in the next paragraph.

Example 4. Find integral of a trigonometric function

Solution. The cosine exponent is odd. Therefore, let's imagine

t= sin x(Then dt=cos x dx ). Then we get

Returning to the old variable, we finally find

Example 5. Find integral of a trigonometric function

.

Solution. The cosine exponent, as in the previous example, is odd, but larger. Let's imagine

and make a change of variable t= sin x(Then dt=cos x dx ). Then we get

Let's open the brackets

and we get

Returning to the old variable, we get the solution

Example 6. Find integral of a trigonometric function

Solution. The exponents of sine and cosine are even. Therefore, we transform the integrand function as follows:

Then we get

In the second integral we make a change of variable, setting t= sin2 x. Then (1/2)dt= cos2 x dx . Hence,

Finally we get

Using the Variable Replacement Method

Variable Replacement Method when integrating trigonometric functions, it can be used in cases where the integrand contains only sine or only cosine, the product of sine and cosine, in which either sine or cosine is in the first degree, tangent or cotangent, as well as the quotient of even powers of sine and cosine of one and the same argument. In this case, it is possible to perform permutations not only sin x = t and sin x = t, but also tg x = t and ctg x = t .

Example 8. Find integral of a trigonometric function

.

Solution. Let's change the variable: , then . The resulting integrand can be easily integrated using the table of integrals:

.

Example 9. Find integral of a trigonometric function

Solution. Let's transform the tangent into the ratio of sine and cosine:

Let's change the variable: , then . The resulting integrand is table integral with a minus sign:

.

Returning to the original variable, we finally get:

.

Example 10. Find integral of a trigonometric function

Solution. Let's change the variable: , then .

Let's transform the integrand to apply the trigonometric identity :

We change the variable, not forgetting to put a minus sign in front of the integral (see above, what is equal to dt). Next, we factor the integrand and integrate according to the table:

Returning to the original variable, we finally get:

.

Find the integral of a trigonometric function yourself, and then look at the solution

Universal trigonometric substitution

Universal trigonometric substitution can be used in cases where the integrand does not fall under the cases discussed in the previous paragraphs. Basically, when sine or cosine (or both) is in the denominator of a fraction. It has been proven that sine and cosine can be replaced by another expression containing the tangent of half the original angle as follows:

But note that universal trigonometric substitution often entails quite complex algebraic transformations, so it is best used when no other method works. Let us look at examples where, together with the universal trigonometric substitution, substitution under the differential sign and the method of indefinite coefficients are used.

Example 12. Find integral of a trigonometric function

.

Solution. Solution. Let's take advantage universal trigonometric substitution. Then
.

We multiply the fractions in the numerator and denominator by , and take out the two and place it in front of the integral sign. Then

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 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; log(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