Methods for calculating uncertain integrals. Integrals for dummies: How to solve, calculation rules, explanation

Is it possible to put a nonlinear function under the differential sign? Yes, if a replacement expression is a product of two multipliers: one factor is a complex function from some nonlinear function, and another factor is derived from this nonlinear function. Consider what is said on the examples.

Find uncertain integrals.

Example 1.. ∫ (2x + 1) (x 2 + x + 2) 5 dx \u003d ∫ (x 2 + x + 2) 5 d (x 2 + x + 2) \u003d (x² + x + 2) 6 : 6 + C.

What is this integrand? The product of the power function from (x 2 + x + 2) and the multiplier (2x + 1), which is equal to the degree derivative: (x 2 + x + 2) "\u003d 2x + 1.

This allowed us to bring (2x + 1) under the sign of the differential:

∫u 5 du \u003d u 6 : 6+ C. (Formula 1). )

Check. (F (x) + c) "\u003d ((x² + x + 2) 6 : 6 + c) '\u003d 1/6 · 6 (x 2 + x + 2) 5 · (x 2 + x + 2) "\u003d

\u003d (x 2 + x + 2) 5 · (2x + 1) \u003d (2x + 1) (x 2 + x + 2) 5 \u003d f (x).

Example 2. ∫ (3x 2 - 2x + 3) (x 3 - x 2 + 3x + 1) 5 dx \u003d ∫ (x 3 - x 2 + 3x + 1) 5 d (x 3 - x 2 + 3x + 1) \u003d

\u003d (x³ x² + 3x + 1) 6 : 6 + C.

And how does this example differ from Example 1? Yes, nothing! The same fifth degree with the base (x 3 - x 2 + 3x + 1) is multiplied by three-shred (3x 2 - 2x + 3), which is the degree of degree derivative: (x 3 - x 2 + 3x + 1) "\u003d 3x 2 - 2x + 3. This is the foundation of the degree we failed under the sign of the differential, from which the value of the integrated expression has not changed, and then applied the same formula 1). ( Integrals)

Example 3.

Here, the derivative of (2 - 3 - 3x) will give (6x 2 - 3), and we have

there is (12x 2 - 6), that is, the expression in 2 The times more, it means that we will bring (2-3 - 3x) under the sign of the differential, and in front of the integral put the multiplier 2 . Apply formula 2) (sheet ).

This is what happens:

Let's check, considering that:

Examples. Find uncertain integrals.

1. ∫ (6x + 5) 3 DX. How will we decide? We look at the sheet and we argue about this: the integrand is a degree, and we have a formula for the degree integral (formula 1) ), but there is a foundation u. and integration variable too u.

And we have a variable integration H., and the foundation of the degree (6x + 5). We will replace the integration variable: instead of DX write D (6x + 5). What changed? Since, what is standing after the Differential Difference Differential, default, is differentiated,

then D (6x + 5) \u003d 6dx, i.e. When replacing the variable x to a variable (6x + 5), the integrated function increased 6 times, so before the integral sign, we put the factor of 1/6. You can write these arguments like this:

So, we solved this example by the introduction of a new variable (the variable x was replaced by a variable 6x + 5). And where did the new variable (6x + 5) recorded? Under the sign of the differential. Therefore, this method The introduction of a new variable is often called method (or in a way ) Saving(new variable ) under the sign of differential.

In the second example, we first received a degree with a negative indicator, and then led to the sign of differential (7x-2) and used the degree integral formula 1) (integrals ).

We will analyze the example of the example 3.

There is a 1/5 coefficient in front of the integral. Why? Since d (5x-2) \u003d 5dx, then the function of the differential function U \u003d 5x-2, we increased the integrand expression 5 times, so that the value of this expression does not change - it was necessary to divide on 5, i.e. . Multiply to 1/5. Next, the formula was used 2) (Integrals) .

All the simplest formulas of the integrals will be viewed:

∫f (x) dx \u003d f (x) + c, moreover, equality should be performed:

(F (x) + c) "\u003d F (x).

Integration formulas can be obtained by referring to the corresponding differentiation formulas.

Really,

Exponent n. Maybe fractional. Often you have to find an indefinite integral from the function y \u003d √h. Calculate the integral from the function f (x) \u003d √x using the formula 1) .

We write this example in the formula 2) .

Since (x + c) "\u003d 1, then ∫dx \u003d x + c.

3) ∫dx \u003d x + c.

Replacing 1 / x² on x -2, calculate the integral from 1 / xx.

And it was possible to obtain this answer by the appeal of a known formula for differentiation:

We write our reasoning in the formula 4).

Multiplying both parts of the equality obtained by 2, we obtain the formula 5).

Find integrals from the main trigonometric functions, knowing their derivatives: (sinx) "\u003d COSX; (COSX)" \u003d - SINX; (TGX) "\u003d 1 / COS²X; (CTGX)" \u003d - 1 / SIN²X. We obtain the integration formulas 6) — 9).

6) ∫COSXDX \u003d SINX + C;

7) ∫sinxdx \u003d -cosx + C;

After studying the indicative and logarithmic functions, add a few more formulas.

The main properties of an uncertain integral.

I. The derivative of an indefinite integral is equal to the integrand .

(∫f (x) dx) "\u003d f (x).

II.The differential of an indefinite integral is equal to the initial expression.

d∫f (x) dx \u003d f (x) dx.

III. The indefinite integral of the differential (derivative) of some function is equal to the sum of this function and an arbitrary constant C.

∫df (x) \u003d f (x) + Cor ∫F "(x) dx \u003d f (x) + c.

Note: In I, II and III properties, the differential and integral signs (integral and differential) "eat" each other!

IV. A permanent multiplier of the integrated expression can be reached by the sign of the integral.

∫kf (x) dx \u003d k · ∫f (x) dx,where k. - A constant value that is not equal to zero.

V.The integral of the algebraic amount of functions is equal to the algebraic amount of integrals from these functions.

∫ (f (x) ± g (x)) dx \u003d ∫f (x) dx ± ∫g (x) dx.

Vi.If f (x) is a primitive for f (x), and k. and b. - permanent values, and k.≠ 0, then (1 / k) · f (kx + b) is a primitive for f (kx + b). Indeed, according to the rule of calculation of the derivative complex function We have:

You can write:

For each mathematical action there is an opposite effect. For differentiation (finding derived functions), there is also reverse action - Integration. By integration, they are found (restored) the function according to its derivative or differential. Found function is called predo-shaped.

Definition. Differential function F (X) called primitive for function f (x) At a given interval, if for all h. Equality is right from this gap: F '(x) \u003d f (x).

Examples. Find primary functions: 1) F (x) \u003d 2x; 2) f (x) \u003d 3cos3x.

1) Since (x²) '\u003d 2x, then, by definition, the function f (x) \u003d x² will be a primitive for the function f (x) \u003d 2x.

2) (SIN3X) '\u003d 3COS3X. If you designate f (x) \u003d 3cos3x and f (x) \u003d sin3x, then, by definition, it is primitive, we have: f '(x) \u003d f (x), and, it means f (x) \u003d sin3x is a primitive for F ( x) \u003d 3cos3x.

Note that and (sin3x +5 )′= 3COS3x., and (sin3x -8,2 )′= 3COS3x., ... in general You can write: (sin3x + S.)′= 3COS3x.where FROM - Some permanent value. These examples indicate the ambiguity of the integration action, in contrast to the differentiation action, when any differentiable function, there is a single derivative.

Definition. If the function F (X) is a primary for function f (x) At some interval, then the set of all the primary this features is:

F (x) + Cwhere C is any valid number.

The combination of all the primitive F (x) + C function f (x) on the interval under consideration is called an uncertain integral and indicated by the symbol ∫ (integral sign). Record: ∫f (x) dx \u003d f (x) + c.

Expression ∫F (x) dx They read: "The integral of EF from X on DE X".

f (x) dx - concrentist,

f (x) - integrated function,

h. - variable integration.

F (X) - Perfect for function f (x),

FROM - Some permanent value.

Now the considered examples can be written as follows:

1) ∫ 2xdx \u003d x² + c. 2) ∫ 3COS3XDX \u003d SIN3X + C.

What does the D sign mean?

d - Differential sign - has a double purpose: first, this sign separates the integrated function from the integration variable; Secondly, everything that stands after this sign is differentially default and multiplied by the integrand function.

Examples. Find integrals: 3) ∫ 2pxdx; 4) ∫ 2pxdp.

3) After the differential icon d. worth it h. H., but r

∫ 2khrdx \u003d Px² + p. Compare with example 1).

Let's check. F '(x) \u003d (px² + c)' \u003d p · (x²) '+ c' \u003d p · 2x \u003d 2px \u003d f (x).

4) After the differential icon d. worth it r. So, the integration variable R, and multiplier h. It should be considered a certain constant value.

∫ 2Hrdr \u003d ² + s. Compare with examples 1) and 3).

Let's check. F '(p) \u003d (p²x + c)' \u003d x · (p²) '+ c' \u003d x · 2p \u003d 2px \u003d f (p).

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The solution of integrals is the task is light, but only for the elect. This article is for those who want to learn to understand the integrals, but does not know anything about them or almost nothing. Integral ... why is it needed? How to calculate it? What is a certain and indefinite integral? If the only integral application known to you is to get a crochet in the form of an integral icon. Something useful from hard to reach places, then welcome! Learn how to solve the integrals and why without it it is impossible to do.

We study the concept of "integral"

Integration was known in ancient Egypt. Of course, not in modern video, but still. Since then, mathematics wrote a lot of books on this topic. Especially distinguished Newton and Leibnits But the essence of things has not changed. How to understand integrals from scratch? In no way! To understand this topic, the basic knowledge of the foundations of mathematical analysis will still need. It is these fundamental information about you will find in our blog.

Uncertain integral

Let us have some kind of function f (x) .

Uncertain integral function f (x) This feature is called F (X) , the derivative of which is equal to the function f (x) .

In other words, the integral is a derivative on the contrary or primitive. By the way, about how to read in our article.

Predictive exists for all continuous functions. Also, the constant sign is often added to the primary, as the derivatives differ in the constant coincide. The process of finding the integral is called integration.

Simple example:

To constantly not to calculate the primitive elementary functions, it is convenient to reduce the table and use the ready-made values:

Certain integral

Having a deal with the concept of integral, we are dealing with infinitely small values. The integral will help calculate the figure of the figure, the mass of the inhomogeneous body, passed under the uneven movement path and much more. It should be remembered that the integral is the amount of infinitely large number Infinitely small terms.

As an example, imagine a schedule of some function. How to find an area of \u200b\u200bfigures limited by a graph of the function?

With the help of the integral! We divide the curvilinear trapezium, limited by the coordinate axes and the graph of the function, on infinitely small segments. Thus, the figure will be divided into thin columns. The sum of the area of \u200b\u200bthe columns will be the area of \u200b\u200bthe trapezoid. But remember that such a calculation will give an exemplary result. However, the smaller the segments will already be, the more accurate will be the calculation. If we reduce them to such an extent that the length will strive for zero, the amount of segments will strive for the area of \u200b\u200bthe figure. This is a specific integral that is written as follows:

Points A and B are called integration limits.

Baria Alibasov and the Group "Integral"

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Rules for calculating integrals for dummies

Properties of an uncertain integral

How to solve an indefinite integral? Here we will consider the properties of an uncertain integral, which will be useful when solving examples.

• The derivative of the integral is equal to the integrand function:

• The constant can be made from the sign of the integral:

• The integral from the amount is equal to the amount of integrals. Also also for difference:

Properties of a specific integral

• Linearity:

• The integral sign changes if the integration limits are swapped:

• For any Points a., b. and from:

We have already found out that a certain integral is the limit of the amount. But how to get a specific value when solving the example? For this, there is a Newton-Leibnic formula:

Examples of solutions of integrals

Below will consider several examples of finding uncertain integrals. We suggest you independently understand the subtleties of the solution, and if something is incomprehensible, ask questions in the comments.

To secure the material, see the video about how integrals are solved in practice. Do not despair if the integral is not given immediately. Ask, and they will tell you about calculating the integrals all that know themselves. With our help of any triple or krivolynoe integral Along the closed surface will become forces.

Integral calculus.

PRINTING FUNCTION.

Definition: FunctionF (x) is called a primitive functionthe functionF (x) on the segment, if at any point of this segment is true equality:

It should be noted that there may be infinitely many for the same function. They will differ from each other for some constant number.

F 1 (x) \u003d f 2 (x) + c.

Uncertain integral.

Definition: Uncertain integralthe functionF (x) is called a set of primitive functions that are determined by the relation:

Record:

The condition for the existence of an indefinite integral on some segment is the continuity of the function on this segment.

Properties:

1.

2.

3.

4.

Example:

Finding the value of an indefinite integral is mainly due to the finding of a primitive function. For some functions, this is a rather complicated task. The following will be considered ways of finding uncertain integrals for basic classes of functions - rational, irrational, trigonometric, indicative, etc.

For convenience, the significance of uncertain integrals of most elementary functions are assembled into special integral tables that are sometimes very voluminous. They include the various most common combinations of functions. But most of the formulas presented in these tables are consequences of each other, so below the table of the main integrals with which you can get the values \u200b\u200bof uncertain integrals of various functions.

Integral		Value	Integral		Value
		lNSINX + C.
		lN.

Integration methods.

Consider three basic integration methods.

Direct integration.

The direct integration method is based on the assumption of a possible value of a primitive function with further verification of this value to differentiation. In general, we note that differentiation is a powerful tool for checking the results of integration.

Consider the use of this method using the example:

Requires the value of the integral . Based on a known differentiation formula
it can be concluded that the desired integral is equal
where C is a constant number. However, on the other hand
. Thus, we can finally conclude:

Note that, in contrast to differentiation, where, for finding a derivative, clear techniques and methods were used, the rules for finding a derivative, finally determining the derivative, for integration such methods are not available. If, when you find the derivative, we used, so to speak, constructive methods that, based on certain rules, led to the result, then when finding a primary one, it is necessary to fully rely on knowledge of the tables of derivatives and primitive.

As for the direct integration method, it is applicable only for some very limited classes of functions. Functions for which it is possible to find a primary very little from the go. Therefore, in most cases, the methods described below are used.

The method of substitution (replacement of variables).

Theorem: If you want to find an integral
But it is difficult to find a primitive, then by replacing x \u003d  (t) anddx \u003d  (t), DtP is:

Evidence : Differentiating the proposed equality:

Upon reviewed by the property number 2 of an indefinite integral:

f.(x.) dX. = f.[  (t.)]  (t.) dt.

what, taking into account the introduced designations and is the initial assumption. Theorem is proved.

Example.Find an indefinite integral
.

We will replace t. = sINX., dt. = cosxdt..

Example.

Replacement
We get:

Below will be considered other examples of the application of the substitution method for various types of functions.

Integration in parts.

The method is based on the well-known formula of the derivative of the work:

(UV)  \u003d uv + Vu

where uiv are some functions from x.

In differential form: D (UV) \u003d UDV + VDU

Integrating, we get:
, and in accordance with the properties of an indefinite integral above:

or
;

Received the integration formula in parts, which allows the integrals of many elementary functions.

Example.

As can be seen, the sequential use of the integration formula in parts allows you to gradually simplify the function and bring the integral to the table.

Example.

It can be seen that as a result of the re-use of integration in parts, the function failed to simplify the table. However, the last resulting integral is no different from the source. Therefore, we move it into the left part of equality.

Thus, the integral is found at all without the use of integral tables.

Before considering in detail the integration methods of different classes of functions, we give a few more examples of finding uncertain integrals by bringing them to tabular.

Example.

Example.

Example.

Example.

Example.

Example.

Example.

Example.

Example.

Example.

Integrating elementary fractions.

Definition: Elementarythe fractions of the following four types are called:

I.
III.

II.
IV.

m, n- integers (m2, n2) ib 2 - 4ac<0.

The first two types of integrals from the elementary fractions are quite simply given to the table substitution T \u003d AX + B.

Consider the integration method of elementary fractions of the type III.

The integral of the fraction of the form III would be presented in the form:

Here, in general, it is shown to bring the integral of a fraction of the form IIIO to two table integrals.

Consider the application of the above formula on the examples.

Example.

Generally speaking, if three-stared AX 2 + BX + ensepassB 2 - 4AC\u003e 0, then the fraction by definition is not elementary, however, it can nevertheless integrate the method specified above.

Example.

Example.

We now consider the methods of integrating the simplest fractions of the IVTP.

First, consider the special case at m \u003d 0, n \u003d 1.

Then the integral of the view
it is possible to present in the database of a complete square in the form of a full square
. Let's make the following transformation:

The second integral entering into this equality will take in parts.

Denote:

For the source integral we get:

The resulting formula is called recurrent.If you apply itN-1 time, then the table integral will be
.

Let us return now to the integral from the elementary fraction of the IVT type of the general case.

In the resulting equality, the first integral by substitution t. = u. 2 + s.located to the table , and the recurrent formula considered above is applied to the second integral.

Despite the seeming complexity of the integration of the elementary fraction of the form IV, it is easy to use enough for fractions with a small degree n., And the versatility and the generality of the approach makes it possible a very simple implementation of this method on a computer.

Example:

Integrating rational functions.

Integrating rational fractions.

In order to integrate the rational fraction, it is necessary to decompose it on elementary fractions.

Theorem: If a
- the correct rational fraction, the denominator (x) of which is represented as a product of linear and quadratic multipliers (we note that any polynomial with valid coefficients can be represented in this form: P.(x.) = (x. - a.)  …(x. - b.)  (x. 2 + px. + q.)  …(x. 2 + rX. + s.)  ), then this fraction can be decomposed on the elementary following scheme:

where A I, B I, M I, N i, R I, S I are some permanent values.

In the integration of rational fractions, it is resorted to the decomposition of the initial fraction on the elementary. To find the magnitude of I, B I, M I, N i, R I, S I, use the so-called method of uncertain coefficientsThe essence of which is that in order for two polynomials to be identically equal, it is necessary and enough to be equal to the coefficients with the same degrees x.

Application of this method Consider on a specific example.

Example.

When leading to a common denominator and equating the corresponding numerals, we get:

Example.

Because The fraction is wrong, then it should be pre-highlight the whole part:

6x 5 - 8x 4 - 25x 3 + 20x 2 - 76x- 7 3x 3 - 4x 2 - 17x + 6

6x 5 - 8x 4 - 34x 3 + 12x 2 2x 2 + 3

9x 3 + 8x 2 - 76x - 7

9x 3 - 12x 2 - 51x +18

20x 2 - 25x - 25

Spread the denominator of the resulting fraction on multipliers. It can be seen that at x \u003d 3 denominator the fraci turns into zero. Then:

3x 3 - 4x 2 - 17x + 6x- 3

3X 3 - 9X 2 3X 2 + 5X- 2

Thus, 3x 3 - 4x 2 - 17x + 6 \u003d (x- 3) (3x 2 + 5x- 2) \u003d (x- 3) (x + 2) (3x-1). Then:

In order to avoid when you find undefined disclosure coefficients, grouping and solving a system of equations (which in some cases it may be quite large) used so-called method of arbitrary values. The essence of the method is that the expression obtained above are alternately somewhat (according to the number of uncertain coefficients) of arbitrary values \u200b\u200bx. To simplify calculations, it is accepted as arbitrary values \u200b\u200bto take points in which the denomoter is zero, i.e. In our case - 3, -2, 1/3. We get:

We finally get:

=

Example.

Find uncertain coefficients:

Then the value of the specified integral:

Integrating some trigonometrics

functions.

Integrals from trigonometric functions may be infinitely a lot. Most of these integrals cannot be calculated analytically, so consider some the main types functions that can always be integrated.

Integral View
.

Here, the R - the designation of some rational function from the variablesSinxAcosx.

Integrals of this species are calculated by substitution
. This substitution allows you to convert trigonometric function to rational.

,

Then

In this way:

The transformation described above is called universal trigonometric substitution.

Example.

The undoubted advantage of this substitution is that it is always possible to convert trigonometric function to rational and calculate the corresponding integral. The disadvantages include the fact that when converting it may turn out a rather complicated rational function, the integration of which will take a lot of time and strength.

However, if it is impossible to apply a more rational replacement of the variable, this method is the only one intensive.

Example.

Integral View
if a

functionR.cOSX.

Despite the possibility of calculating such an integral with a universal trigonometric substitution, more rational to apply substitution t. = sINX..

Function
it may contain as many in even degrees, and, therefore, it can be converted to a rational function of relativesinx.

Example.

Generally speaking, for the use of this method, only the oddness of the function relative to the cosine is needed, and the degree of sine, which is included in the function can be any, both in both the fractional.

Integral View
if a

functionR. is odd aboutsINX..

By analogy with the case discussed above, substitution t. = cOSX.

Example.

Integral View

functionR. even aboutsINX. andcOSX.

For converting the RV function, the substitution is used

t \u003d TGX.

Example.

Integral works of sinuses and cosine

different arguments.

Depending on the type of product, one of three formulas will be applied:

Example.

Example.

Sometimes, when integrating trigonometric functions, it is convenient to use well-known trigonometric formulas to reduce the order of functions.

Example.

Example.

Sometimes some non-standard techniques are applied.

Example.

Integrating some irrational functions.

Not every irrational function may have an integral expressed by elementary functions. To find the integral from the irrational function, apply a substitution that will allow converting a function to rational, the integral of which can always be found.

Consider some techniques for integrating various types of irrational functions.

Integral View
wheren.- natural number.

With the help of substitution
the function is rationalized.

Example.

If the composition of the irrational function includes the roots of various degrees, then as a new variable, rationally take the root of the degree equal to the smallest total multiple degrees of the roots included in the expression.

We will illustrate this on the example.

Example.

Integrating binomine differentials.

Definition: Bininominal differentialcalled expression

x. m. (a. + bX. n. ) p. dX.

where m., n., and p.– rational numbers.

As proved by Academician Chebyshev P.L. (1821-1894), the integral from the binomine differential can be expressed through elementary functions only in the following three cases:

If a r- an integer, then the integral is rationalized by substitution

where - a common denominator m.and n..

There is an overview of the methods for calculating uncertain integrals. The main integration methods that include integration of the amount and difference, making a permanent integral sign, replace the variable, integrating in parts. Special methods and techniques for the integration of fractions, roots, trigonometric and indicative functions.

Pred-like and indefinite integral

The primitive f (x) from the function f (x) is such a function, the derivative of which is equal to F (X):
F '(x) \u003d f (x), x ∈ Δ,
Where Δ - The gap on which this equation is performed.

The totality of all the primordial is called an uncertain integral:
,
where C is a constant, independent of the variable x.

Basic formulas and integration methods

Table integrals

The ultimate goal of calculating uncertain integrals - by transformations, clarify the specified integral to the expression containing the simplest or tabular integrals.
See Table Integrals \u003e\u003e\u003e

The integration rule of the amount (difference)

Making a permanent integral sign

Let c be a constant, independent of x. Then it can be submitted for the integral sign:

Replacing the variable

Let x be a function from the variable t, x \u003d φ (t), then
.
Or vice versa, t \u003d φ (x),
.

By replacing the variable, you can not only calculate simple integrals, but also to simplify the calculation of more complex.

Integration rule in parts

Integration of fractions (rational functions)

We introduce the designation. Let p k (x), q m (x), R n (x) be denoted by the degrees k, m, n, respectively, relative to the variable x.

Consider the integral consisting of fractions of polynomials (the so-called rational function):

If k ≥ n, then you first need to highlight the whole part of the fraci:
.
The integral from the polynomial S k-n (x) is calculated by the integral table.

The integral remains:
where M.< n .
To calculate it, the integrand should be decomposed on the simplest fraction.

To do this, find the roots of the equation:
Q n (x) \u003d 0.
Using the roots obtained, you need to represent the denominator in the form of a work of the factors:
Q n (x) \u003d s (x - a) n a (x - b) n b ... (x 2 + ex + f) n e (x 2 + gx + k) n g ....
Here s is the coefficient at x n, x 2 + ex + f\u003e 0, x 2 + gx + k\u003e 0, ....

After that, decompose the fraction on the simplest:

Integrating, we obtain an expression consisting of simpler integrals.
Integrals of type

The T \u003d x - a is given to the table substation.

Consider the integral:

We transform the numerator:
.
Subject to the integrand, we obtain the expression in which two integral includes:
,
.
The first, substitution T \u003d x 2 + EX + F is given to the table.
The second, according to the formula of bringing:

Located to integral

We give its denominator to the sum of the squares:
.
Then substitution, integral

It is also provided to the table.

Integration of irrational functions

We introduce the designation. Let R (U 1, U 2, ..., U N) mean a rational function from variables U 1, U 2, ..., u n. I.e
,
where p, q is polynomials from variables U 1, U 2, ..., u n.

Linear irrationality

Consider the integrals of the form:
,
where - rational numbers, M 1, N 1, ..., m s, n s are integers.
Let N be a common denominator of the numbers R 1, ..., R s.
Then the integral comes down to the integral from the rational functions of the substitution:
.

Integrals from differential binomes

Consider the integral:
,
where m, n, p is rational numbers, a, b - valid numbers.
Such integrals are reduced to integrals from rational functions in three cases.

1) If P is an integer. The substitution x \u003d t n, where n is the total denominator of the fractions M and N.
2) If - the whole. Substitution A x N + B \u003d T M, where m is the number of numbers p.
3) If - a whole. Substitution A + B X - N \u003d T M, where M is the denominator of the number P.

If none of the three numbers is an integer, then according to the Chebyshev theorem, the integrals of this species cannot be expressed by the final combination of elementary functions.

In some cases, first it is useful to bring the integral to more convenient M and P values. This can be done using formulas:
;
.

Integrals containing square root of square three

Here we consider the integrals of the form:
,

Euler substitutions

Such integrals can be reduced to integrals from rational functions of one of the three substitutions of Euler:
, with a\u003e 0;
, with C\u003e 0;
where x 1 is the root of the equation A x 2 + b x + c \u003d 0. If this equation has valid roots.

Trigonometric and hyperbolic substitutions

Direct methods

In most cases, the substitutions of the Euler lead to longer calculations than direct methods. With direct methods, the integral is given to one of the species listed below.

I type

The integral of the form:
,
where p n (x) is a polynomial degree n.

Such integrals are the method of uncertain coefficients using identity:

Differentiating this equation and equating the left and right parts, we find the coefficients a i.

II type

The integral of the form:
,
where P M (x) is a polynomial degree m.

Substitution T \u003d. (X - α) -1 This integral is driven to the previous type. If m ≥ n, then the fraction should be allocated to the whole part.

III type

The third and most complex type:
.

Here you need to make a substitution:
.
After which the integral will take the form:
.
Next, permanent α, β, you need to choose such that the coefficients at T appealed to zero:
B \u003d 0, b 1 \u003d 0.
Then the integral disintegrates the sum of the integrals of two types:
;
,
which are integrated, respectively, substitutions:
z 2 \u003d a 1 t 2 + C 1;
y 2 \u003d A 1 + C 1 T -2.

General

Integration of transcendental (trigonometric and indicative) functions

We note in advance that the methods that are applicable to trigonometric functions are also applicable for hyperbolic functions. For this reason, we will not consider the integration of hyperbolic functions separately.

Integrating rational trigonometric functions from COS X and SIN X

Consider the integrals from the trigonometric functions of the form:
,
where R is a rational function. This can also include tangents and catanges that should be converted through sinuses and cosines.

When integrating such functions, it is useful to keep in mind the three rules:
1) if R ( cOS X, SIN X) multiplied by -1 from the change of sign in front of one of the values cOS X. or sIN X., it is useful to identify another of them.
2) if R ( cOS X, SIN X) does not change from the change of sign simultaneously before cOS X. and sIN X., it is useful to put tG X \u003d T or cTG X \u003d T.
3) The substitution in all cases leads to an integral from rational fraction. Unfortunately, this substitution leads to longer computing than previous, if they are applicable.

Production of power functions from COS X and SIN X

Consider the integrals of the form:

If M and N are rational numbers, then one of the substitutions T \u003d sIN X. or t \u003d cOS X. The integral is reduced to the integral from the differential binoma.

If M and N are integers, the integrals are calculated by integrating in parts. At the same time, the following formulas are obtained:

;
;
;
.

Integration in parts

The use of the formula Euler

If the integrand is linearly relative to one of the functions
cOS AX. or sin AX.It is convenient to apply the Euler formula:
e IAX \u003d. cOS AX + ISIN AX (where i 2 \u003d - 1 ),
Replacing this feature on e IAX and highlighting valid (when replacing cOS AX.) or imaginary part (when replacing sin AX.) From the result obtained.

References:
N.M. Gunter, R.O. Kuzmin, Collection of tasks on higher mathematics, "Lan", 2003.

Earlier, on a given function, guided by various formulas and rules, we found it derived. The derivative has numerous applications: this is the speed of movement (or, summarizing, the rate of leakage of any process); Corner coefficient tangent to function graphics; Using a derivative, you can explore the function on monotony and extremum; It helps solve optimization tasks.

But along with the task of finding the speed on a well-known law, the inverse problem is also found - the task of restoring the law of movement at known speed. Consider one of these tasks.

Example 1. The material point is moving along the straight, the rate of its movement at time t is given by the formula V \u003d GT. Find the law of movement.
Decision. Let S \u003d S (T) be the desired law of movement. It is known that S "(T) \u003d V (T). So, to solve the problem, it is necessary to select the function S \u003d S (T), the derivative of which is GT. It is not difficult to guess that \\ (s (t) \u003d \\ FRAC (GT ^ 2) (2) \\). In fact
\\ (s "(t) \u003d \\ left (\\ FRAC (GT ^ 2) (2) \\ RIGHT)" \u003d \\ FRAC (G) (2) (T ^ 2) "\u003d \\ FRAC (G) (2) \\ Answer: \\ (S (T) \u003d \\ FRAC (GT ^ 2) (2) \\)
Immediately note that the example is solved right, but incomplete. We got \\ (s (t) \u003d \\ FRAC (GT ^ 2) (2) \\). In fact, the task has infinitely many solutions: any function of the form \\ (S (T) \u003d \\ FRAC (GT ^ 2) (2) + C \\), where C is an arbitrary constant, can serve as a law of movement, because \\ (\\ left (\\ FRAC (GT ^ 2) (2) + C \\ RIGHT) "\u003d GT \\)

For the task of becoming more specific, we needed to fix the initial situation: specify the coordinate of the moving point at some point in time, for example, at t \u003d 0. If, say, S (0) \u003d S 0, then from the equality S (T) \u003d (GT 2) / 2 + C We obtain: S (0) \u003d 0 + s, i.e. c \u003d s 0. Now the law of motion is defined uniquely: S (T) \u003d (GT 2) / 2 + S 0.

In mathematics, mutually reverse operations are assigned different names, invent special designations, for example: construction of a square (x 2) and extraction

Square root (\\ (\\ sqrt (x) \\)), sinus (sin x) and arcsinus (arcsin x), etc. The process of finding a derivative according to a given function called Differentiation , and inverse operation, i.e., the process of finding a function according to a given derivative -integration the term "derivative" itself can justify "Lowering": the function y \u003d f (x) "produces a new function in" \u003d F "(x). The function y \u003d f (x) acts as it were as a "parent", but mathematics, naturally, do not call it "parent" or "manufacturer", they say that this is, with respect to the function, "\u003d F" (x) , Primary image, or primitive..

The function y \u003d f (x) is called a primitive for the function y \u003d f (x) on the period x, if for \\ (x \\ in x \\) the equality f "(x) \u003d f (x)

Definition. In practice, the interval X is usually not indicated, but imply (as a natural field definition area).

We give examples.

{!LANG-a4a0b1ae7e0217a61de9f867a3b655ee!}
1) The function y \u003d x 2 is a primitive to function y \u003d 2x, since for any x equality (x 2) "\u003d 2x
2) The function y \u003d x 3 is a primitive for function y \u003d 3x 2, since for any x equality (x 3) "\u003d 3x 2
3) The function y \u003d sin (x) is a primitive for the function y \u003d cos (x), since for any x equality (sin (x)) "\u003d COS (X)

When finding primary, as well as derivatives, not only formulas are used, but also some rules. They are directly related to the relevant rules for calculating derivatives.

We know that the derivative of the amount is equal to the amount of derivatives. This rule creates the appropriate rule of primary finding.

Rule 1. The first-shaped amount is equal to the amount of primitive.

We know that the permanent multiplier can be reached for a sign of the derivative. This rule creates the appropriate rule of primary finding.

Rule 2. If f (x) is a primitive for f (x), then kf (x) is primitive for kf (x).

Theorem 1. If y \u003d f (x) is a primitive for function y \u003d f (x), then the function \\ (y \u003d \\ Frac (1) (k) f (kx + m) is valid for the function \\ (Y \u003d \\ FRAC (1) (K) F (KX + M) \\)

Theorem 2. If y \u003d f (x) is a primitive for the function y \u003d f (x) on the period x, then the function y \u003d f (x) is infinitely many primitive, and they all have the form y \u003d f (x) + C.

Integration methods

Method for replacing a variable (substitution method)

The method of integrating the substitution is to introduce a new integration variable (that is, substitutions). At the same time, the specified integral is provided to a new integral, which is table or reduced to it. Common methods The selection of substitutions does not exist. The ability to correctly determine the substitution is acquired by practice.
Let us calculate the integral \\ (\\ TextStyle \\ int f (x) dx \\). We will make the substitution \\ (x \u003d \\ varphi (t) \\) where \\ (\\ varphi (t) \\) is a function having a continuous derivative.
Then \\ (dx \u003d \\ varphi "(t) \\ cdot dt \\) and on the basis of the property of the invariance of the integration formula of an indefinite integral, we obtain the integration formula of the substitution:
\\ (\\ int f (x) dx \u003d \\ int f (\\ varphi (t)) \\ Cdot \\ Varphi "(T) DT \\)

Integrating the expressions of the form \\ (\\ TextStyle \\ int \\ sin ^ n x \\ cos ^ M x Dx \\)

If M is odd, M\u003e 0, it is more convenient to make the substitution SIN x \u003d t.
If N is odd, n\u003e 0, it is more convenient to make the cos x \u003d t substitution.
If N and M are read, it is more convenient to make the substitution TG x \u003d t.

Integration in parts

Integration in parts - Application of the following formula for integration:
\\ (\\ TextStyle \\ int u \\ cdot dv \u003d u \\ cdot v - \\ int v \\ cdot du \\)
or:
\\ (\\ textstyle \\ int u \\ cdot v "\\ cdot dx \u003d u \\ cdot v - \\ int v \\ cdot u" \\ CDOT DX \\)

Table of indefinite integrals (primitive) some functions

$$ \\ int 0 \\ cdot dx \u003d c $$$$ \\ int 1 \\ cdot dx \u003d x + c $$$$ \\ int x ^ n dx \u003d \\ FRAC (x ^ (n + 1)) (n + 1 ) + C \\; \\; (N \\ NEQ -1) $$$$ \\ int \\ FRAC (1) (X) DX \u003d \\ ln | x | + C $$$$ \\ int e ^ x dx \u003d e ^ x + c $$$$ \\ int a ^ x dx \u003d \\ FRAC (a ^ x) (\\ ln a) + c \\; \\; (A\u003e 0, \\; \\; A \\ NEQ 1) $$$$ \\ int \\ cos x dx \u003d \\ sin x + c $$$$ \\ int \\ sin x dx \u003d - \\ cos x + C $$$ $ \\ int \\ FRAC (DX) (\\ cos ^ 2 x) \u003d \\ Text (TG) x + C $$$$ \\ int \\ FRAC (DX) (\\ Sin ^ 2 x) \u003d - \\ Text (CTG) x + C $$$$ \\ int \\ FRAC (DX) (\\ SQRT (1-x ^ 2)) \u003d \\ Text (ArcSin) x + C $$$$ \\ int \\ FRAC (DX) (1 + x ^ 2 ) \u003d \\ Text (ArCTG) X + C $$$$ \\ int \\ text (CH) x DX \u003d \\ Text (SH) x + C $$$$ \\ INT \\ TEXT (SH) x DX \u003d \\ TEXT (CH ) x + C $$