Using transformations of expressions, the simplest methods of integration. Integration methods

4.1. SIMPLE INTEGRATION METHODS 4.1.1. The concept of an indefinite integral

In differential calculus, the problem of finding the derivative or differential with respect to a given function was considered y= F(x), i.e. it was necessary to find f(x)= F"(x) or dF(x)= F"(x)dx= f(x)dx. Let us pose the inverse problem: to restore the differentiated function, i.e., knowing the derivative f(x)(or differential f(x)dx), find such a function F(x), to F"(x)= f(x). This task turns out to be much more difficult than the task of differentiation. For example, let the speed of movement of a point be known, but we need to find the law

her movements S= S(t), and To solve such

tasks, new concepts and actions are introduced.

Definition. Differentiable function F(x) called antiderivative for function f(x) on (a; b), If F"(x)= f(x) on (a; b).

For example, for f(x) = x 2 antiderivative because

For f(x) = cos x the antiderivative will be F(x) = sin x, because F"(x) = (sin x)" = cos x, which coincides with f(x).

Does an antiderivative always exist for a given function? f(x)? Yes, if this function is continuous on (a; b). In addition, there are countless numbers of primitives, and they differ from each other only by a constant term. Indeed, sin x+ 2, sin x- 2, sin x+ c- all these functions will be antiderivatives for cos x(the derivative of a constant value is 0) - fig. 4.1.

Definition. Expression F(x)+ C, Where WITH- an arbitrary constant value that defines the set of antiderivatives for the function f(x), called indefinite integral and is indicated by the symbol , i.e. , where the sign is the sign of the indefinite

integral, f(x)- called integrand function, f (x)dx- by the integrand, x- integration variable.

Rice. 4.1. Example of a family of integral curves

Definition. The operation of finding an antiderivative from a given derivative or differential is called integration this function.

Integration is the inverse action of differentiation; it can be verified by differentiation, and differentiation is unique, and integration gives the answer up to a constant. Giving a constant value WITH specific values By-

We get various functions

each of which defines a curve on the coordinate plane called integral. All graphs of integral curves are shifted parallel to each other along the axis Oy. Therefore, a geometrically indefinite integral is a family of integral curves.

So, new concepts (antiderivative and indefinite integral) and a new action (integration) have been introduced, but how do you still find the antiderivative? To easily answer this question, you must first compile and memorize a table of indefinite integrals of basic elementary functions. It is obtained by inverting the corresponding differentiation formulas. For example, if

Typically, the table includes some integrals obtained after applying the simplest integration methods. These formulas are marked in the table. 4.1 with the symbol “*” and are proven in the further presentation of the material.

Table 4.1. Table of basic indefinite integrals

Formula 11 from table. 4.1 may look like
,

because. A similar remark about the form

mules 13:

4.1.2. Properties of indefinite integrals

Let's consider the simplest properties of the indefinite integral, which will allow us to integrate not only the basic elementary functions.

1. The derivative of the indefinite integral is equal to the integrand:

2. The differential of the indefinite integral is equal to the integrand:

3. The indefinite integral of the differential of a function is equal to this function added to an arbitrary constant:

Example 1. Example 2.

4. The constant factor can be taken out of the integral sign: Example 3.

5. The integral of the sum or difference of two functions is equal to the sum or difference of the integrals of these functions:

Example 4.

The integration formula remains valid if the integration variable is a function: if That

An arbitrary function that has a continuous derivative. This property is called invariance.

Example 5. , That's why

Compare with

There is no universal method of integration. Below we will present some methods that allow you to calculate a given integral using properties 1-5 and table. 4.1.

4.1.3.Direct integration

This method consists of direct use of table integrals and properties 4 and 5. Examples.

4.1.4.Decomposition method

This method consists of expanding the integrand into a linear combination of functions with already known integrals.

Examples.

4.1.5. Method of subscribing to the differential sign

To reduce this integral to a tabular one, it is convenient to make differential transformations.

1. Subsuming the differential sign of a linear function

from here
in particular, dx = d(x + b),

the differential does not change if you add to the variable

or subtract a constant value. If the variable increases several times, then the differential is multiplied by its reciprocal value. Examples with solutions.

Let's check formulas 9*, 12* and 14* from the table. 4.1, using the method of subsuming the differential sign:

Q.E.D.

2. Subsuming the basic elementary functions under the differential sign:

Comment. Formulas 15* and 16* can be verified by differentiation (see property 1). For example,

and this is the integrand function from formula 16*.

4.1.6. Method for separating a perfect square from a quadratic trinomial

When integrating expressions like or

separating a perfect square from a quadratic trinomial

ax 2 + bx+ c it is possible to reduce them to tabular 12*, 14*, 15* or 16* (see Table 4.1).

Since in general this operation looks more complicated than it actually is, we will limit ourselves to examples.

Examples.

1.

Solution. Here we extract the perfect square from the quadratic trinomial x 2 + 6x+ 9 = (x 2 + 6x+ 9) - 9 + 5 = (x+ 3) 2 - 4, and then we use the method of subsuming the differential sign.

Using similar reasoning, we can calculate the following integrals:

2. 3.

At the final stage of integration, formula 16* was used.

4.1.7. Basic integration methods

There are two such methods: the method of changing a variable, or substitution, and integration by parts.

Variable Replacement Method

There are two formulas for changing a variable in an indefinite integral:

1) 2)

Here, the essence is monotone differentiable functions

tions of their variables.

The art of applying the method consists mainly in choosing functions so that the new integrals are tabular or reduce to them. The final answer should return to the old variable.

Note that substitution under the differential sign is a special case of variable replacement.

Examples.

Solution.A new variable should be entered heretso as to get rid of the square root. Let's putx+ 1 = t, Then x= t 2+ 1, and dx = 2 tdt:

Solution. Replacing x- 2 per t, we obtain a monomial in the denominator and after term-by-term division the integral is reduced to the tabular integral of the power function:

When passing to a variable x formulas used:

Method of integration by parts

The differential of the product of two functions is determined by the formula

Integrating this equality (see property 3), we find:

From here This is the formula integration by

parts.

Integration by parts involves the subjective representation of the integrand in the form u . dV, and at the same time the integral should be easier than Otherwise application

the method doesn't make sense.

So, the method of integration by parts assumes the ability to isolate factors from the integrand u And dV taking into account the above requirements.

We present a number of typical integrals that can be found by the method of integration by parts. 1. Integrals of the form

Where P(x)- polynomial; k- constant. In this case u= P(x), and dV- all other factors.

Example 1.

2.Integrals of type

Here we put other factors.

Example 2.

Example 3.
Example 4.

Any result can be verified by differentiation. For example, in this case

The result is correct.

3.Integrals of the form

where a, b- const. Behind u should take e ax , sin bx or cos bx.

Example 5.

From here we get Example 6.

From here

Example 7.
Example 8.

Solution.Here you need to first make a change of variable, and then integrate by parts:

Example 9.
Example 10.

Solution. This integral can be found with equal success either by replacing the variable 1 + x 2 = t 2 or by integrating by parts:

Independent work

Perform direct integration (1-10).

Apply simple integration methods (11-46).

Perform integration using change of variable and integration by parts methods (47-74).

A function F(x) differentiable in a given interval X is called antiderivative of the function f(x), or the integral of f(x), if for every x ∈X the following equality holds:

F " (x) = f(x). (8.1)

Finding all antiderivatives for a given function is called its integration. Indefinite integral function f(x) on a given interval X is the set of all antiderivative functions for the function f(x); designation -

If F(x) is some antiderivative of the function f(x), then ∫ f(x)dx = F(x) + C, (8.2)

where C is an arbitrary constant.

Table of integrals

Directly from the definition we obtain the main properties of the indefinite integral and a list of tabular integrals:

1) d∫f(x)dx=f(x)

2)∫df(x)=f(x)+C

3) ∫af(x)dx=a∫f(x)dx (a=const)

4) ∫(f(x)+g(x))dx = ∫f(x)dx+∫g(x)dx

List of tabular integrals

1. ∫x m dx = x m+1 /(m + 1) +C; (m ≠ -1)

3.∫a x dx = a x /ln a + C (a>0, a ≠1)

4.∫e x dx = e x + C

5.∫sin x dx = cosx + C

6.∫cos x dx = - sin x + C

7. = arctan x + C

8. = arcsin x + C

10. = - ctg x + C

Variable replacement

To integrate many functions, use the variable replacement method or substitutions, allowing you to reduce integrals to tabular form.

If the function f(z) is continuous on [α,β], the function z =g(x) has a continuous derivative and α ≤ g(x) ≤ β, then

∫ f(g(x)) g " (x) dx = ∫f(z)dz, (8.3)

Moreover, after integration on the right side, the substitution z=g(x) should be made.

To prove it, it is enough to write the original integral in the form:

∫ f(g(x)) g " (x) dx = ∫ f(g(x)) dg(x).

For example:

Method of integration by parts

Let u = f(x) and v = g(x) be functions that have continuous . Then, according to the work,

d(uv))= udv + vdu or udv = d(uv) - vdu.

For the expression d(uv), the antiderivative will obviously be uv, so the formula holds:

∫ udv = uv - ∫ vdu (8.4.)

This formula expresses the rule integration by parts. It leads the integration of the expression udv=uv"dx to the integration of the expression vdu=vu"dx.

Let, for example, you want to find ∫xcosx dx. Let us put u = x, dv = cosxdx, so du=dx, v=sinx. Then

∫xcosxdx = ∫x d(sin x) = x sin x - ∫sin x dx = x sin x + cosx + C.

The rule of integration by parts has a more limited scope than substitution of variables. But there are whole classes of integrals, for example,

∫x k ln m xdx, ∫x k sinbxdx, ∫ x k cosbxdx, ∫x k e ax and others, which are calculated precisely using integration by parts.

Definite integral

The concept of a definite integral is introduced as follows. Let a function f(x) be defined on an interval. Let us divide the segment [a,b] into n parts by points a= x 0< x 1 <...< x n = b. Из каждого интервала (x i-1 , x i) возьмем произвольную точку ξ i и составим сумму f(ξ i) Δx i где
Δ x i =x i - x i-1. A sum of the form f(ξ i)Δ x i is called integral sum, and its limit at λ = maxΔx i → 0, if it exists and is finite, is called definite integral functions f(x) of a before b and is designated:

F(ξ i)Δx i (8.5).

The function f(x) in this case is called integrable on the interval, numbers a and b are called lower and upper limits of the integral.

The following properties are true for a definite integral:

4), (k = const, k∈R);

5)

6)

7) f(ξ)(b-a) (ξ∈).

The last property is called mean value theorem.

Let f(x) be continuous on . Then on this segment there is an indefinite integral

∫f(x)dx = F(x) + C

and takes place Newton-Leibniz formula, connecting the definite integral with the indefinite integral:

F(b) - F(a). (8.6)

Geometric interpretation: the definite integral is the area of ​​a curvilinear trapezoid bounded from above by the curve y=f(x), straight lines x = a and x = b and a segment of the axis Ox.

Improper integrals

Integrals with infinite limits and integrals of discontinuous (unbounded) functions are called not your own. Improper integrals of the first kind - These are integrals over an infinite interval, defined as follows:

(8.7)

If this limit exists and is finite, then it is called convergent improper integral of f(x) on the interval [a,+ ∞), and the function f(x) is called integrable over an infinite interval[a,+ ∞). Otherwise, the integral is said to be does not exist or diverges.

Improper integrals on the intervals (-∞,b] and (-∞, + ∞) are defined similarly:

Let us define the concept of an integral of an unbounded function. If f(x) is continuous for all values x segment , except for the point c, at which f(x) has an infinite discontinuity, then improper integral of the second kind of f(x) ranging from a to b the amount is called:

if these limits exist and are finite. Designation:

Examples of integral calculations

Example 3.30. Calculate ∫dx/(x+2).

Solution. Let us denote t = x+2, then dx = dt, ∫dx/(x+2) = ∫dt/t = ln|t| + C = ln|x+2| +C.

Example 3.31. Find ∫ tgxdx.

Solution.∫ tgxdx = ∫sinx/cosxdx = - ∫dcosx/cosx. Let t=cosx, then ∫ tgxdx = -∫ dt/t = - ln|t| + C = -ln|cosx|+C.

Example3.32 . Find ∫dx/sinx

Solution.

Example3.33. Find .

Solution. = .

Example3.34 . Find ∫arctgxdx.

Solution. Let's integrate by parts. Let us denote u=arctgx, dv=dx. Then du = dx/(x 2 +1), v=x, whence ∫arctgxdx = xarctgx - ∫ xdx/(x 2 +1) = xarctgx + 1/2 ln(x 2 +1) +C; because
∫xdx/(x 2 +1) = 1/2 ∫d(x 2 +1)/(x 2 +1) = 1/2 ln(x 2 +1) +C.

Example3.35 . Calculate ∫lnxdx.

Solution. Applying the integration by parts formula, we obtain:
u=lnx, dv=dx, du=1/x dx, v=x. Then ∫lnxdx = xlnx - ∫x 1/x dx =
= xlnx - ∫dx + C= xlnx - x + C.

Example3.36 . Calculate ∫e x sinxdx.

Solution. Let us denote u = e x, dv = sinxdx, then du = e x dx, v =∫ sinxdx= - cosx → ∫ e x sinxdx = - e x cosx + ∫ e x cosxdx. We also integrate the integral ∫e x cosxdx by parts: u = e x , dv = cosxdx, du=e x dx, v=sinx. We have:
∫ e x cosxdx = e x sinx - ∫ e x sinxdx. We obtained the relation ∫e x sinxdx = - e x cosx + e x sinx - ∫ e x sinxdx, from which 2∫e x sinx dx = - e x cosx + e x sinx + C.

Example 3.37. Calculate J = ∫cos(lnx)dx/x.

Solution. Since dx/x = dlnx, then J= ∫cos(lnx)d(lnx). Replacing lnx through t, we arrive at the table integral J = ∫ costdt = sint + C = sin(lnx) + C.

Example 3.38 . Calculate J = .

Solution. Considering that = d(lnx), we substitute lnx = t. Then J = .

Example 3.39 . Calculate the integral J = .

Solution. We have: . Therefore =
=
=. entered like this: sqrt(tan(x/2)).

And if in the result window you click on Show steps in the upper right corner, you will get a detailed solution.

Integral calculus.

Antiderivative function.

Definition: The function F(x) is called antiderivative function function f(x) on the segment if the equality is true at any point of this segment:

It should be noted that there can be an infinite number of antiderivatives for the same function. They will differ from each other by some constant number.

F 1 (x) = F 2 (x) + C.

Indefinite integral.

Definition: Indefinite integral function f(x) is a set of antiderivative functions that are defined by the relation:

Write down:

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

Properties:

1.

2.

3.

4.

Example:

Finding the value of the indefinite integral is mainly associated with finding the antiderivative of the function. For some functions this is quite a difficult task. Below we will consider methods for finding indefinite integrals for the main classes of functions - rational, irrational, trigonometric, exponential, etc.

For convenience, the values ​​of the indefinite integrals of most elementary functions are collected in special tables of integrals, which are sometimes quite voluminous. They include various commonly used combinations of functions. But most of the formulas presented in these tables are consequences of each other, so below we present a table of basic integrals, with the help of which you can obtain the values ​​of indefinite integrals of various functions.

Integral		Meaning	Integral		Meaning
		lnsinx+ C
		ln

Integration methods.

Let's consider three main methods of integration.

Direct integration.

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

Let's look at the application of this method using an example:

We need to find the value of the integral . Based on the well-known differentiation formula
we can conclude that the sought integral is equal to
, where C is some constant number. However, on the other hand
. Thus, we can finally conclude:

Note that, in contrast to differentiation, where clear techniques and methods were used to find the derivative, rules for finding the derivative, and finally the definition of the derivative, such methods are not available for integration. If, when finding the derivative, we used, so to speak, constructive methods, which, based on certain rules, led to the result, then when finding the antiderivative we have to rely mainly on the knowledge of tables of derivatives and antiderivatives.

As for the direct integration method, it is applicable only for some very limited classes of functions. There are very few functions for which you can immediately find an antiderivative. Therefore, in most cases, the methods described below are used.

Method of substitution (replacing variables).

Theorem: If you need to find the integral
, but it is difficult to find the antiderivative, then using the replacement x = (t) and dx = (t)dt we get:

Proof : Let us differentiate the proposed equality:

According to property No. 2 of the indefinite integral discussed above:

f(x) dx = f[  (t)]  (t) dt

which, taking into account the introduced notation, is the initial assumption. The theorem has been proven.

Example. Find the indefinite integral
.

Let's make a replacement t = sinx, dt = cosxdt.

Example.

Replacement
We get:

Below we will consider other examples of using the substitution method for various types of functions.

Integration by parts.

The method is based on the well-known formula for the derivative of a product:

(uv) = uv + vu

where u and v are some functions of x.

In differential form: d(uv) = udv + vdu

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

or
;

We have obtained a formula for integration by parts, which allows us to find the integrals of many elementary functions.

Example.

As you can see, consistent application of the integration by parts formula allows you to gradually simplify the function and bring the integral to a tabular one.

Example.

It can be seen that as a result of repeated application of integration by parts, the function could not be simplified to tabular form. However, the last integral obtained is no different from the original one. Therefore, we move it to the left side of the equality.

Thus, the integral was found without using tables of integrals at all.

Before considering in detail the methods of integrating various classes of functions, we give several more examples of finding indefinite integrals by reducing them to tabular ones.

Example.

Example.

Example.

Example.

Example.

Example.

Example.

Example.

Example.

Example.

Integration of elementary fractions.

Definition: Elementary The following four types of fractions are called:

I.
III.

II.
IV.

m, n – natural numbers (m  2, n  2) and b 2 – 4ac<0.

The first two types of integrals of elementary fractions can be quite simply brought to the table by substitution t = ax + b.

Let us consider the method of integrating elementary fractions of type III.

The fraction integral of type III can be represented as:

Here, in general form, the reduction of a fraction integral of type III to two tabular integrals is shown.

Let's look at the application of the above formula using examples.

Example.

Generally speaking, if the trinomial ax 2 + bx + c has the expression b 2 – 4ac >0, then the fraction, by definition, is not elementary, however, it can nevertheless be integrated in the manner indicated above.

Example.

Example.

Let us now consider methods for integrating simple fractions of type IV.

First, let's consider a special case with M = 0, N = 1.

Then the integral of the form
can be represented in the form by selecting the complete square in the denominator
. Let's make the following transformation:

We will take the second integral included in this equality by parts.

Let's denote:

For the original integral we obtain:

The resulting formula is called recurrent. If you apply it n-1 times, you get a table integral
.

Let us now return to the integral of an elementary fraction of type IV in the general case.

In the resulting equality, the first integral using the substitution t = u 2 + s reduced to tabular , and the recurrence formula discussed above is applied to the second integral.

Despite the apparent complexity of integrating an elementary fraction of type IV, in practice it is quite easy to use for fractions with a small degree n, and the universality and generality of the approach makes possible a very simple implementation of this method on a computer.

Example:

Integration of rational functions.

Integrating rational fractions.

In order to integrate a rational fraction, it is necessary to decompose it into elementary fractions.

Theorem: If
- a proper rational fraction, the denominator P(x) of which is represented as a product of linear and quadratic factors (note that any polynomial with real 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 into elementary ones according to the following scheme:

where A i, B i, M i, N i, R i, S i are some constant quantities.

When integrating rational fractions, they resort to decomposing the original fraction into elementary ones. To find the quantities A i, B i, M i, N i, R i, S i, the so-called method of uncertain coefficients, the essence of which is that in order for two polynomials to be identically equal, it is necessary and sufficient that the coefficients at the same powers of x be equal.

Let's consider the use of this method using a specific example.

Example.

Reducing to a common denominator and equating the corresponding numerators, we get:

Example.

Because If the fraction is improper, you must first select its 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

Let's factorize the denominator of the resulting fraction. It can be seen that at x = 3 the denominator of the fraction turns to zero. Then:

3x 3 – 4x 2 – 17x + 6 x - 3

3x 3 – 9x 2 3x 2 + 5x - 2

So 3x 3 – 4x 2 – 17x + 6 = (x – 3)(3x 2 + 5x – 2) = (x – 3)(x + 2)(3x – 1). Then:

In order to avoid opening brackets, grouping and solving a system of equations (which in some cases may turn out to be quite large) when finding uncertain coefficients, the so-called arbitrary value method. The essence of the method is that several (according to the number of undetermined coefficients) arbitrary values ​​of x are substituted into the above expression. To simplify calculations, it is customary to take as arbitrary values ​​points at which the denominator of the fraction is equal to zero, i.e. in our case – 3, -2, 1/3. We get:

Finally we get:

=

Example.

Let's find the undetermined coefficients:

Then the value of the given integral:

Integration of some trigonometrics

functions.

There can be an infinite number of integrals from trigonometric functions. Most of these integrals cannot be calculated analytically at all, so we will consider some of the most important types of functions that can always be integrated.

Integral of the form
.

Here R is the designation of some rational function of the variables sinx and cosx.

Integrals of this type are calculated using substitution
. This substitution allows you to convert a trigonometric function to a rational one.

,

Then

Thus:

The transformation described above is called universal trigonometric substitution.

Example.

The undoubted advantage of this substitution is that with its help you can always transform a trigonometric function into a rational one and calculate the corresponding integral. The disadvantages include the fact that the transformation can result in a rather complex rational function, the integration of which will take a lot of time and effort.

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

Example.

Integral of the form
If

functionRcosx.

Despite the possibility of calculating such an integral using the universal trigonometric substitution, it is more rational to use the substitution t = sinx.

Function
can contain cosx only in even powers, and therefore can be converted into a rational function with respect to sinx.

Example.

Generally speaking, to apply this method, only the oddness of the function relative to the cosine is necessary, and the degree of the sine included in the function can be any, both integer and fractional.

Integral of the form
If

functionRis odd relative tosinx.

By analogy with the case considered above, the substitution is made t = cosx.

Example.

Integral of the form

functionReven relativelysinxAndcosx.

To transform the function R into a rational one, use the substitution

t = tgx.

Example.

Integral of the product of sines and cosines

various arguments.

Depending on the type of work, 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 used.

Example.

Integration of some irrational functions.

Not every irrational function can have an integral expressed by elementary functions. To find the integral of an irrational function, you should use a substitution that will allow you to transform the function into a rational one, the integral of which can always be found, as is always known.

Let's look at some techniques for integrating various types of irrational functions.

Integral of the form
Wheren- natural number.

Using substitution
the function is rationalized.

Example.

If the composition of an irrational function includes roots of various degrees, then as a new variable it is rational to take the root of a degree equal to the least common multiple of the degrees of the roots included in the expression.

Let's illustrate this with an example.

Example.

Integration of binomial differentials.

Definition. The integration method, in which a given integral is reduced to one or more table integrals by means of identical transformations of the integrand (or integrand expression) and applying the properties of the indefinite integral direct integration .

Often, during direct integration, the following differential transformations are used (the operation of “entering under the differential sign”):

For example. 1) ;

When calculating these integrals, we used formulas 1 and 2 of the table of integrals, which is given below.

Table of basic indefinite integrals.

1. Integration method by substitution (variable replacement).

The method of integration by substitution involves introducing a new integration variable. In this case, the given integral is reduced to a new integral, which is tabular or reducible to it.

This integration method is based on the following theorem:

Theorem. Let the function f(x) be represented in the form: f(x)=g(j(x))×j¢(x), then if G(u) is an antiderivative for g(u), then G(j( x)) is antiderivative of g(j(x)). That is, there is equality: .

For example.

1. Method of integration by parts.

Integration by parts consists of representing the integrand of some integral as the product of two factors u and dv, then using the integration by parts formula.

Theorem Let the functions u(x) and v(x) be differentiable, then the formula holds:

Since u¢(x)dx=du, v¢(x)dx=dv, the formula can be rewritten as:

For example.

The integration by parts formula can be used several times during the solution process.

For example

For example

Let's move from the right side of the equality to the left:

Some types of integrals that are convenient to calculate using the integration by parts method:

; ; , where P(x) is a polynomial in x, k is a certain number	u=P(x), dv – other factors
; ; ; ;	dv=P(x)dх, u – all other factors
; , where a and b are some numbers	, dv – other factors

1. Integrating rational fractions.

Definition Rational we will call fractions of the form , where P n (x), Q m (x) are polynomials of the nth and mth degrees, respectively, in x. The simplest rational fractions include fractions of four types:

Where A and a are some real numbers, the simplest fraction first type;

– simple fraction second type;

– simple fraction third type;

– simple fraction fourth type.

Let's consider the integration of fractions of the first three types.

3) Integration of the simplest fraction of the third type is carried out in two stages. Let's look at the integration process using an example.

(we select the derivative of the denominator in the numerator for subsequent entry under the differential sign: (x 2 +2x+3)¢=2x+2)

Definition Rational fractions are called correct if the degree of the polynomial in the numerator is less than the degree of the polynomial in the denominator and wrong if the degree of the polynomial in the numerator is greater than or equal to the degree of the polynomial in the denominator.

In the case of an improper rational fraction, it is possible to isolate the whole part. To do this, the polynomial from the numerator is divided with the remainder by the polynomial in the denominator. The resulting quotient will be the integer part, and the remainder will be the numerator of the new proper rational fraction. For example, let's select the whole part: .

Thus, the integration of rational fractions in both cases comes down to the integration of a proper rational fraction, which is not always the simplest rational fraction of one of the four types.

Let's consider some polynomial Q(x). Let the number a be the root of this polynomial, then Q(x)=(x-a)Q 1 (x), where Q 1 (x) is a polynomial of degree 1 less than the degree of Q(x). The number a can be a root of multiplicity k, then Q(x) = (x-a) to Q 2 (x), where Q 2 (x) is a polynomial of degree k less than the degree of Q(x). In addition, the polynomial Q(x), along with real roots, can have a complex root a+bi, then the complex number a-bi will also be a root of Q(x). In this case, Q(x)=(x 2 +px+q)Q 3 (x), where x 2 +px+q=(x-(a+bi))(x-(a-bi)). If the indicated complex numbers are roots of multiplicity m, then Q(x)=(x 2 +px+q) m Q 4 (x).

Thus, any polynomial Q(x) can be represented as:

Q(x)=(x-a 1) to 1 (x-a 2) to 2 ...(x-a n) k n (x 2 +p 1 x+q 1) m 1 (x 2 +p 2 x+ q 2) m 2 …(x 2 +p s x+q s) m s.

Theorem. Any proper rational fraction can be represented as a sum of the simplest rational fractions of types 1-4.

For example. Let's consider an algorithm for representing a proper rational fraction as a sum of the simplest rational fractions of types 1-4.

Since the denominators of the fractions are equal, obviously the numerators must also be equal, and this equality is possible if the coefficients are equal for the same powers of x. Thus, substituting their values ​​instead of the undetermined coefficients A, B, C: .

For example Find the integral.

The integrand is an improper rational fraction. After dividing the numerator by the denominator with the remainder we get: .

Let us decompose a proper rational fraction into its simplest fractions using the method of indefinite coefficients:

It follows that Solving the resulting system of linear equations, we obtain Then , that is, = ;

We'll find it separately

Thus, .

1. Integration of trigonometric functions.

1. Let it be necessary to find , where R is some function

When finding such integrals, it is often useful to use the universal trigonometric substitution: . With its help, you can always go from the integral of a trigonometric function to the integral of a rational function:

Х=2arctgt, .

2. If the function R(sinx, cosx) is odd relative to sinx, that is, R(-sinx, cosx)=- R(sinx, cosx), then use the substitution cosx=t;

3. If the function R(sinx, cosx) is odd with respect to cosx, that is, R(sinx, -cosx)=- R(sinx, cosx), then use the substitution sinx=t;

4. If the function R(sinx, cosx) is even with respect to sinx and cosx, that is, R(-sinx, -cosx)=R(sinx, cosx), then use the substitution tgx=t; the same substitution applies in the case of .

For example.

For example Find the integral. The integrand is even with respect to sinx, then we use the substitution tgx=t.

5. To find integrals of the form, use the following techniques:

a) if n is an odd positive integer, then use the substitution sinx=t;

b) if m is an odd positive integer, then use the substitution сosx=t;

c) if m and n are non-negative even integers, then order reduction formulas are used; ; ;

d) if m+n is an even negative integer, then use the substitution tgx=t.

For example. .

For example.. ; are reduced to integrals of trigonometric functions using the following substitutions:

a) for the integral, substitution x=a×sint;

b) for the integral, substitution x=a×tgt;

c) for the integral, substitution .

We cannot always calculate antiderivative functions, but the differentiation problem can be solved for any function. That is why there is no single integration method that can be used for any type of calculation.

In this material, we will look at examples of solving problems related to finding the indefinite integral, and see what types of integrands each method is suitable for.

Yandex.RTB R-A-339285-1

Direct integration method

The main method for calculating the antiderivative function is direct integration. This action is based on the properties of the indefinite integral, and for the calculations we need a table of antiderivatives. Other methods can only help bring the original integral to tabular form.

Example 1

Calculate the set of antiderivatives of the function f (x) = 2 x + 3 2 · 5 x + 4 3 .

Solution

First, let's change the form of the function to f (x) = 2 x + 3 2 5 x + 4 3 = 2 x + 3 2 5 x + 4 1 3.

We know that the integral of the sum of functions will be equal to the sum of these integrals, which means:

∫ f (x) d x = ∫ 3 2 5 x + 4 3 = 2 x + 3 2 5 x + 4 1 3 d x = ∫ 3 2 5 x + 4 1 3 d x

We derive the numerical coefficient behind the integral sign:

∫ f (x) d x = ∫ 2 x d x + ∫ 3 2 (5 x + 4) 1 3 d x = = ∫ 2 x d x + 2 3 ∫ (5 x + 4) 1 3 d x

To find the first integral, we will need to refer to the table of antiderivatives. We take from it the value ∫ 2 x d x = 2 x ln 2 + C 1

To find the second integral, you will need a table of antiderivatives for the power function ∫ x p · d x = x p + 1 p + 1 + C , as well as the rule ∫ f k · x + b d x = 1 k · F (k · x + b) + C .

Therefore, ∫ f (x) d x = ∫ 2 x d x + 3 2 ∫ 5 x + 4 1 3 d x = = 2 x ln 2 + C 1 + 3 2 3 20 (5 x + 4) 4 3 + C 2 = = 2 x ln 2 + 9 40 5 x + 4 4 3 + C

We got the following:

∫ f (x) d x = ∫ 2 x d x + 3 2 ∫ 5 x + 4 1 3 d x = = 2 x ln 2 + C 1 + 3 2 3 20 (5 x + 4) 4 3 + C 2 = = 2 x ln 2 + 9 40 5 x + 4 4 3 + C

with C = C 1 + 3 2 C 2

Answer:∫ f (x) d x = 2 x ln 2 + 9 40 5 x + 4 4 3 + C

We devoted a separate article to direct integration using tables of antiderivatives. We recommend that you familiarize yourself with it.

Substitution method

This method of integration consists in expressing the integrand through a new variable introduced specifically for this purpose. As a result, we should get a tabular form of the integral or simply a less complex integral.

This method is very useful when you need to integrate functions with radicals or trigonometric functions.

Example 2

Evaluate the indefinite integral ∫ 1 x 2 x - 9 d x .

Solution

Let's add one more variable z = 2 x - 9 . Now we need to express x in terms of z:

z 2 = 2 x - 9 ⇒ x = z 2 + 9 2 ⇒ d x = d z 2 + 9 2 = z 2 + 9 2 " d z = 1 2 z d z = z d z

∫ d x x 2 x - 9 = ∫ z d z z 2 + 9 2 · z = 2 ∫ d z z 2 + 9

We take the table of antiderivatives and find out that 2 ∫ d z z 2 + 9 = 2 3 a r c t g z 3 + C .

Now we need to return to the variable x and get the answer:

2 3 a r c t g z 3 + C = 2 3 a r c t g 2 x - 9 3 + C

Answer:∫ 1 x 2 x - 9 d x = 2 3 a r c t g 2 x - 9 3 + C .

If we have to integrate functions with irrationality of the form x m (a + b x n) p, where the values ​​m, n, p are rational numbers, then it is important to correctly formulate an expression for introducing a new variable. Read more about this in the article on integrating irrational functions.

As we said above, the substitution method is convenient to use when you need to integrate a trigonometric function. For example, using a universal substitution, you can reduce an expression to a fractionally rational form.

This method explains the integration rule ∫ f (k · x + b) d x = 1 k · F (k · x + b) + C .

We add another variable z = k x + b. We get the following:

x = z k - b k ⇒ d x = d z k - b k = z k - b k " d z = d z k

Now we take the resulting expressions and add them to the integral specified in the condition:

∫ f (k x + b) d x = ∫ f (z) d z k = 1 k ∫ f (z) d z = = 1 k F z + C 1 = F (z) k + C 1 k

If we accept C 1 k = C and return to the original variable x, then we get:

F (z) k + C 1 k = 1 k F k x + b + C

Method of subscribing to the differential sign

This method is based on transforming the integrand into a function of the form f (g (x)) d (g (x)). After this, we perform a substitution by introducing a new variable z = g (x), find an antiderivative for it and return to the original variable.

∫ f (g (x)) d (g (x)) = g (x) = z = ∫ f (z) d (z) = = F (z) + C = z = g (x) = F ( g(x)) + C

To solve problems faster using this method, keep a table of derivatives in the form of differentials and a table of antiderivatives on hand to find the expression to which the integrand will need to be reduced.

Let us analyze a problem in which we need to calculate the set of antiderivatives of the cotangent function.

Example 3

Calculate the indefinite integral ∫ c t g x d x .

Solution

Let's transform the original expression under the integral using basic trigonometric formulas.

c t g x d x = cos s d x sin x

We look at the table of derivatives and see that the numerator can be subsumed under the differential sign cos x d x = d (sin x), which means:

c t g x d x = cos x d x sin x = d sin x sin x, i.e. ∫ c t g x d x = ∫ d sin x sin x .

Let us assume that sin x = z, in this case ∫ d sin x sin x = ∫ d z z. According to the table of antiderivatives, ∫ d z z = ln z + C . Now let's return to the original variable ∫ d z z = ln z + C = ln sin x + C .

The entire solution can be briefly written as follows:

∫ с t g x d x = ∫ cos x d x sin x = ∫ d sin x sin x = s i n x = t = = ∫ d t t = ln t + C = t = sin x = ln sin x + C

Answer: ∫ c t g x d x = ln sin x + C

The method of subscribing to the differential sign is very often used in practice, so we advise you to read a separate article dedicated to it.

Method of integration by parts

This method is based on transforming the integrand into a product of the form f (x) d x = u (x) v " x d x = u (x) d (v (x)), after which the formula ∫ u (x) d ( v (x)) = u (x) · v (x) - ∫ v (x) · d u (x).This is a very convenient and common solution method. Sometimes partial integration in one problem has to be applied several times before obtaining the desired result.

Let us analyze a problem in which we need to calculate the set of antiderivatives of the arctangent.

Example 4

Calculate the indefinite integral ∫ a r c t g (2 x) d x .

Solution

Let's assume that u (x) = a r c t g (2 x), d (v (x)) = d x, in this case:

d (u (x)) = u " (x) d x = a r c t g (2 x) " d x = 2 d x 1 + 4 x 2 v (x) = ∫ d (v (x)) = ∫ d x = x

When we calculate the value of the function v (x), we should not add an arbitrary constant C.

∫ a r c t g (2 x) d x = u (x) v (x) - ∫ v (x) d (u (x)) = = x a r c t g (2 x) - ∫ 2 x d x 1 + 4 x 2

We calculate the resulting integral using the method of subsuming the differential sign.

Since ∫ a r c t g (2 x) d x = u (x) · v (x) - ∫ v (x) d (u (x)) = x · a r c t g (2 x) - ∫ 2 x d x 1 + 4 x 2 , then 2 x d x = 1 4 d (1 + 4 x 2) .

∫ a r c t g (2 x) d x = x · a r c t g (2 x) - ∫ 2 x d x 1 + 4 x 2 = = x · a r c t g (2 x) - 1 4 ln 1 + 4 x 2 + C 1 = = x · a r c t g (2 x) - 1 4 ln 1 + 4 x 2 + C

Answer:∫ a r c t g (2 x) d x = x · a r c t g (2 x) - 1 4 ln 1 + 4 x 2 + C .

The main difficulty in using this method is the need to choose which part to take as the differential and which part as the function u (x). The article on the method of integration by parts provides some advice on this issue that you should familiarize yourself with.

If we need to find the set of antiderivatives of a fractionally rational function, then we must first represent the integrand as a sum of simple fractions, and then integrate the resulting fractions. For more information, see the article on integrating simple fractions.

If we integrate a power expression of the form sin 7 x · d x or d x (x 2 + a 2) 8, then we will benefit from recurrence formulas that can gradually lower the power. They are derived using sequential repeated integration by parts. We recommend reading the article “Integration using recurrence formulas.

Let's summarize. To solve problems, it is very important to know the method of direct integration. Other methods (substitution, substitution, integration by parts) also allow you to simplify the integral and bring it to tabular form.

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