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Calculation of a definite integral by the direct integration method. Integration methods

A review of methods for calculating indefinite integrals is presented. The main methods of integration are considered, which include integrating the sum and difference, placing a constant outside the integral sign, replacing a variable, and integrating by parts. Special methods and techniques for integrating fractions, roots, trigonometric and exponential functions are also discussed.

Content

Rule for integrating sums (differences)

Moving the constant outside the integral sign

Let c be a constant independent of x. Then it can be taken out of the integral sign:

Variable replacement

Let x be a function of the variable t, x = φ(t), then
.
Or vice versa, t = φ(x) ,
.

Using a change of variable, you can not only calculate simple integrals, but also simplify the calculation of more complex ones.

Integration by parts rule

Integration of fractions (rational functions)

Let us introduce the notation. Let P k (x), Q m (x), R n (x) denote polynomials of degrees k, m, n, respectively, with respect to the variable x.

Consider an integral consisting of a fraction of polynomials (the so-called rational function):

If k ≥ n, then you first need to select the whole part of the fraction:
.
The integral of the polynomial S k-n (x) is calculated using the table of integrals.

The integral remains:
, where m< n .
To calculate it, the integrand must be decomposed into simple fractions.

To do this you need to find the roots of the equation:
Q n (x) = 0 .
Using the obtained roots, you need to represent the denominator as a product of factors:
Q n (x) = 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 for x n, x 2 + ex + f > 0, x 2 + gx + k > 0, ....

After this, break down the fraction into its simplest form:

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

are reduced to tabular substitution t = x - a.

Consider the integral:

Let's transform the numerator:
.
Substituting into the integrand, we obtain an expression that includes two integrals:
,
.
The first one, by substitution t = x 2 + ex + f, is reduced to a tabular one.
Second, according to the reduction formula:

is reduced to the integral

Let's reduce its denominator to the sum of squares:
.
Then by substitution, the integral

is also tabulated.

Integration of irrational functions

Let us introduce the notation. Let R(u 1, u 2, ..., u n) mean a rational function of the variables u 1, u 2, ..., u n. That is
,
where P, Q are polynomials in the variables u 1, u 2, ..., u n.

Fractional linear irrationality

Let's consider integrals of the form:
,
where are rational numbers, m 1, n 1, ..., m s, n s are integers.
Let n be the common denominator of the numbers r 1, ..., r s.
Then the integral is reduced to the integral of rational functions by substitution:
.

Integrals from differential binomials

Consider the integral:
,
where m, n, p are rational numbers, a, b are real numbers.
Such integrals reduce to integrals of rational functions in three cases.

1) If p is an integer. Substitution x = t N, where N is the common denominator of the fractions m and n.
2) If - an integer. Substitution a x n + b = t M, where M is the denominator of the number p.
3) If - an integer. Substitution a + b x - n = t M, where M is the denominator of the number p.

If none of the three numbers is an integer, then, according to Chebyshev’s theorem, integrals of this type cannot be expressed by a finite combination of elementary functions.

In some cases, it is first useful to reduce the integral to more convenient values m and p. This can be done using reduction formulas:
;
.

Integrals containing the square root of a square trinomial

Here we consider integrals of the form:
,

Euler substitutions

Such integrals can be reduced to integrals of rational functions of one of three Euler substitutions:
, for a > 0;
, for c > 0 ;
, where x 1 is the root of the equation a x 2 + b x + c = 0. If this equation has real roots.

Trigonometric and hyperbolic substitutions

Direct methods

In most cases, Euler substitutions result in longer calculations than direct methods. Using direct methods, the integral is reduced to one of the forms listed below.

Type I

Integral of the form:
,
where P n (x) is a polynomial of degree n.

Such integrals are found by the method of indefinite coefficients using the identity:

Differentiating this equation and equating the left and right sides, we find the coefficients A i.

Type II

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

Substitution t = (x - α) -1 this integral is reduced to the previous type. If m ≥ n, then the fraction should have an integer 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, the constants α, β must be chosen such that the coefficients for t become zero:
B = 0, B 1 = 0.
Then the integral decomposes into the sum of integrals of two types:
;
,
which are integrated, respectively, by substitutions:
z 2 = A 1 t 2 + C 1 ;
y 2 = A 1 + C 1 t -2 .

General case

Integration of transcendental (trigonometric and exponential) functions

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

Integration of rational trigonometric functions of cos x and sin x

Let's consider integrals of trigonometric functions of the form:
,
where R is a rational function. This may also include tangents and cotangents, which should be converted using sines and cosines.

When integrating such functions, it is useful to keep three rules in mind:
1) if R( cos x, sin x) multiplied by -1 from the sign change before one of the quantities cos x or sin x, then it is useful to denote the other of them by t.
2) if R( cos x, sin x) does not change due to a change in sign at the same time before cos x And sin x, then it is useful to put tg x = t or cot x = t.
3) substitution in all cases leads to the integral of the rational fraction. Unfortunately, this substitution results in longer calculations than the previous ones, if applicable.

Product of power functions of cos x and sin x

Let's consider integrals of the form:

If m and n are rational numbers, then one of the substitutions t = sin x or t = cos x the integral is reduced to the integral of the differential binomial.

If m and n are integers, then the integrals are calculated by integration by parts. This produces the following reduction formulas:

;
;
;
.

Integration by parts

Application of Euler's formula

If the integrand is linear with respect to one of the functions
cos ax or sinax, then it is convenient to apply Euler’s formula:
e iax = cos ax + isin ax(where i 2 = - 1 ),
replacing this function with e iax and highlighting the real one (when replacing cos ax) or imaginary part (when replacing sinax) from the obtained result.

References:
N.M. Gunther, R.O. Kuzmin, Collection of problems in higher mathematics, “Lan”, 2003.

Basic integration methods

There are three main integration method.

1. Direct integration− calculation of integrals using a table of integrals and the basic properties of indefinite integrals.

Example 3. Calculate the integral: ∫ tg 2 xdx.

2. Substitution method . In many cases, the introduction of a new integration variable allows one to reduce the calculation of a given integral to finding a tabular one. This method is also called variable replacement method.

Theorem 3. Let the function x = φ(t) defined, continuous and differentiable on a certain interval T let it go X- the set of values of this function, on it, i.e. on T complex function defined f(φ(t)). Then if ∫ f(x)dx= F(x)+ C, That

∫f(x)dx=∫f(φ(t)) φ ′ (t)dt. (1)

Formula (1) is called formula changing a variable in an indefinite integral.

Comment. After calculating the integral ∫ f(φ(t)) φ ′ (t)dt you need to go back to the variable X.

Example 4. Find the integral: ∫cos 3 x sin xdx.

a) Replace sin xdx on (− d cos x), i.e. we introduce the function cos x under the differential sign. We get

3. Method of integration by parts

Theorem 4. Let the functions u(x) And v(x) defined and differentiable on a certain interval X let it go u′ (x)v(x) has an antiderivative on this interval, i.e. there is an integral ∫ u′( x)v(x)dx.

Then on this interval the function has an antiderivative and u(x)v′ (x) and the formula is correct:

∫u(x)v′( x)dx= u(x)v(x) −∫v(x)u′( x)dx(2)

∫udv= uv−∫vdu.(2′)

Formulas (2) and (2′) are called formulas for integration by parts in the indefinite integral.

Using the method of integration by parts, integrals of the following functions are calculated: P(x)arcsin( ax),P(x)arccos( ax), P(x)arctg( ax), P(x)arcctg( ax),P(x)ln x, P(x)e kx, P(x)sin kx, P(x)cos kx, Here P(x)- polynomial; e ax cos bx, e ax sin bx.

Of course, these functions do not exhaust all the integrals that are calculated using the method of integration by parts.

Example 6. Find the integral: ∫ arctg 3xdx.

Solution. Let's put u= arctg 3x; dv= dx. Then

According to formula (2) we have

The problem of finding an antiderivative function does not always have a solution, while we can differentiate any function. This explains the lack of a universal integration method.

In this article we will look at the basic methods for finding the indefinite integral using examples with detailed solutions. We will also group the types of integrand functions characteristic of each integration method.

Page navigation.

Direct integration.

Undoubtedly, the main method of finding an antiderivative function is direct integration using a table of antiderivatives and the properties of the indefinite integral. All other methods are used only to reduce the original integral to tabular form.

Example.

Find the set of antiderivatives of the function.

Solution.

Let's write the function in the form .

Since the integral of a sum of functions is equal to the sum of integrals, then

The numerical coefficient can be taken out of the integral sign:

The first of the integrals is reduced to tabular form, therefore, from the table of antiderivatives for the exponential function we have .

To find the second integral, we use the table of antiderivatives for the power function and the rule . That is, .

Hence,

Where

Integration by substitution method.

The essence of the method is that we introduce a new variable, express the integrand through this variable, and as a result we arrive at a tabular (or simpler) form of the integral.

Very often, the substitution method comes to the rescue when integrating trigonometric functions and functions with radicals.

Example.

Find the indefinite integral .

Solution.

Let's introduce a new variable. Let's express x through z:

We substitute the resulting expressions into the original integral:

From the table of antiderivatives we have .

It remains to return to the original variable x:

Answer:

Very often the substitution method is used when integrating trigonometric functions. For example, using the universal trigonometric substitution allows you to transform the integrand into a fractionally rational form.

The substitution method allows you to explain the integration rule .

We introduce a new variable, then

We substitute the resulting expressions into the original integral:

If we accept and return to the original variable x, we get

Submitting the differential sign.

The method of subsuming the differential sign is based on reducing the integrand to the form . Next, the substitution method is used: a new variable is introduced and after finding the antiderivative for the new variable, we return to the original variable, that is

For convenience, place it in front of your eyes in the form of differentials to make it easier to convert the integrand, as well as a table of antiderivatives to see what form to convert the integrand to.

For example, let's find the set of antiderivatives of the cotangent function.

Example.

Find the indefinite integral.

Solution.

The integrand can be transformed using trigonometry formulas:

Looking at the table of derivatives, we conclude that the expression in the numerator can be subsumed under the differential sign , That's why

That is .

Let it be then . From the table of antiderivatives we see that . Returning to the original variable .

Without explanation, the solution is written as follows:

Integration by parts.

Integration by parts is based on representing the integrand as a product and then applying the formula. This method is a very powerful integration tool. Depending on the integrand, the method of integration by parts sometimes has to be applied several times in a row before obtaining the result. For example, let's find the set of antiderivatives of the arctangent function.

Example.

Calculate the indefinite integral.

Solution.

Let it be then

It should be noted that when finding the function v(x) do not add an arbitrary constant C.

Now we apply the integration by parts formula:

We calculate the last integral using the method of subsuming it under the differential sign.

Since then . That's why

Hence,

Where .

Answer:

The main difficulties in integrating by parts arise from the choice: which part of the integrand to take as the function u(x) and which part as the differential d(v(x)). However, there are a number of standard recommendations, which we recommend that you familiarize yourself with in the section integration by parts.

When integrating power expressions, for example or , use recurrent formulas that allow you to reduce the degree from step to step. These formulas are obtained by successive repeated integration by parts. We recommend that you familiarize yourself with the section integration using recurrence formulas.

In conclusion, I would like to summarize all the material in this article. The basis of the fundamentals is the method of direct integration. The methods of substitution, substitution under the differential sign and the method of integration by parts make it possible to reduce the original integral to a tabular one.

In this topic we will talk in detail about the properties of the indefinite integral and about finding the integrals themselves using the mentioned properties. We will also work with the table of indefinite integrals. The material presented here is a continuation of the topic "Indefinite integral. Beginning". To be honest, test papers rarely contain integrals that can be taken using typical tables and/or simple properties. These properties can be compared to the alphabet, knowledge and understanding of which is necessary to understand the mechanism for solving integrals in other topics. Often integration using tables of integrals and properties of the indefinite integral is called direct integration.

What I'm getting at: the functions change, but the formula for finding the derivative remains unchanged, unlike the integral, for which we already had to list two methods.

Let's go further. To find the derivative $y=x^(-\frac(1)(2))\cdot(1+x^(\frac(1)(4)))^\frac(1)(3)$ all the same applies the same formula $(u\cdot v)"=u"\cdot v+u\cdot v"$, into which you will have to substitute $u=x^(-\frac(1)(2))$, $v=( 1+x^(\frac(1)(4)))^\frac(1)(3)$. But to find the integral $\int x^(-\frac(1)(2))\cdot( 1+x^(\frac(1)(4)))^\frac(1)(3) dx$ will require the use of a new method - Chebyshev substitutions.

And finally: to find the derivative of the function $y=\sin x\cdot\frac(1)(x)$, the formula $(u\cdot v)"=u"\cdot v+u\cdot v"$ is again applicable, into which instead of $u$ and $v$ we substitute $\sin x$ and $\frac(1)(x)$, respectively. But $\int \sin x\cdot\frac(1)(x) dx$ is not is taken, or more precisely, is not expressed in terms of a finite number of elementary functions.

Let's summarize: where one formula was needed to find the derivative, four were required for the integral (and this is not the limit), and in the latter case the integral refused to be found at all. The function was changed - a new integration method was needed. This is where we have multi-page tables in reference books. The lack of a general method (suitable for solving “manually”) leads to an abundance of private methods that are applicable only for integrating their own, extremely limited class of functions (in further topics we will deal with these methods in detail). Although I cannot help but note the presence of the Risch algorithm (I advise you to read the description on Wikipedia), it is only suitable for program processing of indefinite integrals.

Question #3

But if there are so many of these properties, how can I learn to take integrals? It was easier with derivatives!

For a person, there is only one way so far: to solve as many examples as possible using various integration methods, so that when a new indefinite integral appears, you can choose a solution method for it based on your experience. I understand that the answer is not very reassuring, but there is no other way.

Properties of the indefinite integral

Property No. 1

The derivative of the indefinite integral is equal to the integrand, i.e. $\left(\int f(x) dx\right)"=f(x)$.

This property is quite natural, since the integral and derivative are mutually inverse operations. For example, $\left(\int \sin 3x dx\right)"=\sin 3x$, $\left(\int \left(3x^2+\frac(4)(\arccos x)\right) dx\ right)"=3x^2+\frac(4)(\arccos x)$ and so on.

Property No. 2

The indefinite integral of the differential of some function is equal to this function, i.e. $\int \mathrm d F(x) =F(x)+C$.

Usually this property is perceived as somewhat difficult, since it seems that there is “nothing” under the integral. To avoid this, you can write the indicated property as follows: $\int 1\mathrm d F(x) =F(x)+C$. An example of using this property: $\int \mathrm d(3x^2+e^x+4)=3x^2+e^x+4+C$ or, if you like, in this form: $\int 1\; \mathrm d(3x^2+e^x+4) =3x^2+e^x+4+C$.

Property No. 3

The constant factor can be taken out of the integral sign, i.e. $\int a\cdot f(x) dx=a\cdot\int f(x) dx$ (we assume that $a\neq 0$).

The property is quite simple and, perhaps, does not require comments. Examples: $\int 3x^5 dx=3\cdot \int x^5 dx$, $\int (2x+4e^(7x)) dx=2\cdot\int(x+2e^(7x))dx $, $\int kx^2dx=k\cdot\int x^2dx$ ($k\neq 0$).

Property No. 4

The integral of the sum (difference) of two functions is equal to the sum (difference) of the integrals of these functions:

$$\int(f_1(x)\pm f_2(x))dx=\int f_1(x)dx\pm\int f_2(x)dx$$

Examples: $\int(\cos x+x^2)dx=\int \cos xdx+\int x^2 dx$, $\int(e^x - \sin x)dx=\int e^xdx -\ int \sin x dx$.

In standard tests, properties No. 3 and No. 4 are usually used, so we will dwell on them in more detail.

Example No. 3

Find $\int 3 e^x dx$.

Let's use property No. 3 and take out the constant, i.e. number $3$, for the integral sign: $\int 3 e^x dx=3\cdot\int e^x dx$. Now let's open the table of integrals and substituting $u=x$ into formula No. 4 we get: $\int e^x dx=e^x+C$. It follows that $\int 3 e^x dx=3\cdot\int e^x dx=3e^x+C$. I assume that the reader will immediately have a question, so I will formulate this question separately:

Question #4

If $\int e^x dx=e^x+C$, then $\int 3 e^x dx=3\cdot\int e^x dx=3\cdot\left(e^x+C\right) =3e^x+3C$! Why did they just write $3e^x+C$ instead of $3e^x+3C$?

The question is completely reasonable. The point is that the integral constant (i.e. that same number $C$) can be represented in the form of any expression: the main thing is that this expression “runs through” the entire set of real numbers, i.e. varied from $-\infty$ to $+\infty$. For example, if $-\infty≤ C ≤ +\infty$, then $-\infty≤ \frac(C)(3) ≤ +\infty$, so the constant $C$ can be represented in the form $\frac(C)( 3)$. We can write that $\int e^x dx=e^x+\frac(C)(3)$ and then $\int 3 e^x dx=3\cdot\int e^x dx=3\cdot\left (e^x+\frac(C)(3)\right)=3e^x+C$. As you can see, there is no contradiction here, but you need to be careful when changing the form of the integral constant. For example, representing the constant $C$ as $C^2$ would be an error. The point is that $C^2 ≥ 0$, i.e. $C^2$ does not change from $-\infty$ to $+\infty$ and does not “run through” all real numbers. Likewise, it would be a mistake to represent a constant as $\sin C$, because $-1≤ \sin C ≤ 1$, i.e. $\sin C$ does not "run" through all values of the real axis. In what follows, we will not discuss this issue in detail, but will simply write the constant $C$ for each indefinite integral.

Example No. 4

Find $\int\left(4\sin x-\frac(17)(x^2+9)-8x^3 \right)dx$.

Let's use property No. 4:

$$\int\left(4\sin x-\frac(17)(x^2+9)-8x^3 \right) dx=\int 4\sin x dx-\int\frac(17)(x ^2+9)dx-\int8x^3dx$$

Now let’s take the constants (numbers) outside the integral signs:

$$\int 4\sin x dx-\int\frac(17)(x^2+9)dx-\int8x^3dx=4\int \sin x dx-17\int\frac(dx)(x^ 2+9)-8\int x^3dx$$

Next, we will work with each obtained integral separately. The first integral, i.e. $\int \sin x dx$, can be easily found in the table of integrals under No. 5. Substituting $u=x$ into formula No. 5 we get: $\int \sin x dx=-\cos x+C$.

To find the second integral $\int\frac(dx)(x^2+9)$ you need to apply formula No. 11 from the table of integrals. Substituting $u=x$ and $a=3$ into it we get: $\int\frac(dx)(x^2+9)=\frac(1)(3)\cdot \arctg\frac(x)( 3)+C$.

And finally, to find $\int x^3dx$ we use formula No. 1 from the table, substituting $u=x$ and $\alpha=3$ into it: $\int x^3dx=\frac(x^(3 +1))(3+1)+C=\frac(x^4)(4)+C$.

All integrals included in the expression $4\int \sin x dx-17\int\frac(dx)(x^2+9)-8\int x^3dx$ have been found. All that remains is to substitute them:

$$4\int \sin x dx-17\int\frac(dx)(x^2+9)-8\int x^3dx=4\cdot(-\cos x)-17\cdot\frac(1) (3)\cdot\arctg\frac(x)(3)-8\cdot\frac(x^4)(4)+C=\\ =-4\cdot\cos x-\frac(17)(3 )\cdot\arctg\frac(x)(3)-2\cdot x^4+C.$$

The problem is solved, the answer is: $\int\left(4\sin x-\frac(17)(x^2+9)-8x^3 \right)dx=-4\cdot\cos x-\frac(17 )(3)\cdot\arctg\frac(x)(3)-2\cdot x^4+C$. I will add one small note to this problem:

Just a small note

Perhaps no one will need this insert, but I’ll still mention that $\frac(1)(x^2+9)\cdot dx=\frac(dx)(x^2+9)$. Those. $\int\frac(17)(x^2+9)dx=17\cdot\int\frac(1)(x^2+9)dx=17\cdot\int\frac(dx)(x^2 +9)$.

Let's look at an example in which we use formula No. 1 from the table of integrals to interpose irrationalities (roots, in other words).

Example No. 5

Find $\int\left(5\cdot\sqrt(x^4)-\frac(14)(\sqrt(x^6))\right)dx$.

To begin with, we will do the same actions as in example No. 3, namely: we will decompose the integral into two and move the constants beyond the signs of the integrals:

$$\int\left(5\cdot\sqrt(x^4)-\frac(14)(\sqrt(x^6)) \right)dx=\int\left(5\cdot\sqrt(x^ 4) \right)dx-\int\frac(14)(\sqrt(x^6)) dx=\\ =5\cdot\int\sqrt(x^4) dx-14\cdot\int\frac( dx)(\sqrt(x^6)) $$

Since $\sqrt(x^4)=x^(\frac(4)(7))$, then $\int\sqrt(x^4) dx=\int x^(\frac(4)(7 ))dx$. To find this integral, we apply formula No. 1, substituting $u=x$ and $\alpha=\frac(4)(7)$ into it: $\int x^(\frac(4)(7))dx=\ frac(x^(\frac(4)(7)+1))(\frac(4)(7)+1)+C=\frac(x^(\frac(11)(7)))(\ frac(11)(7))+C=\frac(7\cdot\sqrt(x^(11)))(11)+C$. If you wish, you can represent $\sqrt(x^(11))$ as $x\cdot\sqrt(x^(4))$, but this is not necessary.

Let us now turn to the second integral, i.e. $\int\frac(dx)(\sqrt(x^6))$. Since $\frac(1)(\sqrt(x^6))=\frac(1)(x^(\frac(6)(11)))=x^(-\frac(6)(11) )$, then the integral under consideration can be represented in the following form: $\int\frac(dx)(\sqrt(x^6))=\int x^(-\frac(6)(11))dx$. To find the resulting integral, we apply formula No. 1 from the table of integrals, substituting $u=x$ and $\alpha=-\frac(6)(11)$ into it: $\int x^(-\frac(6)(11) ))dx=\frac(x^(-\frac(6)(11)+1))(-\frac(6)(11)+1)+C=\frac(x^(\frac(5) (11)))(\frac(5)(11))+C=\frac(11\cdot\sqrt(x^(5)))(5)+C$.

Substituting the results obtained, we get the answer:

$$5\cdot\int\sqrt(x^4) dx-14\cdot\int\frac(dx)(\sqrt(x^6))= 5\cdot\frac(7\cdot\sqrt(x^( 11)))(11)-14\cdot\frac(11\cdot\sqrt(x^(5)))(5)+C= \frac(35\cdot\sqrt(x^(11)))( 11)-\frac(154\cdot\sqrt(x^(5)))(5)+C. $$

Answer: $\int\left(5\cdot\sqrt(x^4)-\frac(14)(\sqrt(x^6))\right)dx=\frac(35\cdot\sqrt(x^(11 )))(11)-\frac(154\cdot\sqrt(x^(5)))(5)+C$.

And finally, let’s take the integral that falls under formula No. 9 of the table of integrals. Example No. 6, which we will now move on to, could be solved in another way, but this will be discussed in subsequent topics. For now, we will remain within the framework of using the table.

Example No. 6

Find $\int\frac(12)(\sqrt(15-7x^2))dx$.

First, let's do the same operation as before: moving the constant (the number $12$) outside the integral sign:

$$ \int\frac(12)(\sqrt(15-7x^2))dx=12\cdot\int\frac(1)(\sqrt(15-7x^2))dx=12\cdot\int \frac(dx)(\sqrt(15-7x^2)) $$

The resulting integral $\int\frac(dx)(\sqrt(15-7x^2))$ is already close to the tabular one $\int\frac(du)(\sqrt(a^2-u^2))$ (formula No. 9 table of integrals). The difference in our integral is that before $x^2$ under the root there is a coefficient $7$, which the table integral does not allow. Therefore, we need to get rid of this seven by moving it beyond the root sign:

$$ 12\cdot\int\frac(dx)(\sqrt(15-7x^2))=12\cdot\int\frac(dx)(\sqrt(7\cdot\left(\frac(15)() 7)-x^2\right)))= 12\cdot\int\frac(dx)(\sqrt(7)\cdot\sqrt(\frac(15)(7)-x^2))=\frac (12)(\sqrt(7))\cdot\int\frac(dx)(\sqrt(\frac(15)(7)-x^2)) $$

If we compare the table integral $\int\frac(du)(\sqrt(a^2-u^2))$ and $\int\frac(dx)(\sqrt(\frac(15)(7)-x^ 2))$ it becomes clear that they have the same structure. Only in the integral $\int\frac(dx)(\sqrt(\frac(15)(7)-x^2))$ instead of $u$ there is $x$, and instead of $a^2$ there is $\frac (15)(7)$. Well, if $a^2=\frac(15)(7)$, then $a=\sqrt(\frac(15)(7))$. Substituting $u=x$ and $a=\sqrt(\frac(15)(7))$ into the formula $\int\frac(du)(\sqrt(a^2-u^2))=\arcsin\ frac(u)(a)+C$, we get the following result:

$$ \frac(12)(\sqrt(7))\cdot\int\frac(dx)(\sqrt(\frac(15)(7)-x^2))= \frac(12)(\sqrt (7))\cdot\arcsin\frac(x)(\sqrt(\frac(15)(7)))+C $$

If we take into account that $\sqrt(\frac(15)(7))=\frac(\sqrt(15))(\sqrt(7))$, then the result can be rewritten without the “three-story” fractions:

$$ \frac(12)(\sqrt(7))\cdot\arcsin\frac(x)(\sqrt(\frac(15)(7)))+C=\frac(12)(\sqrt(7 ))\cdot\arcsin\frac(x)(\frac(\sqrt(15))(\sqrt(7)))+C= \frac(12)(\sqrt(7))\cdot\arcsin\frac (\sqrt(7)\;x)(\sqrt(15))+C $$

The problem is solved, the answer is received.

Answer: $\int\frac(12)(\sqrt(15-7x^2))dx=\frac(12)(\sqrt(7))\cdot\arcsin\frac(\sqrt(7)\;x) (\sqrt(15))+C$.

Example No. 7

Find $\int\tg^2xdx$.

There are methods for integrating trigonometric functions. However, in this case, you can get by with knowledge of simple trigonometric formulas. Since $\tg x=\frac(\sin x)(\cos x)$, then $\left(\tg x\right)^2=\left(\frac(\sin x)(\cos x) \right)^2=\frac(\sin^2x)(\cos^2x)$. Considering $\sin^2x=1-\cos^2x$, we get:

$$ \frac(\sin^2x)(\cos^2x)=\frac(1-\cos^2x)(\cos^2x)=\frac(1)(\cos^2x)-\frac(\ cos^2x)(\cos^2x)=\frac(1)(\cos^2x)-1 $$

Thus, $\int\tg^2xdx=\int\left(\frac(1)(\cos^2x)-1\right)dx$. Expanding the resulting integral into the sum of integrals and applying tabular formulas, we will have:

$$ \int\left(\frac(1)(\cos^2x)-1\right)dx=\int\frac(dx)(\cos^2x)-\int 1dx=\tg x-x+C . $$

Answer: $\int\tg^2xdx=\tg x-x+C$.

Direct integration is understood as a method of integration in which a given integral is reduced to one or more table integrals by means of identical transformations of the integrand and the application of the properties of the indefinite integral.

Example 1. Find.

 Dividing the numerator by the denominator, we get:

=
.

Note that there is no need to put an arbitrary constant after each term, because their sum is also an arbitrary constant, which we write at the end.

Example 2. Find
.

 We transform the integrand as follows:

Applying table integral 1, we obtain:

.

Example 3.

Example 4.

Example 5.

=
.

In some cases, finding integrals is simplified by using artificial techniques.

Example 6. Find
.

 Multiply the integrand by
we find

=
.

Example 7.

Example 8 .

2. Integration by change of variable method

It is not always possible to calculate a given integral by direct integration, and sometimes this is associated with great difficulties. In these cases, other techniques are used. One of the most effective is the variable replacement method. Its essence lies in the fact that by introducing a new integration variable it is possible to reduce a given integral to a new one, which is relatively easy to take directly. There are two variations of this method.

a) Method of subsuming a function under the differential sign

By definition of the differential of the function
.

The transition in this equality from left to right is called “summarizing the factor”
under the differential sign."

Theorem on the invariance of integration formulas

Any integration formula retains its form when replacing the independent variable with any differentiable function from it, i.e., if

, then
,

Where
- any differentiable function of x. Its values must belong to the interval in which the function defined and continuous.

Proof:

From what
, should
. Let us now take the function
. For its differential, due to the property of invariance of the form of the first differential of the function , we have

Let it be necessary to calculate the integral
. Let us assume that there are a differentiable function
and function
such that the integrand
can be written as

those. integral calculation
reduces to calculating the integral
and subsequent substitution
.

Example 1. .

Example 2. .

Example 3 . .

Example 4 . .

Example 5 .
.

Example 6 . .

Example 7 . .

Example 8. .

Example 9. .

Example 10 . .

Example 11.

Example 12 . FindI=
(0).

 Let us represent the integrand function in the form:

Hence,

Thus,
.

Example 12a. Find I=
,
.

 Since
,

hence I= . 

Example 13. Find
(0).

 In order to reduce this integral to a tabular one, we divide the numerator and denominator of the integrand by :

We have placed a constant factor under the differential sign. Considering it as a new variable, we get:

.

Let us also calculate the integral, which is important when integrating irrational functions.

Example 14. FindI=
( X A,A0).

 We have
.

So,

( X A,A0).

The examples presented illustrate the importance of the ability to present a given

differential expression
to mind
, Where there is some function from x And g– a function simpler to integrate than f.

In these examples, differential transformations such as

Where b– constant value

often used in finding integrals.

In the table of basic integrals it was assumed that x there is an independent variable. However, this table, as follows from the above, fully retains its meaning if under x understand any continuously differentiable function of an independent variable. Let us generalize a number of formulas from the table of basic integrals.

3a.
.

4.
.

5.
=
.

6.
=
.

7.
=
.

8.
( X A,A0).

9.
(A0).

Operation of summarizing a function
under the differential sign is equivalent to changing the variable X to a new variable
. The following examples illustrate this point.

Example 15. FindI=
.

 Let’s replace the variable using the formula
, Then
, i.e.
andI=
.

Replacing u his expression
, we finally get

I=
.

The transformation performed is equivalent to subsuming the differential sign of the function
.

Example 16. Find
.

 Let's put
, Then
, where
. Hence,

Example 17. Find
.

 Let
, Then
, or
. Hence,

In conclusion, we note that different ways of integrating the same function sometimes lead to functions that are different in appearance. This apparent contradiction can be eliminated if we show that the difference between the obtained functions is a constant value (see the theorem proven in Lecture 1).

Examples:

The results differ by a constant amount, which means both answers are correct.

b) I=
.

It is easy to verify that any of the answers differ from each other only by a constant amount.

b) Substitution method (method of introducing a new variable)

Let the integral
(
- continuous) cannot be directly converted to tabular form. Let's make a substitution
, Where
- a function that has a continuous derivative. Then
,
And

. (3)

Formula (3) is called the change of variable formula in the indefinite integral.

How to choose the right substitution? This is achieved through practice in integration. But it is possible to establish a number of general rules and some techniques for special cases of integration.

The rule for integration by substitution is as follows.

Determine which table integral this integral is reduced to (after first transforming the integrand, if necessary).

Determine which part of the integrand to replace with a new variable, and write down this replacement.

Find the differentials of both parts of the record and express the differential of the old variable (or an expression containing this differential) in terms of the differential of the new variable.

Make a substitution under the integral.

Find the resulting integral.

A reverse replacement is made, i.e. go to the old variable.

Let's illustrate the rule with examples.

Example 18. Find
.

Example 19. Find
.

=
.

We find this integral by summing
under the differential sign.

=.

Example 20. Find
(
).

, i.e.
, or
. From here
, i.e.
.

Thus we have
. Replacing its expression through x, we finally find the integral, which plays an important role in the integration of irrational functions:
(
).

Students nicknamed this integral the “long logarithm.”

Sometimes instead of substitution
it is better to perform a variable replacement of the form
.

Example 21. Find
.

Example 22. Find
.

 Let’s use the substitution
. Then
,
,
.

Therefore, .

In a number of cases, finding the integral is based on using the methods of direct integration and subsuming functions under the differential sign at the same time (see example 12).

Let us illustrate this combined approach to calculating the integral, which plays an important role in the integration of trigonometric functions.

Example 23. Find
.

=
.

So,
.

Another approach to calculating this integral:

Example 24. Find
.

Note that choosing a successful substitution is usually difficult. To overcome them, you need to master the technique of differentiation and have a good knowledge of table integrals.

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