Method of changing a variable in an indefinite integral. Examples of solutions. Change of variable and integration by parts in the indefinite integral
A ways to reduce integrals to tabular ones We have listed for you:
variable replacement method;
method of integration by parts;
Direct integration method
methods of representing indefinite integrals through tabular ones for integrals of rational fractions;
methods for representing indefinite integrals through table integrals for integrals of irrational expressions;
ways to express indefinite integrals through tabular ones for integrals of trigonometric functions.
Indefinite integral of a power function
Indefinite integral of the exponential function
But the indefinite integral of the logarithm is not a tabular integral; instead, the formula is tabular:
Indefinite integrals of trigonometric functions: Integrals of sine, cosine and tangent
Indefinite integrals with inverse trigonometric functions
Reduction to tabular form or direct integration method. Using identical transformations of the integrand, the integral is reduced to an integral to which the basic rules of integration are applicable and it is possible to use a table of basic integrals.
Example
Exercise. Find the integral
Solution. Let's use the properties of the integral and reduce this integral to tabular form.
Answer. 
Technically variable replacement method in the indefinite integral is implemented in two ways:
– Subsuming a function under the differential sign. – Actually changing the variable.
Subsuming a function under the differential sign
Example 2
Find the indefinite integral. Perform check.
Let's analyze the integrand function. Here we have a fraction, and the denominator is a linear function (with “x” to the first power). We look at the table of integrals and find the most similar thing: .
We bring the function under the differential sign: 
Those who find it difficult to immediately figure out which fraction to multiply by can quickly reveal the differential in a draft: . Yeah, it turns out that this means that in order for nothing to change, I need to multiply the integral by . Next we use the tabular formula:
Examination: 
The original integrand function has been obtained, which means that the integral has been found correctly.
Variable change method in indefinite integral
Example 5
Find the indefinite integral.
As an example, I took the integral that we looked at at the very beginning of the lesson. As we have already said, to solve the integral we liked the tabular formula 
, and I would like to reduce the whole matter to her.
The idea behind the replacement method is to replace a complex expression (or some function) with a single letter. In this case, it suggests itself: The second most popular letter for replacement is the letter . In principle, you can use other letters, but we will still adhere to traditions.
So: 
But when we replace it, we are left with ! Probably, many guessed that if a transition is made to a new variable, then in the new integral everything should be expressed through the letter, and there is no place for a differential there at all. The logical conclusion is that it is necessary turn into some expression that depends only on.
The action is as follows. After we have selected a replacement, in this example, we need to find the differential. With differentials, I think everyone has already established friendship.
Since then
After sorting out the differential, I recommend rewriting the final result as briefly as possible: Now, according to the rules of proportion, we express the one we need:
Eventually: 
Thus: 
And this is already the most tabular integral 
(the table of integrals is, of course, also valid for the variable ).
Finally, all that remains is to carry out the reverse replacement. Let us remember that.
Ready.
The final design of the example considered should look something like this:
“
Let's replace: 
“
The icon does not have any mathematical meaning; it means that we have interrupted the solution for intermediate explanations.
When preparing an example in a notebook, it is better to mark the reverse substitution with a simple pencil.
Attention! In the following examples, finding the differential will not be described in detail.
And now it’s time to remember the first solution: 
What is the difference? There is no fundamental difference. It's actually the same thing. But from the point of view of designing the task, the method of subsuming a function under the differential sign is much shorter. The question arises. If the first method is shorter, then why use the replacement method? The fact is that for a number of integrals it is not so easy to “fit” the function to the sign of the differential.
Integration by parts. Examples of solutions
Integrals of logarithms
Example 1
Find the indefinite integral.
Classic. From time to time this integral can be found in tables, but it is not advisable to use a ready-made answer, since the teacher has spring vitamin deficiency and will swear heavily. Because the integral under consideration is by no means tabular - it is taken in parts. We decide:
We interrupt the solution for intermediate explanations.
We use the integration by parts formula: 
The formula is applied from left to right
We look at the left side: . Obviously, in our example (and in all the others that we will consider), something needs to be designated as , and something as .
In integrals of the type under consideration foralways denoted by logarithm.
Technically, the design of the solution is implemented as follows; we write in the column:
That is, we denoted the logarithm as, and the remaining part integrand expression.
Next stage: find the differential:
A differential is almost the same as a derivative; we have already discussed how to find it in previous lessons.
Now we find the function. In order to find the function you need to integrate right side lower equality:
Now we open our solution and construct the right side of the formula: . By the way, here is a sample of the final solution with small notes.
If the function x=φ(t) has a continuous derivative, then in a given indefinite integral ∫f(x)dx you can always go to a new variable t using the formula
∫f(x)dx=∫f(φ(t))φ"(t)dt
Then find the integral from the right side and return to the original variable. In this case, the integral on the right side of this equality may turn out to be simpler than the integral on the left side of this equality, or even tabular. This method of finding the integral is called the change of variable method.
Example 7. ∫x√x-5dx
To get rid of the root, we set √x-5=t. Hence x=t 2 +5 and therefore dx=2tdt. Making the substitution, we consistently have:
∫x√x-5dx=∫(t 2 +5) 2tdt=∫(2t 4 +10t 2)dt=2∫t 4 dt+10∫t 2 dt=
III. Method of integration by parts
The integration by parts method is based on the following formula:
∫udv=uv-∫vdu
where u(x),v(x) are continuously differentiable functions. The formula is called the integration by parts formula. This formula shows that the integral ∫udv leads to the integral ∫vdu, which may turn out to be simpler than the original one, or even tabular.
Example 12. Find the indefinite integral ∫xe -2x dx
Let's use the method of integration by parts. Let's put u=x, dv=e -2x dx. Then du=dx, v=∫xe -2x dx=-e -2x +C Therefore, according to the formula we have: ∫xe -2x dx=x(-e -2x)-∫- -2 dx=-e -2x -e -2x +C
23 . Rational fraction is a fraction whose numerator and denominator are polynomials.
Rational fractions. The simplest rational fractions and their integration
Any rational function can be represented as a rational fraction, that is, as a ratio of two polynomials:
If the degree of the numerator is lower than the degree of the denominator, then the fraction is called correct, otherwise the fraction is called wrong.
If the fraction is improper, then by dividing the numerator by the denominator (according to the rule for dividing polynomials), you can represent this fraction as the sum of a polynomial and some proper fraction: 
, Where M(x)- a polynomial, but a proper fraction.
Example: Let us be given an improper rational fraction.
Then 
, since when dividing by a corner we get a remainder (4x-6).
Since the integration of polynomials does not present any fundamental difficulties, the main difficulty in integrating rational fractions lies in the integration of proper rational fractions.
There are several types of rational fractions:
II. Type: (k-positive integer ³2).
IY. View: 
(k-integer³2).
Let's consider integrals of the simplest rational fractions.
I. 
.
II. 
=A 
.
24 .Integrating rational fractions
Let the integrand be a rational fraction where and are polynomials (polynomials) of degrees k And n respectively. Without loss of generality, we can assume that k < n, since otherwise you can always present the numerator in the form P(x) = Q(x)R(x) + S(x) where R(x) and S(x) are polynomials, usually called, as in the case real numbers, quotient and remainder, and the degree of the polynomial S(x) is less n. Then
,
(1.1)
and we can calculate the integral of the polynomial R(x). Let us show by example how to obtain expansion (1.1). Let P(x) = x 7 + 3x 6 + 3x 5 – 3x 3 + 4x 2 + x -2, Q(x) = x + 3x 2 + x-2. Let's divide the polynomial P(x) by the polynomial Q(x) in the same way as we divide real numbers (we obtain the solution using a column division calculator). Thus, we received the whole part of the fraction (the quotient from the division of the polynomial P by the polynomial Q) R(x) = x 4 + 2x 2 – 4x + 7 and the remainder S(x) = 9x 2 – 14x +12 from this division. According to the basic theorem of algebra, any polynomial can be decomposed into simple factors, that is, represented in the form , where are the roots of the polynomial Q(x) repeated as many times as their multiplicity. Let the polynomial Q(x) have n different roots. Then a proper rational fraction can be represented as 
, where are numbers to be determined. If is a root of multiplicity α, then in the decomposition into simple fractions there corresponds α terms 
. If x j is the complex root of the multiplicity of a polynomial with real coefficients, then the complex conjugate number is also the root of the multiplicity α of this polynomial. In order not to deal with complex numbers when integrating rational fractions, the terms in the expansion of a proper rational fraction corresponding to pairs of complex conjugate roots are combined and written as one term of the form , if - roots of multiplicity one. If are roots of multiplicity , then they correspond to terms and the corresponding expansion has the form
Thus, the integration of proper rational fractions has been reduced to the integration of the simplest fractions, of which the tabular ones can be found using the recurrence formula, which is obtained by integration by parts. Integrals, in the case where the denominator has complex roots (discriminant), are reduced, using the selection of a complete square, to integrals, by replacement. One way to find coefficients in the expansion of a proper rational fraction is as follows. The right-hand side of the resulting expansion with undetermined coefficients is reduced to a common denominator. Since the denominators of the right and left sides are equal, the numerators, which are polynomials, must also be equal. Equating the coefficients at the same degrees (since polynomials are equal if the coefficients at the same degrees are equal), we obtain a system of linear equations for determining these coefficients.
25. Integration of irrational functions - The general principle of integrating irrational expressions is to replace the variable, which allows you to get rid of the roots in the integrand. For some function classes this goal is achieved using standard substitutions.
Integrals of the form 
.
Integrals of the form 
are calculated by replacing or .
Integrals of the form 
are calculated by replacing or .
26 . Integration of irrational functions - The general principle of integrating irrational expressions is to replace the variable, which allows you to get rid of the roots in the integrand. For some function classes this goal is achieved using standard substitutions.
Integrals of the form 
, where is the rational function of its arguments, are calculated by replacing 
.
Integrals of the form 
are calculated by replacing or .
Integrals of the form 
are calculated by replacing or . Integrals of the form 
are calculated by replacing or .
Integration by substitution (variable replacement). Suppose you need to calculate an integral that is not tabular. The essence of the substitution method is that in the integral the variable x is replaced by the variable t according to the formula x = q(t), from which dx = q"(t)dt.
Theorem. Let the function x=t(t) be defined and differentiable on a certain set T and let X be the set of values of this function on which the function f(x) is defined. Then if on the set X the function f(x) has an antiderivative, then on the set T the formula is valid:
Formula (1) is called the change of variable formula in the indefinite integral.
Integration by parts. The method of integration by parts follows from the formula for the differential of the product of two functions. Let u(x) and v(x) be two differentiable functions of the variable x. Then:
d(uv)=udv+vdu. - (3)
Integrating both sides of equality (3), we obtain:
But since then:
Relation (4) is called the integration by parts formula. Using this formula, find the integral. It is advisable to use it when the integral on the right side of formula (4) is simpler to calculate than the original one.
In formula (4) there is no arbitrary constant C, since on the right side of this formula there is an indefinite integral containing an arbitrary constant.
We present some frequently encountered types of integrals calculated by the method of integration by parts.
I. Integrals of the form, (P n (x) is a polynomial of degree n, k is a certain number). To find these integrals, it is enough to set u=P n (x) and apply formula (4) n times.
II. Integrals of the form, (Pn(x) is a polynomial of degree n with respect to x). They can be found using frequencies, taking for u a function that is a multiplier for P n (x).
Direct integration
Basic integration formulas
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Calculation of integrals by direct use of the table of simple integrals and the basic properties of indefinite integrals is called direct integration.
Example 1.
Example 2.
Example 3.
This is the most common method of integrating a complex function, consisting of transforming the integral by moving to another integration variable.
If it is difficult to reduce the integral to a tabular one using elementary transformations, then in this case the substitution method is used. The essence of this method is that by introducing a new variable it is possible to reduce this integral to a new integral, which is relatively easy to take directly.
To integrate by substitution method, use the solution scheme:
2) find the differential from both replacement parts;
3) express the entire integrand through a new variable (after which a table integral should be obtained);
4) find the resulting table integral;
5) perform a reverse replacement.
Find the integrals:
Example 1 . Substitution:cosx=t,-sinxdx=dt,
Solution:
Example 2.∫e -x3 x 2 dx Substitution:-x 3 =t, -3x 2 dx=dt, Solution:∫e -x3 x 2 dx=∫e t (-1/3)dt=-1/3e t +C=-1/3e -x3 +C
Example 3.
Substitution: 1+sinx=t , cosxdx=dt ,
Solution: .
SECTION 1.5. Definite integral, methods of its calculation.
item 1 The concept of a definite integral
Task. Find the increment of a function that is antiderivative of a function f(x), when passing the argument x from the value a to value b.
Solution. Let us assume that integration has found: ∫ (x)dx = F(x)+C.
Then F(x)+C 1, Where C 1- any given number will be one of the antiderivative functions for this function f(x). Let's find its increment when the argument moves from the value a to value b. We get:
x=b - x=a =F(b) +C 1 - F(a) -C 1 =F(b)-F(a)
As we see, in the expression for the increment of the antiderivative function F(x)+C 1 no constant value C 1. And since under C 1 any given number was implied, the result obtained leads to the following conclusion: on argument transition x from the value x=a to value x=b all functions F(x)+C, antiderivatives for a given function f(x), have the same increment equal to F(b)-F(a).
This increment is usually called the definite integral and denoted by the symbol: and reads: integral of A before b from the function f(x) over dх or, in short, the integral of A before b from f(x)dx.
Number A called lower limit integration, number b - top; segment a ≤ x ≤ b – segment of integration. It is assumed that the integrand function f(x) continuous for all values x, satisfying the conditions: a x b
Definition. Increment of antiderivative functions F(x)+C on argument transition x from the value x=a to value x=b, equal to the difference F(b)-F(a), is called a definite integral and is denoted by the symbol: so that if ∫ (x)dx = F(x)+C, then = F(b)-F(a) - given the equality is called the Newton-Leibniz formula.
item 2 Basic properties of the definite integral
All properties are formulated in the proposition that the functions under consideration are integrable in the corresponding intervals.
item 3 Direct calculation of the definite integral
To calculate the definite integral, when you can find the corresponding indefinite integral, use the Newton–Leibniz formula
those. the definite integral is equal to the difference between the values of any antiderivative function at the upper and lower limits of integration.
This formula shows the procedure for calculating a definite integral:
1) find the indefinite integral of this function;
2) into the resulting antiderivative, substitute first the upper and then the lower limit of the integral instead of the argument;
3) subtract the result of substituting the lower limit from the result of substituting the upper limit.
Example 1: 
Calculate the integral:
Example 2: Calculate the integral:
p.4 Calculation of a definite integral by substitution method
The calculation of the definite integral by the substitution method is as follows:
1) replace part of the integrand with a new variable;
2) find new limits of the definite integral;
3) find the differential from both replacement parts;
4) express the entire integrand through a new variable (after which a table integral should be obtained); 5) calculate the resulting definite integral.
Example 1: Calculate the integral:
Substitution: 1+cosx=t,-sinxdx=dt, 
SECTION 1.6. Geometric meaning of a definite integral.
Area of a curved trapezoid:
It is known that a definite integral on a segment represents the area of a curvilinear trapezoid bounded by the graph of the function f(x).
The area of a figure bounded by certain lines can be found using certain integrals if the equations of these lines are known.
Let on the segment [a; b] a continuous function is given y = ƒ(x) ≥ 0. Let us find the area of this trapezoid.
Area of the figure bounded by axis 0 x, two vertical straight lines x = a, x = b and the graph of the function y = ƒ(x) (figure), determined by the formula:
This is the geometric meaning of the definite integral.
Example 1:
Calculate the area of the figure bounded by the lines: y=x2.+2, y=0, x= -2, x=1.
Solution: Let's make a drawing (note that the equation y=0 defines the Ox axis).
Answer: S = 9 units 2
Example 2:
Calculate the area of the figure bounded by the lines: y= - e x, x=1 and coordinate axes.
Solution: Let's make a drawing.
If a curved trapezoid completely located under the Ox axis, then its area can be found using the formula:
In this case: 
Attention! If you are asked to find the area of a figure using a definite integral, then the area is always positive! That is why the minus appears in the formula just discussed.
SECTION 1.7. Application of the definite integral
p.1 Calculation of the volume of a body of rotation
If a curved trapezoid is adjacent to the Ox axis, and straight lines y=a, y=b and the graph of the function y= F(x) (Fig. 1), then the volume of the body of revolution is determined by a formula containing an integral.
The volume of the body of revolution is equal to:
Example:
Find the volume of the body limited by the surface of rotation of the line around the Ox axis at 0≤ x ≤4.
Solution: V
units 3. Answer: unit 3.
SECTION 3.1. Ordinary differential equations
item 1 The concept of a differential equation
Definition. Differential equation is an equation containing a function of a set of variables and their derivatives.
The general form of such an equation 
=0, where F is a known function of its arguments, specified in a fixed area; x - independent variable (variable by which it is differentiated); y - dependent variable (the one from which derivatives are taken and the one to be determined); - derivative of the dependent variable y with respect to the independent variable x.
item 2 Basic concepts of differential equation
In order of a differential equation is called the order of the highest derivative included in it.
For example:
A second order equation is a first order equation.
Any function that connects variables and turns a differential equation into a true equality is called decision differential equation.
General solution of a first-order differential equation is a function of and an arbitrary constant C that turns this equation into an identity in .
The general solution, written in the implicit form =0, is called general integral.
Private decision equation =0 is a solution obtained from the general solution for a fixed value - a fixed number.
The problem of finding a particular solution to a differential equation of the nth order (n= 1,2,3,...), satisfying the initial conditions of the form
called Cauchy problem.
item 3 First order differential equations with separable variables
A first order differential equation is called a separable equation if it can be represented as 
can be rewritten in the form 
. If 
. Let's integrate: 
.
To solve an equation of this type you need:
1. Separate variables;
2. By integrating the equation with separated variables, find the general solution of this equation;
3. Find a particular solution that satisfies the initial conditions (if they are given).
Example 1. Solve the equation. Find a particular solution that satisfies the condition y=4 at x=-2.
Solution: This is a separated variable equation. Integrating, we find the general solution to the equation: . To obtain a simpler general solution, we represent the constant term on the right side in the form C/2. We have or is a general solution. Substituting the values y=4 and x=-2 into the general solution, we get 16=4+C, from which C=12.
So, a particular solution of the equation that satisfies this condition has the form
Example 2. Find a particular solution to the equation if .
Solution: 
, 
, 
, 
, 
, 
common decision.
We substitute the values of x and y into the particular solution: , , 
private solution.
Example 3. Find the general solution to the equation . Solution: ,, 
, 
- common decision.
item 4 Differential equations of order higher than the first
An equation of the form or is solved by double integration: , , whence . Having integrated this function, we obtain a new function of f(x), which we denote by F(x). Thus, ; 
. Let's integrate again: or y=Ф(x). We obtained a general solution to the equation containing two arbitrary constants and .
Example 1. Solve the equation.
Solution:, 
, ,
Example 2. Solve the equation . Solution: 
, , 
.
SECTION 3.2. Number series, its members
Definition 1.Number series is called an expression of the form ++…++…, (1)
Where , , …, , … - numbers belonging to some specific number system.
Thus, we can talk about real series for which R, about complex series for which C,i= 1, 2, …, n, ...
Previously, given a given function, guided by various formulas and rules, we found its derivative. The derivative has numerous uses: it is the speed of movement (or, more generally, the speed of any process); the angular coefficient of the tangent to the graph of the function; using the derivative, you can examine the function for monotonicity and extrema; it helps solve optimization problems.
But along with the problem of finding the speed according to a known law of motion, there is also an inverse problem - the problem of restoring the law of motion according to a known speed. Let's consider one of these problems.
Example 1. A material point moves in a straight line, its speed at time t is given by the formula v=gt. Find the law of motion.
Solution. Let s = s(t) be the desired law of motion. It is known that s"(t) = v(t). This means that to solve the problem you need to select a function s = s(t), the derivative of which is equal to gt. It is not difficult to guess that \(s(t) = \frac(gt^ 2)(2)\).
\(s"(t) = \left(\frac(gt^2)(2) \right)" = \frac(g)(2)(t^2)" = \frac(g)(2) \ cdot 2t = gt\)
Answer: \(s(t) = \frac(gt^2)(2) \)
Let us immediately note that the example is solved correctly, but incompletely. We got \(s(t) = \frac(gt^2)(2) \). In fact, the problem has infinitely many solutions: any function of the form \(s(t) = \frac(gt^2)(2) + C\), where C is an arbitrary constant, can serve as a law of motion, since \(\left (\frac(gt^2)(2) +C \right)" = gt \)
To make the problem more specific, we had to fix the initial situation: indicate the coordinate of a moving point at some point in time, for example at t = 0. If, say, s(0) = s 0, then from the equality s(t) = (gt 2)/2 + C we get: s(0) = 0 + C, i.e. C = s 0. Now the law of motion is uniquely defined: s(t) = (gt 2)/2 + s 0.
In mathematics, mutually inverse operations are given different names, special notations are invented, for example: squaring (x 2) and square root (\(\sqrt(x) \)), sine (sin x) and arcsine (arcsin x) and etc. The process of finding the derivative of a given function is called differentiation, and the inverse operation, i.e. the process of finding a function from a given derivative, is integration.
The term “derivative” itself can be justified “in everyday terms”: the function y = f(x) “gives birth” to a new function y" = f"(x). The function y = f(x) acts as a “parent”, but mathematicians, naturally, do not call it a “parent” or “producer”; they say that it is, in relation to the function y" = f"(x) , primary image, or primitive.
Definition. The function y = F(x) is called antiderivative for the function y = f(x) on the interval X if the equality F"(x) = f(x) holds for \(x \in X\)
In practice, the interval X is usually not specified, but is implied (as the natural domain of definition of the function).
Let's give examples.
1) The function y = x 2 is antiderivative for the function y = 2x, since for any x the equality (x 2)" = 2x is true
2) The function y = x 3 is antiderivative for the function y = 3x 2, since for any x the equality (x 3)" = 3x 2 is true
3) The function y = sin(x) is antiderivative for the function y = cos(x), since for any x the equality (sin(x))" = cos(x) is true
When finding antiderivatives, as well as derivatives, not only formulas are used, but also some rules. They are directly related to the corresponding rules for calculating derivatives.
We know that the derivative of a sum is equal to the sum of its derivatives. This rule generates the corresponding rule for finding antiderivatives.
Rule 1. The antiderivative of a sum is equal to the sum of the antiderivatives.
We know that the constant factor can be taken out of the sign of the derivative. This rule generates the corresponding rule for finding antiderivatives.
Rule 2. If F(x) is an antiderivative for f(x), then kF(x) is an antiderivative for kf(x).
Theorem 1. If y = F(x) is an antiderivative for the function y = f(x), then the antiderivative for the function y = f(kx + m) is the function \(y=\frac(1)(k)F(kx+m) \)
Theorem 2. If y = F(x) is an antiderivative for the function y = f(x) on the interval X, then the function y = f(x) has infinitely many antiderivatives, and they all have the form y = F(x) + C.
Integration methods
Variable replacement method (substitution method)
The method of integration by substitution involves introducing a new integration variable (that is, substitution). In this case, the given integral is reduced to a new integral, which is tabular or reducible to it. There are no general methods for selecting substitutions. The ability to correctly determine substitution is acquired through practice.
Let it be necessary to calculate the integral \(\textstyle \int F(x)dx \). Let's make the substitution \(x= \varphi(t) \) where \(\varphi(t) \) is a function that has a continuous derivative.
Then \(dx = \varphi " (t) \cdot dt \) and based on the invariance property of the integration formula for the indefinite integral, we obtain the integration formula by substitution:
\(\int F(x) dx = \int F(\varphi(t)) \cdot \varphi " (t) dt \)
Integration of expressions of the form \(\textstyle \int \sin^n x \cos^m x dx \)
If m is odd, m > 0, then it is more convenient to make the substitution sin x = t.
If n is odd, n > 0, then it is more convenient to make the substitution cos x = t.
If n and m are even, then it is more convenient to make the substitution tg x = t.
Integration by parts
Integration by parts - applying the following formula for integration:
\(\textstyle \int u \cdot dv = u \cdot v - \int v \cdot du \)
or:
\(\textstyle \int u \cdot v" \cdot dx = u \cdot v - \int v \cdot u" \cdot dx \)
