Change of variable to an integral theorem. Variable change in indefinite integral
In this lesson, we will get acquainted with one of the most important and most common techniques that is used in the course of solving indefinite integrals - the variable change method. To successfully master the material, initial knowledge and skills of integration are required. If there is a feeling of an empty full teapot in integral calculus, then first you should familiarize yourself with the material where I explained in an accessible form what an integral is and analyzed in detail the basic examples for beginners.
Technically, the method of changing a variable in an indefinite integral is implemented in two ways:
- Bringing the function under the differential sign;
- The actual replacement of the variable.
In fact, they are one and the same, but the design of the solution looks different.
Let's start with a simpler case.
Assigning a function under the differential sign
In the classroom Indefinite integral. Examples of solutions we learned how to open the differential, I remind the example that I gave:
That is, opening the differential is formally almost the same as finding the derivative.
Example 1
Check.
We look at the table of integrals and find a similar formula: 
... But the problem is that under the sine we have not just the letter "X", but a complex expression. What to do?
We bring the function under the differential sign:
By opening the differential, it is easy to check that: 
In fact and 
Is a record of the same thing.
But, nevertheless, the question remained, how did we come to the idea that at the first step we need to write our integral exactly like this: 
? Why so and not otherwise?
Formula 
(and all other tabular formulas) are valid and applicable NOT ONLY for a variable, but for any complex expression ONLY A FUNCTION ARGUMENT(- in our example) AND THE EXPRESSION UNDER THE SIGN OF THE DIFFERENTIAL WERE IDEAL
.
Therefore, the mental reasoning when solving should be something like this: “I need to solve the integral. I looked in the table and found a similar formula 
... But I have a complicated argument and I can't immediately use the formula. However, if I manage to get it under the differential sign, then everything will be fine. If I write, then. But in the original integral there is no triple factor, therefore, so that the integrand does not change, I need to multiply it by. " In the course of approximately such mental reasoning, a record is born:
Now you can use the tabular formula 
:
Ready
The only difference is that we do not have the letter "X", but a complex expression.
Let's check. We open the table of derivatives and differentiate the answer: 
The original integrand is obtained, which means that the integral is found correctly.
Please note that during the test we used the rule for differentiating a complex function 
. As a matter of fact, bringing the function under the sign of the differential and 
Are two mutually inverse rules.
Example 2
We analyze the integrand. Here we have a fraction, and the denominator is a linear function (with "x" in the first degree). 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 can quickly reveal the differential on a draft:. Yeah, it turns out, so that nothing changes, I need to multiply the integral by.
Next, we use the tabular formula 
:
Examination: 
The original integrand is obtained, which means that the integral is found correctly.
Example 3
Find the indefinite integral. Check.
Example 4
Find the indefinite integral. Check.
This is an example for a do-it-yourself solution. The answer is at the end of the lesson.
With some experience solving integrals, such examples will seem easy, and will click like nuts:
At the end of this section, I would also like to dwell on the "free" case when a variable enters in a linear function with a unit coefficient, for example:
Strictly speaking, the solution should look like this:
As you can see, putting the function under the differential sign was "painless", without any multiplication. Therefore, in practice, such long decision is often neglected and immediately recorded that 
... But be prepared, if necessary, to explain to the teacher how you decided! Since there is no integral in the table at all.
Variable change method in indefinite integral
We turn to the consideration of the general case - the method of changing variables in an indefinite integral.
Example 5
Find the indefinite integral.
As an example, I took the integral that we considered at the very beginning of the lesson. As we 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 one letter.
In this case, it begs:
The second most popular replacement letter is the letter.
In principle, other letters can be used, but we will still stick to tradition.
So: 
But when replacing, we still have! Probably, many have guessed that if a transition to a new variable is carried out, then in the new integral everything should be expressed through a letter, and there is no place for the differential at all.
It follows a logical conclusion that you need turn into some expression that depends only on.
The action is as follows. After we have found a replacement, in this example, we need to find the differential. With differentials, I think everyone has already established a friendship.
Since then
After the showdown with the differential, I recommend rewriting the final result as short 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 naturally also valid for a variable).
In conclusion, it remains to carry out the reverse replacement. Remember that.
Ready.
The final layout of the considered example should look something like this:
“
Let's replace: 
“
The icon does not carry any mathematical meaning, it means that we have interrupted the solution for intermediate explanations.
When writing an example in a notebook, it is better to superscript the reverse replacement with a simple pencil.
Attention! In the following examples, finding the differential will not be described in detail.
Now is the time to remember the first solution: 
What is the difference? There is no fundamental difference. They are actually the same thing. But from the point of view of the design of the task, the method of bringing the function under the differential sign is much shorter.
The question arises. If the first way is shorter, then why use the replace method? The point is that for a number of integrals it is not so easy to "fit" the function under the sign of the differential.
Example 6
Find the indefinite integral. 
Let's make a replacement: (it's hard to think of another replacement here) 
As you can see, as a result of the replacement, the original integral has become much simpler - reduced to an ordinary power function. This is the purpose of the replacement - to simplify the integral.
Lazy advanced people can easily solve this integral by putting a function under the differential sign: 
Another thing is that such a solution is not obvious for all students. In addition, already in this example, the use of the method of bringing a function under the differential sign significantly increases the risk of confusion in the solution.
Example 7
Find the indefinite integral. Check.
Example 8
Find the indefinite integral. 
Replacement:
It remains to find out what will become 
Okay, we expressed it, but what to do with the “x” remaining in the numerator ?!
From time to time, in the course of solving integrals, the following trick occurs: we express from the same substitution! 
Example 9
Find the indefinite integral.
This is an example for a do-it-yourself solution. The answer is at the end of the lesson.
Example 10
Find the indefinite integral.
Surely some have noticed that there is no variable replacement rule in my lookup table. This was done deliberately. The rule would confuse explanation and understanding, since it does not appear explicitly in the examples above.
Now is the time to talk about the basic premise of using the variable replacement method: the integrand must contain some function and its derivative:(functions may not be in the work)
In this regard, when finding integrals, one often has to look into the table of derivatives.
In this example, note that the degree of the numerator is one less than the degree of the denominator. In the table of derivatives, we find the formula that just lowers the degree by one. And, therefore, if you designate for the denominator, then the chances are great that the numerator will turn into something good.
By changing the variable, you can calculate simple integrals and, in some cases, simplify the calculation of more complex ones.
The method of changing a variable consists in the fact that we from the initial variable of integration, let it be x, go to another variable, which we denote as t. In this case, we assume that the variables x and t are related by some relation x = x (t), or t = t (x)... For example, x = ln t, x = sin t, t = 2 x + 1, etc. Our task is to choose such a relationship between x and t so that the original integral either reduces to a tabular one, or becomes simpler.
Basic Variable Replacement Formula
Consider the expression under the integral sign. It consists of the product of the integrand, which we denote as f (x) and differential dx:. Let us pass to a new variable t, choosing some relation x = x (t)... Then we have to express the function f (x) and the differential dx in terms of the variable t.
To express the integrand f (x) through the variable t, you just need to substitute the selected ratio x = x instead of the variable x (t).
Differential conversion is done like this:
.
That is, the differential dx is equal to the product of the derivative of x with respect to t and the differential dt.
Then
.
In practice, the most common case is when we perform a replacement, choosing a new variable as a function of the old one: t = t (x)... If we guessed that the integrand can be represented as
,
where t ′ (x) is the derivative of t with respect to x, then
.
So, the basic formula for changing a variable can be represented in two forms.
(1)
,
where x is a function of t.
(2)
,
where t is a function of x.
Important note
In tables of integrals, the variable of integration is most often denoted as x. However, it should be noted that the variable of integration can be denoted by any letter. Moreover, an expression can be used as a variable of integration.
As an example, consider the tabular integral
.
Here x can be replaced with any other variable or function of the variable. Here are examples of possible options:
;
;
.
IN last example it should be borne in mind that when passing to the variable of integration x, the differential is transformed as follows:
.
Then
.
This example is the essence of integration by substitution. That is, we have to guess that
.
Then the integral is reduced to a tabular one.
.
You can calculate this integral by changing the variable using the formula (2)
... Put t = x 2 + x... Then
;
;
.
Examples of Integration by Variable Change
1)
We calculate the integral
.
Note that (sin x) ′ = cos x... Then
.
Here we have applied the substitution t = sin x.
2)
We calculate the integral
.
Note that. Then
.
Here we performed integration by changing the variable t = arctg x.
3)
Let's integrate
.
Note that. Then
... Here, during integration, we changed the variable t = x 2 + 1
.
Linear substitutions
Linear substitutions are perhaps the most common. This is a replacement for a variable of the form
t = ax + b,
where a and b are constants. With such a replacement, the differentials are related by the relation
.
Examples of Integration by Linear Substitutions
A) Calculate the integral
.
Solution.
.
B) Find the integral
.
Solution.
Let's use the properties of the exponential function.
.
ln 2 is constant. We calculate the integral.
.
C) Calculate the integral
.
Solution.
Let us reduce the square polynomial in the denominator of the fraction to the sum of squares.
.
We calculate the integral.
.
D) Find the integral
.
Solution.
We transform the polynomial at the root.
.
We integrate using the variable change method.
.
Earlier we got the formula
.
From here
.
Substituting this expression, we get the final answer.
E) Calculate the integral
.
Solution.
Let's apply the formula for the product of sine and cosine.
;
.
We integrate and make substitutions.
.
References:
N.M. Gunther, R.O. Kuzmin, Collection of problems in higher mathematics, "Lan", 2003.
Pemona and its properties
The antiderivative of a function f (x) on the interval (a; b) is a function F (x) such that equality holds for any x from a given interval.
If we take into account the fact that the derivative of the constant С is equal to zero, then the equality 
... Thus, the function f (x) has a set of antiderivatives F (x) + C, for an arbitrary constant C, and these antiderivatives differ from each other by an arbitrary constant value.
Antiderivative properties.
If the function F (x) is the antiderivative for the function f (x) on the interval X, then the function f (x) + C, where C is an arbitrary constant, will also be the antiderivative for f (x) on this interval.
If the function F (x) is some antiderivative for the function f (x) on the interval X = (a, b), then any other antiderivative F1 (x) can be represented as F1 (x) = F (x) + C, where C is a constant function on X.
2 Definition of an indefinite integral.
The whole set of antiderivatives of a function f (x) is called the indefinite integral of this function and is denoted 
.
The expression is called the integrand, and f (x) is called the integrand. The integrand is the differential of the function f (x).
The action of finding an unknown function by its given differential is called indefinite integration, because the result of integration is not one function F (x), but the set of its antiderivatives F (x) + C.
properties of an indefinite integral (properties of an antiderivative).
The derivative of the result of integration is equal to the integrand.
The indefinite integral of the differential of a function is equal to the sum of the function itself and an arbitrary constant.
where k is an arbitrary constant. The coefficient can be taken outside the sign of the indefinite integral.
The indefinite integral of the sum / difference of functions is equal to the sum / difference of the indefinite integrals of the functions.
Variable change in indefinite integral
Variable replacement in the indefinite integral is performed using substitutions of two types:
a) where is a monotone, continuously differentiable function of the new variable t. Variable replacement formula in this case:
Where U is the new variable. The formula for changing a variable with this substitution:
Integration piece by piece
Finding the integral by the formula 
Is called integration by parts. Here U = U (x), υ = υ (x) are continuously differentiable functions of x. With the help of this formula, finding the integral is reduced to finding another integral; its application is expedient in those cases when the last integral is either simpler than the original one, or similar to it.
In this case, for υ is taken such a function that is simplified during differentiation, and for dU - that part of the integrand, the integral of which is known or can be found.
Newton-Leibniz formula
Continuity of the definite integral as a function of the upper limit
If the function y = f (x) is integrable on an interval, then, obviously, it is also integrable on an arbitrary interval [a, x] embedded in. Function 
,
where x Î is called an integral with a variable upper limit. The value of the function Ф (х) at the point x is equal to the area S (x) under the curve y = f (x) on the segment [a, x]. This is the geometric meaning of an integral with a variable upper limit.
Theorem. If the function f (x) is continuous on an interval then the function Φ (x) is also continuous on [a, b].
Let Δx be such that x + Δx Î. We have
By the mean value theorem, there is a value with Î [x, x + Δ x] such that Since c Î, and the function f (x) is bounded, then passing to the limit as Δ x → 0, we obtain
1st order SDT
What is the difference between homogeneous differential equations and other types of differential equations? The easiest way to explain this right away is with a concrete example.
Solve differential equation
What should be analyzed first when solving any first-order differential equation? First of all, it is necessary to check, is it possible to immediately separate the variables with the help of "school" actions? Usually, this analysis is done mentally or trying to separate the variables on a draft.
In this example, the variables cannot be divided (you can try to swap terms from part to part, take factors out of parentheses, etc.). By the way, in this example, the fact that variables cannot be divided is quite obvious due to the presence of a factor
The question arises - how to solve this diffusion?
It is necessary to check if the given equation is not homogeneous? The verification is simple, and the verification algorithm itself can be formulated as follows:
To the original equation:
Instead of x, substitute for y, substitute the derivative without touching: 
The letter lambda is some abstract numeric parameter, the point is not in the lambdas themselves, and not in their values, but the point is this:
If, as a result of the transformations, it is possible to reduce ALL "lambdas" (ie to obtain the original equation), then this differential equation is homogeneous.
Obviously, lambdas are immediately reduced in exponent: 
Now we put the lambda outside the brackets on the right side: 
Both sides of the equation can be reduced to this same lambda: As a result, all lambdas disappeared like a dream, like morning fog, and we got the original equation.
Conclusion: This equation is homogeneous
LOU.General properties of solutions
that is, it is linear with respect to the unknown function y and its derivatives and. The coefficients and and the right-hand side of this equation are continuous.
If the right side of the equation, then the equation is called linear inhomogeneous. If, then the equation has the form
 | (9) |
and is called linear homogeneous.
Let and be any particular solutions of equation (9), that is, do not contain arbitrary constants.
Theorem 1. If and are two particular solutions of a linear homogeneous equation of the second order, then it is also a solution of this equation.
Since and are solutions of equation (9), they turn this equation into identity, that is
 and  | (10) |
Substitute in equation (9). Then we have:
By virtue of (10). Hence, the solution to the equation.
Theorem 2. If is a solution to a linear homogeneous second-order equation, and C- constant, then it is also a solution to this equation.
Proof. Substitute in equation (9). We get: that is, the solution to the equation.
Consequence. If and are solutions to equation (9), then it is also its solution by virtue of theorems (1) and (2).
Definition. Two solutions and equations (9) are called linearly dependent (on an interval) if it is possible to choose such numbers and that are not equal to zero at the same time such that the linear combination of these solutions is identically zero on, that is, if.
If such numbers cannot be found, then the solutions are called linearly independent (on a segment).
Obviously, decisions will be linearly dependent if and only if their ratio is constant, that is (or vice versa).
Indeed, if and are linearly dependent, then where at least one constant or nonzero. For example, let. Then,, Denoting we get, that is, the relation is constant.
Conversely, if then 
... Here the coefficient at, that is, nonzero, which by definition means that they are linearly dependent.
Comment. From the definition of linearly independent solutions and the reasoning above, we can conclude that if and are linearly independent, then their ratio cannot be constant.
For example, the functions and at are linearly independent, since 
, as . And here are the 5 functions x and x- linearly dependent, as their relationship.
Theorem. If and are linearly independent particular solutions of a linear homogeneous second order equation, then their linear combination, where and are arbitrary constants, is a general solution of this equation.
Proof. By virtue of Theorems 1 and 2 (and their corollaries), is a solution to Eq. (9) for any choice of constants and.
If solutions and are linearly independent, then is a general solution, since this solution contains two arbitrary constants that cannot be reduced to one.
At the same time, if they were linearly dependent solutions, they would no longer be a general solution. In this case, where α -constant. Then where is constant. cannot be a general solution of a second-order differential equation, since it depends only on one constant.
So, the general solution to equation (9):
19. The concept of a linearly independent system of functions. Vronsky determinant. sufficient condition for linear independence. the concept of a fundamental system of function. Examples. A necessary and sufficient condition for the nonzero Wronski determinant on the segment [a, b]
The concept of a linearly independent system of functions 
Functions 
are called linearly dependent on if one of them is a linear combination of the others. In other words, the functions are called linearly dependent on if there exist numbers, of which at least one is not equal to zero, such that
If identity (4) holds only in the case when all, then the functions are called linearly independent on.
System of solutions linearly independent on an interval
of a homogeneous differential equation of the th order (3) with coefficients continuous on is called the fundamental system of solutions of this equation.
To solve a linear homogeneous differential equation of the th order (3) with continuous coefficients, it is necessary to find its fundamental system of solutions.
According to Theorem 1, an arbitrary linear combination of solutions, i.e., the sum
, (5)
where are arbitrary numbers, is, in turn, the solution of equation (3) on. But it turns out that, and vice versa, every solution to the differential equation (3) on an interval is a certain linear combination of the indicated (mutually independent) particular solutions of it (see Theorem 4 below), which form a fundamental system of solutions.
Thus, the general solution of the homogeneous differential equation (3) has the form (5), where are arbitrary constants, and are particular solutions (3) that form a fundamental system of solutions to the homogeneous equation.
Note that the general solution to the inhomogeneous equation (1) is the sum of some of its particular solutions and general solution homogeneous equation
. (6)
Indeed,
.
On the other hand, if there is an arbitrary solution to equation (1), then
and, therefore, there is a solution to the homogeneous equation; but then there are numbers such that
,
i.e., for these numbers, equality (6) holds.
Vronsky's determinant.
Theorem 2. If functions are linearly dependent on and have derivatives up to the th order, then the determinant
. (7)
I
Determinant (7) is called the Wronsky determinant or Wronskian and is denoted by the symbol 
.
Proof. Since the functions are linearly dependent on, then there exist not all numbers equal to zero for which identity (4) is satisfied on. Differentiating it once, we obtain the system of equations
By hypothesis, this homogeneous system has a nontrivial solution (i.e., at least one) for. The latter is possible when the determinant of the system, which is the Wronski determinant, is identically zero. The theorem is proved.
Comment. From Theorem 2 it follows that if at least at one point, then the functions are linearly independent on.
Example 2. Functions are linearly independent on any, since
.
Example 3. Functions are linearly independent on any, if - different numbers (real or complex).
Indeed.
,
since the last determinant is the Vandermonde determinant, which for different is not equal to zero.
Example 4. Functions 
are linearly independent on any.
Since and
then the linear independence of these functions follows from the second example.
Theorem 3. For solutions 
linear differential homogeneous equations with continuous coefficients were linearly independent on, it is necessary and sufficient that for all.
Proof. 1) If on, then the functions are linearly independent regardless of whether they are solutions to the equation or not (see remark).
2) Let are linearly independent functions on and are solutions of the equation.
Let us prove that everywhere on. Let us assume the contrary, that there is a point at which. Let us choose numbers that are not simultaneously equal to zero, so that they are solutions of the system
(8)
This can be done since the determinant of system (8) exists. Then, by Theorem 1, the function 
will be a solution to the equation with zero initial conditions (by (8))
But the trivial solution also satisfies the same conditions. By virtue of the existence and uniqueness theorem, there can be only one solution satisfying these initial conditions, therefore, on, i.e., the functions are linearly dependent on, which was not assumed. The theorem is proved.
If are discontinuous functions in the interval where we are looking for a solution, then the equation may have more than one solution that satisfies the initial conditions, and then it is possible that on.
Example 5. It is easy to check that the functions
are linearly independent on and for them on.
This is because the function 
is a general solution to the equation
,
where 
is discontinuous at a point. The existence and uniqueness theorem does not hold for this equation (in a neighborhood of the point). Not only a function, but also a function is a solution to a differential equation that satisfies the conditions and at.
General solution structure.
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Theorem 4. If are linearly independent on solutions of a linear homogeneous differential equation of the th order with continuous coefficients 
, then the function
, (9)
where are arbitrary constants, is the general solution of the equation, i.e., the sum (9) for any, is a solution of this equation and, conversely, any solution of this equation can be represented as a sum (9) for the corresponding values.
Proof. We already know that the sum (9) for any is a solution to the equation. Let, conversely, be an arbitrary solution to this equation. We put
For the numbers obtained 
we compose a linear system of equations for unknown numbers:, it is enough to find some - real constants. To find a general solution to equation (8), we proceed as follows. We compose the characteristic equation for equation (8):. Using the initial conditions, we define
Consider the linear differential equation n-Th order
y (n) + a n -1 (x)y (n- 1) + ... + a 1 (x)y" + a 0 (x)y = f(x).
with continuous coefficients a n -1 (x), a n -2 (x), ..., a 1 (x), a 0 (x) and continuous right-hand side f(x).
Superposition principle based on the following properties of solutions of linear differential equations.
1. If y 1 (x) and y 2 (x) are two solutions to the linear homogeneous differential equation
y (n) + a n -1 (x)y (n- 1) + ... + a 1 (x)y" + a 0 (x)y = 0
then any linear combination of them y(x) = C 1 y 1 (x) + C 2 y 2 (x) is a solution to this homogeneous equation.
2. If y 1 (x) and y 2 (x) are two solutions to the linear inhomogeneous equation L(y) = f(x), then their difference y(x) = y 1 (x) − y 2 (x) is a solution to the homogeneous equation L(y) = 0 .
3. Any solution to an inhomogeneous linear equation L(y) = f(x) is the sum of any fixed (particular) solution of the inhomogeneous equation and some solution of the homogeneous equation.
4. If y 1 (x) and y 2 (x) - solutions of linear inhomogeneous equations L(y) = f 1 (x) and L(y) = f 2 (x), respectively, then their sum y(x) =y 1 (x) + y 2 (x) is a solution to the inhomogeneous equation L(y) = f 1 (x) + f 2 (x).
Usually this last statement is called superposition principle.
Constant variation method
Consider the th-order inhomogeneous equation
where the coefficients and the right-hand side are given continuous functions on the interval.
Let us assume that we know the fundamental system of solutions corresponding homogeneous equation
As we showed in § 1.15 (formula (6)), the general solution to equation (1) is equal to the sum of the general solution to equation (2) and some solution to equation (1).
The inhomogeneous equation (1) can be solved by
Integration by substitution (variable replacement). Suppose it is required 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), whence dx = q "(t) dt.
Theorem. Let the function x = q (t) be defined and differentiable on some 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 following formula is valid:
Formula (1) is called the variable change 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 formula for integration by parts. Finding the integral using this formula. It is advisable to use it when the integral on the right-hand side of formula (4) is easier to calculate than the original one.
In formula (4) there is no arbitrary constant C, since on the right-hand side of this formula there is an indefinite integral containing an arbitrary constant.
Here are some common 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 some 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 by frequent, taking for u a function that is a factor of P n (x).
Variable change in an indefinite integral is used to find integrals in which one of the functions is the derivative of another function. Let there be an integral $ \ int f (x) dx $, make the substitution $ x = \ phi (t) $. Note that the function $ \ phi (t) $ is differentiable, so we can find $ dx = \ phi "(t) dt $.
Now we substitute $ \ begin (vmatrix) x = \ phi (t) \\ dx = \ phi "(t) dt \ end (vmatrix) $ in the integral and we get that:
$$ \ int f (x) dx = \ int f (\ phi (t)) \ cdot \ phi "(t) dt $$
This is variable change formula in indefinite integral.
Variable Replacement Algorithm
Thus, if an integral of the form is specified in the problem: $$ \ int f (\ phi (x)) \ cdot \ phi "(x) dx $$ It is advisable to replace the variable with a new one: $$ t = \ phi (x) $ $ $$ dt = \ phi "(t) dt $$
After that, the integral will be presented in a form that can be easily taken by the main integration methods: $$ \ int f (\ phi (x)) \ cdot \ phi "(x) dx = \ int f (t) dt $$
Don't forget to revert the replaced variable back to $ x $ as well.
Examples of solutions
| Example 1 |
|
Find indefinite integral by substitution of variable: $$ \ int e ^ (3x) dx $$ |
| Solution |
|
We replace the variable in the integral with $ t = 3x, dt = 3dx $: $$ \ int e ^ (3x) dx = \ int e ^ t \ frac (dt) (3) = \ frac (1) (3) \ int e ^ t dt = $$ The integral of the exponent is still the same according to the integration table, although instead of $ x $ it is written $ t $: $$ = \ frac (1) (3) e ^ t + C = \ frac (1) (3) e ^ (3x) + C $$ If you can't solve your problem, then send it to us. We will provide a detailed solution. You will be able to familiarize yourself with the course of the calculation and get information. This will help you get credit from your teacher in a timely manner! |
| Answer |
| $$ \ int e ^ (3x) dx = \ frac (1) (3) e ^ (3x) + C $$ |
