The method of changing a variable in an indefinite integral. Solution examples

When solving some types of integrals, a transformation is performed, as they say insertion under the differential sign. This is done in order to obtain a tabular integral and to take it easily. To do this, apply the formula: $$ f"(x) dx = d(f(x)) $$

I would like to note such an important nuance that students think about. How does this method differ from the method of replacing a variable (substitution)? It's the same thing, only it looks different in the recordings. Both are correct.

Formula

If the product of two functions is traced in the integrand, one of which is the differential of the other, then put under the sign of the differential desired function. It looks like this:

$$ \int f(\varphi(x)) \varphi"(x) dx = \int f(\varphi(x)) d(\varphi(x))=\int f(u) du $$ $$ u=\varphi(x) $$

Summing up the main functions

In order to successfully use this method of solving, you need to know the tables of derivatives and integration. The following formulas follow from them:

Solution examples

So, in the article we analyzed how some types of integrals are solved by entering under the differential sign. We recalled the differentials of frequently used elementary functions. If it is not possible or there is not enough time to solve the tasks of examinations on your own, then we will provide you with our assistance as soon as possible. Just fill out the order form and we will contact you.

The integrals that we will consider are similar to the integrals of the previous paragraph, they have the form: or

(coefficients a, b and f are not equal to zero).

That is, we have a linear function in the numerator. How to solve such integrals?

Example 14

Find the indefinite integral

Please be careful, now we will consider a typical algorithm.

1) When an integral of the form

Or

(where the coefficients a, b and f are not equal to zero), then the first thing we do is ... take a draft. The fact is that now we have to perform a small selection.

2) Let us form the numerator of the integrand by identical transformations (we express the numerator in terms of the denominator). To do this, for now, we simply enclose the expression that is in the denominator in this example (it does not matter - under the root or without the root), under the differential sign: .

3) Opening the differential:

Let's look at the numerator of our integral:

Slightly different things turned out .... And now we need to choose a factor for the differential, such that when it is opened, it turns out at least 3 x. AT this case With a suitable multiplier, you get:

4) For self-control, we again open our differential:

Let's look at the numerator of our integral again:

Already closer, but we got not “that” term (+2), but another: (+3/2).

5) To our differential

we attribute the term that we originally had in the integrand:

- Subtract ( in this case - subtract, sometimes it is necessary, on the contrary, to add)

our "not that" term:

- We take both constants in brackets and assign the differential icon to the right:

- Subtract (in some examples you need to add) constants:

.

6) We check:

We got exactly the numerator of the integrand, which means that the selection was successful.

The clean design of the solution looks something like this:

(1) We select the numerator on the draft according to the above algorithm. Be sure to check whether the selection is correct. With a certain experience in solving integrals, the selection is not difficult to perform in the mind.

(2) Divide the numerator by the denominator term by term. In practical problem solving, this step can be omitted

(3) Using the linearity property, we separate the integrals. It is expedient to take out all the constants outside the signs of the integrals.

(4) The first integral is actually tabular, we use the formula (constant C we will add later, when we take the second integral). In the second integral, we single out the full square (we considered this type of integrals in the previous paragraph). The rest is a matter of technique.

And, for a snack, a couple of examples for an independent solution - one is easier, the other is more difficult.

Example 15

Find the indefinite integral

Example 16

Find the indefinite integral

To solve Examples 15 and 16, a special case of integrating a power function, which is not in our lookup table, will be useful:

.

Example 15: Solution:

Example 16: Solution:

.

§ 5. Integrals and their applications

.

5.1. Basic definitions and formulas. Function F(x) is antiderivative function f(x), if on some set X equality F (x)= f(x). The collection of all primitives for f(x) called indefinite integral and is denoted. At the same time, if F(x) - any of the originals f(x), then
, constant C runs through the whole set of real numbers. Table 2 shows the main formulas in which u= u(x).

table 2

1) 2) 3) 4) 5) 6) 7) 8) , 9)

10) 11) 12) 13) 14) 15) 16)

If a f(x) is a function continuous on the interval [ a; b], then exists definite integral from this function, which can be calculated from Newton-Leibniz formula:

, (5.1)

where F(x) - any prototype f(x). Unlike the indefinite integral (which is a set of functions), the definite integral is a number.

Both indefinite and definite integrals have the property linearity(the integral of the sum of functions is equal to the sum of the integrals, and the constant factor can be taken out of the integral sign):

.

Example 5.1. Find: a)
; b)
.

Solution. In the task a) we first simplify the integrand by dividing term by term each term from the numerator by the denominator, then we use the property linearity and "table" formulas 1)-3):

In the task b) besides linearity and "table" formulas 3), 9), 1), use the Newton-Leibniz formula (5.1):

5.2. Inserting under the sign of the differential and changing the variable. It can be seen that sometimes a part of the integrand forms a differential of some expression, which allows the use of tabular formulas.

Example 5.2 Find: a)
; b)
.

Solution. In the example a) it can be noticed that
and then use the formula 5) at u=ln x:

When b)
, and therefore due to 11) at
we get:

Remark 1. When introducing under the differential sign, it is useful, along with those used above, to take into account the following relations:

;
;
; ; ;

;
;
;
.

Remark 2. Integrals from example 5.2. could also be found by changing the variable. In this case, in a certain integral, the limits of integration should also be changed. Conversions to 5.2.b) would look like this, for example:

In the general case, the choice of replacement is determined by the form of the integrand. In some cases, special replacements are recommended. For example, if the expression contains an irrationality of the form
, then we can put
or
.

Example 5.3 Find: a)
; b)
.

Solution. When a) we have

(after replacement, the tabular formula was applied 11 )).

When deciding b) we necessarily change the limits of integration.

5.3. Integration by parts. In some cases, the "integration by parts formula" helps. For the indefinite integral, it has the form

, (5.2)

for a certain

, (5.3)

It is important to take into account the following.

1) If the integrand contains the product of a polynomial of x on functions
, then as u a polynomial is chosen, and the expression remaining under the integral sign refers to dv.

2) If the integrand contains inverse trigonometric ( ) or logarithmic (
) function, then as u one of them is selected.

Example 5.4. Find: a)
; b)
.

Solution. When a) apply the formula (5.2) and second rule. Exactly, we suppose
. Then
. Further,
, and therefore
. Consequently, . In the resulting integral, we select the integer part of the integrand (this is done when the degree of the numerator is not less than the degree of the denominator):

.

The final solution looks like this:

In the example b) use (5.3) and the first of the rules.

5.4. Integration of expressions containing a square trinomial. The main ideas are to single out a full square in a square trinomial and to carry out a linear replacement, which makes it possible to reduce the original integral to a tabular form 10 )-16 ).

Example 5.5. Find: a)
; b)
; in)
.

Solution. When a) we act as follows:

therefore (taking into account 13) )

When solving the example b) additional transformations are required due to the presence of a variable in the numerator of the integrand. Selecting the full square in the denominator (), we get:

For the second of the integrals, due to 11) (Table 2) we have:
. In the first integral, we introduce under the sign of the differential:

Thus, putting everything together and returning to the variable x, we get:

In the example in) we also pre-select the full square:

5.5. Integration of the simplest trigonometric functions. When integrating expressions of the form
(where m and n are natural numbers), it is recommended to take into account the following rules.

1) If both degrees are even, then the “decrease in degree” formulas are applied: ; .

2) Suppose that any of the numbers m and n- odd. For example, n=2 k+1. In this case, one of the powers of the function cosx “split off” to bring under the differential sign (because ). In the remaining expression
using the basic trigonometric identity
express through
(). After transforming the integrand (and taking into account the linearity property), we obtain an algebraic sum of integrals of the form
, each of which can be found using the formula 2) from table 2:
.

In addition, in some cases the formulas are also useful

Example 5.6. Find: a)
; b)
; in)
.

Solution. a) The integrand includes an odd (5th) power sinx, so we act on second rule, given that .

In the example b) use the formula (5.4 ), linearity indefinite integral, equality
and tabular formula 4):

When in) successively lower the degree, we take into account linearity, the possibility of introducing a constant under the differential sign and the necessary tabular formulas:

5.6. Applications of a definite integral. As is known, a curvilinear trapezoid corresponding to a non-negative and continuous on the segment [ a; b] functions f(x), called the area bounded by the graph of the function y= f(x), axis OX and two vertical lines x= a, x= b. Briefly, this can be written as follows: fig.3). and where

First, let's talk a little about the problem statement in general view, and then move on to examples of integration by substitution. Suppose we have some integral $\int g(x) \; dx$. However, there is no required formula in the table of integrals, and it is not possible to split the given integral into several tabular integrals (i.e., direct integration is no longer necessary). However, the problem will be solved if we manage to find some substitution $u=\varphi(x)$ that reduces our integral $\int g(x) \; dx$ to some table integral $\int f(u) \; du=F(u)+C$. After applying the formula $\int f(u) \; du=F(u)+C$ we only have to return the variable $x$. Formally, this can be written as follows:

$$\int g(x) \; dx=|u=\varphi(x)|=\int f(u) \; du=F(u)+C=F(\varphi(x))+C.$$

The problem is how to choose such a substitution $u$. This will require knowledge, firstly, of the table of derivatives and the ability to use it to differentiate complex functions, and secondly, tables of indefinite integrals. In addition, we will desperately need a formula, which I will write down below. If $y=f(x)$ then:

\begin(equation)dy=y"dx\end(equation)

Those. the differential of some function is equal to the derivative of this function multiplied by the differential of the independent variable. This rule is very important, and it is it that will allow you to use the substitution method. Here we indicate a couple of special cases that are obtained from formula (1). Let $y=x+C$, where $C$ is some constant (a number, to put it simply). Then, substituting the expression $x+C$ in formula (1) instead of $y$, we get the following:

$$ d(x+C)=(x+C)" dx $$

Since $(x+C)"=x"+C"=1+0=1$, the above formula becomes:

$$ d(x+C)=(x+C)" dx=1\cdot dx=dx.$$

Let us write down the obtained result separately, i.e.

\begin(equation)dx=d(x+C)\end(equation)

The resulting formula means that adding a constant under a differential does not change this differential, i.e. $dx=d(x+10)$, $dx=d(x-587)$ and so on.

Consider one more special case for formula (1). Let $y=Cx$, where $C$, again, is some constant. Let us find the differential of this function by substituting the expression $Cx$ instead of $y$ into formula (1):

$$ d(Cx)=(Cx)"dx $$

Since $(Cx)"=C\cdot (x)"=C\cdot 1=C$, the above formula $d(Cx)=(Cx)"dx$ becomes $d(Cx)=Cdx $ If we divide both parts of this formula by $C$ (assuming $C\neq 0$), we get $\frac(d(Cx))(C)=dx$ This result can be rewritten in a slightly different form:

\begin(equation)dx=\frac(1)(C)\cdot d(Cx)\;\;\;(C\neq 0)\end(equation)

The resulting formula says that the multiplication of the expression under the differential by a certain non-zero constant requires the introduction of an appropriate multiplier that compensates for such multiplication. For example, $dx=\frac(1)(5) d(5x)$, $dx=-\frac(1)(19) d(-19x)$.

In examples No. 1 and No. 2, formulas (2) and (3) will be considered in detail.

A note about formulas

In this topic, both formulas 1-3 will be used, as well as formulas from the table of indefinite integrals, which also have their own numbers. To avoid confusion, let's agree on the following: if the topic contains the text "we use formula No. 1", then it means literally the following "we use formula No. 1, located on this page". If we need a formula from the table of integrals, then we will specify this each time separately. For example, like this: "we use formula No. 1 from the table of integrals."

And one more small note

Before starting to work with examples, it is recommended that you familiarize yourself with the material presented in previous topics on the concept of an indefinite integral and . The presentation of the material in this topic is based on the information specified in the mentioned topics.

Example #1

Find $\int \frac(dx)(x+4)$.

If we turn to , we cannot find a formula that exactly matches the integral $\int \frac(dx)(x+4)$. The formula No. 2 of the table of integrals is closest to this integral, i.e. $\int \frac(du)(u)=\ln|u|+C$. The problem is this: the formula $\int \frac(du)(u)=\ln|u|+C$ assumes that in the integral $\int \frac(du)(u)$ the expressions in the denominator and under the differential must be are the same (both there and there is one letter $u$). In our case, in $\int \frac(dx)(x+4)$, the letter $x$ is under the differential, and the expression $x+4$ is in the denominator, i.e. there is a clear discrepancy with the tabular formula. Let's try to "adjust" our integral to the table one. What happens if $x+4$ is substituted for the differential instead of $x$? To answer this question, we use , substituting the expression $x+4$ instead of $y$ into it:

$$ d(x+4)=(x+4)"dx $$

Since $(x+4)"=x"+(4)"=1+0=1$, the equality $ d(x+4)=(x+4)"dx $ becomes:

$$ d(x+4)=1\cdot dx=dx $$

So $dx=d(x+4)$. To be honest, the same result could have been obtained by simply substituting the number $4$ instead of the constant $C$. In the future, we will do just that, but for the first time we analyzed the procedure for obtaining the equality $dx=d(x+4)$ in detail. But what does the equality $dx=d(x+4)$ give us?

And it gives us the following conclusion: if $dx=d(x+4)$, then $d(x+4)$ can be substituted into the integral $\int \frac(dx)(x+4)$ instead of $dx$ , and the integral does not change from this:

$$ \int \frac(dx)(x+4)=\int \frac(d(x+4))(x+4)$$

We made this transformation only so that the resulting integral would completely correspond to the tabular formula $\int \frac(du)(u)=\ln|u|+C$. To make this correspondence quite explicit, we replace the expression $x+4$ with the letter $u$ (i.e., we make substitution$u=x+4$):

$$ \int \frac(dx)(x+4)=\int \frac(d(x+4))(x+4)=|u=x+4|=\int \frac(du)(u )=\ln|u|+C.$$

In fact, the problem has already been solved. It remains only to return the variable $x$. Remembering that $u=x+4$, we get: $\ln|u|+C=\ln|x+4|+C$. Complete Solution without explanation it looks like this:

$$ \int \frac(dx)(x+4)=\int \frac(d(x+4))(x+4)=|u=x+4|=\int \frac(du)(u )=\ln|u|+C=\ln|x+4|+C.$$

Answer: $\int \frac(dx)(x+4)=\ln|x+4|+C$.

Example #2

Find $\int e^(3x) dx$.

If we turn to the table of indefinite integrals, we will not be able to find a formula that exactly corresponds to the integral $\int e^(3x) dx$. Formula No. 4 from the table of integrals is closest to this integral, i.e. $\int e^u du=e^u+C$. The problem is this: the formula $\int e^u du=e^u+C$ assumes that in the integral $\int e^u du$ the expressions in the power of $e$ and under the differential must be the same (both there and there there is one letter $u$). In our case, in $\int e^(3x) dx$, the letter $x$ is under the differential, and the expression $3x$ is in the power of $e$, i.e. there is a clear discrepancy with the tabular formula. Let's try to "adjust" our integral to the table one. What happens if $3x$ is substituted for the differential instead of $x$? To answer this question, we use , substituting the expression $3x$ instead of $y$ into it:

$$ d(3x)=(3x)"dx $$

Since $(3x)"=3\cdot (x)"=3\cdot 1=3$, the equality $d(3x)=(3x)"dx$ becomes:

$$ d(3x)=3dx $$

Dividing both sides of the resulting equality by $3$, we get: $\frac(d(3x))(3)=dx$, i.e. $dx=\frac(1)(3)\cdot d(3x)$. Actually, the equality $dx=\frac(1)(3)\cdot d(3x)$ could be obtained by simply substituting the number $3$ instead of the constant $C$. In the future, we will do just that, but for the first time we analyzed the procedure for obtaining the equality $dx=\frac(1)(3)\cdot d(3x)$ in detail.

What did the resulting equality $dx=\frac(1)(3)\cdot d(3x)$ give us? It means that $\frac(1)(3)\cdot d(3x)$ can be substituted into the integral $\int e^(3x) dx$ instead of $dx$ without changing the integral:

$$ \int e^(3x) dx= \int e^(3x) \cdot\frac(1)(3) d(3x) $$

We take the constant $\frac(1)(3)$ out of the integral sign and replace the expression $3x$ with the letter $u$ (i.e., we make substitution$u=3x$), after which we apply the tabular formula $\int e^u du=e^u+C$:

$$ \int e^(3x) dx= \int e^(3x) \cdot\frac(1)(3) d(3x)=\frac(1)(3)\cdot \int e^(3x) d(3x)=|u=3x|=\frac(1)(3)\cdot\int e^u du=\frac(1)(3)\cdot e^u+C.$$

As in the previous example, you need to return the original variable $x$. Since $u=3x$, then $\frac(1)(3)\cdot e^u+C=\frac(1)(3)\cdot e^(3x)+C$. The complete solution without comments looks like this:

$$ \int e^(3x) dx= \int e^(3x) \cdot\frac(1)(3) d(3x)=\frac(1)(3)\cdot \int e^(3x) d(3x)=|u=3x|=\frac(1)(3)\cdot\int e^u du=\frac(1)(3)\cdot e^u+C=\frac(1)( 3)\cdot e^(3x)+C.$$

Answer: $ \int e^(3x) dx= \frac(1)(3)\cdot e^(3x)+C$.

Example #3

Find $\int (3x+2)^2 dx$.

There are two ways to find this integral. The first way is to expand the brackets and directly integrate. The second way is to use the substitution method.

First way

Since $(3x+2)^2=9x^2+12x+4$, then $\int (3x+2)^2 dx=\int (9x^2+12x+4)dx$. Representing the integral $\int (9x^2+12x+4)dx$ as a sum of three integrals and taking the constants out of the signs of the corresponding integrals, we get:

$$ \int (9x^2+12x+4)dx=\int 9x^2 dx+\int 12x dx+\int 4 dx=9\cdot \int x^2 dx+12\cdot \int x dx+4\ cdot \int 1 dx $$

To find $\int x^2 dx$, let's substitute $u=x$ and $\alpha=2$ into formula #1 of the integral table: $\int x^2 dx=\frac(x^(2+1))( 2+1)+C=\frac(x^3)(3)+C$. Similarly, substituting $u=x$ and $\alpha=1$ into the same formula from the table, we get: $\int x^1 dx=\frac(x^(1+1))(1+1)+ C=\frac(x^2)(2)+C$. Since $\int 1 dx=x+C$, then:

$$ 9\cdot \int x^2 dx+12\cdot \int x dx+4\cdot \int 1 dx=9\cdot\frac(x^3)(3)+12\cdot \frac(x^ 2)(2)+4\cdot x+C=3x^3+6x^2+4x+C. $$

$$ \int (9x^2+12x+4)dx=\int 9x^2 dx+\int 12x dx+\int 4 dx=9\cdot \int x^2 dx+12\cdot \int x dx+4\ cdot \int 1 dx=\\ =9\cdot\frac(x^3)(3)+12\cdot \frac(x^2)(2)+4\cdot x+C=3x^3+6x^ 2+4x+C. $$

Second way

We will not open the brackets. Let's try to make the expression $3x+2$ appear under the differential instead of $x$. This will allow you to enter a new variable and apply the spreadsheet formula. We need the factor $3$ to appear under the differential, therefore, substituting $C=3$ into the value, we get $d(x)=\frac(1)(3)d(3x)$. In addition, the term $2$ is missing under the differential. According to the addition of a constant under the sign of the differential does not change this differential, i.e. $\frac(1)(3)d(3x)=\frac(1)(3)d(3x+2)$. From the conditions $d(x)=\frac(1)(3)d(3x)$ and $\frac(1)(3)d(3x)=\frac(1)(3)d(3x+2) $ we have: $dx=\frac(1)(3)d(3x+2)$.

I note that the equality $dx=\frac(1)(3)d(3x+2)$ can be obtained in another way:

$$ d(3x+2)=(3x+2)"dx=((3x)"+(2)")dx=(3\cdot x"+0)dx=3\cdot 1 dx=3dx;\ \dx=\frac(1)(3)d(3x+2). $$

We use the resulting equality $dx=\frac(1)(3)d(3x+2)$, substituting into the integral $\int (3x+2)^2 dx$ the expression $\frac(1)(3)d(3x +2)$ instead of $dx$. We take the constant $\frac(1)(3)$ out of the sign of the resulting integral:

$$ \int (3x+2)^2 dx=\int (3x+2)^2 \cdot \frac(1)(3)d(3x+2)=\frac(1)(3)\cdot \ int(3x+2)^2 d(3x+2). $$

The further solution is to perform the substitution $u=3x+2$ and apply formula No. 1 from the table of integrals:

$$ \frac(1)(3)\cdot \int (3x+2)^2 d(3x+2)=|u=3x+2|=\frac(1)(3)\cdot \int u^ 2 du=\frac(1)(3)\cdot \frac(u^(2+1))(2+1)+C=\frac(u^3)(9)+C. $$

Returning the expression $3x+2$ instead of $u$, we get:

$$ \frac(u^3)(9)+C=\frac((3x+2)^3)(9)+C. $$

The complete solution without explanation is this:

$$ \int (3x+2)^2 dx=\frac(1)(3)\cdot \int (3x+2)^2 d(3x+2)=|u=3x+2|=\\ = \frac(1)(3)\cdot \int u^2 du=\frac(u^3)(9)+C=\frac((3x+2)^3)(9)+C. $$

I foresee a couple of questions, so I'll try to formulate them and give answers.

Question #1

Something doesn't add up here. When we solved in the first way, we got that $\int (9x^2+12x+4)dx=3x^3+6x^2+4x+C$. When solving the second way, the answer became: $\int (3x+2)^2 dx=\frac((3x+2)^3)(9)+C$. However, the transition from the second answer to the first does not work! If we open the brackets, we get the following:

$$ \frac((3x+2)^3)(9)+C=\frac(27x^3+54x^2+36x+8)(9)+C=\frac(27x^3)(9) +\frac(54x^2)(9)+\frac(36x)(9)+\frac(8)(9)+C=3x^3+6x^2+4x+\frac(8)(9)+ C. $$

The answers don't match! Where did the extra fraction $\frac(8)(9)$ come from?

This question suggests that you should refer to the previous topics. Read the topic about the concept of an indefinite integral (giving Special attention question #2 at the end of the page) and direct integration (pay attention to question #4). In these topics, this issue is covered in detail. In short, the integral constant $C$ can be represented in different forms. For example, in our case, renaming $C_1=C+\frac(8)(9)$, we get:

$$ 3x^3+6x^2+4x+\frac(8)(9)+C=3x^3+6x^2+4x+C_1. $$

Therefore, there is no contradiction, the answer can be written both in the form $3x^3+6x^2+4x+C$, and in the form $\frac((3x+2)^3)(9)+C$.

Question #2

Why was it decided the second way? This is just too much complication! Why use a bunch of extra formulas to find an answer that can be obtained in a couple of steps in the first way? All that was needed was to open the brackets by applying the school formula.

Well, firstly, this is not such a complication. When you understand the substitution method, you will begin to solve such examples in one line: $\int (3x+2)^2 dx=\frac(1)(3)\cdot \int (3x+2)^2 d( 3x+2)=\frac((3x+2)^3)(9)+C$. However, let's look at this example in a different way. Imagine that you need to calculate not $\int (3x+2)^2 dx$, but $\int (3x+2)^(200) dx$. When solving the second way, you only have to slightly tweak the degrees and the answer will be ready:

$$ \int (3x+2)^(200) dx=\frac(1)(3)\cdot \int (3x+2)^(200) d(3x+2)=|u=3x+2| =\\ =\frac(1)(3)\cdot \int u^(200) du=\frac(u^(201))(603)+C=\frac((3x+2)^(201) )(603)+C. $$

Now imagine that the same integral $\int (3x+2)^(200) dx$ needs to be taken in the first way. First, you will need to open the bracket $(3x+2)^(200)$, thus obtaining the sum of two hundred and one terms! And then each term will also have to be integrated. Therefore, the conclusion here is this: for large degrees, the method of direct integration is not suitable. The second method, despite the apparent complexity, is more practical.

Example #4

Find $\int \sin2x dx$.

We will solve this example in three different ways.

First way

Let's look at the table of integrals. Formula No. 5 from this table is closest to our example, i.e. $\int \sin u du=-\cos u+C$. To fit the integral $\int \sin2x dx$ to the form $\int \sin u du$, we use , introducing the factor $2$ under the differential sign. Actually, we already did this in example No. 2, so we can do without detailed comments:

$$ \int \sin 2x dx=\left|dx=\frac(1)(2)\cdot d(2x) \right|=\int \sin 2x \cdot\frac(1)(2)d(2x )=\\ =\frac(1)(2) \int \sin 2x d(2x)=|u=2x|=\frac(1)(2) \int \sin u du=-\frac(1) (2)\cos u+C=-\frac(1)(2)\cos 2x+C. $$

Answer: $\int \sin2x dx=-\frac(1)(2)\cos 2x+C$.

Second way

To solve the second way, we apply a simple trigonometric formula: $\sin 2x=2\sin x\cos x$. Let us substitute the expression $2 \sin x \cos x$ instead of $\sin 2x$, taking the constant $2$ out of the integral sign:

What is the purpose of such a transformation? There is no integral $\int \sin x\cos x dx$ in the table, but we can slightly transform $\int \sin x\cos x dx$ to make it look more like a table. To do this, find $d(\cos x)$ using . Let's substitute $\cos x$ instead of $y$ into the mentioned formula:

$$ d(\cos x)=(\cos x)"dx=-\sin x dx. $$

Since $d(\cos x)=-\sin x dx$, then $\sin x dx=-d(\cos x)$. Since $\sin x dx=-d(\cos x)$, we can substitute $-d(\cos x)$ instead of $\sin x dx$ in $\int \sin x\cos x dx$. The value of the integral will not change:

$$ 2\cdot\int \sin x\cos x dx=2\cdot\int \cos x \cdot (-d(\cos x))=-2\int\cos x d(\cos x) $$

In other words, we brought under the differential$\cosx$. Now, by substituting $u=\cos x$, we can apply formula #1 from the table of integrals:

$$ -2\int\cos x d(\cos x)=|u=\cos x|=-2\int u du=-2\cdot \frac(u^2)(2)+C=-u^ 2+C=-\cos^2x+C. $$

Answer received. In general, you can omit the letter $u$. When you acquire sufficient skill in solving this kind of integrals, then the need for additional notation will disappear. The complete solution without explanation is this:

$$ \int \sin 2x dx=2\cdot\int \sin x\cos x dx=|\sin x dx=-d(\cos x)|=-2\int\cos x d(\cos x)= |u=\cos x|=\\ =-2\int u du=-2\cdot \frac(u^2)(2)+C=-u^2+C=-\cos^2x+C. $$

Answer: $\int \sin2x dx=-\cos^2x+C$.

Third way

To solve it in the third way, we apply the same trigonometric formula: $\sin 2x=2\sin x\cos x$. Let us substitute the expression $2 \sin x \cos x$ instead of $\sin 2x$, taking the constant $2$ out of the integral sign:

$$ \int \sin 2x dx=\int 2 \sin x\cos x dx=2\cdot\int \sin x\cos x dx $$

Find $d(\sin x)$ using . Let's substitute $\sin x$ instead of $y$ into the mentioned formula:

$$ d(\sin x)=(\sin x)"dx=\cos x dx. $$

So $d(\sin x)=\cos x dx$. It follows from the resulting equality that we can substitute $d(\sin x)$ instead of $\cos x dx$ in $\int \sin x\cos x dx$. The value of the integral will not change:

$$ 2\cdot\int \sin x\cos x dx=2\cdot\int \sin x \cdot d(\sin x) $$

In other words, we brought under the differential$\sinx$. Now, by substituting $u=\sin x$, we can apply formula #1 from the table of integrals:

$$ 2\int\sin x d(\sin x)=|u=\sin x|=2\int u du=2\cdot \frac(u^2)(2)+C=u^2+C= \sin^2x+C. $$

Answer received. The complete solution without explanation is:

$$ \int \sin 2x dx=2\cdot\int \sin x\cos x dx=|\cos x dx=d(\sin x)|=2\cdot\int \sin x \cdot d(\sin x)=|u=\sin x|=\\ =2\int u du=2\cdot \frac(u^2)(2)+C=u^2+C=\sin^2x+C. $$

Answer: $\int \sin2x dx=\sin^2x+C$.

It is possible that after reading this example, especially the three different (at first glance) answers, a question will arise. Let's consider it.

Question #3

Wait. The answers should match, but they are different! In example #3, the difference was only in the constant $\frac(8)(9)$, but here even outwardly the answers are not similar: $-\frac(1)(2)\cos 2x+C$, $-\ cos^2x+C$, $\sin^2x+C$. Is it really all about the integral constant $C$ again?

Yes, it's in this constant. Let's reduce all the answers to one form, after which this difference in constants will become quite clear. Let's start with $-\frac(1)(2)\cos 2x+C$. We use a simple trigonometric equation: $\cos 2x=1-2\sin^2 x$. Then the expression $-\frac(1)(2)\cos 2x+C$ becomes:

$$ -\frac(1)(2)\cos 2x+C=-\frac(1)(2)\cdot(1-2\sin^2 x)+C=-\frac(1)(2) +\frac(1)(2)\cdot 2\sin^2x+C=\sin^2x+C-\frac(1)(2). $$

Now let's work with the second answer, i.e. $-\cos^2x+C$. Since $\cos^2 x=1-\sin^2x$, then:

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

The three answers we got in example #4 became: $\sin^2 x+C-\frac(1)(2)$, $\sin^2x+C-1$, $\sin^2x+ C$. I think it is now clear that they differ from each other only by a certain number. Those. the matter again turned out to be in the integral constant. As you can see, a small difference in the integral constant can, in principle, greatly change appearance answer, - but this does not stop the answer from being correct. What I'm leading to: if you see an answer in the collection of tasks that does not match yours, then this does not mean at all that your answer is wrong. It is possible that you simply came to the answer in a different way than the author of the problem intended. And to make sure the answer is correct, a check based on the definition of an indefinite integral will help. For example, if the integral $\int \sin2x dx=-\frac(1)(2)\cos 2x+C$ is found correctly, then the equality $\left(-\frac(1)(2)\cos 2x+ C\right)"=\sin 2x$. Let's check whether it is true that the derivative of $\left(-\frac(1)(2)\cos 2x+C\right)$ is equal to the integrand $\sin 2x $:

$$ \left(-\frac(1)(2)\cos 2x+C\right)"=\left(-\frac(1)(2)\cos 2x\right)"+C"=-\frac (1)(2)\cdot(\cos 2x)"+0=\\ =-\frac(1)(2)\cdot (-\sin 2x)\cdot (2x)"=-\frac(1) (2)\cdot (-\sin 2x)\cdot 2=\sin 2x.$$

Verification completed successfully. The equality $\left(-\frac(1)(2)\cos 2x+C\right)"=\sin 2x$ holds, so the formula $\int \sin2x dx=-\frac(1)(2)\cos 2x+C$ is correct In example #5 we will also check the result to make sure it is correct. control work the requirement to check the result is present.

The method of subsuming under the differential sign is rarely given in the literature, so we will first show why it is beneficial.

Often in the integrand one can see 2 fragments, one of which similar to derivative another. For example,

a) in the integral the numerator x similar to the derivative of :
;

b) integral
can be imagined as
, where
;

c) function
in the integral
- this is
.

Such integrals are often proposed to be found by replacing the new variable a function whose derivative discovered. So, for the indicated integrals

what if
, then
, then
and
, where

b) because
, then
, then
and
, that's why

The replacement method is described in more detail in § 4.

However, the calculation of the 3rd integral using the replacement is already associated with difficulties. Let, noticing that
, we have replaced
.

Then
and
. express
through t it is possible like this:

(
, that's why
). Substitute:

As a result of cumbersome actions, almost everything was reduced and a simple tabular integral was obtained. The question arises whether it was possible to arrive at it faster if almost no expression was needed.

Indeed, there is a shorter solution:

then, replacing
, we immediately obtain the integral

In the same way, one could find the integrals

Here the steps are shown in great detail, and half of them can be skipped. The following will make the solution especially short.

Table of main differentials

Examples of subsuming under the differential sign

3) ;

PD1. Find Integrals

1) a)
; b)
; in)
; G)
; e)
;

e)
; and)
; h)
; and)
; to)
;

2) a)
; b)
; in)
; G)
; e)
;

e)
; and)
; h)
; and)
; to)
;

3) a)
; b)
; in)
; G)
; e)

e)
; and)
; h)
; and)
; to)
;

4) a)
; b)
; in)
; G)
; e)
;

e)
; and)
; h)
; and)
; to)
;

5) a)
; b)
; in)
; G)
; e)
;

e)
; and)
; h)
; and)
; to)
.

§ 3. Integrals of functions containing a quadratic expression

When integrating functions containing the expression
, the formula will help
. For example,

b)
;

It is convenient to denote the resulting bracket by a new letter and pass to the integral over this variable (the differentials of the new and old variables will coincide).

The coefficient in front of the square is better to take out of the bracket:

,

and then, if possible, for the integral sign. So,

The purpose of the replacement is to pass to an integral without a linear term
, since integrals containing only
, are easier, and often - according to the table. At the same time, it is important to remember that
,
, etc.

Namely (see § 2),

where a- any number, and the number
. In addition, at

where
.

Remark 1. After the replacement, integrals often appear
,
or
. They can be found like this:

similarly in the 2nd and 3rd cases.

However, integrals of the form
are quite complex. Use ready-made formulas

(check by differentiation that this is indeed the case).

CI1. Find using equality
and replacements
:

Example 1(to be short
labeled as
.

When searching
and
took into account that
and
respectively, and applied the basic rule of tabular integration.

CI2. Find the integrals by expanding each into a sum of integrals, one of which is tabular, and the other is similar to those found in task KI1:

Example 2 Let's find the integral
, expanding into the sum of two:

Answer:(the module is not needed, because always
).

Example 3 Let us take in the same way the integral
:

The most rational way to find integrals is as follows:

where did you learn that
;

Then where
.

Answer: .

Remark 2. In the future, you will often have to break the integral into 2 or 3 integrals, in each of which a constant appears (
, etc.). For brevity, we will mean (but not indicate) the constants in each individual auxiliary integral (or indicate, but not accompany with a number), and we will write only the general constant C in the answer. At the same time, always C is some linear combination.

CI3. Having obtained a full square in the denominator and making a replacement, find

Example 4
Noticing that

replace
, then
and.

Substitute in the integral:

Example 5

Since , we can make the substitution
, with which
and
. Substitute:

Example 6

Here, we replace
, where
and
. Substitute:

where
. Let's break the integral into two:

.

Just like in the previous examples,

and the 2nd integral is tabular:
.

So, where
. Thereby

Example 7

Now, replacement
, that's why
and
.

We pass to the integral of the new variable:

where
.

We will find separately

in)
(table integral).

Multiply the 2nd result by 7, the 3rd by 10, collect similar terms and return to the old variable:

CI4. Find integrals of irrational functions:

Example 8 Let's find
. A similar integral without a root has already been found above (Example 6), and it is enough to add a root at the appropriate step:

,

where
. Breaking down

and find

b)
.

Thus, where
.

Answer: .

Example 9
It is convenient to get a full square like this:

where
. Then

.

Let's replace
. Wherein
and
:

We act in the same way as in example 8:

Answer: .

Remark 3. It is impossible to take out the “-” sign or any negative common factor from under the root:
;, etc. Example 9 shows the only possible The right way actions.

Example 10 Let's see what will change if we put a square in example 9: we find
. Now, after the same substitutions, it turns out that

As usual,

and the 2nd and 3rd integrals are found in the same way as in Example 9:

;

.

According to the instructions on page 19, the 1st integral can be converted as follows:

where again
, a

The new integral is found either by trigonometric substitution
, or by repeated integration by parts, taking
and
. Let's use the formula
(page 19):

Multiply all integrals by their respective coefficients and put together:

in the answer we give like terms.