Online calculator. Calculate the indefinite integral (antiderivative). Basic integration methods

Complex integrals

This article concludes the topic of indefinite integrals, and includes integrals that I find quite complex. The lesson was created at the repeated requests of visitors who expressed their wish that more difficult examples be analyzed on the site.

It is assumed that the reader of this text is well prepared and knows how to apply basic integration techniques. Dummies and people who are not very confident in integrals should refer to the very first lesson - Indefinite integral. Examples of solutions, where you can master the topic almost from scratch. More experienced students can become familiar with techniques and methods of integration that have not yet been encountered in my articles.

What integrals will be considered?

First we will consider integrals with roots, for the solution of which we successively use variable replacement And integration by parts. That is, in one example two techniques are combined at once. And even more.

Then we will get acquainted with interesting and original method of reducing the integral to itself. Quite a few integrals are solved this way.

The third issue of the program will be integrals of complex fractions, which flew past the cash desk in previous articles.

Fourthly, additional integrals from trigonometric functions will be analyzed. In particular, there are methods that avoid time-consuming universal trigonometric substitution.

(2) In the integrand function, we divide the numerator by the denominator term by term.

(3) We use the linearity property of the indefinite integral. In the last integral immediately put the function under the differential sign.

(4) We take the remaining integrals. Note that in a logarithm you can use parentheses rather than a modulus, since .

(5) We carry out a reverse replacement, expressing “te” from the direct replacement:

Masochistic students can differentiate the answer and get the original integrand, as I just did. No, no, I did the check in the right sense =)

As you can see, during the solution we had to use even more than two solution methods, so to deal with such integrals you need confident integration skills and quite a bit of experience.

In practice, of course, the square root is more common; here are three examples for solving it yourself:

Example 2

Find indefinite integral

Example 3

Find the indefinite integral

Example 4

Find the indefinite integral

These examples are of the same type, so the complete solution at the end of the article will only be for Example 2; Examples 3-4 have the same answers. Which replacement to use at the beginning of decisions, I think, is obvious. Why did I choose examples of the same type? Often found in their role. More often, perhaps, just something like .

But not always, when under the arctangent, sine, cosine, exponential and other functions there is a root of linear function, you have to use several methods at once. In a number of cases, it is possible to “get off easy,” that is, immediately after the replacement, a simple integral is obtained, which can be easily taken. The easiest of the tasks proposed above is Example 4, in which, after replacement, a relatively simple integral is obtained.

By reducing the integral to itself

A witty and beautiful method. Let's take a look at the classics of the genre:

Example 5

Find the indefinite integral

Under the root is a quadratic binomial, and when trying to integrate this example the kettle can suffer for hours. Such an integral is taken in parts and reduced to itself. In principle, it’s not difficult. If you know how.

Let us denote the integral under consideration by a Latin letter and begin the solution:

Let's integrate by parts:

(1) Prepare the integrand function for term-by-term division.

(2) We divide the integrand function term by term. It may not be clear to everyone, but I’ll describe it in more detail:

(3) We use the linearity property of the indefinite integral.

(4) Take the last integral (“long” logarithm).

Now let's look at the very beginning of the solution:

And at the end:

What happened? As a result of our manipulations, the integral was reduced to itself!

Let's equate the beginning and the end:

Move to the left side with a change of sign:

And we take the deuce to right side. As a result:

The constant, strictly speaking, should have been added earlier, but I added it at the end. I strongly recommend reading what the rigor is here:

Note: More strictly, the final stage of the solution looks like this:

Thus:

The constant can be redesignated by . Why can it be redesignated? Because he still accepts it any values, and in this sense there is no difference between constants and.
As a result:

A similar trick with constant renotation is widely used in differential equations. And there I will be strict. And here I allow such freedom only in order not to confuse you with unnecessary things and to focus attention precisely on the integration method itself.

Example 6

Find the indefinite integral

Another typical integral for independent solution. Full solution and answer at the end of the lesson. There will be a difference with the answer in the previous example!

If under square root is a quadratic trinomial, then the solution in any case reduces to two analyzed examples.

For example, consider the integral . All you need to do is first select a complete square:
.
Next, a linear replacement is carried out, which does “without any consequences”:
, resulting in the integral . Something familiar, right?

Or this example, with a quadratic binomial:
Select a complete square:
And, after linear replacement, we obtain the integral, which is also solved using the algorithm already discussed.

Let's look at two more typical examples of how to reduce an integral to itself:
– integral of the exponential multiplied by sine;
– integral of the exponential multiplied by the cosine.

In the listed integrals by parts you will have to integrate twice:

Example 7

Find the indefinite integral

The integrand is the exponential multiplied by the sine.

We integrate by parts twice and reduce the integral to itself:

As a result of double integration by parts, the integral was reduced to itself. We equate the beginning and end of the solution:

We move it to the left side with a change of sign and express our integral:

Ready. At the same time, it is advisable to comb the right side, i.e. take the exponent out of brackets, and place the sine and cosine in brackets in a “beautiful” order.

Now let's go back to the beginning of the example, or more precisely, to integration by parts:

We designated the exponent as. The question arises: is it the exponent that should always be denoted by ? Not necessary. In fact, in the considered integral fundamentally doesn't matter, what do we mean by , we could have gone the other way:

Why is this possible? Because the exponential turns into itself (both during differentiation and integration), sine and cosine mutually turn into each other (again, both during differentiation and integration).

That is, we can also denote a trigonometric function. But, in the example considered, this is less rational, since fractions will appear. If you wish, you can try to solve this example using the second method; the answers must match.

Example 8

Find the indefinite integral

This is an example for you to solve on your own. Before you decide, think about what is more advantageous in this case to designate as , an exponential or a trigonometric function? Full solution and answer at the end of the lesson.

And, of course, do not forget that most of the answers in this lesson are quite easy to check by differentiation!

The examples considered were not the most complex. In practice, integrals are more common where the constant is both in the exponent and in the argument of the trigonometric function, for example: . Many people will get confused in such an integral, and I often get confused myself. The fact is that there is a high probability of fractions appearing in the solution, and it is very easy to lose something through carelessness. In addition, there is a high probability of an error in the signs; note that the exponent has a minus sign, and this introduces additional difficulty.

At the final stage, the result is often something like this:

Even at the end of the solution, you should be extremely careful and correctly understand the fractions:

Integrating Complex Fractions

We are slowly approaching the equator of the lesson and begin to consider integrals of fractions. Again, not all of them are super complex, it’s just that for one reason or another the examples were a little “off topic” in other articles.

Continuing the theme of roots

Example 9

Find the indefinite integral

In the denominator under the root there is a quadratic trinomial plus an “appendage” in the form of an “X” outside the root. An integral of this type can be solved using a standard substitution.

We decide:

The replacement here is simple:

Let's look at life after replacement:

(1) After substitution, we reduce the terms under the root to a common denominator.
(2) We take it out from under the root.
(3) The numerator and denominator are reduced by . At the same time, under the root, I rearranged the terms in a convenient order. With some experience, steps (1), (2) can be skipped by performing the commented actions orally.
(4) The resulting integral, as you remember from the lesson Integrating Some Fractions, is being decided complete square extraction method. Select a complete square.
(5) By integration we obtain an ordinary “long” logarithm.
(6) We carry out the reverse replacement. If initially , then back: .
(7) The final action is aimed at straightening the result: under the root we again bring the terms to a common denominator and take them out from under the root.

Example 10

Find the indefinite integral

This is an example for you to solve on your own. Here a constant is added to the lone “X”, and the replacement is almost the same:

The only thing you need to do additionally is to express the “x” from the replacement being carried out:

Full solution and answer at the end of the lesson.

Sometimes in such an integral there may be a quadratic binomial under the root, this does not change the method of solution, it will be even simpler. Feel the difference:

Example 11

Find the indefinite integral

Example 12

Find the indefinite integral

Brief solutions and answers at the end of the lesson. It should be noted that Example 11 is exactly binomial integral, the solution method of which was discussed in class Integrals of irrational functions.

Integral of an indecomposable polynomial of the 2nd degree to the power

(polynomial in denominator)

More rare, but nevertheless found in practical examples type of integral.

Example 13

Find the indefinite integral

But let’s return to the example with lucky number 13 (honestly, I didn’t guess correctly). This integral is also one of those that can be quite frustrating if you don’t know how to solve.

The solution starts with an artificial transformation:

I think everyone already understands how to divide the numerator by the denominator term by term.

The resulting integral is taken in parts:

For an integral of the form ( – natural number) withdrawn recurrent reduction formula:
, Where – integral of a degree lower.

Let us verify the validity of this formula for the solved integral.
In this case: , , we use the formula:

As you can see, the answers are the same.

Example 14

Find the indefinite integral

This is an example for you to solve on your own. The sample solution uses the above formula twice in succession.

If under the degree is indivisible square trinomial, then the solution is reduced to a binomial by isolating the perfect square, for example:

What if there is an additional polynomial in the numerator? In this case, the method of indefinite coefficients is used, and the integrand function is expanded into a sum of fractions. But in my practice there is such an example never met so I missed it this case in the article Integrals of fractional-rational functions, I'll skip it now. If you still encounter such an integral, look at the textbook - everything is simple there. I don’t think it’s advisable to include material (even simple ones), the probability of encountering which tends to zero.

Integrating complex trigonometric functions

The adjective “complex” for most examples is again largely conditional. Let's start with tangents and cotangents in high degrees. From the point of view of the solving methods used, tangent and cotangent are almost the same thing, so I will talk more about tangent, implying that the demonstrated method for solving the integral is valid for cotangent too.

In the above lesson we looked at universal trigonometric substitution for solutions certain type integrals of trigonometric functions. The disadvantage of universal trigonometric substitution is that its use often results in cumbersome integrals with difficult calculations. And in some cases, universal trigonometric substitution can be avoided!

Let's consider another canonical example, the integral of one divided by sine:

Example 17

Find the indefinite integral

Here you can use universal trigonometric substitution and get the answer, but there is a more rational way. I will provide the complete solution with comments for each step:

(1) We use the trigonometric formula for the sine of a double angle.
(2) We carry out an artificial transformation: Divide in the denominator and multiply by .
(3) Using the well-known formula in the denominator, we transform the fraction into a tangent.
(4) We bring the function under the differential sign.
(5) Take the integral.

Pair simple examples for independent solution:

Example 18

Find the indefinite integral

Note: The very first step should be to use the reduction formula and carefully carry out actions similar to the previous example.

Example 19

Find the indefinite integral

Well, this is a very simple example.

Complete solutions and answers at the end of the lesson.

I think now no one will have problems with integrals:
and so on.

What is the idea of ​​the method? The idea is that, using transformations, trigonometric formulas organize only tangents and the derivative of tangent in the integrand. That is, we're talking about about replacement: . In Examples 17-19 we actually used this replacement, but the integrals were so simple that we got by with an equivalent action - subsuming the function under the differential sign.

Similar reasoning, as I already mentioned, can be carried out for the cotangent.

There is also a formal prerequisite for applying the above replacement:

The sum of the powers of cosine and sine is a negative integer EVEN number, For example:

for the integral – a negative integer EVEN number.

! Note : if the integrand contains ONLY a sine or ONLY a cosine, then the integral is also taken for a negative odd degree (the simplest cases are in Examples No. 17, 18).

Let's look at a couple of more meaningful tasks based on this rule:

Example 20

Find the indefinite integral

The sum of the powers of sine and cosine: 2 – 6 = –4 is a negative integer EVEN number, which means that the integral can be reduced to tangents and its derivative:

(1) Let's transform the denominator.
(2) Using the well-known formula, we obtain .
(3) Let's transform the denominator.
(4) We use the formula .
(5) We bring the function under the differential sign.
(6) We carry out replacement. More experienced students may not carry out the replacement, but it is still better to replace the tangent with one letter - there is less risk of getting confused.

Example 21

Find the indefinite integral

This is an example for you to solve on your own.

Hang in there, the championship rounds are about to begin =)

Often the integrand contains a “hodgepodge”:

Example 22

Find the indefinite integral

This integral initially contains a tangent, which immediately leads to an already familiar thought:

I will leave the artificial transformation at the very beginning and the remaining steps without comment, since everything has already been discussed above.

A couple of creative examples for your own solution:

Example 23

Find the indefinite integral

Example 24

Find the indefinite integral

Yes, in them, of course, you can lower the powers of sine and cosine, and use a universal trigonometric substitution, but the solution will be much more efficient and shorter if it is carried out through tangents. Full solution and answers at the end of the lesson

By a definite integral from a continuous function f(x) on the final segment [ a, b] (where ) is the increment of some of its antiderivatives on this segment. (In general, understanding will be noticeably easier if you repeat the topic of the indefinite integral) In this case, the notation is used

As can be seen in the graphs below (the increment of the antiderivative function is indicated by ), the definite integral can be either positive or negative number (It is calculated as the difference between the value of the antiderivative in the upper limit and its value in the lower limit, i.e. as F(b) - F(a)).

Numbers a And b are called the lower and upper limits of integration, respectively, and the segment [ a, b] – segment of integration.

Thus, if F(x) – some antiderivative function for f(x), then, according to the definition,

(38)

Equality (38) is called Newton-Leibniz formula . Difference F(b) – F(a) is briefly written as follows:

Therefore, we will write the Newton-Leibniz formula like this:

(39)

Let us prove that the definite integral does not depend on which antiderivative of the integrand is taken when calculating it. Let F(x) and F( X) are arbitrary antiderivatives of the integrand. Since these are antiderivatives of the same function, they differ by a constant term: Ф( X) = F(x) + C. That's why

This establishes that on the segment [ a, b] increments of all antiderivatives of the function f(x) match up.

Thus, to calculate a definite integral, it is necessary to find any antiderivative of the integrand, i.e. First you need to find the indefinite integral. Constant WITH excluded from subsequent calculations. Then the Newton-Leibniz formula is applied: in antiderivative function the value of the upper limit is substituted b , further - the value of the lower limit a and the difference is calculated F(b) - F(a) . The resulting number will be a definite integral..

At a = b by definition accepted

Example 1.

Solution. First, let's find the indefinite integral:

Applying the Newton-Leibniz formula to the antiderivative

(at WITH= 0), we get

However, when calculating a definite integral, it is better not to find the antiderivative separately, but to immediately write the integral in the form (39).

Example 2. Calculate definite integral

Solution. Using formula

Properties of the definite integral

Theorem 2.The value of the definite integral does not depend on the designation of the integration variable, i.e.

(40)

Let F(x) – antiderivative for f(x). For f(t) the antiderivative is the same function F(t), in which the independent variable is only designated differently. Hence,

Based on formula (39), the last equality means the equality of the integrals

Theorem 3.The constant factor can be taken out of the sign of the definite integral, i.e.

(41)

Theorem 4.The definite integral of the algebraic sum of a finite number of functions is equal to algebraic sum definite integrals of these functions, i.e.

(42)

Theorem 5.If a segment of integration is divided into parts, then the definite integral over the entire segment is equal to the sum of definite integrals over its parts, i.e. If

(43)

Theorem 6.When rearranging the limits of integration, the absolute value of the definite integral does not change, but only its sign changes, i.e.

(44)

Theorem 7(mean value theorem). A definite integral is equal to the product of the length of the integration segment and the value of the integrand at some point inside it, i.e.

(45)

Theorem 8.If the upper limit of integration is greater than the lower one and the integrand is non-negative (positive), then the definite integral is also non-negative (positive), i.e. If

Theorem 9.If the upper limit of integration is greater than the lower one and the functions and are continuous, then the inequality

can be integrated term by term, i.e.

(46)

The properties of the definite integral make it possible to simplify the direct calculation of integrals.

Example 5. Calculate definite integral

Using Theorems 4 and 3, and when finding antiderivatives - table integrals (7) and (6), we obtain

Definite integral with variable upper limit

Let f(x) – continuous on the segment [ a, b] function, and F(x) is its antiderivative. Consider the definite integral

(47)

and through t the integration variable is designated so as not to confuse it with the upper bound. When it changes X the definite integral (47) also changes, i.e. it is a function of the upper limit of integration X, which we denote by F(X), i.e.

(48)

Let us prove that the function F(X) is an antiderivative for f(x) = f(t). Indeed, differentiating F(X), we get

because F(x) – antiderivative for f(x), A F(a) is a constant value.

Function F(X) – one of the infinite number of antiderivatives for f(x), namely the one that x = a goes to zero. This statement is obtained if in equality (48) we put x = a and use Theorem 1 of the previous paragraph.

Calculation of definite integrals by the method of integration by parts and the method of change of variable

where, by definition, F(x) – antiderivative for f(x). If we change the variable in the integrand

then, in accordance with formula (16), we can write

In this expression

antiderivative function for

In fact, its derivative, according to rule of differentiation of complex functions, is equal

Let α and β be the values ​​of the variable t, for which the function

takes values ​​accordingly a And b, i.e.

But, according to the Newton-Leibniz formula, the difference F(b) – F(a) There is

The integration by parts formula looks like:
.

The method of integration by parts consists of applying this formula. At practical application It is worth noting that u and v are functions of the integration variable. Let the integration variable be designated as x (the symbol after the differential sign d at the end of the integral notation). Then u and v are functions of x: u(x) and v(x) .
Then
, .
And the formula for integration by parts takes the form:
.

That is, the integrand function must consist of the product of two functions:
,
one of which we denote as u: g(x) = u, and for the other the integral must be calculated (more precisely, the antiderivative must be found):
, then dv = f(x) dx .

In some cases f(x) = 1 . That is, in the integral
,
we can put g(x) = u, x = v .

Summary

So, in this method, the integration by parts formula should be remembered and applied in two forms:
;
.

Integrals calculated by integration by parts

Integrals containing logarithms and inverse trigonometric (hyperbolic) functions

Integrals containing logarithms and inverse trigonometric or hyperbolic functions are often integrated by parts. In this case, the part that contains the logarithm or inverse trigonometric (hyperbolic) functions is denoted by u, the remaining part by dv.

Here are examples of such integrals, which are calculated by the method of integration by parts:
, , , , , , .

Integrals containing the product of a polynomial and sin x, cos x or e x

Using the integration by parts formula, integrals of the form are found:
, , ,
where P(x) is a polynomial in x. When integrating, the polynomial P(x) is denoted by u, and e ax dx, cos ax dx or sin ax dx- via dv.

Here are examples of such integrals:
, , .

Examples of calculating integrals using the method of integration by parts

Examples of integrals containing logarithms and inverse trigonometric functions

Example

Calculate the integral:

Detailed solution

Here the integrand contains a logarithm. Making substitutions
u = ln x,
dv = x 2 dx.
Then
,
.

We calculate the remaining integral:
.
Then
.
At the end of the calculations, it is necessary to add the constant C, since the indefinite integral is the set of all antiderivatives. It could also be added in intermediate calculations, but this would only clutter up the calculations.

Shorter solution

You can imagine a solution in more short version. To do this, you do not need to make substitutions with u and v, but you can group the factors and apply the integration by parts formula in the second form.

.

Answer

Examples of integrals containing the product of a polynomial and sin x, cos x or ex

Example

Calculate the integral:
.

Solution

Let us introduce the exponent under the differential sign:
e - x dx = - e - x d(-x) = - d(e - x).

Let's integrate by parts.
.
We also use the method of integration by parts.
.
.
.
Finally we have.

In this topic we will talk in detail about the calculation of indefinite integrals using the so-called “integration by parts formula”. We will need a table of indefinite integrals and a table of derivatives. In the first part, standard examples will be discussed, which are mostly found in standard calculations and tests. More complex examples discussed in the second part.

The problem statement in the standard case is as follows. Let's say that under the integral we have two functions of different nature : polynomial and trigonometric function, polynomial and logarithm, polynomial and inverse trigonometric function and so on. In this situation, it is advantageous to separate one function from another. Roughly speaking, it makes sense to break the integrand into parts - and deal with each part separately. Hence the name: “integration by parts.” The application of this method is based on the following theorem:

Let the functions $u(x)$ and $v(x)$ be differentiable on some interval, and on this interval there exists an integral $\int v \; du$. Then on the same interval there also exists the integral $\int u \; dv$, and the following equality is true:

\begin(equation) \int u \; dv=u\cdot v-\int v\; du \end(equation)

Formula (1) is called the “integration by parts formula.” Sometimes, when applying the above theorem, they talk about using the “method of integration by parts.” The essence of this method will be important to us, which we will consider using examples. There are several standard cases in which formula (1) clearly applies. It is these cases that will become the topic of this page. Let $P_n(x)$ - nth polynomial degrees. Let's introduce two rules:

Rule #1

For integrals of the form $\int P_n(x) \ln x \;dx$, $\int P_n(x) \arcsin x \;dx$, $\int P_n(x) \arccos x \;dx$, $\ int P_n(x)\arctg x \;dx$, $\int P_n(x) \arcctg x \;dx$ we take $dv=P_n(x)dx$.

Rule #2

For integrals of the form $\int P_n(x) a^x \;dx$ ($a$ is some positive number), $\int P_n(x) \sin x \;dx$, $\int P_n(x) \cos x \;dx$, $\int P_n(x)ch x \;dx$, $\int P_n (x) sh x \;dx$ we take $u=P_n(x)$.

Let me immediately note that the above entries should not be taken literally. For example, in integrals of the form $\int P_n(x) \ln x \;dx$ there will not necessarily be exactly $\ln x$. Both $\ln 5x$ and $\ln (10x^2+14x-5)$ can be located there. Those. the notation $\ln x$ should be taken as a kind of generalization.

One more thing. It happens that the integration by parts formula has to be applied several times. Let's talk about this in more detail in examples No. 4 and No. 5. Now let's move on directly to solving typical problems. Solving problems whose level is slightly higher than standard is discussed in the second part.

Example No. 1

Find $\int (3x+4) \cos (2x-1) \; dx$.

Below the integral is the polynomial $3x+4$ and the trigonometric function $\cos (2x-1)$. This is a classic case for applying the formula, so let’s take the given integral by parts. The formula requires that the integral $\int (3x+4) \cos (2x-1)\; dx$ was represented in the form $\int u\; dv$. We need to choose expressions for $u$ and for $dv$. We can take $3x+4$ as $u$, then $dv=\cos (2x-1)dx$. We can take $u=\cos (2x-1)$, then $dv=(3x+4)dx$. To do right choice let's turn to . Given integral $\int (3x+4) \cos (2x-1)\; dx$ falls under the form $\int P_n(x) \cos x \;dx$ (the polynomial $P_n(x)$ in our integral has the form $3x+4$). According to, you need to choose $u=P_n(x)$, i.e. in our case $u=3x+4$. Since $u=3x+4$, then $dv=\cos(2x-1)dx$.

However, simply choosing $u$ and $dv$ is not enough. We will also need the values ​​of $du$ and $v$. Since $u=3x+4$, then:

$$ du=d(3x+4)=(3x+4)"dx=3dx.$$

Now let's look at the function $v$. Since $dv=\cos(2x-1)dx$, then according to the definition of the indefinite integral we have: $ v=\int \cos(2x-1)\; dx$. To find the required integral, we apply the following to the differential sign:

$$ v=\int \cos(2x-1)\; dx=\frac(1)(2)\cdot \int \cos(2x-1)d(2x-1)=\frac(1)(2)\cdot \sin(2x-1)+C=\frac (\sin(2x-1))(2)+C. $$

However, we do not need the entire infinite set of functions $v$, which is described by the formula $\frac(\sin(2x-1))(2)+C$. We need some one function from this set. To get the required function, you need to substitute some number instead of $C$. The easiest way, of course, is to substitute $C=0$, thereby obtaining $v=\frac(\sin(2x-1))(2)$.

So, let's put all of the above together. We have: $u=3x+4$, $du=3dx$, $dv=\cos(2x-1)dx$, $v=\frac(\sin(2x-1))(2)$. Substituting all this into the right side of the formula we have:

$$ \int (3x+4) \cos (2x-1) \; dx=(3x+4)\cdot\frac(\sin(2x-1))(2)-\int \frac(\sin(2x-1))(2)\cdot 3dx. $$

All that remains, in fact, is to find $\int\frac(\sin(2x-1))(2)\cdot 3dx$. Taking the constant (i.e. $\frac(3)(2)$) outside the integral sign and applying the method of introducing it under the differential sign, we obtain:

$$ (3x+4)\cdot \frac(\sin(2x-1))(2)-\int \frac(\sin(2x-1))(2)\cdot 3dx= \frac((3x+ 4)\cdot\sin(2x-1))(2)-\frac(3)(2)\int \sin(2x-1) \;dx= \\ =\frac((3x+4)\cdot \sin(2x-1))(2)-\frac(3)(4)\int \sin(2x-1) \;d(2x-1)= \frac((3x+4)\cdot\sin (2x-1))(2)-\frac(3)(4)\cdot (-\cos (2x-1))+C=\\ =\frac((3x+4)\cdot\sin(2x -1))(2)+\frac(3)(4)\cdot \cos (2x-1)+C. $$

So $\int (3x+4) \cos (2x-1) \; dx=\frac((3x+4)\cdot\sin(2x-1))(2)+\frac(3)(4)\cdot \cos (2x-1)+C$. In abbreviated form, the solution process is written as follows:

$$ \int (3x+4) \cos (2x-1) \; dx=\left | \begin(aligned) & u=3x+4; \; du=3xdx.\\ & dv=\cos(2x-1)dx; \; v=\frac(\sin(2x-1))(2). \end(aligned) \right |=\\ =(3x+4)\cdot\frac(\sin(2x-1))(2)-\int \frac(\sin(2x-1))(2) \cdot 3dx= \frac((3x+4)\cdot\sin(2x-1))(2)-\frac(3)(2)\int \sin(2x-1) \;dx=\\ = \frac((3x+4)\cdot\sin(2x-1))(2)-\frac(3)(4)\cdot (-\cos (2x-1))+C= \frac((3x +4)\cdot\sin(2x-1))(2)+\frac(3)(4)\cdot\cos (2x-1)+C. $$

The indefinite integral has been found by parts; all that remains is to write down the answer.

Answer: $\int (3x+4) \cos (2x-1) \; dx=\frac((3x+4)\cdot\sin(2x-1))(2)+\frac(3)(4)\cdot \cos (2x-1)+C$.

I believe there is a question here, so I’ll try to formulate it and give an answer.

Why did we take exactly $u=3x+4$ and $dv=\cos(2x-1)dx$? Yes, the integral has been solved. But maybe if we took $u=\cos (2x-1)$ and $dv=(3x+4)dx$ the integral would also be found!

No, if we take $u=\cos (2x-1)$ and $dv=(3x+4)dx$, then nothing good will come of it - the integral will not be simplified. Judge for yourself: if $u=\cos(2x-1)$, then $du=(\cos(2x-1))"dx=-2\sin(2x-1)dx$. Moreover, since $ dv=(3x+4)dx$, then:

$$ v=\int (3x+4) \; dx=\frac(3x^2)(2)+4x+C.$$

Taking $C=0$, we get $v=\frac(3x^2)(2)+4x$. Let us now substitute the found values ​​of $u$, $du$, $v$ and $dv$ into the formula:

$$ \int (3x+4) \cos (2x-1) \; dx=\cos (2x-1)\cdot \left(\frac(3x^2)(2)+4x \right) - \int \left(\frac(3x^2)(2)+4x \right) \cdot (-2\sin(2x-1)dx)=\\ =\cos (2x-1)\cdot \left(\frac(3x^2)(2)+4x \right) +2\cdot\ int \left(\frac(3x^2)(2)+4x \right) \sin(2x-1)\;dx $$

And what have we come to? We came to the integral $\int \left(\frac(3x^2)(2)+4x \right) \sin(2x-1)\;dx$, which is clearly more complicated than the original integral $\int (3x+4 ) \cos (2x-1) \; dx$. This suggests that the choice of $u$ and $dv$ was made poorly. After applying the integration by parts formula, the resulting integral should be simpler than the original one. When finding the indefinite integral by parts, we must simplify it, not complicate it, so if after applying formula (1) the integral becomes more complicated, then the choice of $u$ and $dv$ was made incorrectly.

Example No. 2

Find $\int (3x^4+4x-1) \ln 5x \; dx$.

Below the integral is a polynomial (i.e. $3x^4+4x-1$) and $\ln 5x$. This case falls under , so let's take the integral by parts. The given integral has the same structure as the integral $\int P_n(x) \ln x\; dx$. Again, as in example No. 1, we need to select some part of the integrand $(3x^4+4x-1) \ln 5x \; dx$ as $u$, and some part as $dv$. According to , you need to choose $dv=P_n(x)dx$, i.e. in our case $dv=(3x^4+4x-1)dx$. If from the expression $(3x^4+4x-1) \ln 5x \; dx$ "remove" $dv=(3x^4+4x-1)dx$, then $\ln 5x$ will remain - this will be the function $u$. So, $dv=(3x^4+4x-1)dx$, $u=\ln 5x$. To apply the formula, we also need $du$ and $v$. Since $u=\ln 5x$, then:

$$ du=d(\ln 5x)=(\ln 5x)"dx=\frac(1)(5x)\cdot 5 dx=\frac(1)(x)dx. $$

Now let's find the function $v$. Since $dv=(3x^4+4x-1)dx$, then:

$$ v=\int(3x^4+4x-1)\; dx=\frac(3x^5)(5)+2x^2-x+C. $$

From the entire found infinite set of functions $\frac(3x^5)(5)+2x^2-x+C$ we need to choose one. And the easiest way to do this is by taking $C=0$, i.e. $v=\frac(3x^5)(5)+2x^2-x$. Everything is ready to apply the formula. Let us substitute the values ​​$u=\ln 5x$, $du=\frac(1)(x)dx$, $v=\frac(3x^5)(5)+2x^2-x$ and $dv=(3x^4+4x-1)dx$ we will have:

$$ \int (3x^4+4x-1) \ln 5x \; dx=\left | \begin(aligned) & u=\ln 5x; \; du=\frac(1)(x)dx.\\ & dv=(3x^4+4x-1)dx; \; v=\frac(3x^5)(5)+2x^2-x. \end(aligned) \right |=\\ =\ln 5x \cdot \left (\frac(3x^5)(5)+2x^2-x \right)-\int \left (\frac(3x^ 5)(5)+2x^2-x \right)\cdot \frac(1)(x)dx=\\ =\left (\frac(3x^5)(5)+2x^2-x \right )\cdot\ln 5x -\int \left (\frac(3x^4)(5)+2x-1 \right)dx=\\ =\left (\frac(3x^5)(5)+2x^ 2-x \right)\cdot\ln 5x - \left (\frac(3x^5)(25)+x^2-x \right)+C=\\ =\left (\frac(3x^5) (5)+2x^2-x \right)\cdot\ln 5x - \frac(3x^5)(25)-x^2+x+C. $$

Answer: $\int (3x^4+4x-1) \ln 5x \; dx=\left (\frac(3x^5)(5)+2x^2-x \right)\cdot\ln 5x - \frac(3x^5)(25)-x^2+x+C$.

Example No. 3

Find $\int \arccos x\; dx$.

This integral has the structure $\int P_n(x) \arccos x \;dx$, falling under . I understand that a reasonable question will immediately arise: “where in the given integral $\int\arccos x \; dx$ did they hide the polynomial $P_n(x)$? There is no polynomial there, only arccosine and that’s it!” However, in fact, not only the arc cosine is located under the integral. I will present the integral $\int arccos x\; dx$ in this form: $\int 1\cdot\arccos x \; dx$. Agree that multiplying by one will not change the integrand. This unit is $P_n(x)$. Those. $dv=1\cdot dx=dx$. And as $u$ (according to ) we take $\arccos x$, i.e. $u=\arccos x$. We find the values ​​$du$ and $v$, which are involved in the formula, in the same way as in the previous examples:

$$ du=(\arccos x)"dx=-\frac(1)(\sqrt(1-x^2))dx;\\ v=\int 1\; dx=x+C. $$

As in previous examples, assuming $C=0$ we get $v=x$. Substituting all the found parameters into the formula, we will have the following:

$$ \int \arccos x \; dx=\left | \begin(aligned) & u=\arccos x; \; du=-\frac(1)(\sqrt(1-x^2))dx.\\ & dv=dx; \; v=x. \end(aligned) \right |=\\ =\arccos x \cdot x-\int x\cdot \left(-\frac(1)(\sqrt(1-x^2))dx \right)= \ arccos x \cdot x+\int \frac(xdx)(\sqrt(1-x^2))=\\ =x\cdot\arccos x-\frac(1)(2)\cdot\int (1-x ^2)^(-\frac(1)(2))d(1-x^2)= =x\cdot\arccos x-\frac(1)(2)\cdot\frac((1-x^ 2)^(\frac(1)(2)))(\frac(1)(2))+C=\\ =x\cdot\arccos x-\sqrt(1-x^2)+C. $$

Answer: $\int\arccos x\; dx=x\cdot\arccos x-\sqrt(1-x^2)+C$.

Example No. 4

Find $\int (3x^2+x) e^(7x) \; dx$.

In this example, the integration by parts formula will have to be applied twice. Integral $\int (3x^2+x) e^(7x) \; dx$ has the structure $\int P_n(x) a^x \;dx$. In our case, $P_n(x)=3x^2+x$, $a=e$. According to we have: $u=3x^2+x$. Accordingly, $dv=e^(7x)dx$.

$$ du=(3x^2+x)"=(6x+1)dx;\\ v=\int e^(7x)\;dx=\frac(1)(7)\cdot \int e^( 7x)\;d(7x)=\frac(1)(7)\cdot e^(7x)+C=\frac(e^(7x))(7)+C. $$

Again, as in previous examples, assuming $C=0$, we have: $v=\frac(e^(7x))(7)$.

$$ \int (3x^2+x) e^(7x) \; dx=\left | \begin(aligned) & u=3x^2+x; \; du=(6x+1)dx.\\ & dv=e^(7x)dx; \; v=\frac(e^(7x))(7). \end(aligned) \right |=\\ =(3x^2+x)\cdot\frac(e^(7x))(7)-\int \frac(e^(7x))(7)\cdot (6x+1)dx= \frac((3x^2+x)e^(7x))(7)-\frac(1)(7)\cdot \int (6x+1) e^(7x)\ ;dx. $$

We have arrived at the integral $\int (6x+1) e^(7x)\;dx$, which again must be taken in parts. Taking $u=6x+1$ and $dv=e^(7x)dx$ we have:

$$ \frac((3x^2+x)e^(7x))(7)-\frac(1)(7)\cdot \int (6x+1) e^(7x)\;dx=\left | \begin(aligned) & u=6x+1; \; du=6dx.\\ & dv=e^(7x)dx; \; v=\frac(e^(7x))(7). \end(aligned) \right |=\\ =\frac((3x^2+x)e^(7x))(7)-\frac(1)(7)\cdot \left ((6x+1) \cdot\frac(e^(7x))(7) - \int\frac(e^(7x))(7)\cdot 6\;dx \right)=\\ =\frac((3x^2+ x)e^(7x))(7) -\frac((6x+1)e^(7x))(49) +\frac(6)(49)\cdot\int\ e^(7x)\; dx=\\ =\frac((3x^2+x)e^(7x))(7) -\frac((6x+1)e^(7x))(49) +\frac(6)(49 )\cdot\frac(e^(7x))(7)+C=\\ =\frac((3x^2+x)e^(7x))(7) -\frac((6x+1)e ^(7x))(49) +\frac(6\; e^(7x))(343)+C. $$

The resulting answer can be simplified by opening the brackets and rearranging the terms:

$$ \frac((3x^2+x)e^(7x))(7) -\frac((6x+1)e^(7x))(49) +\frac(6\; e^(7x ))(343)+C=e^(7x)\cdot \left(\frac(3x^2)(7)+\frac(x)(49)-\frac(1)(343) \right)+ C. $$

Answer: $\int (3x^2+x) e^(7x) \; dx=e^(7x)\cdot \left(\frac(3x^2)(7)+\frac(x)(49)-\frac(1)(343) \right)+C$.

Example No. 5

Find $\int (x^2+5)\sin(3x+1) \; dx$.

Here, as in the previous example, integration by parts is applied twice. Detailed explanations were given earlier, so I will give only the solution:

$$ \int (x^2+5)\sin(3x+1) \; dx=\left | \begin(aligned) & u=x^2+5; \; du=2xdx.\\ & dv=\sin(3x+1)dx; \; v=-\frac(\cos(3x+1))(3). \end(aligned) \right |=\\ =(x^2+5)\cdot \left(-\frac(\cos(3x+1))(3) \right)-\int\left(-\ frac(\cos(3x+1))(3) \right)\cdot 2xdx=\\ = -\frac((x^2+5)\cdot\cos(3x+1))(3) +\frac (2)(3)\int x\cos(3x+1)dx= \left | \begin(aligned) & u=x; \; du=dx.\\ & dv=\cos(3x+1)dx; \; v=\frac(\sin(3x+1))(3). \end(aligned) \right |=\\ =-\frac((x^2+5)\cdot\cos(3x+1))(3) +\frac(2)(3)\cdot \left( x\cdot\frac(\sin(3x+1))(3)-\int\frac(\sin(3x+1))(3)dx \right)=\\ =-\frac((x^2 +5)\cdot\cos(3x+1))(3) +\frac(2x\sin(3x+1))(9)-\frac(2)(9)\cdot\int\sin(3x+ 1)dx=\\ =-\frac((x^2+5)\cdot\cos(3x+1))(3) +\frac(2x\sin(3x+1))(9)-\frac (2)(9)\cdot \left(-\frac(\cos(3x+1))(3)\right)+C=\\ = -\frac((x^2+5)\cdot\cos (3x+1))(3) +\frac(2x\sin(3x+1))(9)+\frac(2\cos(3x+1))(27)+C=\\ =-\frac (x^2\cdot\cos(3x+1))(3)-\frac(5\cdot\cos(3x+1))(3) +\frac(2x\sin(3x+1))(9 )+\frac(2\cos(3x+1))(27)+C=\\ =-\frac(x^2\cdot\cos(3x+1))(3) +\frac(2x\sin (3x+1))(9)-\frac(43\cos(3x+1))(27)+C. $$

Answer: $\int (x^2+5)\sin(3x+1) \; dx=-\frac(x^2\cdot\cos(3x+1))(3) +\frac(2x\sin(3x+1))(9)-\frac(43\cos(3x+1) )(27)+C$.

The application of the method of integration by parts in somewhat non-standard cases that are not subject to rules No. 1 and No. 2 will be given in

Integration by parts- a method used to solve definite and indefinite integrals, when one of the integrands is easily integrable and the other is differentiable. A fairly common method for finding integrals, both indefinite and definite. Main sign, when you need to use it, it is a certain function consisting of the product of two functions that cannot be integrated point-blank.

Formula

In order to successfully use this method it is necessary to disassemble and learn the formulas.

Formula for integration by parts in not definite integral:

$$ \int udv = uv - \int vdu $$

Formula for integration by parts in a definite integral:

$$ \int \limits_(a)^(b) udv = uv \bigg |_(a)^(b) - \int \limits_(a)^(b) vdu $$

Examples of solutions

Let us consider in practice examples of solutions to integration by parts, which are often proposed by teachers during tests. Please note that under the integral symbol there is a product of two functions. This is a sign that this method is suitable for the solution.

Example 1

Find the integral $ \int xe^xdx $

Solution

We see that the integrand consists of two functions, one of which, upon differentiation, instantly turns into unity, and the other is easily integrated. To solve the integral, we use the method of integration by parts. Let us assume that $ u = x \rightarrow du=dx $ and $ dv = e^x dx \rightarrow v=e^x $ We substitute the found values ​​into the first integration formula and get: $$ \int xe^x dx = xe^x - \int e^x dx = xe^x - e^x + C $$ If you cannot solve your problem, then send it to us. We will provide detailed solution. You will be able to view the progress of the calculation and gain information. This will help you get your grade from your teacher in a timely manner!

Answer

$$ \int xe^x dx = xe^x - e^x + C $$

Example 4

Calculate the integral $ \int \limits_0 ^1 (x+5) 3^x dx $

Solution

By analogy with the previous solved examples, we will figure out which function to integrate without problems, which to differentiate. Please note that if we differentiate $ (x+5) $, then this expression will be automatically converted to unity, which will be to our advantage. So we do this: $$ u=x+5 \rightarrow du=dx, dv=3^x dx \rightarrow v=\frac(3^x)(ln3) $$ Now all unknown functions have been found and can be put into the second formula for integration by parts for a definite integral. $$ \int \limits_0 ^1 (x+5) 3^x dx = (x+5) \frac(3^x)(\ln 3) \bigg |_0 ^1 - \int \limits_0 ^1 \frac (3^x dx)(\ln 3) = $$ $$ = \frac(18)(\ln 3) - \frac(5)(\ln 3) - \frac(3^x)(\ln^2 3)\bigg| _0 ^1 = \frac(13)(\ln 3) - \frac(3)(\ln^2 3)+\frac(1)(\ln^2 3) = \frac(13)(\ln 3 )-\frac(4)(\ln^2 3) $$

Answer

$$ \int\limits_0 ^1 (x+5)3^x dx = \frac(13)(\ln 3)-\frac(4)(\ln^2 3) $$