Complex integrals

This article completes the topic of indefinite integrals, and it includes integrals that I consider quite difficult. The lesson was created at the repeated request 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 the basic techniques of integration. Dummies and people who are not very confident in integrals should refer to the very first lesson - Indefinite integral. Solution examples where you can learn the topic almost from scratch. More experienced students can get acquainted with the techniques and methods of integration, which have not yet been encountered in my articles.

What integrals will be considered?

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

Then we will get acquainted with an interesting and original method of reducing the integral to itself. Not so few integrals are solved in this way.

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

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

(2) B integrand divide the numerator by the denominator term by term.

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

(4) We take the remaining integrals. Note that you can use brackets in the logarithm and not the modulus, because .

(5) We carry out the reverse substitution, expressing from the direct substitution "te":

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, in the course of the solution, even more than two solution methods had to be used, so to deal with such integrals, you need confident integration skills and not the least experience.

In practice, of course, the square root is more common, here are three examples for independent solution:

Example 2

Find the 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 be only for Example 2, in Examples 3-4 - one answer. Which replacement to use at the beginning of decisions, I think, is obvious. Why did I choose the same type of examples? Often found in their roles. More often, perhaps, just something like .

But not always, when under the arc tangent, sine, cosine, exponent, and other functions there is a root of linear function, it is necessary to apply 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 is taken elementarily. The easiest of the tasks proposed above is Example 4, in which, after the replacement, a relatively simple integral is obtained.

The method of reducing the integral to itself

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

Example 5

Find the indefinite integral

There is a square binomial under the root, and when trying to integrate this example, the teapot can suffer for hours. Such an integral is taken by parts and reduces to itself. In principle, it is not difficult. If you know how.

Let us denote the considered integral by a Latin letter and start the solution:

Integrating by parts:

(1) We prepare the integrand for term-by-term division.

(2) We divide the integrand term by term. Perhaps not everyone understands, I will write in more detail:

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

(4) We take the last integral ("long" logarithm).

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

And for the ending:

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

Equate the beginning and end:

We transfer to the left side with a change of sign:

And we demolish the deuce to the 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 is the severity here:

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

Thus:

The constant can be re-named with . Why can you rename? Because it still takes any values, and in this sense there is no difference between constants and.
As a result:

A similar trick with constant renaming is widely used in differential equations. And there I will be strict. And here such liberties are allowed by me only in order not to confuse you with unnecessary things and focus on the very method of integration.

Example 6

Find the indefinite integral

Another typical integral for independent solution. Complete Solution and the answer at the end of the lesson. The difference with the answer of the previous example will be!

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

For example, consider the integral . All you need to do is in advance select a full square:
.
Next, a linear replacement is carried out, which manages "without any consequences":
, resulting in an integral . Something familiar, right?

Or this example, with a square binomial:
Selecting a full square:
And, after a linear replacement , we get the integral , which is also solved by the already considered algorithm.

Consider two more typical examples of how to reduce an integral to itself:
is the integral of the exponent multiplied by the sine;
is the integral of the exponent multiplied by the cosine.

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

Example 7

Find the indefinite integral

The integrand is the exponent 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 is reduced to itself. Equate the beginning and end of the solution:

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

Ready. Along the way, it is desirable 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 rather, to integration by parts:

For we have designated the exhibitor. The question arises, it is the exponent that should always be denoted by ? Not necessary. In fact, in the considered integral fundamentally doesn't matter, what to denote for, one could go the other way:

Why is this possible? Because the exponent turns into itself (when differentiating and integrating), the sine and cosine mutually turn into each other (again, both when differentiating and integrating).

That is, the trigonometric function can be denoted as well. But, in the considered example, this is less rational, since fractions will appear. If you wish, you can try to solve this example in the second way, the answers must be the same.

Example 8

Find the indefinite integral

This is a do-it-yourself example. Before deciding, think about what is more profitable in this case to designate for, exponential or trigonometric function? Full solution and answer at the end of the lesson.

And, of course, don't forget that most of the answers in this lesson are fairly easy to check by differentiation!

The examples were considered not the most difficult. 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 have to get confused in such an integral, and I myself often get confused. The fact is that in the solution there is a high probability of the appearance of fractions, and it is very easy to lose something due to inattention. In addition, there is a high probability of error in signs, note that there is a minus sign in the exponent, and this introduces additional difficulty.

At the final stage, it often turns out something like this:

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

Integration of 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, just 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 square trinomial plus outside the root "appendage" in the form of "x". An integral of this form is solved using a standard substitution.

We decide:

The replacement here is simple:

Looking 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) We reduce the numerator and denominator 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 Integration of some fractions, is solved extraction method full square . Select a full 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 hairdressing 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 a do-it-yourself example. Here, a constant is added to the lone x, and the replacement is almost the same:

The only thing that needs to be done additionally is to express the "x" from the replacement:

Full solution and answer at the end of the lesson.

Sometimes in such an integral there may be a square binomial under the root, this does not change the way the solution is solved, it will even 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 considered in the lesson Integrals of irrational functions.

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

(polynomial in denominator)

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

Example 13

Find the indefinite integral

But back to the example with the lucky number 13 ( honestly, did not guess). This integral is also from the category of those with which you can pretty much suffer 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) derived recurrent downgrading formula:
, Where is an integral of a lower degree.

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 a do-it-yourself example. The sample solution uses the above formula twice in succession.

If under the degree is indecomposable square trinomial, then the solution is reduced to a binomial by extracting the full square, for example:

What if there is an additional polynomial in the numerator? In this case, the method of indeterminate coefficients is used, and the integrand is expanded into a sum of fractions. But in my practice of such an example never met, so I skipped this case in the article Integrals of a fractional-rational function, I'll skip it now. If such an integral still occurs, see the textbook - everything is simple there. I do not consider it expedient to include material (even simple), the probability of meeting with which tends to zero.

Integration of complex trigonometric functions

The adjective "difficult" for most examples is again largely conditional. Let's start with tangents and cotangents in high degrees. From the point of view of the methods used to solve the tangent and cotangent are almost the same, so I will talk more about the tangent, meaning that the demonstrated method of solving the integral is also valid for the cotangent.

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

Consider another canonical example, the integral of unity divided by the sine:

Example 17

Find the indefinite integral

Here you can use the universal trigonometric substitution and get the answer, but there is a more rational way. I will provide a 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: In the denominator we divide and multiply by .
(3) According to the well-known formula in the denominator, we turn the fraction into a tangent.
(4) We bring the function under the sign of the differential.
(5) We take the integral.

Pair simple examples for independent solution:

Example 18

Find the indefinite integral

Hint: The very first step is 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 behind the method? The idea is that with the help of transformations, trigonometric formulas organize in the integrand only the tangents and the derivative of the tangent. That is, we are talking about replacing: . In Examples 17-19, we actually used this replacement, but the integrals were so simple that it was done with an equivalent action - bringing the function under the differential sign.

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

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

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

for an integral, an integer negative EVEN number.

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

Consider a couple of more meaningful tasks for this rule:

Example 20

Find the indefinite integral

The sum of the degrees of sine and cosine: 2 - 6 \u003d -4 - 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) According to 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 the replacement. More experienced students may not carry out the replacement, but still it is better to replace the tangent with one letter - there is less risk of confusion.

Example 21

Find the indefinite integral

This is a do-it-yourself example.

Hold on, the championship rounds begin =)

Often in the integrand there is a "hodgepodge":

Example 22

Find the indefinite integral

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

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

A couple of creative examples for an independent solution:

Example 23

Find the indefinite integral

Example 24

Find the indefinite integral

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

It is shown that the integral of the product of power functions of sin x and cos x can be reduced to an integral of the differential binomial. For integer values ​​of the exponents, such integrals are easily calculated in parts or using reduction formulas. The derivation of reduction formulas is given. An example of the calculation of such an integral is given.

Content

See also:
Table of indefinite integrals

Reduction to the integral of the differential binomial

Consider integrals of the form:

Such integrals are reduced to the integral of the differential binomial of one of the substitutions t = sin x or t= cos x.

Let's demonstrate this by substituting
t = sin x.
Then
dt = (sin x)′ dx = cos x dx;
cos 2 x \u003d 1 - sin 2 x \u003d 1 - t 2;

If m and n are rational numbers, then differential binomial integration methods should be applied.

Integration with integers m and n

Next, consider the case where m and n are integers (not necessarily positive). In this case, the integrand is a rational function of sin x And cos x. Therefore, the rules presented in the section "Integration of trigonometric rational functions" can be applied.

However, taking into account specific features, it is easier to use reduction formulas, which are easily obtained by integration by parts.

Cast formulas

Reduction Formulas for the Integral

look like:

;
;
;
.

They do not need to be memorized, since they are easily obtained by integration by parts.

Proof of reduction formulas

We integrate by parts.


Multiplying by m + n, we get the first formula:

Similarly, we obtain the second formula.

We integrate by parts.


Multiplying by m + n, we get the second formula:

Third formula.

We integrate by parts.


Multiplying by n + 1 , we get the third formula:

Similarly, for the fourth formula.

We integrate by parts.


Multiplying by m + 1 , we get the fourth formula:

Example

Let's calculate the integral:

Let's transform:

Here m = 10, n = - 4.

We apply the reduction formula:

For m = 10, n = - 4:

For m = 8, n = - 2:

We apply the reduction formula:

For m = 6, n = - 0:

For m = 4, n = - 0:

For m = 2, n = - 0:

We calculate the remaining integral:

We collect intermediate results in one formula.

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

See also:

Principal Integrals Every Student Should Know

The listed integrals are the basis, the basis of the foundations. These formulas, of course, should be remembered. When calculating more complex integrals, you will have to use them constantly.

Pay Special attention to formulas (5), (7), (9), (12), (13), (17), and (19). Do not forget to add an arbitrary constant C to the answer when integrating!

Integral of a constant

∫ A d x = A x + C (1)

Power function integration

In fact, one could confine oneself to formulas (5) and (7), but the rest of the integrals from this group are so common that it is worth paying a little attention to them.

∫ x d x = x 2 2 + C (2)
∫ x 2 d x = x 3 3 + C (3)
∫ 1 x d x = 2 x + C (4)
∫ 1 x d x = log | x | +C(5)
∫ 1 x 2 d x = − 1 x + C (6)
∫ x n d x = x n + 1 n + 1 + C (n ≠ − 1) (7)

Integrals of the exponential function and of hyperbolic functions

Of course, formula (8) (perhaps the most convenient to remember) can be considered as a special case of formula (9). Formulas (10) and (11) for the integrals of the hyperbolic sine and hyperbolic cosine are easily derived from formula (8), but it is better to just remember these relationships.

∫ e x d x = e x + C (8)
∫ a x d x = a x log a + C (a > 0, a ≠ 1) (9)
∫ s h x d x = c h x + C (10)
∫ c h x d x = s h x + C (11)

Basic integrals of trigonometric functions

A mistake that students often make: they confuse the signs in formulas (12) and (13). Remembering that the derivative of the sine is equal to the cosine, for some reason many people believe that the integral of the sinx function is equal to cosx. This is not true! The integral of sine is "minus cosine", but the integral of cosx is "just sine":

∫ sin x d x = − cos x + C (12)
∫ cos x d x = sin x + C (13)
∫ 1 cos 2 x d x = t g x + C (14)
∫ 1 sin 2 x d x = − c t g x + C (15)

Integrals Reducing to Inverse Trigonometric Functions

Formula (16), which leads to the arc tangent, is naturally a special case of formula (17) for a=1. Similarly, (18) is a special case of (19).

∫ 1 1 + x 2 d x = a r c t g x + C = − a r c c t g x + C (16)
∫ 1 x 2 + a 2 = 1 a a r c t g x a + C (a ≠ 0) (17)
∫ 1 1 − x 2 d x = arcsin x + C = − arccos x + C (18)
∫ 1 a 2 − x 2 d x = arcsin x a + C = − arccos x a + C (a > 0) (19)

More complex integrals

These formulas are also desirable to remember. They are also used quite often, and their output is rather tedious.

∫ 1 x 2 + a 2 d x = ln | x + x2 + a2 | +C(20)
∫ 1 x 2 − a 2 d x = ln | x + x 2 − a 2 | +C(21)
∫ a 2 − x 2 d x = x 2 a 2 − x 2 + a 2 2 arcsin x a + C (a > 0) (22)
∫ x 2 + a 2 d x = x 2 x 2 + a 2 + a 2 2 ln | x + x2 + a2 | + C (a > 0) (23)
∫ x 2 − a 2 d x = x 2 x 2 − a 2 − a 2 2 ln | x + x 2 − a 2 | + C (a > 0) (24)

General integration rules

1) The integral of the sum of two functions is equal to the sum of the corresponding integrals: ∫ (f (x) + g (x)) d x = ∫ f (x) d x + ∫ g (x) d x (25)

2) The integral of the difference of two functions is equal to the difference of the corresponding integrals: ∫ (f (x) − g (x)) d x = ∫ f (x) d x − ∫ g (x) d x (26)

3) The constant can be taken out of the integral sign: ∫ C f (x) d x = C ∫ f (x) d x (27)

It is easy to see that property (26) is simply a combination of properties (25) and (27).

4) Integral of a complex function if the inner function is linear: ∫ f (A x + B) d x = 1 A F (A x + B) + C (A ≠ 0) (28)

Here F(x) is the antiderivative for the function f(x). Note that this formula only works when the inner function is Ax + B.

Important: there is no universal formula for the integral of the product of two functions, as well as for the integral of a fraction:

∫ f (x) g (x) d x = ? ∫ f (x) g (x) d x = ? (thirty)

This does not mean, of course, that a fraction or a product cannot be integrated. It's just that every time you see an integral like (30), you have to invent a way to "fight" with it. In some cases, integration by parts will help you, somewhere you will have to make a change of variable, and sometimes even "school" formulas of algebra or trigonometry can help.

A simple example for calculating the indefinite integral

Example 1. Find the integral: ∫ (3 x 2 + 2 sin x − 7 e x + 12) d x

We use formulas (25) and (26) (the integral of the sum or difference of functions is equal to the sum or difference of the corresponding integrals. We get: ∫ 3 x 2 d x + ∫ 2 sin x d x − ∫ 7 e x d x + ∫ 12 d x

Recall that the constant can be taken out of the integral sign (formula (27)). The expression is converted to the form

3 ∫ x 2 d x + 2 ∫ sin x d x − 7 ∫ e ​​x d x + 12 ∫ 1 d x

Now let's just use the table of basic integrals. We will need to apply formulas (3), (12), (8) and (1). Let's integrate the power function, sine, exponent and constant 1. Don't forget to add an arbitrary constant C at the end:

3 x 3 3 - 2 cos x - 7 e x + 12 x + C

After elementary transformations, we get the final answer:

X 3 − 2 cos x − 7 e x + 12 x + C

Test yourself with differentiation: take the derivative of the resulting function and make sure that it is equal to the original integrand.

Summary table of integrals

∫ A d x = A x + C

∫ x d x = x 2 2 + C

∫ x 2 d x = x 3 3 + C

∫ 1 x d x = 2 x + C

∫ 1 x d x = log | x | + C

∫ 1 x 2 d x = − 1 x + C

∫ x n d x = x n + 1 n + 1 + C (n ≠ − 1)

∫ e x d x = e x + C

∫ a x d x = a x ln a + C (a > 0, a ≠ 1)

∫ s h x d x = c h x + C

∫ c h x d x = s h x + C

∫ sin x d x = − cos x + C

∫ cos x d x = sin x + C

∫ 1 cos 2 x d x = t g x + C

∫ 1 sin 2 x d x = − c t g x + C

∫ 1 1 + x 2 d x = a r c t g x + C = − a r c c t g x + C

∫ 1 x 2 + a 2 = 1 a a r c t g x a + C (a ≠ 0)

∫ 1 1 − x 2 d x = arcsin x + C = − arccos x + C

∫ 1 a 2 − x 2 d x = arcsin x a + C = − arccos x a + C (a > 0)

∫ 1 x 2 + a 2 d x = ln | x + x2 + a2 | + C

∫ 1 x 2 − a 2 d x = ln | x + x 2 − a 2 | + C

∫ a 2 − x 2 d x = x 2 a 2 − x 2 + a 2 2 arcsin x a + C (a > 0)

∫ x 2 + a 2 d x = x 2 x 2 + a 2 + a 2 2 ln | x + x2 + a2 | + C (a > 0)

∫ x 2 − a 2 d x = x 2 x 2 − a 2 − a 2 2 ln | x + x 2 − a 2 | + C (a > 0)

Download the table of integrals (part II) from this link

If you are studying at a university, if you have any difficulties with higher mathematics ( mathematical analysis, linear algebra, probability theory, statistics), if you need the services of a qualified teacher, go to the page of a tutor in higher mathematics. Let's solve your problems together!

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Hello again, friends!

As I promised, from this lesson we will begin to surf the endless expanses of the poetic world of integrals and begin to solve a wide variety of (sometimes very beautiful) examples. :)

In order to competently navigate the entire integral variety and not get lost, we need only four things:

1) Table of integrals. All details about her . How exactly to work with her - in this.

2) Properties of linearity of the indefinite integral (integral of the sum/difference and product by a constant).

3) Table of derivatives and differentiation rules.

Yes, don't be surprised! Without the ability to count derivatives, there is absolutely nothing to catch in integration. Agree, it makes no sense, for example, to learn division without knowing how to multiply. :) And very soon you will see that without perfect differentiation skills, you cannot calculate any serious integral that goes beyond the scope of elementary tabular ones.

4) Methods of integration.

There are very, very many of them. For a specific class of functions - its own. But among all their rich diversity, three basic ones stand out:

– ,

– ,

– .

About each of them - in separate lessons.

And now, finally, let's start solving the long-awaited examples. In order not to jump from section to section, I will duplicate once more the entire gentleman's set, which will come in handy for our further work. Keep all the tools at hand.)

First of all, this table of integrals:

In addition, we need the basic properties of the indefinite integral (linearity properties):

Well, the necessary equipment is prepared. Time to go! :)

Direct application of the table

In this section, the most simple and harmless examples will be considered. The algorithm here is simple to horror:

1) We look at the table and look for the desired formula (formulas);

2) Apply the properties of linearity (where required);

3) We carry out the transformation according to tabular formulas and add a constant at the end WITH (don't forget!) ;

4) Write down the answer.

So let's go.)

Example 1

There is no such function in our table. But there is an integral of the power function in general view(second group). In our case n=5. So we substitute the five instead of n and carefully calculate the result:

Ready. :)

Of course, this example is quite primitive. Purely for acquaintance.) But the ability to integrate degrees makes it easy to calculate integrals from any polynomials and other power structures.

Example 2

Under the integral sum. Well, okay. We have linearity properties for this case. :) We divide our integral into three separate ones, take all the constants out of the signs of the integrals and count each according to the table (group 1-2):

Please note: constant WITH appears at the very moment when ALL signs of the integral disappear! Of course, after that you have to constantly carry it with you. So what to do…

Of course, it is usually not necessary to paint in such detail. This is purely for understanding. To get the point.)

For example, very soon, without much hesitation, you will mentally give an answer to monsters like:

Polynomials are the most free functions in integrals.) And in diffurs, in physics, in strength of materials and other serious disciplines, polynomials will have to be integrated constantly. Get used to it.)

The next example will be a little trickier.

Example 3

I hope everyone understands that our integrand can be written like this:

The integrand is separate, and the multiplier dx (differential icon)- separately.

Comment: in this lesson the multiplier dx in the process of integration Bye does not participate in any way, and for the time being we are mentally "hammering" him. :) We work only with integrand. But let's not forget about him. Very soon, literally next lesson dedicated, we will remember him. And we will feel the importance and power of this icon in full force!)

In the meantime, our gaze is turned to the integrand function

Doesn't look much like a power function, but that's it. :) If we recall the school properties of roots and degrees, then it is quite possible to transform our function:

And x to the power of minus two-thirds is already table function! The second group n=-2/3. And the constant 1/2 is not a hindrance to us. We take it outside, beyond the integral sign, and directly according to the formula we consider:

In this example, we were helped elementary properties degrees. And this is how it should be done in most cases, when there are single roots or fractions under the integral. Therefore, a couple practical advice when integrating power structures:

We replace fractions with powers with negative exponents;

We replace the roots with powers with fractional exponents.

But in the final answer, the transition from degrees back to fractions and roots is a matter of taste. Personally, I turn back - it's more aesthetically pleasing, or something.

And please, carefully count all the fractions! We carefully follow the signs and what goes where - what is the numerator and what is the denominator.

What? Tired of already boring power functions? OK! We take the bull by the horns!

Example 4

If now we bring everything under the integral to a common denominator, then we can get stuck on this example seriously and for a long time.) But, looking more closely at the integrand, you can see that our difference consists of two tabular functions. So let's not pervert, but instead expand our integral into two:

The first integral is an ordinary power function, (2nd group, n=-1): 1/x = x -1 .

Our traditional formula for the antiderivative power function

Doesn't work here, but for us n=-1 there is a worthy alternative - a formula with a natural logarithm. This one:

Then, according to this formula, the first fraction will be integrated as follows:

And the second fraction also a table function! Learned? Yes! This seventh formula with "high" logarithm:

The constant "a" in this formula is equal to two: a=2.

Important note: Please note the constantWITH with intermediate integration I nowhere I do not attribute! Why? Because she will go to the final answer the whole example. This is quite enough.) Strictly speaking, the constant must be written after each individual integration - at least intermediate, at least final: so the indefinite integral requires ...)

For example, after the first integration, I would have to write:

After the second integration:

But the whole point is that the sum / difference of arbitrary constants is also some constant! In our case, for the final answer, we need from the first integral subtract second. Then we will succeed difference two intermediate constants:

C 1 -C 2

And we have every right to replace this very difference of constants one constant! And just redesignate it with the letter "C" familiar to us. Like this:

C 1 -C 2 \u003d C

So we attribute this same constant WITH to the final result and get the answer:

Yes, they are fractions! Multi-storey logarithms when they are integrated are the most common thing. We also get used to it.)

Remember:

With intermediate integration of several terms, the constant WITH after each of them you can not write. It is enough to include it in the final answer of the entire example. In the end.

The next example is also with a fraction. For warm up.)

Example 5

In the table, of course, there is no such function. But there is similar function:

This is the latest eighth formula. With arctangent. :)

This one:

And God himself ordered us to adjust our integral to this formula! But there is one problem: in the tabular formula before x 2 there is no coefficient, but we have a nine. We can't use the formula directly yet. But in our case, the problem is completely solvable. Let's first take this nine out of brackets, and then we will generally take it out of the limits of our fraction.)

And the new fraction is the tabular function we need at number 8! Here a 2 \u003d 4/9. Or a=2/3.

All. We take 1/9 out of the integral sign and use the eighth formula:

Here is the answer. This example, with a coefficient before x 2, I chose it that way. To make it clear what to do in such cases. :) If before x 2 there is no coefficient, then such fractions will also be integrated in the mind.

For example:

Here a 2 = 5, so "a" itself would be "the root of five". In general, you understand.)

And now we will slightly modify our function: we will write the denominator under the root.) Now we will take such an integral:

Example 6

The denominator has a root. Naturally, the corresponding formula for integration has also changed, yes.) Again we climb into the table and look for the right one. We have roots in the formulas of the 5th and 6th groups. But in the sixth group, there is only a difference under the roots. And we have the sum. So we are working on fifth formula, with a "long" logarithm:

Number A we have five. Substitute in the formula and get:

And all things. This is the answer. Yes, yes, it's that simple!

If doubts creep in, then it is always possible (and necessary) to check the result by reverse differentiation. Let's check? And then all of a sudden, some kind of crap?

We differentiate (we do not pay attention to the module and perceive it as ordinary brackets):

Everything is fair. :)

By the way, if in the integrand under the root we change the sign from plus to minus, then the formula for integration will remain the same. It is no coincidence that in the table under the root is plus/minus. :)

For example:

Important! In case of minus first the place under the root should be exactly x 2, and on second – number. If under the root everything is the opposite, then the corresponding tabular formula will be already another!

Example 7

Under the root again minus, but x 2 with five changed places. It looks similar, but not the same... Our table also has a formula for this case.) Formula number six, we have not worked with it yet:

And now - carefully. In the previous example, our five acted as a number A . Here the five will act as a number and 2!

Therefore, for the correct application of the formula, do not forget to take the root of the five:

And now the example is solved in one step. :)

That's it! Only the terms under the root have changed places, and the result of integration has changed significantly! Logarithm and arcsine... so please do not confuse these two formulas! Although the integrands are very similar...

Bonus:

In tabular formulas 7-8, there are coefficients before the logarithm and arc tangent 1/(2а) And 1/a respectively. And in an alarming combat situation, when writing these formulas, even nerds hardened by studies often get confused where 1/a, And where 1/(2а). Here's a simple trick for you to remember.

In formula number 7

The denominator of the integrand is difference of squares x 2 - a 2. Which, according to the scary school formula, is decomposed as (x-a)(x+a). On two multiplier. Keyword - two. And these two when integrating, brackets go to the logarithm: with a minus up, with a plus - down.) And the coefficient in front of the logarithm is also 1/( 2 A).

But in the formula number 8

The denominator of the fraction is sum of squares. But the sum of squares x2 +a2 indecomposable into simpler factors. Therefore, whatever one may say, it will remain in the denominator one factor. And the coefficient in front of the arc tangent will also be 1/a.

And now, for a change, let's integrate something from trigonometry.)

Example 8

The example is simple. So simple that people, without even looking at the table, immediately joyfully write the answer and ... they arrived. :)

We follow the signs! This is the most common mistake when integrating sines/cosines. Do not confuse with derivatives!

Yes, (sin x)" = cos x And (cos x)’ = - sin x.

But!

Since people usually remember the derivatives at the very least, in order not to get confused in the signs, the technique for remembering the integrals here is very simple:

Integral of sine/cosine = minus derivative of the same sine/cosine.

For example, we know from school that the derivative of the sine is equal to the cosine:

(sin x)" = cos x.

Then for integral from the same sine will be true:

And that's it.) With the cosine the same thing.

Let's fix our example:

Preliminary elementary transformations of the integrand

Up to this point, there have been the simplest examples. To get a feel for how the table works and not make mistakes in choosing a formula.)

Of course, we did some simple transformations - we took out factors, broke them into terms. But the answer still lay on the surface one way or another.) However ... If the calculation of integrals was limited only to the direct use of the table, then there would be a complete freebie around and life would become boring.)

Now let's look at more solid examples. Those where directly, it seems, nothing is decided. But it is worth remembering literally a couple of elementary school formulas or transformations, as the road to the answer becomes simple and understandable. :)

Application of trigonometry formulas

Let's continue to have fun with trigonometry.

Example 9

There is no such function in the table. But in school trigonometry there is this little-known identity:

We now express the square of the tangent we need from it and insert it under the integral:

Why is this done? And then, that after such a transformation, our integral will be reduced to two tabular ones and will be taken in mind!

See:

Now let's analyze our actions. At first glance, everything seems to be simple. But let's think about this. If we had a task differentiate the same function, then we would exactly knew exactly what to do - to apply formula derivative of a complex function:

And that's it. Simple and trouble-free technology. It always works and is guaranteed to lead to success.

But what about the integral? And here we had to dig into trigonometry, dig up some obscure formula in the hope that it would somehow help us get out and reduce the integral to a tabular one. And it’s not a fact that it would help us, it’s not a fact at all ... That is why integration is a more creative process than differentiation. Art, I would even say. :) And this is not the most difficult example. It's only the beginning!

Example 10

What inspires? The table of integrals is still powerless, yes. But, if you look again into our treasury of trigonometric formulas, you can dig out a very, very useful double angle cosine formula:

So we apply this formula to our integrand. In the role of "alpha" we have x / 2.

We get:

The effect is amazing, right?

These two examples clearly show that the pre-transformation of the function before integration quite acceptable and sometimes makes life tremendously easier! And in integration this procedure (transformation of the integrand) is an order of magnitude more justified than in differentiation. You will see later.)

Let's take a look at a couple more typical transformations.

Abbreviated multiplication formulas, parentheses expansion, reduction of likes and term division method.

Usual banal school transformations. But sometimes only they save, yes.)

Example 11

If we considered the derivative, then no problem: the formula for the derivative of the product and - forward. But the standard formula for integral from the work does not exist. And the only way out here is to open all the brackets so that a polynomial is obtained under the integral. And we will somehow integrate the polynomial.) But we will also open the brackets wisely: the formulas for abbreviated multiplication are a powerful thing!

(x 2 - 1) 2 (x 2 + 1) 2 = ((x 2 - 1)(x 2 + 1)) 2 = ((x 2) 2 - 1 2) 2 = (x 4 - 1) 2 = x 8 - 2x 4 + 1

And now we consider:

And all things.)

Example 12

Again, the standard formula for fraction integral does not exist. However, the denominator of the integrand contains lonely x. This radically changes the situation.) Let's divide the numerator by the denominator term by term, reducing our terrible fraction to a harmless sum of tabular power functions:

I will not comment specifically on the procedure for integrating degrees: they are not small anymore.)

We integrate the sum of power functions. By plate.)

That's all.) By the way, if the denominator was not x, but, say, x+1, like this:

Then this trick with term-by-member division would not have gone so easily. It is because of the presence of the root in the numerator and one in the denominator. I would have to get rid of the root. But such integrals are much more complicated. About them - in other lessons.

See! One has only to slightly modify the function - the approach to its integration immediately changes. Sometimes dramatically!) There is no clear standard scheme. Each function has its own approach. Sometimes even unique.

In some cases, conversions in fractions are even more tricky.

Example 13

And here, how can the integral be reduced to a set of tabular ones? Here you can deftly dodge by adding and subtracting the expression x2 in the numerator of a fraction followed by term division. Very skillful reception in integrals! Watch the master class! :)

And now, if we replace the original fraction with the difference of two fractions, then our integral breaks up into two tabular ones - the already familiar power function and arc tangent (formula 8):

Well, what can I say? Wow!

This add/subtract numerator trick is very popular in integrating rational fractions. Very! I recommend taking note.

Example 14

Here, too, the same technology rules. You only need to add / subtract one in order to select the expression in the denominator from the numerator:

Generally speaking, rational fractions (with polynomials in the numerator and denominator) are a separate very extensive topic. The thing is that rational fractions are one of the very few classes of functions for which a universal way of integrating exists. The method of decomposition into simple fractions, coupled with . But this method is very time consuming and is usually used as heavy artillery. More than one lesson will be devoted to him. In the meantime, we are training and getting our hands on simple functions.

Let's summarize today's lesson.

Today we examined in detail how to use the table, with all the nuances, analyzed many examples (and not the most trivial ones) and got acquainted with the simplest methods of reducing integrals to tabular ones. And so we will now do Always. Whatever terrible function is under the integral, with the help of a wide variety of transformations, we will ensure that, sooner or later, our integral, one way or another, is reduced to a set of tabular ones.

A few practical tips.

1) If under the integral there is a fraction, in the numerator of which is the sum of degrees (roots), and in the denominator - lone x, then we use the term-by-term division of the numerator by the denominator. We replace the roots with powers fractional indicators and work according to formulas 1-2.

2) In trigonometric constructions, first of all, we try the basic formulas of trigonometry - double / triple angle,

It can be very lucky. Or maybe not…

3) Where necessary (especially in polynomials and fractions), we useabbreviated multiplication formulas:

(a+b) 2 = a 2 +2ab+b 2

(a-b) 2 = a 2 -2ab+b 2

(a-b)(a+b) = a 2 -b 2

4) When integrating fractions with polynomials, we try to artificially highlight the expression (s) in the numerator in the denominator. Very often, the fraction is simplified and the integral is reduced to a combination of tabular ones.

Well, friends? I see you're starting to like integrals. :) Then we fill our hand and solve the examples on our own.) Today's material is quite enough to successfully deal with them.

What? Do not know, ? Yes! We have not gone through this yet.) But here they do not need to be directly integrated. And may the school course help you!)

Answers (in disarray):

For better results, I strongly recommend purchasing a collection of problems on G.N. matan. Berman. Cool thing!

And that's all I have for today. Good luck!