Definite integral of a power function. Integrating the product of power functions of sin x and cos x

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 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 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 a 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 trying to integrate this example can give the teapot a headache 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 move the two 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 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 the square root there is a square trinomial, then the solution in any case comes down 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 necessarily. 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)

A more rare type of integral, but nevertheless encountered in practical examples.

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) we derive 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 is expanded into a sum of fractions. But in my practice there is such an example never met, so I missed 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 powers. 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 solving a certain type of 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.

A couple of simple examples for you to solve on your own:

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:
etc.

What is the idea of the method? The idea is to use transformations and trigonometric formulas to organize only tangents and the tangent derivative into the integrand. That is, we are talking about replacing: . 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 the 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

Principal integrals that every student should know

The listed integrals are the basis, the basis of the fundamentals. These formulas should definitely 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). Don't forget to add an arbitrary constant C to your answer when integrating!

Integral of a constant

∫ A d x = A x + C (1)

Integrating a Power Function

In fact, it was possible to limit ourselves to only formulas (5) and (7), but the rest of the integrals from this group occur so often 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 = ln | 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 exponential functions and hyperbolic functions

Of course, formula (8) (perhaps the most convenient for memorization) 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 simply remember these relations.

∫ e x d x = e x + C (8)
∫ a x d x = a x ln 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 is that 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 function sinx is equal to cosx. This is not true! The integral of sine is equal to “minus cosine”, but the integral of cosx is equal to “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 that reduce to inverse trigonometric functions

Formula (16), leading to the arctangent, 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

It is also advisable to remember these formulas. They are also used quite often, and their output is quite tedious.

∫ 1 x 2 + a 2 d x = ln | x + x 2 + a 2 | +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 + x 2 + a 2 | + 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 rules of integration

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 an antiderivative for the function f(x). Please note: 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 = ? (30)

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

A simple example of calculating the indefinite integral

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

Let us 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 obtain: ∫ 3 x 2 d x + ∫ 2 sin x d x − ∫ 7 e x d x + ∫ 12 d x

Let us remember 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, exponential 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 by 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 = ln | 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 + x 2 + a 2 | +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 + x 2 + a 2 | + 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 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 higher mathematics tutor. We will solve your problems together!

Reduction to the integral of a differential binomial

Let's 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 performing the substitution
t = sin x.
Then
dt = (sin x)′ dx = cos x dx;
cos 2 x = 1 - sin 2 x = 1 - t 2;

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

Integration with integers m and n

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

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

Reduction formulas

Reduction formulas for the integral

have the form:

;
;
;
.

There is no need to memorize them, since they are easily obtained by integrating by parts.

Proof of reduction formulas

Let's integrate by parts.

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

We similarly obtain the second formula.

Let's integrate by parts.

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

Third formula.

Let's integrate by parts.

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

Similarly, for the fourth formula.

Let's integrate by parts.

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

Example

Let's calculate the integral:

Let's convert:

Here m = 10, n = - 4.

We apply the reduction formula:

When m = 10, n = - 4:

When m = 8, n = - 2:

We apply the reduction formula:

When m = 6, n = - 0:

When m = 4, n = - 0:

When m = 2, n = - 0:

We calculate the remaining integral:

We collect intermediate results into one formula.

Used literature:
N.M. Gunther, R.O. Kuzmin, Collection of problems in higher mathematics, “Lan”, 2003.

See also:

Hello again, friends!

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

To navigate competently in all the integral diversity and not get lost, we need only four things:

1) Table of integrals. All the details about her - . This is how exactly to work with her.

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

3) Table of derivatives and differentiation rules.

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

4) Integration methods.

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

– ,

– .

Each of them will be discussed in separate lessons.

And now, finally, let's get down to solving the long-awaited examples. In order not to jump from section to section, I will duplicate once again the entire gentleman’s set, which will be useful for our further work. Let all the tools be at hand.)

First of all, this table of integrals:

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

Well, the necessary equipment is prepared. It's time to go! :)

Direct application of the table

This paragraph will consider the simplest and most harmless examples. The algorithm here is terribly simple:

1) Look at the table and look for the required formula(s);

2) Apply linearity properties (where required);

3) We carry out the transformation using 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 a power function in general form (second group). In our case n = 5. So we substitute the five for n and carefully calculate the result:

Ready. :)

Of course, this example is completely primitive. Purely for acquaintance.) But the ability to integrate powers makes it easy to calculate integrals of any polynomials and other power constructions.

Example 2

Below the integral is the sum. Oh well. We have linearity properties for this case. :) We split our integral into three separate ones, take all the constants out of the signs of the integrals and count each one according to the table (group 1-2):

Please note: constant WITH appears exactly at the moment when ALL integral signs disappear! Of course, after that you have to constantly carry it around with you. What to do...

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

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

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

The next example will be a little cooler.

Example 3

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

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

Comment: in this lesson multiplier dx in the process of integration Bye doesn’t participate in any way, and we are mentally “forgetting” about him for now. :) We only work with integrand function. But let's not forget about him. Very soon, literally in the next lesson dedicated to, we will remember about it. And we will feel the importance and power of this icon in full force!)

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

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

And x to the power minus two-thirds is already a tabular function! 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 calculate directly using the formula:

In this example, we were helped by the elementary properties of degrees. And this should be done in most cases when there are lonely roots or fractions under the integral. Therefore, a couple of practical tips when integrating power constructions:

We replace fractions with powers with negative exponents;

We replace roots with powers with fractional exponents.

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

And please, count all fractions carefully! We carefully monitor the signs and what goes where – what is in the numerator and what is the denominator.

What? Tired of boring power functions already? OK! Let's take the bull by the horns!

Example 4

If we now bring everything under the integral to a common denominator, then we can get stuck on this example seriously and for a long time.) But, taking a closer look at the integrand, we can see that our difference consists of two tabular functions. So let's not get perverted, but instead decompose 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 of a 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 like this:

And the second fraction is also a table function! Did you find out? 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 don't attribute it! Why? Because she will go to the final answer whole example. This is quite enough.) Strictly speaking, the constant must be written after each individual integration - be it intermediate or final: that’s what the indefinite integral requires...)

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

After the second integration:

But the trick 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 can do it difference two intermediate constants:

C 1 -C 2

And we have every right to replace this very difference in constants one constant! And simply redesignate it with the letter “C” that is familiar to us. Like this:

C 1 -C 2 = C

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

Yes, yes, they are fractions! Multistory logarithms when integrated are the most common thing. We're getting used to it too.)

Remember:

During intermediate integration of several terms, the constant WITH After each of them you don’t have to write. It is enough to include it in the final answer of the entire example. At the very end.

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

Example 5

The table, of course, does not have such a function. But there is similar function:

This is the very last one 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 cannot yet use the formula directly. But in our case the problem is completely solvable. Let’s first take this nine out of brackets, and then take it outside of our fraction altogether.)

And the new fraction is the table function we already need, number 8! Here and 2 =4/9. Or a=2/3.

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

This is the answer. This example, with a coefficient in front x 2, I chose it that way on purpose. 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 will be “root of five”. In general, you understand.)

Now let’s slightly modify our function: we’ll write the denominator under the root.) Now we’ll take this integral:

Example 6

The denominator now has a root. Naturally, the corresponding formula for integration has also changed, yes.) Again we go into the table and look for the appropriate 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 amount. So, we are working on fifth formula, with a "long" logarithm:

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

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

If doubts creep in, you can (and should) always check the result by reverse differentiation. Shall we check? What if it’s some kind of screw-up?

Let’s differentiate (we don’t pay attention to the module and treat it as ordinary brackets):

Everything is fair. :)

By the way, if in the integrand under the root you 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 there is plus/minus. :)

For example:

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

Example 7

Under the root again minus, but x 2 with the five we switched places. It’s similar, but not the same thing... For this case, our table also has a formula.) Formula number six, we haven’t worked with it yet:

But now - carefully. In the previous example, we used five as a number A . Here five will act as a number a 2!

Therefore, to apply the formula correctly, do not forget to extract the root of five:

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

Just like that! Just the terms under the root were swapped, and the result of integration changed significantly! Logarithm and arcsine... So please do not confuse these two formulas! Although the integrand functions are very similar...

Bonus:

In tabular formulas 7-8 there are coefficients before the logarithm and arctangent 1/(2a) And 1/a respectively. And in an alarming combat situation, when writing down these formulas, even nerds seasoned by their studies often get confused, where is it simple 1/a, and where 1/(2a). Here's a simple trick to remember.

In formula No. 7

The denominator of the integrand contains difference of squares x 2 – a 2. Which, according to the fearful school formula, breaks down as (x-a)(x+a). On two multiplier Keyword – two. And these two when integrating, the 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 formula No. 8

The denominator of the fraction contains sum of squares. But the sum of squares x 2 +a 2 cannot be decomposable into simpler factors. Therefore, whatever one may say, the denominator will remain so one factor. And the coefficient in front of the arctangent will also be 1/a.

Now let’s integrate some trigonometry for a change.)

Example 8

The example is simple. So simple that people, without even looking at the table, immediately happily write the answer and... we've arrived. :)

Let's 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 derivatives at the very least, in order not to get confused in the signs, the technique for remembering integrals is very simple:

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

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

(sin x)" = cos x.

Then for integral from the same sine it will be true:

That's all.) It's the same with cosine.

Let's now fix our example:

Preliminary elementary transformations of the integrand

Up to this point there were 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 the factors and divided 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 application of the table, then there would be a lot of freebies around and life would become boring.)

Now let's take a closer look at the examples. The kind where nothing seems to be decided directly. But it’s worth remembering just a couple of elementary school formulas or transformations, and the road to the answer becomes simple and clear. :)

Application of trigonometry formulas

Let's continue to have fun with trigonometry.

Example 9

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

Now we express from it the squared tangent we need and insert it under the integral:

Why was this done? And then, 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 simpler than ever. But let's think about this. If we were faced with a task differentiate the same function, then we would exactly knew exactly what to do - apply formula derivative of a complex function:

That's all. Simple and trouble-free technology. It always works and is guaranteed to lead to success.

What about the integral? But here we had to rummage through 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. Or else there will be more!

Example 10

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

So we apply this formula to our integrand function. In the “alpha” role we have x/2.

We get:

The effect is amazing, isn't it?

These two examples clearly show that pre-transforming a function before integration It’s completely acceptable and sometimes makes life enormously easier! And in integration this procedure (transformation of the integrand) is an order of magnitude more justified than in differentiation. You'll see everything later.)

Let's look at a couple more typical transformations.

Formulas for abbreviated multiplication, opening parentheses, bringing similar ones and the method of term-by-term division.

The usual banal school transformations. But sometimes they are the only ones who save, yes.)

Example 11

If we were calculating the derivative, then there would be no problem: the formula for the derivative of the product and - go ahead. But the standard formula for integral does not exist from the work. And the only way out here is to open all the brackets so that under the integral we get a polynomial. And we’ll somehow integrate the polynomial.) But we’ll also open the brackets wisely: abbreviated multiplication formulas are powerful things!

(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

Now we count:

And that's all.)

Example 12

Again, the standard formula for integral of a fraction 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 tabulated power functions:

I won’t comment specifically on the procedure for integrating the degrees: they’re not small anymore.)

Let's integrate the sum of power functions. According to the sign.)

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

This trick with term-by-term division would not have worked so easily. It is precisely because of the presence of a root in the numerator and a unit 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 to fractions are even more tricky.

Example 13

And here, how can you reduce the integral to a set of tabular ones? Here you can cleverly dodge by adding and subtracting the expression x 2 in the numerator of the fraction followed by term-by-term division. A very clever trick in integrals! Watch the master class! :)

And now, if we replace the original fraction with the difference of two fractions, then our integral splits into two tabular ones - the power function that is already familiar to us and the arctangent (formula 8):

Well, what can we say? Wow!

This trick of adding/subtracting terms in the numerator is very popular in integrating rational fractions. Very! I recommend taking note.

Example 14

The same technology rules here too. You just need to add/subtract one to extract the expression in the denominator from the numerator:

Generally speaking, rational fractions (with polynomials in the numerator and denominator) are a separate, very broad topic. The point is that rational fractions are one of the very few classes of functions for which a universal method of integration exists. The method of decomposition into simple fractions, coupled with . But this method is very labor-intensive and is usually used as heavy artillery. More than one lesson will be dedicated to him. In the meantime, we are training and getting better at simple functions.

Let's summarize today's lesson.

Today we looked in detail at exactly how to use the table, with all the nuances, analyzed many examples (and not the most trivial ones) and got acquainted with the simplest techniques for reducing integrals to tabular ones. And this is how we will do it now Always. No matter what 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.

Some practical tips.

1) If the integral is a fraction, the numerator of which is the sum of powers (roots), and the denominator is lonely x power, then we use term-by-term division of the numerator by the denominator. Replace roots with powers of c 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,

You might get 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 isolate the expression(s) in the denominator in the numerator. 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 get better at solving the examples ourselves.) Today’s material is quite enough to successfully cope with them.