Calculation of the simplest indefinite integrals. Integrating the product of power functions of sin x and cos x Integral of a complex power function

Hello again, friends!

As I promised, with this lesson we will begin to explore the endless expanses poetic world integrals and start solving 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 power function in general terms (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 done purely for understanding. 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 table function! Second group n=-2/3. And the constant 1/2 is not a hindrance for 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 practical advice 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 for a long time.) But, taking a closer look at the integrand, we can see that our difference consists of two table 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 a suitable 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?

We differentiate (we don’t pay attention to the module and perceive 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 swapped 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 joyfully 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 look at more solid 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 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 best complex 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 trigonometric formulas, then 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 – open all the brackets so that under the integral you 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) is a separate, very extensive 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 examined in detail 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 methods of reducing integrals to tabular ones. And this is how we will do it now Always. Whatever terrible function may be 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.

What? Don't know? Yes! We haven’t gone through this yet.) But there is no need to directly integrate them here. And may the school course help you!)

Answers (in disarray):

For better results, I strongly recommend purchasing a collection of problems based on G.N. Mathan. Berman. Cool stuff!

That's all I have for today. Good luck!

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

Content

See also:
Table of indefinite integrals

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:

At m = 10, n = - 4:

At m = 8, n = - 2:

We apply the reduction formula:

At m = 6, n = - 0:

At m = 4, n = - 0:

At m = 2, n = - 0:

We calculate the remaining integral:

We collect intermediate results into one formula.

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

See also:

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 integrals of 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 reducing 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) Integral of the difference of two functions equal to the difference 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: does not exist 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!

You might also be interested in

On this page you will find:

1. Actually, the table of antiderivatives - it can be downloaded in PDF format and printed;

2. Video on how to use this table;

3. A bunch of examples of calculating the antiderivative from various textbooks and tests.

In the video itself, we will analyze many problems where you need to calculate antiderivatives of functions, often quite complex, but most importantly, they are not power functions. All functions summarized in the table proposed above must be known by heart, like derivatives. Without them, further study of integrals and their application to solve practical problems is impossible.

Today we continue to study antiderivatives and move on to a little more complex topic. If last time we looked at antiderivatives only of power functions and slightly more complex constructions, today we will look at trigonometry and much more.

As I said in the last lesson, antiderivatives, unlike derivatives, are never solved “right away” using any standard rules. Moreover, the bad news is that, unlike the derivative, the antiderivative may not be considered at all. If we write absolutely random function and we try to find its derivative, then with a very high probability we will succeed, but the antiderivative will almost never be calculated in this case. But there is good news: there is a fairly large class of functions called elementary functions, the antiderivatives of which are very easy to calculate. And all the other more complex structures that are given on all kinds of tests, independent tests and exams, in fact, are made up of these elementary functions through addition, subtraction and other simple actions. The prototypes of such functions have long been calculated and compiled into special tables. It is these functions and tables that we will work with today.

But we will start, as always, with a repetition: let’s remember what an antiderivative is, why there are infinitely many of them and how to define them general view. To do this, I picked up two simple problems.

Solving easy examples

Example #1

Let us immediately note that $\frac(\text( )\!\!\pi\!\!\text( ))(6)$ and in general the presence of $\text( )\!\!\pi\!\!\ text( )$ immediately hints to us that the required antiderivative of the function is related to trigonometry. And, indeed, if we look at the table, we will find that $\frac(1)(1+((x)^(2)))$ is nothing more than $\text(arctg)x$. So let's write it down:

In order to find, you need to write down the following:

\[\frac(\pi )(6)=\text(arctg)\sqrt(3)+C\]

\[\frac(\text( )\!\!\pi\!\!\text( ))(6)=\frac(\text( )\!\!\pi\!\!\text( )) (3)+C\]

Example No. 2

Here also we're talking about O trigonometric functions. If we look at the table, then, indeed, this is what happens:

We need to find among the entire set of antiderivatives the one that passes through the indicated point:

\[\text( )\!\!\pi\!\!\text( )=\arcsin \frac(1)(2)+C\]

\[\text( )\!\!\pi\!\!\text( )=\frac(\text( )\!\!\pi\!\!\text( ))(6)+C\]

Let's finally write it down:

It's that simple. The only problem is that in order to count the antiderivatives simple functions, you need to learn the table of antiderivatives. However, after studying the derivative table for you, I think this will not be a problem.

Solving problems containing an exponential function

To begin with, let's write the following formulas:

\[((e)^(x))\to ((e)^(x))\]

\[((a)^(x))\to \frac(((a)^(x)))(\ln a)\]

Let's see how this all works in practice.

Example #1

If we look at the contents of the brackets, we will notice that in the table of antiderivatives there is no such expression for $((e)^(x))$ to be in a square, so this square must be expanded. To do this, we use the abbreviated multiplication formulas:

Let's find the antiderivative for each of the terms:

\[((e)^(2x))=((\left(((e)^(2)) \right))^(x))\to \frac(((\left(((e)^ (2)) \right))^(x)))(\ln ((e)^(2)))=\frac(((e)^(2x)))(2)\]

\[((e)^(-2x))=((\left(((e)^(-2)) \right))^(x))\to \frac(((\left(((e )^(-2)) \right))^(x)))(\ln ((e)^(-2)))=\frac(1)(-2((e)^(2x))) \]

Now let’s collect all the terms into a single expression and get the general antiderivative:

Example No. 2

This time the degree is larger, so the abbreviated multiplication formula will be quite complex. So let's open the brackets:

Now let’s try to take the antiderivative of our formula from this construction:

As you can see, there is nothing complicated or supernatural in the antiderivatives of the exponential function. All of them are calculated through tables, but attentive students will probably notice that the antiderivative $((e)^(2x))$ is much closer to simply $((e)^(x))$ than to $((a)^(x ))$. So maybe there is some more special rule, which allows, knowing the antiderivative $((e)^(x))$, to find $((e)^(2x))$? Yes, such a rule exists. And, moreover, it is an integral part of working with the table of antiderivatives. We will now analyze it using the same expressions that we just worked with as an example.

Rules for working with the table of antiderivatives

Let's write our function again:

In the previous case, we used the following formula to solve:

\[((a)^(x))\to \frac(((a)^(x)))(\operatorname(lna))\]

But now let’s do it a little differently: let’s remember on what basis $((e)^(x))\to ((e)^(x))$. As I already said, because the derivative $((e)^(x))$ is nothing more than $((e)^(x))$, therefore its antiderivative will be equal to the same $((e) ^(x))$. But the problem is that we have $((e)^(2x))$ and $((e)^(-2x))$. Now let's try to find the derivative of $((e)^(2x))$:

\[((\left(((e)^(2x)) \right))^(\prime ))=((e)^(2x))\cdot ((\left(2x \right))^( \prime ))=2\cdot ((e)^(2x))\]

Let's rewrite our construction again:

\[((\left(((e)^(2x)) \right))^(\prime ))=2\cdot ((e)^(2x))\]

\[((e)^(2x))=((\left(\frac(((e)^(2x)))(2) \right))^(\prime ))\]

This means that when we find the antiderivative $((e)^(2x))$ we get the following:

\[((e)^(2x))\to \frac(((e)^(2x)))(2)\]

As you can see, we got the same result as before, but we did not use the formula to find $((a)^(x))$. Now this may seem stupid: why complicate the calculations when there is a standard formula? However, in slightly more complex expressions you will find that this technique is very effective, i.e. using derivatives to find antiderivatives.

As a warm-up, let's find the antiderivative of $((e)^(2x))$ in a similar way:

\[((\left(((e)^(-2x)) \right))^(\prime ))=((e)^(-2x))\cdot \left(-2 \right)\]

\[((e)^(-2x))=((\left(\frac(((e)^(-2x)))(-2) \right))^(\prime ))\]

When calculating, our construction will be written as follows:

\[((e)^(-2x))\to -\frac(((e)^(-2x)))(2)\]

\[((e)^(-2x))\to -\frac(1)(2\cdot ((e)^(2x)))\]

We got exactly the same result, but took a different path. It is this path, which now seems a little more complicated to us, that in the future will turn out to be more effective for calculating more complex antiderivatives and using tables.

Pay attention! This is very important point: antiderivatives, like derivatives, can be considered a set in various ways. However, if all calculations and calculations are equal, then the answer will be the same. We have just seen this with the example of $((e)^(-2x))$ - on the one hand, we calculated this antiderivative “right through”, using the definition and calculating it using transformations, on the other hand, we remembered that $ ((e)^(-2x))$ can be represented as $((\left(((e)^(-2)) \right))^(x))$ and only then we used the antiderivative for the function $( (a)^(x))$. However, after all the transformations, the result was the same, as expected.

And now that we understand all this, it’s time to move on to something more significant. Now we will analyze two simple constructions, but the technique that will be used when solving them is a more powerful and useful tool than simply “running” between neighboring antiderivatives from the table.

Problem solving: finding the antiderivative of a function

Example #1

Let's break down the amount that is in the numerators into three separate fractions:

This is a fairly natural and understandable transition - most students do not have problems with it. Let's rewrite our expression as follows:

Now let's remember this formula:

In our case we will get the following:

To get rid of all these three-story fractions, I suggest doing the following:

Example No. 2

Unlike the previous fraction, the denominator is not a product, but a sum. In this case, we can no longer divide our fraction into the sum of several simple fractions, but we must somehow try to make sure that the numerator contains approximately the same expression as the denominator. In this case, it's quite simple to do it:

This notation, which in mathematical language is called “adding a zero,” will allow us to again divide the fraction into two pieces:

Now let's find what we were looking for:

That's all the calculations. Despite the apparent greater complexity than in the previous problem, the amount of calculations turned out to be even smaller.

Nuances of the solution

And this is where the main difficulty of working with tabular antiderivatives lies, this is especially noticeable in the second task. The fact is that in order to select some elements that are easily calculated through the table, we need to know what exactly we are looking for, and it is in the search for these elements that the entire calculation of antiderivatives consists.

In other words, it is not enough just to memorize the table of antiderivatives - you need to be able to see something that does not yet exist, but what the author and compiler of this problem meant. That is why many mathematicians, teachers and professors constantly argue: “What is taking antiderivatives or integration - is it just a tool or is it a real art?” In fact, in my personal opinion, integration is not an art at all - there is nothing sublime in it, it is just practice and more practice. And to practice, let's solve three more serious examples.

We train in integration in practice

Task No. 1

Let's write the following formulas:

\[((x)^(n))\to \frac(((x)^(n+1)))(n+1)\]

\[\frac(1)(x)\to \ln x\]

\[\frac(1)(1+((x)^(2)))\to \text(arctg)x\]

Let's write the following:

Problem No. 2

Let's rewrite it as follows:

The total antiderivative will be equal to:

Problem No. 3

The difficulty of this problem is that, unlike the previous functions above, there is no variable $x$ at all, i.e. We don’t understand what to add or subtract to get at least something similar to what is below. However, in fact, this expression is considered even simpler than any expression from the previous constructions, because this function can be rewritten as follows:

You may now ask: why are these functions equal? Let's check:

Let's rewrite it again:

Let's transform our expression a little:

And when I explain all this to my students, almost always the same problem arises: with the first function everything is more or less clear, with the second you can also figure it out with luck or practice, but what kind of alternative consciousness do you need to have in order to solve the third example? Actually, don't be scared. The technique that we used when calculating the last antiderivative is called “decomposition of a function into its simplest”, and this is a very serious technique, and a separate video lesson will be devoted to it.

In the meantime, I propose to return to what we just studied, namely, to exponential functions and somewhat complicate the problems with their content.

More complex problems for solving antiderivative exponential functions

Task No. 1

Let's note the following:

\[((2)^(x))\cdot ((5)^(x))=((\left(2\cdot 5 \right))^(x))=((10)^(x) )\]

To find the antiderivative of this expression, simply use the standard formula - $((a)^(x))\to \frac(((a)^(x)))(\ln a)$.

In our case, the antiderivative will be like this:

Of course, compared to the design we just solved, this one looks simpler.

Problem No. 2

Again, it's easy to see that this function can easily be divided into two separate terms - two separate fractions. Let's rewrite:

It remains to find the antiderivative of each of these terms using the formula described above:

Despite the apparent great complexity exponential functions Compared to power ones, the overall volume of calculations and calculations turned out to be much simpler.

Of course, for knowledgeable students, what we have just discussed (especially against the backdrop of what we have analyzed before) may seem like elementary expressions. However, when choosing these two problems for today's video lesson, I did not set myself the goal of telling you another complex and sophisticated technique - all I wanted to show you is that you should not be afraid to use standard algebra techniques to transform original functions.

Using a "secret" technique

In conclusion, I would like to look at another interesting technique, which, on the one hand, goes beyond what we mainly discussed today, but, on the other hand, it is, firstly, not at all complicated, i.e. even beginner students can master it, and, secondly, it is quite often found on all kinds of tests and tests. independent work, i.e. knowledge of it will be very useful in addition to knowledge of the table of antiderivatives.

Task No. 1

Obviously, we have something very similar to a power function. What should we do in this case? Let's think about it: $x-5$ is not that much different from $x$ - they just added $-5$. Let's write it like this:

\[((x)^(4))\to \frac(((x)^(5)))(5)\]

\[((\left(\frac(((x)^(5)))(5) \right))^(\prime ))=\frac(5\cdot ((x)^(4))) (5)=((x)^(4))\]

Let's try to find the derivative of $((\left(x-5 \right))^(5))$:

\[((\left(((\left(x-5 \right))^(5)) \right))^(\prime ))=5\cdot ((\left(x-5 \right)) ^(4))\cdot ((\left(x-5 \right))^(\prime ))=5\cdot ((\left(x-5 \right))^(4))\]

It follows from this:

\[((\left(x-5 \right))^(4))=((\left(\frac(((\left(x-5 \right))^(5)))(5) \ right))^(\prime ))\]

There is no such value in the table, so we have now derived this formula ourselves using the standard antiderivative formula for a power function. Let's write the answer like this:

Problem No. 2

Many students who look at the first solution may think that everything is very simple: just replace $x$ in the power function with a linear expression, and everything will fall into place. Unfortunately, everything is not so simple, and now we will see this.

By analogy with the first expression, we write the following:

\[((x)^(9))\to \frac(((x)^(10)))(10)\]

\[((\left(((\left(4-3x \right))^(10)) \right))^(\prime ))=10\cdot ((\left(4-3x \right)) ^(9))\cdot ((\left(4-3x \right))^(\prime ))=\]

\[=10\cdot ((\left(4-3x \right))^(9))\cdot \left(-3 \right)=-30\cdot ((\left(4-3x \right)) ^(9))\]

Returning to our derivative, we can write:

\[((\left(((\left(4-3x \right))^(10)) \right))^(\prime ))=-30\cdot ((\left(4-3x \right) )^(9))\]

\[((\left(4-3x \right))^(9))=((\left(\frac(((\left(4-3x \right))^(10)))(-30) \right))^(\prime ))\]

This immediately follows:

Nuances of the solution

Please note: if nothing essentially changed last time, then in the second case, instead of $-10$, $-30$ appeared. What is the difference between $-10$ and $-30$? Obviously, by a factor of $-3$. Question: where did it come from? If you look closely, you can see that it was taken as a result of calculating the derivative of a complex function - the coefficient that stood at $x$ appears in the antiderivative below. This is very important rule, which I initially did not plan to discuss at all in today’s video tutorial, but without it the presentation of tabular antiderivatives would be incomplete.

So let's do it again. Let there be our main power function:

\[((x)^(n))\to \frac(((x)^(n+1)))(n+1)\]

Now, instead of $x$, let's substitute the expression $kx+b$. What will happen then? We need to find the following:

\[((\left(kx+b \right))^(n))\to \frac(((\left(kx+b \right))^(n+1)))(\left(n+ 1\right)\cdot k)\]

On what basis do we claim this? Very simple. Let's find the derivative of the construction written above:

\[((\left(\frac(((\left(kx+b \right))^(n+1)))(\left(n+1 \right)\cdot k) \right))^( \prime ))=\frac(1)(\left(n+1 \right)\cdot k)\cdot \left(n+1 \right)\cdot ((\left(kx+b \right))^ (n))\cdot k=((\left(kx+b \right))^(n))\]

This is the same expression that originally existed. Thus, this formula is also correct, and it can be used to supplement the table of antiderivatives, or it is better to simply memorize the entire table.

Conclusions from the “secret: technique:

• Both functions that we just looked at can, in fact, be reduced to the antiderivatives indicated in the table by expanding the degrees, but if we can more or less somehow cope with the fourth degree, then I wouldn’t even consider the ninth degree dared to reveal.
• If we were to expand the powers, we would get such a volume of calculations that simple task would borrow from us inadequately large number time.
• That is why such tasks, within which there are linear expressions, there is no need to solve it head on. As soon as you come across an antiderivative that differs from the one in the table only by the presence of the expression $kx+b$ inside, immediately remember the formula written above, substitute it into your table antiderivative, and everything will turn out much faster and easier.

Naturally, due to the complexity and seriousness of this technique, we will return to its consideration many times in future video lessons, but that’s all for today. I hope this lesson will really help those students who want to understand antiderivatives and integration.