Leading 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 your answer when integrating!

Integral of a constant

∫ A d x = A x + C (1)

Power function integration

In fact, it was possible to restrict oneself 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 function 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: they confuse the signs in formulas (12) and (13). Remembering that the derivative of the sine is equal to the cosine, many for some reason believe that the integral of the sinx function is equal to cosx. This is not true! The integral of the sine is equal to "minus the cosine", but the integral of the 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), which leads to the arctangent, is naturally a special case of formula (17) with 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 outside 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 compound 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). Please note: this formula is only suitable for the case 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 "deal" with it. In some cases, integration by parts will help you, somewhere you will have to change a variable, and sometimes even "school" algebra or trigonometry formulas can help.

A simple example for calculating an 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 obtain: ∫ 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 outside 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 need to apply formulas (3), (12), (8) and (1). Let's integrate the power function, sine, exponent and constant 1. Let's not 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 differentiating: take the derivative of the resulting function and make sure that it is equal to the original integrand.

Pivot 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

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Complex integrals

This article completes the topic of indefinite integrals, and includes integrals that I find quite difficult. The lesson was created at the repeated requests of visitors who expressed their wishes that more difficult examples were also 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 about integrals should refer to the very first lesson - Indefinite integral. Examples of solutions, where you can master the topic practically from scratch. More experienced students can get acquainted with the techniques and methods of integration that have not yet been met in my articles.

What integrals will be considered?

First, we will consider integrals with roots, for which solution is consistently used 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 an interesting and original the method of reducing the integral to itself... Not so few integrals are solved in this way.

The third number of the program will go to integrals of complex fractions, which flew past the box office in previous articles.

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

(2) In the integrand, 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 we bring the function under the differential sign.

(4) Take the remaining integrals. Note that parentheses can be used in the logarithm, not modulus, since.

(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 correct sense =)

As you can see, during the solution, even more than two solution methods had to be used, thus, to deal with such integrals, you need confident integration skills and not the smallest experience.

In practice, of course, the square root is more common, here are three examples for an 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 substitution to use at the beginning of the solutions, I think, is obvious. Why did I pick up examples of the same type? They often meet in their role. More often, perhaps, just something like .

But not always, when the root of a linear function is found under the arctangent, sine, cosine, exponent, and other functions, several methods have to be applied at once. In a number of cases, it is possible to "get off easily", 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 replacing, a relatively simple integral is obtained.

By reducing the integral to itself

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

Example 5

Find the indefinite integral

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

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

We integrate piece by piece:

(1) Prepare an integrand function for term division.

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

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

(4) Take the last integral ("long" logarithm).

Now we look at the very beginning of the solution:

And at the end:

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

Let's equate the beginning and the end:

Transferring to left side with sign change:

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

The constant, strictly speaking, should have been added earlier, but added it at the end. I strongly recommend that you read what is strict here:

Note: More strictly The final stage the solution looks like this:

Thus:

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

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

Example 6

Find the indefinite integral

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

If there is a square trinomial under the square root, then the solution in any case is reduced 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 is dispensed with "without any consequences":
, resulting in an integral. Something familiar, right?

Or such an example, with a square binomial:
Select a complete square:
And, after a linear replacement, we get an integral, which is also solved according to the already considered algorithm.

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

In the listed integrals by parts, we will have to integrate two times 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 reduced to itself. Let's equate the beginning and the end of the solution:

Move to the left with a sign change and express our integral:

Ready. Along the way, it is advisable to comb the right side, i.e. put the exponent outside the brackets, and in the brackets arrange the sine and cosine in a "nice" 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, exactly the exponent should always be denoted by? Not necessary. In fact, in the considered integral fundamentally does not matter What to denote for, it was possible to go the other way:

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

That is, you can also designate a trigonometric function. 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 an example for a do-it-yourself solution. Before deciding, think about what is more profitable in this case to designate for, exponent or trigonometric function? Complete solution and answer at the end of the tutorial.

And of course, don't forget that most of the answers in this lesson are easy enough to differentiate!

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 lost 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 by inattention. In addition, there is a high probability of error in signs, note that the exponent has a minus sign, and this introduces additional difficulty.

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

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

Integration of compound fractions

We are slowly getting closer to the equator of the lesson and begin to consider integrals of fractions. Again, not all of them are super complicated, 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 is the square trinomial plus outside the root "appendage" in the form of "x". An integral of this kind is solved using a standard substitution.

We decide:

The replacement is simple:

We look at life after replacement:

(1) After substitution, we bring the terms under the root to a common denominator.
(2) We take out from under the root.
(3) 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 verbally.
(4) The resulting integral, as you remember from the lesson Integration of some fractions, solved full square selection method... Select a complete square.
(5) By integration we get an ordinary "long" logarithm.
(6) We carry out the reverse replacement. If initially, then back:.
(7) The final action is aimed at the hairstyle of 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 a do-it-yourself solution. Here, a constant has been added to the lonely 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:

Complete solution and answer at the end of the tutorial.

Sometimes in such an integral there may be a square binomial under the root, this does not change the 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 considered in the lesson Integrals of Irrational Functions.

Integral of an indecomposable polynomial of degree 2 in degree

(polynomial in the denominator)

More rare, but nevertheless found in practical examples integral form.

Example 13

Find the indefinite integral

But let's go back to the example with lucky number 13 (honestly, I didn't guess right). This integral is also from the category of those with which you can pretty much torment yourself if you don't know how to solve it.

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 piece by piece:

For an integral of the form (is a natural number), we have derived recurrent Degree 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 a do-it-yourself solution. The sample solution uses the above formula twice in succession.

If under the degree there is indecomposable square trinomial, then the solution is reduced to a binomial by selecting a complete square, for example:

What if there is an additional polynomial in the numerator? In this case, the method of undefined coefficients is used, and the integrand is expanded into the 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 will skip it now. If such an integral still occurs, see the textbook - everything is simple there. I do not consider it appropriate to include material (even simple ones), the probability of meeting with which tends to zero.

Integration of complex trigonometric functions

The adjective "difficult" for most examples again carries a lot of conditional character... Let's start with tangents and cotangents in high degrees... From the point of view of the methods used for solving the tangent and the cotangent, they are almost the same, so I will talk more about the tangent, implying that the demonstrated method for solving the integral is valid for the cotangent too.

In the above lesson, we looked at universal trigonometric substitution for solving a certain kind of integrals of trigonometric functions. The disadvantage of the universal trigonometric substitution is that when using it, cumbersome integrals with difficult calculations often arise. And in some cases, universal trigonometric substitution can be avoided!

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

Example 17

Find the indefinite integral

Here you can use generic 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 double angle sine trigonometric formula.
(2) We carry out an artificial transformation: In the denominator, divide and multiply by.
(3) According to the well-known formula in the denominator, we transform the fraction into a tangent.
(4) We bring the function under the sign of the differential.
(5) Take the integral.

Pair simple examples for an independent solution:

Example 18

Find the indefinite integral

Note: The very first step is to use the cast formula and carefully carry out the 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 behind the method? The idea is to organize only the tangents and the derivative of the tangent in the integrand using transformations, trigonometric formulas. I.e, it comes about replacement: ... In Examples 17-19, we actually applied this replacement, but the integrals were so simple that the matter was treated 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 replacement:

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

for an 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).

Consider a couple of more meaningful tasks for this rule:

Example 20

Find the indefinite integral

The sum of the powers of the 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) Transform the denominator.
(2) According to the well-known formula, we obtain.
(3) Transform the denominator.
(4) We use the formula .
(5) We bring the function under the sign of the differential.
(6) We carry out a 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 confusion.

Example 21

Find the indefinite integral

This is an example for a do-it-yourself solution.

Hold on, champion rounds begin =)

Often in the integrand there is a "hodgepodge":

Example 22

Find the indefinite integral

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

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

A couple of creative examples for self-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 carried out through the tangents. Complete solution and answers at the end of the lesson