Complex integrals

This article completes the subject of uncertain integrals, and in it the integrals that I consider quite complicated are included. The lesson was created on the repeated requests of visitors who expressed the wishes so that more difficult examples are dismantled on the site.

It is assumed that the reader of this text is well prepared and knows how to apply the main techniques of integration. Teapots and people who are not very confidently dealt with integrals should be referred to the first lesson - Uncertain integral. Examples of solutionswhere you can master the topic with almost zero. More experienced students can familiarize themselves with the techniques and methods of integration, which in my articles have not yet met.

What integrals will be considered?

First, we will consider integrals with roots, to solve which is consistently used replacing the variable and integration in parts. That is, in one example, two receptions are combined. And even more.

Then we will get acquainted with interesting and original method information integral to yourself. This method is solved not so few integrals.

The third number of the program will go integrals from complex fractions that flew past cash registers in previous articles.

Fourthly, additional integrals from trigonometric functions will be disassembled. In particular, there are methods that allow you to avoid the time consuming of a universal trigonometric substitution.

(2) In the integrand function, the numerator on the denominator.

(3) Use the linearity property of an indefinite integral. In the last integral immediately sweep the function under the sign of the differential.

(4) Take the remaining integrals. Please note that in logarithm you can use brackets, not a module, since.

(5) We hold a replacement, expressing from the direct replacement "TE":

Masochian students can indifferentiate the answer and get the original integrand function as I just did. No, no, I fulfilled the verification in the right sense \u003d)

As you can see, during the decision I had to use even more than two decisions of the solution, so for reprisals with similar 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 an indefinite integral

Example 3.

Find an indefinite integral

Example 4.

Find an indefinite integral

These examples of the same type, so the complete solution at the end of the article will be only for Example 2, in the examples 3-4 - one answers. What replacement to apply at the beginning of the decisions, I think obviously. Why did I pick up the same type of examples? Often found in your role. More often, perhaps, just something like .

But not always, when under Arctgennes, sinus, cosine, exponential, etc. features are the root of a linear function, several methods have to be applied. In some cases, it is possible to "get rid of", that is, immediately after the replacement, a simple integral is obtained, which is elementary takes. The easiest of the proposed tasks is example 4, in it after replacement it turns out a relatively simple integral.

Method information integral to yourself

A witty and beautiful method. Immediately consider the classics of the genre:

Example 5.

Find an indefinite integral

Under the root there is a square biccoon, and when trying to integrate this example, the kettle can suffer for hours. Such an integral is taken in parts and comes down to itself. In principle, it is not difficult. If you know how.

Denote by the considered integral of the Latin letter and begin the solution:

We integrate in parts:

(1) We prepare a replacement function for the soil division.

(2) We divide the replacement function. Perhaps not to all clearly, I will write in more detail:

(3) Use the linearity property of an indefinite integral.

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

Now we look at the very beginning of the decision:

And on the end:

What happened? As a result of our manipulations, the integral got to himself!

We equate the beginning and end:

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

And demo demolose to the right side. As a result:

The constant, strictly speaking, had to be added earlier, but attributed it at the end. I strongly recommend reading what is here for a rigor:

Note: A more strict final stage of the solution looks like this:

In this way:

Constant can be reused through. Why can you reissue? Because it still takes any Values, and in this sense between constants and there is no difference.
As a result:

Such a trick with reissued constant is widely used in differential equations. And there I will be strict. And here such a liberty is allowed by me only in order not to confuse you with superfluous things and focus on the integration method itself.

Example 6.

Find an indefinite integral

Another typical integral for self-decisions. Complete solution and answer at the end of the lesson. The difference with the response of the previous example will be!

If the square root is a square triple, then the solution in any case is reduced to two disassembled examples.

For example, consider the integral . All you need to do is pre- select full square:
.
Next, a linear replacement is carried out, which costs "without any consequences":
As a result, the integral is obtained. Something familiar, right?

Or such an example, with square bounced:
We highlight a full square:
And, after a linear replacement, we get an integral, which is also solved by the algorithm already considered.

Consider two more typical examples on the receipt of information integral to yourself:
- integral from the exhibitory multiplied by sinus;
- Integral from the exhibitory multiplied by cosine.

In the listed integrals in parts will have to be integrated twice:

Example 7.

Find an indefinite integral

The integrand function is an exhibitor multiplied by sinus.

We integrate twice in parts and bring the integral to yourself:

As a result of two-time integration in parts, the integral has gotten to itself. We equate the beginning and ending solutions:

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

Ready. Also, it is desirable to combat the right side, i.e. To make an exponent for brackets, and in brackets to lay sinus with cosine in the "beautiful" order.

Now let's go back to the beginning of the example, or rather - to integration in parts:

For we designated the exhibitor. The question arises, it is always necessary to refer to the exhibitor for? Not necessary. In fact, in the examined integral principle no differenceWhat to refer to, it was possible to go to another way:

Why is it possible? Because the exhibitor turns into itself (and during differentiation, and during integration), the sinus with cosine is mutually becoming each other (again - both during differentiation, and during integration).

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

Example 8.

Find an indefinite integral

This is an example for an independent solution. Before deciding, think about it is more profitable in this case to designate, exponent or trigonometric function? Complete solution and answer at the end of the lesson.

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

Examples were not considered not the most difficult. In practice, integrals are more often found, where there is a constant in the exponent indicator and in the argument of a trigonometric function, for example:. Thought in a similar integral will have to make many, I often confuse me. The fact is that in solving the probability of the appearance of fractions, and is very simply something intense to lose. In addition, the likelihood of errors in signs is great, please note that in the exponent's indicator there is a minus sign, and this makes additional difficulty.

At the final stage, approximately the following is often obtained:

Even at the end of the decision should be extremely attentive and competently deal with fractions:

Integrating complex fractions

Slowly we get to the lesson equator and begin to consider integrals from fractions. Again, not all of them are superswit, just for one reason or another examples were a bit "not in the topic" in other articles.

We continue the topic of roots

Example 9.

Find an indefinite integral

In the denominator, under the root there is a square three-stale plus outside the root "Improve" in the form of "IKSA". The integral of this type is solved using standard replacement.

We decide:

Replacement here is simple:

We look at life after replacement:

(1) After the substitution, we give to the overall denominator terms under the root.
(2) We endure from the root.
(3) Numerator and denominator reducing on. At the same time, under the root, I rearranged the components in a comfortable order. With a certain experiment, steps (1), (2) can be skipped by performing commented actions orally.
(4) the resulting integral, as you remember from the lesson Integrating some fractions, decides method of allocation of a full square. Select a full square.
(5) Integration we get a utmost "long" logarithm.
(6) Conduct a replacement. If initially, then back :.
(7) The final action is aimed at the hairstyle of the result: under the root, they again bring the components to the overall denominator and endure from the root.

Example 10.

Find an indefinite integral

This is an example for an independent solution. Here the constant has been added to the lonely "ICSU", and the replacement is almost the same:

The only thing you need to additionally do is express "X" from the replacement:

Complete solution and answer at the end of the lesson.

Sometimes in such an integral under the root there may be a square bicker, it does not change the solution to solve, it will be even easier. Feel the difference:

Example 11.

Find an indefinite integral

Example 12.

Find an indefinite integral

Brief decisions and answers at the end of the lesson. It should be noted that example 11 is exactly binomial integral, whose decision was considered in the lesson Integrals from irrational functions.

Integral from an indecomposable polynomial of a 2nd degree to degree

(polynomial in the denominator)

More rare, but, nevertheless, in practical examples, the view of the integral.

Example 13.

Find an indefinite integral

But let's return for example with a happy number 13 (honestly, did not fit). This integral is also from the category of those with which you can be pretty enough if you do not know how to solve.

The decision begins with artificial transformation:

How to divide the numerator to the denominator, I think everything is understood.

The resulting integral is taken in parts:

For the view integral (- natural number) removed recurrent Degree reduction formula:
where - Integral degree lower.

I will be convinced of the justice of this formula for the propheted integral.
In this case,:, we use the formula:

As you can see, the answers coincide.

Example 14.

Find an indefinite integral

This is an example for an independent solution. In the sample of the solution, the aforementioned formula was twice.

If under the degree is located independent on multipliers Square threefold, then the solution comes down to bicked by highlighting a complete square, for example:

What if you are additionally in the numerator there is a polynomial? In this case, the method of indefinite coefficients is used, and the integrated function is described in the amount of fractions. But in my practice of such an example i did not meet, so I missed this case in the article Integrals from fractional rational functionI miss and now. If such an integral still meets, see the textbook - everything is simple there. I do not consider it expedient to include the material (even simple), the probability of meeting with which she strives for zero.

Integration of complex trigonometric functions

The adjective "complex" for most examples is in many ways conditional. Let's start with Tangents and Kotangenes in high degrees. From the point of view of the methods of solving Tangent and Kotangent, almost the same thing, so I will talk more about Tangent, implying that the demonstrated reception of the solution of the integral is fair and for Cotangent too.

On the above lesson, we considered universal trigonometric substitution To solve a specific type of integrals from trigonometric functions. The lack of a universal trigonometric substitution is that when it is used, bulky integrals with difficult calculations often occur. And in some cases of a universal trigonometric substitution can be avoided!

Consider another canonical example, the integral from the unit divided into sinus:

Example 17.

Find an indefinite integral

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

(1) Use the trigonometric formula of the dual angle sine.
(2) We carry out an artificial transformation: in the denominator we divide and multiply on.
(3) According to the known formula in the denominator, we turn fraction in the tangent.
(4) Sweep the function under the sign of the differential.
(5) Take the integral.

A couple of simple examples for an independent solution:

Example 18.

Find an indefinite integral

Note: The most first action should be used by the formula And carefully carry out similar to the previous example of action.

Example 19.

Find an indefinite integral

Well, this is a very simple example.

Full solutions and answers at the end of the lesson.

I think now no one has problems with integrals:
etc.

What is the idea of \u200b\u200bthe method? The idea is that with the help of transformations, trigonometric formulas to organize in the integrand only tangents and a tangent derivative. That is, it's about replacing: . In Examples 17-19, we actually applied this replacement, but the integrals were so simple that it cost an equivalent effect - to summarize the function under the sign of the differential.

Similar arguments, as I have already stipulated, you can spend for Cotangent.

There is a formal prerequisite for the use of the above replacement:

The sum of the degrees of cosine and sinus is a whole negative number, eg:

for the integral - a whole negative number.

! Note : If the integrand function contains only sinus or only cosine, then the integral is taken at a negative odd degree (the simplest cases in Examples No. 11, 18).

Consider a couple of more informative tasks for this rule:

Example 20.

Find an indefinite integral

The sum of the degrees of sinus and cosine: 2 - 6 \u003d -4 is a whole negative number, which means that the integral can be reduced to the tangents and its derivative:

(1) We transform the denominator.
(2) According to the famous formula, we get.
(3) We transform the denominator.
(4) We use the formula .
(5) Surrender the function under the sign of the differential.
(6) We replace. More experienced students can not be replaced, but still it is better to replace the tangent with one letter - less risk is confused.

Example 21.

Find an indefinite integral

This is an example for an independent solution.

Hold on, champion's rounds begin \u003d)

Often in the integrand function is "Solyanka":

Example 22.

Find an indefinite integral

In this integral, the Tangent is initially present, which immediately pursues at the already familiar thought:

Artificial transformation at the very beginning and remaining the remaining steps without comment, since everything was mentioned above.

A pair of creative examples for an independent solution:

Example 23.

Find an indefinite integral

Example 24.

Find an indefinite integral

Yes, in them, of course, it is possible to lower the degree of sinus, cosine, to use a universal trigonometric substitution, but the decision 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

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Definition of a primitive function

• Function y \u003d f (x)called primitive for function y \u003d f (x) At a given interval X,if for all h. ∈ H. Equality is performed: F '(x) \u003d f (x)

You can read in two ways:

1. f. derived function F.
2. F. Perfect for function f.

Primitive property

• If a F (X)- Perfect for function f (x) At a given gap, the function f (x) has infinitely many primitive, and all these primitive can be written as F (x) + withwhere C is an arbitrary constant.

Geometric interpretation

• Graphs of all primitive this feature. F (X) obtained from the graph of any one primitive parallel transfer along the axis about w..

The rules for calculating the primary

1. The first amount is equal to the sum of the primordial. If a F (X) - Pred-like for f (x), and G (x) is a primitive for g (x)T. F (x) + g (x) - Pred-like for f (x) + g (x).
2. Permanent multiplier can be made for a derivative mark. If a F (X) - Pred-like for f (x), I. k. - constant, then k · f (x) - Pred-like for k · f (x).
3. If a F (X) - Pred-like for f (x), I. k, B. - constant, and k ≠ 0T. 1 / k · f (kx + b) - Pred-like for f (KX + B).

Remember!

Any feature F (x) \u003d x 2 + where C is an arbitrary constant, and only such a function is a primitive for function f (x) \u003d 2x.

• For example:

  F "(x) \u003d (x 2 + 1)" \u003d 2x \u003d f (x);

  f (x) \u003d 2x, Because F "(x) \u003d (x 2 - 1)" \u003d 2x \u003d f (x);

  f (x) \u003d 2x, Because F "(x) \u003d (x 2 -3)" \u003d 2x \u003d f (x);

The connection between the graphs of the function and its primary:

1. If the graph is function f (X)\u003e 0 on the interval, then the schedule is its primitive F (X) Increases at this interval.
2. If the graph is function f (x) on the interval, then the graph of its primitive F (X) decreases at this interval.
3. If a f (x) \u003d 0then the graph of her primitive F (X) At this point changes with an increasing decrease (or vice versa).

To designate, the sign of an undefined integral is used, that is, the integral without specifying the integration limits.

Uncertain integral

Definition:

• An uncertain integral from the function f (x) is the expression F (x) + C, that is, the combination of all the primary functions of the F (X). Denotes an indefinite integral as follows: \\ int f (x) dx \u003d f (x) + c

• f (x)- refer to the integrated function;
• f (x) dx- are called a concintive expression;
• x. - call integration variable;
• F (X) - one of the primitive functions f (x);
• FROM - Arbitrary constant.

Properties of an indefinite integral

1. The derivative of an indefinite integral is equal to the integrand function: (\\ int f (x) dx) \\ prime \u003d f (x).
2. A permanent multiplier of the integrated expression can be made for an integral sign: \\ int k \\ cdot f (x) dx \u003d k \\ cdot \\ int f (x) dx.
3. The integral from the amount (difference) of functions is equal to the amount (difference) of the integrals from these functions: \\ int (f (x) \\ pm g (x)) dx \u003d \\ int f (x) dx \\ pm \\ int g (x) dx.
4. If a k, B.- constant, and k ≠ 0, then \\ int f (kx + b) dx \u003d \\ frac (1) (k) \\ cdot f (kx + b) + C.

Table of primary and uncertain integrals

Function f (x)	PRINTING F (x) + C	Uncertain integrals \\ int f (x) dx \u003d f (x) + c
0	C.	\\ int 0 dx \u003d c
f (x) \u003d k	F (x) \u003d kx + c	\\ int kdx \u003d kx + c
f (x) \u003d x ^ m, m \\ not \u003d -1	F (x) \u003d \\ frac (x ^ (m + 1)) (m + 1) + c	\\ int x (^ m) dx \u003d \\ frac (x ^ (m + 1)) (M + 1) + C
f (x) \u003d \\ FRAC (1) (x)	F (x) \u003d L n \\ LVERT X \\ RVERT + C	\\ int \\ FRAC (DX) (X) \u003d L N \\ LVERT X \\ RVERT + C
f (x) \u003d E ^ x	F (x) \u003d e ^ x + c	\\ int e (^ x) dx \u003d e ^ x + c
f (x) \u003d a ^ x	F (x) \u003d \\ frac (a ^ x) (L Na) + C	\\ int a (^ x) dx \u003d \\ frac (a ^ x) (L Na) + C
f (x) \u003d \\ sin x	F (x) \u003d - \\ cos x + c	\\ int \\ sin x dx \u003d - \\ cos x + c
f (x) \u003d \\ cos x	F (x) \u003d \\ sin x + c	\\ int \\ cos x dx \u003d \\ sin x + c
f (x) \u003d \\ FRAC (1) (\\ sin (^ 2) x)	F (x) \u003d - \\ ctg x + c	\\ int \\ FRAC (DX) (\\ sin (^ 2) x) \u003d - \\ CTG X + C
f (x) \u003d \\ FRAC (1) (\\ cos (^ 2) x)	F (x) \u003d \\ tg x + c	\\ int \\ FRAC (DX) (\\ sin (^ 2) x) \u003d \\ TG x + C
f (x) \u003d \\ sqrt (x)	F (x) \u003d \\ FRAC (2x \\ SQRT (X)) (3) + C
f (x) \u003d \\ FRAC (1) (\\ SQRT (X))	F (x) \u003d 2 \\ sqrt (x) + c
f (x) \u003d \\ FRAC (1) (\\ sqrt (1-x ^ 2))	F (x) \u003d \\ arcsin x + c	\\ int \\ FRAC (DX) (\\ SQRT (1-X ^ 2)) \u003d \\ ARCSIN X + C
f (x) \u003d \\ FRAC (1) (\\ sqrt (1 + x ^ 2))	F (x) \u003d \\ arctg x + c	\\ int \\ FRAC (DX) (\\ SQRT (1 + x ^ 2)) \u003d \\ arctg x + c
f (x) \u003d \\ FRAC (1) (\\ sqrt (a ^ 2-x ^ 2))	F (x) \u003d \\ arcsin \\ FRAC (X) (A) + C	\\ int \\ FRAC (DX) (\\ sqrt (a ^ 2-x ^ 2)) \u003d \\ arcsin \\ FRAC (X) (A) + C
f (x) \u003d \\ FRAC (1) (\\ sqrt (a ^ 2 + x ^ 2))	F (x) \u003d \\ arctg \\ FRAC (X) (A) + C	\\ int \\ FRAC (DX) (\\ SQRT (A ^ 2 + x ^ 2)) \u003d \\ FRAC (1) (a) \\ arctg \\ FRAC (X) (A) + C
f (x) \u003d \\ FRAC (1) (1 + x ^ 2)	F (x) \u003d \\ arctg + c	\\ int \\ FRAC (DX) (1 + x ^ 2) \u003d \\ arctg + c
f (x) \u003d \\ FRAC (1) (\\ sqrt (x ^ 2-a ^ 2)) (a \\ not \u003d 0)	F (x) \u003d \\ FRAC (1) (2a) L n \\ LVERT \\ FRAC (X-A) (X + A) \\ RVERT + C	\\ int \\ FRAC (DX) (\\ SQRT (X ^ 2-A ^ 2)) \u003d \\ FRAC (1) (2a) L N \\ LVERT \\ FRAC (X-A) (X + A) \\ RVERT + C
f (x) \u003d \\ tg x	F (x) \u003d - L n \\ LVERT \\ COS X \\ RVERT + C	\\ int \\ tg x dx \u003d - L n \\ LVERT \\ COS X \\ RVERT + C
f (x) \u003d \\ ctg x	F (x) \u003d L n \\ LVERT \\ SIN X \\ RVERT + C	\\ int \\ CTG X DX \u003d L N \\ LVERT \\ SIN X \\ RVERT + C
f (x) \u003d \\ FRAC (1) (\\ sin x)	F (x) \u003d L n \\ LVERT \\ TG \\ FRAC (X) (2) \\ RVERT + C	\\ int \\ FRAC (DX) (\\ Sin x) \u003d L n \\ LVERT \\ TG \\ FRAC (X) (2) \\ RVERT + C
f (x) \u003d \\ FRAC (1) (\\ COS X)	F (x) \u003d L n \\ LVERT \\ TG (\\ FRAC (X) (2) + \\ FRAC (\\ PI) (4)) \\ RVERT + C	\\ int \\ FRAC (DX) (\\ COS X) \u003d L N \\ LVERT \\ TG (\\ FRAC (X) (2) + \\ FRAC (\\ PI) (4)) \\ RVERT + C

Formula Newton Labitsa

Let be f (x) This feature, F. Her arbitrary primitive.

\\ int_ (a) ^ (b) f (x) dx \u003d f (x) | _ (a) ^ (b)\u003d F (B) - F (a)

where F (X) - Pred-like for f (x)

That is, the integral function f (X) The interval is equal to the difference in the sights at points b. and a..

Square of curvilinear trapezium

Curvilinear trapezium called a figure limited by a non-negative and continuous schedule on a segment of the function f., Ox axis and straight x \u003d A. and x \u003d B..

The area of \u200b\u200bthe curvilinear trapezium is found according to Newton Labitsa formula:

S \u003d \\ int_ (a) ^ (b) f (x) dx