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Formula for the integral of a complex exponential function. Complex integrals |
Principal integrals that every student should knowThe 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 FunctionIn 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) Integrals of exponential functions and hyperbolic functionsOf course, formula (8) (perhaps the most convenient for memorization) can be considered as a special case of formula (9). Formulas (10) and (11) for the integrals of the hyperbolic sine and hyperbolic cosine are easily derived from formula (8), but it is better to simply remember these relations. ∫ e x d x = e x + C (8) Basic integrals of trigonometric functionsA 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) Integrals that reduce to inverse trigonometric functionsFormula (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) More complex integralsIt 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) General rules of integration1) The integral of the sum of two functions is equal to the sum of the corresponding integrals: ∫ (f (x) + g (x)) d x = ∫ f (x) d x + ∫ g (x) d x (25) 2) The integral of the difference of two functions is equal to the difference of the corresponding integrals: ∫ (f (x) − g (x)) d x = ∫ f (x) d x − ∫ g (x) d x (26) 3) The constant can be taken out of the integral sign: ∫ C f (x) d x = C ∫ f (x) d x (27) It is easy to see that property (26) is simply a combination of properties (25) and (27). 4) Integral of complex function, if the inner function is linear: ∫ f (A x + B) d x = 1 A F (A x + B) + C (A ≠ 0) (28) Here F(x) is an antiderivative for the function f(x). Please note: this formula only works when the inner function is Ax + B. Important: there is no universal formula for the integral of the product of two functions, as well as for the integral of a fraction: ∫ f (x) g (x) d x = ? ∫ f (x) g (x) d x = ? (thirty) 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 integralExample 1. Find the integral: ∫ (3 x 2 + 2 sin x − 7 e x + 12) d xLet 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
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 Complex integralsThis article concludes the topic of indefinite integrals, and includes integrals that I find quite complex. The lesson was created at the repeated requests of visitors who expressed their wish that more difficult examples be analyzed on the site. It is assumed that the reader of this text is well prepared and knows how to apply basic integration techniques. Dummies and people who are not very confident in integrals should refer to the very first lesson - Indefinite integral. Examples of solutions, where you can master the topic almost from scratch. More experienced students can become familiar with techniques and methods of integration that have not yet been encountered in my articles. What integrals will be considered? First we will consider integrals with roots, for the solution of which we successively use variable replacement And integration by parts. That is, in one example two techniques are combined at once. And even more. Then we will get acquainted with interesting and original method of reducing the integral to itself. Quite a few integrals are solved this way. The third issue of the program will be integrals of complex fractions, which flew past the cash desk in previous articles. Fourthly, additional integrals from trigonometric functions will be analyzed. In particular, there are methods that avoid time-consuming universal trigonometric substitution. (2) In the integrand function, we divide the numerator by the denominator term by term. (3) We use the linearity property of the indefinite integral. In the last integral immediately put the function under the differential sign. (4) We take the remaining integrals. Note that in a logarithm you can use parentheses rather than a modulus, since . (5) We carry out a reverse replacement, expressing “te” from the direct replacement: Masochistic students can differentiate the answer and get the original integrand, as I just did. No, no, I did the check in the right sense =) As you can see, during the solution we had to use even more than two solution methods, so to deal with such integrals you need confident integration skills and quite a bit of experience. In practice, of course, the square root is more common; here are three examples for solving it yourself: Example 2 Find the indefinite integral Example 3 Find the indefinite integral Example 4 Find the indefinite integral These examples are of the same type, so the complete solution at the end of the article will only be for Example 2; Examples 3-4 have the same answers. Which replacement to use at the beginning of decisions, I think, is obvious. Why did I choose examples of the same type? Often found in their role. More often, perhaps, just something like But not always, when under the arctangent, sine, cosine, exponential and other functions there is a root of a linear function, you have to use several methods at once. In a number of cases, it is possible to “get off easy,” that is, immediately after the replacement, a simple integral is obtained, which can be easily taken. The easiest of the tasks proposed above is Example 4, in which, after replacement, a relatively simple integral is obtained. By reducing the integral to itselfA witty and beautiful method. Let's take a look at the classics of the genre: Example 5 Find the indefinite integral Under the root is a quadratic binomial, and when trying to integrate this example the kettle can suffer for hours. Such an integral is taken in parts and reduced to itself. In principle, it’s not difficult. If you know how. Let us denote the integral under consideration by a Latin letter and begin the solution: Let's integrate by parts: (1) Prepare the integrand function for term-by-term division. (2) We divide the integrand function term by term. It may not be clear to everyone, but I’ll describe it in more detail: (3) We use the linearity property of the indefinite integral. (4) Take the last integral (“long” logarithm). Now let's look at the very beginning of the solution: What happened? As a result of our manipulations, the integral was reduced to itself! Let's equate the beginning and the end: Move to the left side with a change of sign: And we move the two to the right side. As a result: The constant, strictly speaking, should have been added earlier, but I added it at the end. I strongly recommend reading what the rigor is here: Note:
More strictly The final stage the solution looks like this: A similar trick with constant renotation is widely used in differential equations. And there I will be strict. And here I allow such freedom only in order not to confuse you with unnecessary things and to focus attention precisely on the integration method itself. Example 6 Find the indefinite integral Another typical integral for independent solution. Full solution and answer at the end of the lesson. There will be a difference with the answer in the previous example! If under the square root there is a square trinomial, then the solution in any case comes down to two analyzed examples. For example, consider the integral Or this example, with a quadratic binomial: Let's look at two more typical examples to reduce the integral to itself: In the listed integrals by parts you will have to integrate twice: Example 7 Find the indefinite integral The integrand is the exponential multiplied by the sine. We integrate by parts twice and reduce the integral to itself:
We move it to the left side with a change of sign and express our integral: Ready. At the same time, it is advisable to comb the right side, i.e. take the exponent out of brackets, and place the sine and cosine in brackets in a “beautiful” order. Now let's go back to the beginning of the example, or more precisely, to integration by parts: We designated the exponent as. The question arises: is it the exponent that should always be denoted by ? Not necessary. In fact, in the considered integral fundamentally doesn't matter, what do we mean by , we could have gone the other way: Why is this possible? Because the exponential turns into itself (both during differentiation and integration), sine and cosine mutually turn into each other (again, both during differentiation and integration). That is, we can also denote a trigonometric function. But, in the example considered, this is less rational, since fractions will appear. If you wish, you can try to solve this example using the second method; the answers must match. Example 8 Find the indefinite integral This is an example for you to solve on your own. Before you decide, think about what is more advantageous in this case to designate as , an exponential or a trigonometric function? Full solution and answer at the end of the lesson. And, of course, do not forget that most of the answers in this lesson are quite easy to check by differentiation! The examples considered were not the most complex. In practice, integrals are more common where the constant is both in the exponent and in the argument of the trigonometric function, for example: . Many people will get confused in such an integral, and I often get confused myself. The fact is that there is a high probability of fractions appearing in the solution, and it is very easy to lose something through carelessness. In addition, there is a high probability of an error in the signs; note that the exponent has a minus sign, and this introduces additional difficulty. At the final stage, the result is often something like this: Even at the end of the solution, you should be extremely careful and correctly understand the fractions: Integrating Complex FractionsWe are slowly approaching the equator of the lesson and begin to consider integrals of fractions. Again, not all of them are super complex, it’s just that for one reason or another the examples were a little “off topic” in other articles. Continuing the theme of roots Example 9 Find the indefinite integral In the denominator under the root there is a quadratic trinomial plus an “appendage” in the form of an “X” outside the root. An integral of this type can be solved using a standard substitution. We decide: The replacement here is simple: Let's look at life after replacement: (1) After substitution, we reduce the terms under the root to a common denominator. Example 10 Find the indefinite integral This is an example for you to solve on your own. Here a constant is added to the lone “X”, and the replacement is almost the same: Full solution and answer at the end of the lesson. Sometimes in such an integral there may be a quadratic binomial under the root, this does not change the method of solution, it will be even simpler. Feel the difference: Example 11 Find the indefinite integral Example 12 Find the indefinite integral Brief solutions and answers at the end of the lesson. It should be noted that Example 11 is exactly binomial integral, the solution method of which was discussed in class Integrals of irrational functions. Integral of an indecomposable polynomial of the 2nd degree to the power(polynomial in denominator) A more rare type of integral, but nevertheless encountered in practical examples. Example 13 Find the indefinite integral But let’s return to the example with lucky number 13 (honestly, I didn’t guess correctly). This integral is also one of those that can be quite frustrating if you don’t know how to solve. The solution starts with an artificial transformation: I think everyone already understands how to divide the numerator by the denominator term by term. The resulting integral is taken in parts: For an integral of the form ( – natural number) we derive recurrent reduction formula: Let us verify the validity of this formula for the solved integral. As you can see, the answers are the same. Example 14 Find the indefinite integral This is an example for you to solve on your own. The sample solution uses the above formula twice in succession. If under the degree is indivisible square trinomial, then the solution is reduced to a binomial by isolating the perfect square, for example: What if there is an additional polynomial in the numerator? In this case, the method of indefinite coefficients is used, and the integrand function is expanded into a sum of fractions. But in my practice there is such an example never met, so I missed this case in the article Integrals of fractional-rational functions, I'll skip it now. If you still encounter such an integral, look at the textbook - everything is simple there. I don’t think it’s advisable to include material (even simple ones), the probability of encountering which tends to zero. Integrating complex trigonometric functionsThe adjective “complex” for most examples is again largely conditional. Let's start with tangents and cotangents in high powers. From the point of view of the solving methods used, tangent and cotangent are almost the same thing, so I will talk more about tangent, implying that the demonstrated method for solving the integral is valid for cotangent too. In the above lesson we looked at universal trigonometric substitution for solutions certain type integrals of trigonometric functions. The disadvantage of universal trigonometric substitution is that its use often results in cumbersome integrals with difficult calculations. And in some cases, universal trigonometric substitution can be avoided! Let's consider another canonical example, the integral of one divided by sine: Example 17 Find the indefinite integral Here you can use universal trigonometric substitution and get the answer, but there is a more rational way. I will provide the complete solution with comments for each step: (1) We use the trigonometric formula for the sine of a double angle. Pair simple examples for independent solution: Example 18 Find the indefinite integral Note: The very first step should be to use the reduction formula 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: What is the idea of the method? The idea is to use transformations and trigonometric formulas to organize only tangents and the tangent derivative into the integrand. That is, we are talking about replacing: Similar reasoning, as I already mentioned, can be carried out for the cotangent. There is also a formal prerequisite for applying the above replacement: The sum of the powers of cosine and sine is a negative integer EVEN number, For example: for the integral – a negative integer EVEN number. ! Note : if the integrand contains ONLY a sine or ONLY a cosine, then the integral is also taken for a negative odd degree (the simplest cases are in Examples No. 17, 18). Let's look at a couple of more meaningful tasks based on this rule: Example 20 Find the indefinite integral The sum of the powers of sine and cosine: 2 – 6 = –4 is a negative integer EVEN number, which means that the integral can be reduced to tangents and its derivative: (1) Let's transform the denominator. Example 21 Find the indefinite integral This is an example for you to solve on your own. Hang in there, the championship rounds are about to begin =) Often the integrand contains a “hodgepodge”: Example 22 Find the indefinite integral This integral initially contains a tangent, which immediately leads to an already familiar thought: I will leave the artificial transformation at the very beginning and the remaining steps without comment, since everything has already been discussed above. A couple of creative examples for your own solution: Example 23 Find the indefinite integral Example 24 Find the indefinite integral Yes, in them, of course, you can lower the powers of sine and cosine, and use a universal trigonometric substitution, but the solution will be much more efficient and shorter if it is carried out through tangents. Full solution and answers at the end of the lesson |
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