Complex integrals
This article concludes the topic of indefinite integrals, and includes integrals that I find quite complex. The lesson was created at the repeated requests of visitors who expressed their wish that more difficult examples be analyzed on the site.
It is assumed that the reader of this text is well prepared and knows how to apply basic integration techniques. Dummies and people who are not very confident in integrals should refer to the very first lesson - Indefinite integral. Examples of solutions, where you can master the topic almost from scratch. More experienced students can become familiar with techniques and methods of integration that have not yet been encountered in my articles.
What integrals will be considered?
First we will consider integrals with roots, for the solution of which we successively use variable replacement And integration by parts. That is, in one example two techniques are combined at once. And even more.
Then we will get acquainted with interesting and original method of reducing the integral to itself. Quite a few integrals are solved this way.
The third issue of the program will be integrals of complex fractions, which flew past the cash desk in previous articles.
Fourthly, additional integrals from trigonometric functions will be analyzed. In particular, there are methods that avoid time-consuming universal trigonometric substitution.
(2) In the integrand function, we divide the numerator by the denominator term by term.
(3) We use the linearity property Not definite 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 linear function, you have to use several methods at once. In a number of cases, it is possible to “get off easy,” that is, immediately after the replacement, a simple integral is obtained, which can be easily taken. The easiest of the tasks proposed above is Example 4, in which, after replacement, a relatively simple integral is obtained.
By reducing the integral to itself
Witty and nice 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: 
And at the end: 
What happened? As a result of our manipulations, the integral was reduced to itself!
Let's equate the beginning and the end: 
Move to left side with sign change: 
And we move the two to the right side. As a result: 
The constant, strictly speaking, should have been added earlier, but I added it at the end. I strongly recommend reading what the rigor is here:
Note:
More strictly, the final stage of the solution looks like this:
Thus:
The constant can be redesignated by . Why can it be redesignated? Because he still accepts it any values, and in this sense there is no difference between constants and.
As a result:
A similar trick with constant renotation is widely used in differential equations. And there I will be strict. And here I allow such freedom only in order not to confuse you with unnecessary things and to focus attention precisely on the integration method itself.
Example 6
Find the indefinite integral
Another typical integral for independent solution. Full solution and answer at the end of the lesson. There will be a difference with the answer in the previous example!
If under square root is a quadratic trinomial, then the solution in any case reduces to two analyzed examples.
For example, consider the integral 
. All you need to do is first select a complete square:
.
Next, a linear replacement is carried out, which does “without any consequences”:
, resulting in the integral . Something familiar, right?
Or this example, with a quadratic binomial:
Select a complete square:
And, after linear replacement, we obtain the integral, which is also solved using the algorithm already discussed.
Let's look at two more typical examples of how to reduce an integral to itself:
– integral of the exponential multiplied by sine;
– integral of the exponential multiplied by the cosine.
In the listed integrals by parts you will have to integrate twice:
Example 7
Find the indefinite integral
The integrand is the exponential multiplied by the sine.
We integrate by parts twice and reduce the integral to itself: 
As a result of double integration by parts, the integral was reduced to itself. We equate the beginning and end of the solution:
We move it to the left side with a change of sign and express our integral: 
Ready. At the same time, it is advisable to comb the right side, i.e. take the exponent out of brackets, and place the sine and cosine in brackets in a “beautiful” order.
Now let's go back to the beginning of the example, or more precisely, to integration by parts: 
We designated the exponent as. The question arises: is it the exponent that should always be denoted by ? Not 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 Fractions
We are slowly approaching the equator of the lesson and begin to consider integrals of fractions. Again, not all of them are super complex, it’s just that for one reason or another the examples were a little “off topic” in other articles.
Continuing the theme of roots
Example 9
Find the indefinite integral
In the denominator under the root there is a quadratic trinomial plus an “appendage” in the form of an “X” outside the root. An integral of this type can be solved using a standard substitution.
We decide: 
The replacement here is simple: 
Let's look at life after replacement: 
(1) After substitution, we reduce the terms under the root to a common denominator.
(2) We take it out from under the root.
(3) The numerator and denominator are reduced by . At the same time, under the root, I rearranged the terms in a convenient order. With some experience, steps (1), (2) can be skipped by performing the commented actions orally.
(4) The resulting integral, as you remember from the lesson Integrating Some Fractions, is being decided complete square extraction method. Select a complete square.
(5) By integration we obtain an ordinary “long” logarithm.
(6) We carry out the reverse replacement. If initially , then back: .
(7) The final action is aimed at straightening the result: under the root we again bring the terms to a common denominator and take them out from under the root.
Example 10
Find the indefinite integral 
This is an example for you to solve on your own. Here a constant is added to the lone “X”, and the replacement is almost the same: 
The only thing you need to do additionally is to express the “x” from the replacement being carried out: 
Full solution and answer at the end of the lesson.
Sometimes in such an integral there may be a quadratic binomial under the root, this does not change the method of solution, it will be even simpler. Feel the difference:
Example 11
Find the indefinite integral
Example 12
Find the indefinite integral
Brief solutions and answers at the end of the lesson. It should be noted that Example 11 is exactly binomial integral, the solution method of which was discussed in class Integrals of irrational functions.
Integral of an indecomposable polynomial of the 2nd degree to the power
(polynomial in denominator)
A more rare type of integral, but nevertheless encountered in practical examples.
Example 13
Find the indefinite integral
But let’s return to the example with lucky number 13 (honestly, I didn’t guess correctly). This integral is also one of those that can be quite frustrating if you don’t know how to solve.
The solution starts with an artificial transformation: 
I think everyone already understands how to divide the numerator by the denominator term by term.
The resulting integral is taken in parts: 
For an integral of the form ( – natural number) withdrawn recurrent reduction formula:
, Where 
– integral of a degree lower.
Let us verify the validity of this formula for the solved integral.
In this case: , , we use the formula: 
As you can see, the answers are the same.
Example 14
Find the indefinite integral
This is an example for you to solve on your own. The sample solution uses the above formula twice in succession.
If under the degree is indivisible square trinomial, then the solution is reduced to a binomial by isolating the perfect square, for example:
What if there is an additional polynomial in the numerator? In this case, the method of indefinite coefficients is used, and the integrand 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 functions
The adjective “complex” for most examples is again largely conditional. Let's start with tangents and cotangents in high powers. From the point of view of the solving methods used, tangent and cotangent are almost the same thing, so I will talk more about tangent, implying that the demonstrated method for solving the integral is valid for cotangent too.
In the above lesson we looked at universal trigonometric substitution to solve a certain type of integrals from trigonometric functions. The disadvantage of universal trigonometric substitution is that its use often results in cumbersome integrals with difficult calculations. And in some cases, universal trigonometric substitution can be avoided!
Let's consider another canonical example, the integral of one divided by sine:
Example 17
Find the indefinite integral
Here you can use universal trigonometric substitution and get the answer, but there is a more rational way. I will provide the complete solution with comments for each step:
(1) We use the trigonometric formula for the sine of a double angle.
(2) We carry out an artificial transformation: Divide in the denominator and multiply by .
(3) Using the well-known formula in the denominator, we transform the fraction into a tangent.
(4) We bring the function under the differential sign.
(5) Take the integral.
Pair simple examples for independent solution:
Example 18
Find the indefinite integral
Note: The very first step should be to use the reduction formula 
and carefully carry out actions similar to the previous example.
Example 19
Find the indefinite integral
Well, this is a very simple example.
Complete solutions and answers at the end of the lesson.
I think now no one will have problems with integrals: 
and so on.
What is the idea of the method? The idea is that, using transformations, trigonometric formulas organize only tangents and the derivative of tangent in the integrand. That is, we're talking about about replacement: 
. In Examples 17-19 we actually used this replacement, but the integrals were so simple that we got by with an equivalent action - subsuming the function under the differential sign.
Similar reasoning, as I already mentioned, can be carried out for the cotangent.
There is also a formal prerequisite for applying the above replacement:
The sum of the powers of cosine and sine is a negative integer Even number , For example:
for the integral – a negative integer EVEN number.
! Note : if the integrand contains ONLY a sine or ONLY a cosine, then the integral is also taken for a negative odd degree (the simplest cases are in Examples No. 17, 18).
Let's look at a couple of more meaningful tasks based on this rule:
Example 20
Find the indefinite integral
The sum of the powers of sine and cosine: 2 – 6 = –4 is a negative integer EVEN number, which means that the integral can be reduced to tangents and its derivative: 
(1) Let's transform the denominator.
(2) Using the well-known formula, we obtain .
(3) Let's transform the denominator.
(4) We use the formula 
.
(5) We bring the function under the differential sign.
(6) We carry out replacement. More experienced students may not carry out the replacement, but it is still better to replace the tangent with one letter - there is less risk of getting confused.
Example 21
Find the indefinite integral
This is an example for you to solve on your own.
Hang in there, the championship rounds are about to begin =)
Often the integrand contains a “hodgepodge”:
Example 22
Find the indefinite integral 
This integral initially contains a tangent, which immediately leads to an already familiar thought:
I will leave the artificial transformation at the very beginning and the remaining steps without comment, since everything has already been discussed above.
Pair creative examples for independent 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
A function F(x) differentiable in a given interval X is called antiderivative of the function f(x), or the integral of f(x), if for every x ∈X the following equality holds:
F " (x) = f(x). (8.1)
Finding all antiderivatives for a given function is called its integration. Indefinite integral function f(x) on a given interval X is the set of all antiderivative functions for the function f(x); designation -
If F(x) is some antiderivative of the function f(x), then ∫ f(x)dx = F(x) + C, (8.2)
where C is an arbitrary constant.
Table of integrals
Directly from the definition we obtain the main properties of the indefinite integral and a list of tabular integrals:
1) d∫f(x)dx=f(x)
2)∫df(x)=f(x)+C
3) ∫af(x)dx=a∫f(x)dx (a=const)
4) ∫(f(x)+g(x))dx = ∫f(x)dx+∫g(x)dx
List of tabular integrals
1. ∫x m dx = x m+1 /(m + 1) +C; (m ≠ -1)
3.∫a x dx = a x /ln a + C (a>0, a ≠1)
4.∫e x dx = e x + C
5.∫sin x dx = cosx + C
6.∫cos x dx = - sin x + C
7. = arctan x + C
8. = arcsin x + C
10. = - ctg x + C
Variable replacement
To integrate many functions, use the variable replacement method or substitutions, allowing you to reduce integrals to tabular form.
If the function f(z) is continuous on [α,β], the function z =g(x) has a continuous derivative and α ≤ g(x) ≤ β, then
∫ f(g(x)) g " (x) dx = ∫f(z)dz, (8.3)
Moreover, after integration on the right side, the substitution z=g(x) should be made.
To prove it, it is enough to write the original integral in the form:
∫ f(g(x)) g " (x) dx = ∫ f(g(x)) dg(x).
For example:
Method of integration by parts
Let u = f(x) and v = g(x) be functions that have continuous . Then, according to the work,
d(uv))= udv + vdu or udv = d(uv) - vdu.
For the expression d(uv), the antiderivative will obviously be uv, so the formula holds:
∫ udv = uv - ∫ vdu (8.4.)
This formula expresses the rule integration by parts. It leads the integration of the expression udv=uv"dx to the integration of the expression vdu=vu"dx.
Let, for example, you want to find ∫xcosx dx. Let us put u = x, dv = cosxdx, so du=dx, v=sinx. Then
∫xcosxdx = ∫x d(sin x) = x sin x - ∫sin x dx = x sin x + cosx + C.
The rule of integration by parts has a more limited scope than substitution of variables. But there are whole classes of integrals, for example,
∫x k ln m xdx, ∫x k sinbxdx, ∫ x k cosbxdx, ∫x k e ax and others, which are calculated precisely using integration by parts.
Definite integral
The concept of a definite integral is introduced as follows. Let a function f(x) be defined on an interval. Let us divide the segment [a,b] into n parts by points a= x 0< x 1 <...< x n = b. Из каждого интервала (x i-1 ,
x i) возьмем произвольную точку ξ i и составим сумму f(ξ i)
Δx i где
Δ x i =x i - x i-1. A sum of the form f(ξ i)Δ x i is called integral sum, and its limit at λ = maxΔx i → 0, if it exists and is finite, is called definite integral functions f(x) of a before b and is designated:
F(ξ i)Δx i (8.5).
The function f(x) in this case is called integrable on the interval, numbers a and b are called lower and upper limits of the integral.
The following properties are true for a definite integral:
4), (k = const, k∈R);
5)
6)
7) f(ξ)(b-a) (ξ∈).
The last property is called mean value theorem.
Let f(x) be continuous on . Then on this segment there is an indefinite integral
∫f(x)dx = F(x) + C
and takes place Newton-Leibniz formula, connecting the definite integral with the indefinite integral:
F(b) - F(a). (8.6)
Geometric interpretation: the definite integral is the area of a curvilinear trapezoid bounded from above by the curve y=f(x), straight lines x = a and x = b and a segment of the axis Ox.
Improper integrals
Integrals with infinite limits and integrals of discontinuous (unbounded) functions are called not your own. Improper integrals of the first kind - These are integrals over an infinite interval, defined as follows:
(8.7)
If this limit exists and is finite, then it is called convergent improper integral of f(x) on the interval [a,+ ∞), and the function f(x) is called integrable over an infinite interval[a,+ ∞). Otherwise, the integral is said to be does not exist or diverges.
Improper integrals on the intervals (-∞,b] and (-∞, + ∞) are defined similarly:
Let us define the concept of an integral of an unbounded function. If f(x) is continuous for all values x segment , except for the point c, at which f(x) has an infinite discontinuity, then improper integral of the second kind of f(x) ranging from a to b the amount is called:
if these limits exist and are finite. Designation:
Examples of integral calculations
Example 3.30. Calculate ∫dx/(x+2).
Solution. Let us denote t = x+2, then dx = dt, ∫dx/(x+2) = ∫dt/t = ln|t| + C = ln|x+2| +C.
Example 3.31. Find ∫ tgxdx.
Solution.∫ tgxdx = ∫sinx/cosxdx = - ∫dcosx/cosx. Let t=cosx, then ∫ tgxdx = -∫ dt/t = - ln|t| + C = -ln|cosx|+C.
Example3.32 . Find ∫dx/sinxSolution.
Example3.33. Find .
Solution. = .
Example3.34 . Find ∫arctgxdx.
Solution. Let's integrate by parts. Let us denote u=arctgx, dv=dx. Then du = dx/(x 2 +1), v=x, whence ∫arctgxdx = xarctgx - ∫ xdx/(x 2 +1) = xarctgx + 1/2 ln(x 2 +1) +C; because
∫xdx/(x 2 +1) = 1/2 ∫d(x 2 +1)/(x 2 +1) = 1/2 ln(x 2 +1) +C.
Example3.35 . Calculate ∫lnxdx.
Solution. Applying the integration by parts formula, we obtain:
u=lnx, dv=dx, du=1/x dx, v=x. Then ∫lnxdx = xlnx - ∫x 1/x dx =
= xlnx - ∫dx + C= xlnx - x + C.
Example3.36 . Calculate ∫e x sinxdx.
Solution. Let us denote u = e x, dv = sinxdx, then du = e x dx, v =∫ sinxdx= - cosx → ∫ e x sinxdx = - e x cosx + ∫ e x cosxdx. We also integrate the integral ∫e x cosxdx by parts: u = e x , dv = cosxdx, du=e x dx, v=sinx. We have:
∫ e x cosxdx = e x sinx - ∫ e x sinxdx. We obtained the relation ∫e x sinxdx = - e x cosx + e x sinx - ∫ e x sinxdx, from which 2∫e x sinx dx = - e x cosx + e x sinx + C.
Example 3.37. Calculate J = ∫cos(lnx)dx/x.
Solution. Since dx/x = dlnx, then J= ∫cos(lnx)d(lnx). Replacing lnx through t, we arrive at the table integral J = ∫ costdt = sint + C = sin(lnx) + C.
Example 3.38 . Calculate J = .
Solution. Considering that = d(lnx), we substitute lnx = t. Then J = 
.
Example 3.39
. Calculate the integral J = 
.
Solution. We have: 
. Therefore =
=
=. entered like this: sqrt(tan(x/2)).
And if in the result window you click on Show steps in the upper right corner, you will get a detailed solution.
The integration by parts formula looks like:
.
The method of integration by parts consists of applying this formula. In practical application, it is worth noting that u and v are functions of the integration variable. Let the integration variable be designated as x (the symbol after the differential sign d at the end of the integral notation). Then u and v are functions of x: u(x) and v(x) .
Then
, .
And the formula for integration by parts takes the form:
.
That is, the integrand function must consist of the product of two functions:
,
one of which we denote as u: g(x) = u, and for the other the integral must be calculated (more precisely, the antiderivative must be found):
, then dv = f(x) dx .
In some cases f(x) = 1
. That is, in the integral
,
we can put g(x) = u, x = v .
Summary
So, in this method, the integration by parts formula should be remembered and applied in two forms:
;
.
Integrals calculated by integration by parts
Integrals containing logarithms and inverse trigonometric (hyperbolic) functions
Integrals containing logarithms and inverse trigonometric or hyperbolic functions are often integrated by parts. In this case, the part that contains the logarithm or inverse trigonometric (hyperbolic) functions is denoted by u, the remaining part by dv.
Here are examples of such integrals, which are calculated by the method of integration by parts:
, , , , , , .
Integrals containing the product of a polynomial and sin x, cos x or e x
Using the integration by parts formula, integrals of the form are found:
, , ,
where P(x) is a polynomial in x. When integrating, the polynomial P(x) is denoted by u, and e ax dx, cos ax dx or sin ax dx- via dv.
Here are examples of such integrals:
, , .
Examples of calculating integrals using the method of integration by parts
Examples of integrals containing logarithms and inverse trigonometric functions
Example
Calculate the integral:
Detailed solution
Here the integrand contains a logarithm. Making substitutions
u = ln x,
dv = x 2 dx.
Then
,
.
We calculate the remaining integral:
.
Then
.
At the end of the calculations, it is necessary to add the constant C, since the indefinite integral is the set of all antiderivatives. It could also be added in intermediate calculations, but this would only clutter up the calculations.
Shorter solution
You can present the solution in a shorter version. To do this, you do not need to make substitutions with u and v, but you can group the factors and apply the integration by parts formula in the second form.
.Answer
Examples of integrals containing the product of a polynomial and sin x, cos x or ex
Example
Calculate the integral:
.
Solution
Let us introduce the exponent under the differential sign:
e - x dx = - e - x d(-x) = - d(e - x).
Let's integrate by parts.
.
We also use the method of integration by parts.
.
.
.
Finally we have.
In this topic we will talk in detail about the calculation of indefinite integrals using the so-called “integration by parts formula”. We will need a table of indefinite integrals and a table of derivatives. In the first part, standard examples will be analyzed, which are mostly found in standard calculations and tests. More complex examples are discussed in the second part.
The problem statement in the standard case is as follows. Let's say that under the integral we have two functions of different nature: polynomial and trigonometric function, polynomial and logarithm, polynomial and inverse trigonometric function and so on. In this situation, it is advantageous to separate one function from another. Roughly speaking, it makes sense to break the integrand into parts - and deal with each part separately. Hence the name: “integration by parts.” The application of this method is based on the following theorem:
Let the functions $u(x)$ and $v(x)$ be differentiable on some interval, and on this interval there exists an integral $\int v \; du$. Then on the same interval there also exists the integral $\int u \; dv$, and the following equality is true:
\begin(equation) \int u \; dv=u\cdot v-\int v\; du \end(equation)
Formula (1) is called the “integration by parts formula.” Sometimes, when applying the above theorem, they talk about using the “method of integration by parts.” The essence of this method will be important to us, which we will consider using examples. There are several standard cases in which formula (1) clearly applies. It is these cases that will become the topic of this page. Let $P_n(x)$ be a polynomial of nth degree. Let's introduce two rules:
Rule #1
For integrals of the form $\int P_n(x) \ln x \;dx$, $\int P_n(x) \arcsin x \;dx$, $\int P_n(x) \arccos x \;dx$, $\ int P_n(x)\arctg x \;dx$, $\int P_n(x) \arcctg x \;dx$ we take $dv=P_n(x)dx$.
Rule #2
For integrals of the form $\int P_n(x) a^x \;dx$ ($a$ is some positive number), $\int P_n(x) \sin x \;dx$, $\int P_n(x) \ cos x \;dx$, $\int P_n(x)ch x \;dx$, $\int P_n(x) sh x \;dx$ we take $u=P_n(x)$.
Let me immediately note that the above entries should not be taken literally. For example, in integrals of the form $\int P_n(x) \ln x \;dx$ there will not necessarily be exactly $\ln x$. Both $\ln 5x$ and $\ln (10x^2+14x-5)$ can be located there. Those. the notation $\ln x$ should be taken as a kind of generalization.
One more thing. It happens that the integration by parts formula has to be applied several times. Let's talk about this in more detail in examples No. 4 and No. 5. Now let's move on directly to solving typical problems. Solving problems whose level is slightly higher than standard is discussed in the second part.
Example No. 1
Find $\int (3x+4) \cos (2x-1) \; dx$.
Below the integral is the polynomial $3x+4$ and the trigonometric function $\cos (2x-1)$. This is a classic case for applying the formula, so let’s take the given integral by parts. The formula requires that the integral $\int (3x+4) \cos (2x-1)\; dx$ was represented in the form $\int u\; dv$. We need to choose expressions for $u$ and for $dv$. We can take $3x+4$ as $u$, then $dv=\cos (2x-1)dx$. We can take $u=\cos (2x-1)$, then $dv=(3x+4)dx$. To make the right choice, let's turn to. Given integral $\int (3x+4) \cos (2x-1)\; dx$ falls under the form $\int P_n(x) \cos x \;dx$ (the polynomial $P_n(x)$ in our integral has the form $3x+4$). According to, you need to choose $u=P_n(x)$, i.e. in our case $u=3x+4$. Since $u=3x+4$, then $dv=\cos(2x-1)dx$.
However, simply choosing $u$ and $dv$ is not enough. We will also need the values of $du$ and $v$. Since $u=3x+4$, then:
$$ du=d(3x+4)=(3x+4)"dx=3dx.$$
Now let's look at the function $v$. Since $dv=\cos(2x-1)dx$, then according to the definition of the indefinite integral we have: $ v=\int \cos(2x-1)\; dx$. To find the required integral, we apply the following to the differential sign:
$$ v=\int \cos(2x-1)\; dx=\frac(1)(2)\cdot \int \cos(2x-1)d(2x-1)=\frac(1)(2)\cdot \sin(2x-1)+C=\frac (\sin(2x-1))(2)+C. $$
However, we do not need the entire infinite set of functions $v$, which is described by the formula $\frac(\sin(2x-1))(2)+C$. We need some one function from this set. To get the required function, you need to substitute some number instead of $C$. The easiest way, of course, is to substitute $C=0$, thereby obtaining $v=\frac(\sin(2x-1))(2)$.
So, let's put all of the above together. We have: $u=3x+4$, $du=3dx$, $dv=\cos(2x-1)dx$, $v=\frac(\sin(2x-1))(2)$. Substituting all this into the right side of the formula we have:
$$ \int (3x+4) \cos (2x-1) \; dx=(3x+4)\cdot\frac(\sin(2x-1))(2)-\int \frac(\sin(2x-1))(2)\cdot 3dx. $$
All that remains, in fact, is to find $\int\frac(\sin(2x-1))(2)\cdot 3dx$. Taking the constant (i.e. $\frac(3)(2)$) outside the integral sign and applying the method of introducing it under the differential sign, we obtain:
$$ (3x+4)\cdot \frac(\sin(2x-1))(2)-\int \frac(\sin(2x-1))(2)\cdot 3dx= \frac((3x+ 4)\cdot\sin(2x-1))(2)-\frac(3)(2)\int \sin(2x-1) \;dx= \\ =\frac((3x+4)\cdot \sin(2x-1))(2)-\frac(3)(4)\int \sin(2x-1) \;d(2x-1)= \frac((3x+4)\cdot\sin (2x-1))(2)-\frac(3)(4)\cdot (-\cos (2x-1))+C=\\ =\frac((3x+4)\cdot\sin(2x -1))(2)+\frac(3)(4)\cdot \cos (2x-1)+C. $$
So $\int (3x+4) \cos (2x-1) \; dx=\frac((3x+4)\cdot\sin(2x-1))(2)+\frac(3)(4)\cdot \cos (2x-1)+C$. In abbreviated form, the solution process is written as follows:
$$ \int (3x+4) \cos (2x-1) \; dx=\left | \begin(aligned) & u=3x+4; \; du=3xdx.\\ & dv=\cos(2x-1)dx; \; v=\frac(\sin(2x-1))(2). \end(aligned) \right |=\\ =(3x+4)\cdot\frac(\sin(2x-1))(2)-\int \frac(\sin(2x-1))(2) \cdot 3dx= \frac((3x+4)\cdot\sin(2x-1))(2)-\frac(3)(2)\int \sin(2x-1) \;dx=\\ = \frac((3x+4)\cdot\sin(2x-1))(2)-\frac(3)(4)\cdot (-\cos (2x-1))+C= \frac((3x +4)\cdot\sin(2x-1))(2)+\frac(3)(4)\cdot\cos (2x-1)+C. $$
The indefinite integral has been found by parts; all that remains is to write down the answer.
Answer: $\int (3x+4) \cos (2x-1) \; dx=\frac((3x+4)\cdot\sin(2x-1))(2)+\frac(3)(4)\cdot \cos (2x-1)+C$.
I believe there is a question here, so I’ll try to formulate it and give an answer.
Why did we take exactly $u=3x+4$ and $dv=\cos(2x-1)dx$? Yes, the integral has been solved. But maybe if we took $u=\cos (2x-1)$ and $dv=(3x+4)dx$ the integral would also be found!
No, if we take $u=\cos (2x-1)$ and $dv=(3x+4)dx$, then nothing good will come of it - the integral will not be simplified. Judge for yourself: if $u=\cos(2x-1)$, then $du=(\cos(2x-1))"dx=-2\sin(2x-1)dx$. Moreover, since $ dv=(3x+4)dx$, then:
$$ v=\int (3x+4) \; dx=\frac(3x^2)(2)+4x+C.$$
Taking $C=0$, we get $v=\frac(3x^2)(2)+4x$. Let us now substitute the found values of $u$, $du$, $v$ and $dv$ into the formula:
$$ \int (3x+4) \cos (2x-1) \; dx=\cos (2x-1)\cdot \left(\frac(3x^2)(2)+4x \right) - \int \left(\frac(3x^2)(2)+4x \right) \cdot (-2\sin(2x-1)dx)=\\ =\cos (2x-1)\cdot \left(\frac(3x^2)(2)+4x \right) +2\cdot\ int \left(\frac(3x^2)(2)+4x \right) \sin(2x-1)\;dx $$
And what have we come to? We came to the integral $\int \left(\frac(3x^2)(2)+4x \right) \sin(2x-1)\;dx$, which is clearly more complicated than the original integral $\int (3x+4 ) \cos (2x-1) \; dx$. This suggests that the choice of $u$ and $dv$ was made poorly. After applying the integration by parts formula, the resulting integral should be simpler than the original one. When finding the indefinite integral by parts, we must simplify it, not complicate it, so if after applying formula (1) the integral becomes more complicated, then the choice of $u$ and $dv$ was made incorrectly.
Example No. 2
Find $\int (3x^4+4x-1) \ln 5x \; dx$.
Below the integral is a polynomial (i.e. $3x^4+4x-1$) and $\ln 5x$. This case falls under , so let's take the integral by parts. The given integral has the same structure as the integral $\int P_n(x) \ln x\; dx$. Again, as in example No. 1, we need to select some part of the integrand $(3x^4+4x-1) \ln 5x \; dx$ as $u$, and some part as $dv$. According to , you need to choose $dv=P_n(x)dx$, i.e. in our case $dv=(3x^4+4x-1)dx$. If from the expression $(3x^4+4x-1) \ln 5x \; dx$ "remove" $dv=(3x^4+4x-1)dx$, then $\ln 5x$ will remain - this will be the function $u$. So, $dv=(3x^4+4x-1)dx$, $u=\ln 5x$. To apply the formula, we also need $du$ and $v$. Since $u=\ln 5x$, then:
$$ du=d(\ln 5x)=(\ln 5x)"dx=\frac(1)(5x)\cdot 5 dx=\frac(1)(x)dx. $$
Now let's find the function $v$. Since $dv=(3x^4+4x-1)dx$, then:
$$ v=\int(3x^4+4x-1)\; dx=\frac(3x^5)(5)+2x^2-x+C. $$
From the entire found infinite set of functions $\frac(3x^5)(5)+2x^2-x+C$ we need to choose one. And the easiest way to do this is by taking $C=0$, i.e. $v=\frac(3x^5)(5)+2x^2-x$. Everything is ready to apply the formula. Let us substitute the values $u=\ln 5x$, $du=\frac(1)(x)dx$, $v=\frac(3x^5)(5)+2x^2-x$ and $dv=(3x^4+4x-1)dx$ we will have:
$$ \int (3x^4+4x-1) \ln 5x \; dx=\left | \begin(aligned) & u=\ln 5x; \; du=\frac(1)(x)dx.\\ & dv=(3x^4+4x-1)dx; \; v=\frac(3x^5)(5)+2x^2-x. \end(aligned) \right |=\\ =\ln 5x \cdot \left (\frac(3x^5)(5)+2x^2-x \right)-\int \left (\frac(3x^ 5)(5)+2x^2-x \right)\cdot \frac(1)(x)dx=\\ =\left (\frac(3x^5)(5)+2x^2-x \right )\cdot\ln 5x -\int \left (\frac(3x^4)(5)+2x-1 \right)dx=\\ =\left (\frac(3x^5)(5)+2x^ 2-x \right)\cdot\ln 5x - \left (\frac(3x^5)(25)+x^2-x \right)+C=\\ =\left (\frac(3x^5) (5)+2x^2-x \right)\cdot\ln 5x - \frac(3x^5)(25)-x^2+x+C. $$
Answer: $\int (3x^4+4x-1) \ln 5x \; dx=\left (\frac(3x^5)(5)+2x^2-x \right)\cdot\ln 5x - \frac(3x^5)(25)-x^2+x+C$.
Example No. 3
Find $\int \arccos x\; dx$.
This integral has the structure $\int P_n(x) \arccos x \;dx$, falling under . I understand that a reasonable question will immediately arise: “where in the given integral $\int\arccos x \; dx$ did they hide the polynomial $P_n(x)$? There is no polynomial there, only arccosine and that’s it!” However, in fact, not only the arc cosine is located under the integral. I will present the integral $\int arccos x\; dx$ in this form: $\int 1\cdot\arccos x \; dx$. Agree that multiplying by one will not change the integrand. This unit is $P_n(x)$. Those. $dv=1\cdot dx=dx$. And as $u$ (according to ) we take $\arccos x$, i.e. $u=\arccos x$. We find the values $du$ and $v$, which are involved in the formula, in the same way as in the previous examples:
$$ du=(\arccos x)"dx=-\frac(1)(\sqrt(1-x^2))dx;\\ v=\int 1\; dx=x+C. $$
As in previous examples, assuming $C=0$ we get $v=x$. Substituting all the found parameters into the formula, we will have the following:
$$ \int \arccos x \; dx=\left | \begin(aligned) & u=\arccos x; \; du=-\frac(1)(\sqrt(1-x^2))dx.\\ & dv=dx; \; v=x. \end(aligned) \right |=\\ =\arccos x \cdot x-\int x\cdot \left(-\frac(1)(\sqrt(1-x^2))dx \right)= \ arccos x \cdot x+\int \frac(xdx)(\sqrt(1-x^2))=\\ =x\cdot\arccos x-\frac(1)(2)\cdot\int (1-x ^2)^(-\frac(1)(2))d(1-x^2)= =x\cdot\arccos x-\frac(1)(2)\cdot\frac((1-x^ 2)^(\frac(1)(2)))(\frac(1)(2))+C=\\ =x\cdot\arccos x-\sqrt(1-x^2)+C. $$
Answer: $\int\arccos x\; dx=x\cdot\arccos x-\sqrt(1-x^2)+C$.
Example No. 4
Find $\int (3x^2+x) e^(7x) \; dx$.
In this example, the integration by parts formula will have to be applied twice. Integral $\int (3x^2+x) e^(7x) \; dx$ has the structure $\int P_n(x) a^x \;dx$. In our case, $P_n(x)=3x^2+x$, $a=e$. According to we have: $u=3x^2+x$. Accordingly, $dv=e^(7x)dx$.
$$ du=(3x^2+x)"=(6x+1)dx;\\ v=\int e^(7x)\;dx=\frac(1)(7)\cdot \int e^( 7x)\;d(7x)=\frac(1)(7)\cdot e^(7x)+C=\frac(e^(7x))(7)+C. $$
Again, as in previous examples, assuming $C=0$, we have: $v=\frac(e^(7x))(7)$.
$$ \int (3x^2+x) e^(7x) \; dx=\left | \begin(aligned) & u=3x^2+x; \; du=(6x+1)dx.\\ & dv=e^(7x)dx; \; v=\frac(e^(7x))(7). \end(aligned) \right |=\\ =(3x^2+x)\cdot\frac(e^(7x))(7)-\int \frac(e^(7x))(7)\cdot (6x+1)dx= \frac((3x^2+x)e^(7x))(7)-\frac(1)(7)\cdot \int (6x+1) e^(7x)\ ;dx. $$
We have arrived at the integral $\int (6x+1) e^(7x)\;dx$, which again must be taken in parts. Taking $u=6x+1$ and $dv=e^(7x)dx$ we have:
$$ \frac((3x^2+x)e^(7x))(7)-\frac(1)(7)\cdot \int (6x+1) e^(7x)\;dx=\left | \begin(aligned) & u=6x+1; \; du=6dx.\\ & dv=e^(7x)dx; \; v=\frac(e^(7x))(7). \end(aligned) \right |=\\ =\frac((3x^2+x)e^(7x))(7)-\frac(1)(7)\cdot \left ((6x+1) \cdot\frac(e^(7x))(7) - \int\frac(e^(7x))(7)\cdot 6\;dx \right)=\\ =\frac((3x^2+ x)e^(7x))(7) -\frac((6x+1)e^(7x))(49) +\frac(6)(49)\cdot\int\ e^(7x)\; dx=\\ =\frac((3x^2+x)e^(7x))(7) -\frac((6x+1)e^(7x))(49) +\frac(6)(49 )\cdot\frac(e^(7x))(7)+C=\\ =\frac((3x^2+x)e^(7x))(7) -\frac((6x+1)e ^(7x))(49) +\frac(6\; e^(7x))(343)+C. $$
The resulting answer can be simplified by opening the brackets and rearranging the terms:
$$ \frac((3x^2+x)e^(7x))(7) -\frac((6x+1)e^(7x))(49) +\frac(6\; e^(7x ))(343)+C=e^(7x)\cdot \left(\frac(3x^2)(7)+\frac(x)(49)-\frac(1)(343) \right)+ C. $$
Answer: $\int (3x^2+x) e^(7x) \; dx=e^(7x)\cdot \left(\frac(3x^2)(7)+\frac(x)(49)-\frac(1)(343) \right)+C$.
Example No. 5
Find $\int (x^2+5)\sin(3x+1) \; dx$.
Here, as in the previous example, integration by parts is applied twice. Detailed explanations were given earlier, so I will give only the solution:
$$ \int (x^2+5)\sin(3x+1) \; dx=\left | \begin(aligned) & u=x^2+5; \; du=2xdx.\\ & dv=\sin(3x+1)dx; \; v=-\frac(\cos(3x+1))(3). \end(aligned) \right |=\\ =(x^2+5)\cdot \left(-\frac(\cos(3x+1))(3) \right)-\int\left(-\ frac(\cos(3x+1))(3) \right)\cdot 2xdx=\\ = -\frac((x^2+5)\cdot\cos(3x+1))(3) +\frac (2)(3)\int x\cos(3x+1)dx= \left | \begin(aligned) & u=x; \; du=dx.\\ & dv=\cos(3x+1)dx; \; v=\frac(\sin(3x+1))(3). \end(aligned) \right |=\\ =-\frac((x^2+5)\cdot\cos(3x+1))(3) +\frac(2)(3)\cdot \left( x\cdot\frac(\sin(3x+1))(3)-\int\frac(\sin(3x+1))(3)dx \right)=\\ =-\frac((x^2 +5)\cdot\cos(3x+1))(3) +\frac(2x\sin(3x+1))(9)-\frac(2)(9)\cdot\int\sin(3x+ 1)dx=\\ =-\frac((x^2+5)\cdot\cos(3x+1))(3) +\frac(2x\sin(3x+1))(9)-\frac (2)(9)\cdot \left(-\frac(\cos(3x+1))(3)\right)+C=\\ = -\frac((x^2+5)\cdot\cos (3x+1))(3) +\frac(2x\sin(3x+1))(9)+\frac(2\cos(3x+1))(27)+C=\\ =-\frac (x^2\cdot\cos(3x+1))(3)-\frac(5\cdot\cos(3x+1))(3) +\frac(2x\sin(3x+1))(9 )+\frac(2\cos(3x+1))(27)+C=\\ =-\frac(x^2\cdot\cos(3x+1))(3) +\frac(2x\sin (3x+1))(9)-\frac(43\cos(3x+1))(27)+C. $$
Answer: $\int (x^2+5)\sin(3x+1) \; dx=-\frac(x^2\cdot\cos(3x+1))(3) +\frac(2x\sin(3x+1))(9)-\frac(43\cos(3x+1) )(27)+C$.
The application of the method of integration by parts in somewhat non-standard cases that are not subject to rules No. 1 and No. 2 will be given in
Integration by parts- a method used to solve definite and indefinite integrals, when one of the integrands is easily integrable and the other is differentiable. A fairly common method for finding integrals, both indefinite and definite. The main sign when you need to use it is a certain function consisting of the product of two functions that cannot be integrated point-blank.
Formula
In order to successfully use this method, you need to understand and learn the formulas.
Formula for integration by parts in the indefinite integral:
$$ \int udv = uv - \int vdu $$
Formula for integration by parts in a definite integral:
$$ \int \limits_(a)^(b) udv = uv \bigg |_(a)^(b) - \int \limits_(a)^(b) vdu $$
Examples of solutions
Let us consider in practice examples of solutions to integration by parts, which are often proposed by teachers during tests. Please note that under the integral symbol there is a product of two functions. This is a sign that this method is suitable for the solution.
| Example 1 |
| Find the integral $ \int xe^xdx $ |
| Solution |
|
We see that the integrand consists of two functions, one of which, upon differentiation, instantly turns into unity, and the other is easily integrated. To solve the integral, we use the method of integration by parts. Let us assume that $ u = x \rightarrow du=dx $ and $ dv = e^x dx \rightarrow v=e^x $ We substitute the found values into the first integration formula and get: $$ \int xe^x dx = xe^x - \int e^x dx = xe^x - e^x + C $$ If you cannot solve your problem, then send it to us. We will provide detailed solution. You will be able to view the progress of the calculation and gain information. This will help you get your grade from your teacher in a timely manner! |
| Answer |
|
$$ \int xe^x dx = xe^x - e^x + C $$ |
| Example 4 |
| Calculate the integral $ \int \limits_0 ^1 (x+5) 3^x dx $ |
| Solution |
|
By analogy with the previous solved examples, we will figure out which function to integrate without problems, which to differentiate. Please note that if we differentiate $ (x+5) $, then this expression will be automatically converted to unity, which will be to our advantage. So we do this: $$ u=x+5 \rightarrow du=dx, dv=3^x dx \rightarrow v=\frac(3^x)(ln3) $$ Now all unknown functions have been found and can be put into the second formula for integration by parts for a definite integral. $$ \int \limits_0 ^1 (x+5) 3^x dx = (x+5) \frac(3^x)(\ln 3) \bigg |_0 ^1 - \int \limits_0 ^1 \frac (3^x dx)(\ln 3) = $$ $$ = \frac(18)(\ln 3) - \frac(5)(\ln 3) - \frac(3^x)(\ln^2 3)\bigg| _0 ^1 = \frac(13)(\ln 3) - \frac(3)(\ln^2 3)+\frac(1)(\ln^2 3) = \frac(13)(\ln 3 )-\frac(4)(\ln^2 3) $$ |
| Answer |
| $$ \int\limits_0 ^1 (x+5)3^x dx = \frac(13)(\ln 3)-\frac(4)(\ln^2 3) $$ |