Integration by parts for dummies. Complex integrals
Examples of integration by parts of a similar composition are given to 1st and 2nd year students. These tasks were asked during the test at LNU named after. I. Frank. So that the formulas in the problems and answers do not repeat, we will not describe the problems. According to the conditions of the tasks, you need to either “Find the integral” or “Calculate the integral”.
Example 8. We find the integral using the rule of integration by parts int(u*dv)=u*v-int(v*du).
The main thing here is to choose the right functions for the rule. (For yourself, remember that for dv, if possible, choose periodic functions or those that, when differentiated up to a factor, give themselves - an exponential). In this integral you need to add the sine under the differential 
Further integration is quite simple and we will not dwell on the details. 
Example 9. Again you need to apply the rule of integration by parts u*dv. Here we have the product of a periodic function and an exponential, so it’s up to you to choose what is best to include under the differential. You can use either an exponential or a cosine (in each option we get a recurrent formula). 
We reapply integration by parts
We arrived at a recurrent formula. If we write down the integral we were looking for and the result of the calculations, we get two similar terms
We group them and find the required integral 
Example 10. We have a ready-made entry for the integral under the u*dv rule. 
Find du and perform integration 
We reduce the second integral to a tabular formula and calculate it 
Example 11. Let us denote cos(ln(x))=u as a new variable and find du , then enter it under the differential
We reapply the rule of integration by parts to the integral
We arrived at the recurrent formula 
with which we calculate the unknown integral
Example 12. To find the integral, we select a complete square in the denominator. Next, reducing the denominator to the well-known integration formula, we obtain the arctangent 
Remember well the order of alternating multipliers. The unit is divided by the root of the free term appears in front of the arctangent, and this factor is also present in the arctangent in front of the variable.
If you cannot solve the integral yourself, then ask for help.
The method of integration by parts is used mainly when the integrand consists of the product of two factors of a certain type. The integration by parts formula looks like:
It makes it possible to reduce the calculation of a given integral 
to the calculation of the integral 
, which turns out to be simpler than this one.
Most of the integrals calculated by the method of integration by parts can be divided into three groups:
1. Integrals of the form 
,
,
, Where 
– polynomial, 
– a number not equal to zero
In this case, through 
denote a polynomial 
.
2. Integrals of the form 
,
,
,
,
, Where 
– polynomial.
In this case, through 
denote 
, and the rest of the integrand through 
:
3. Integrals of the form 
,
, Where 
– numbers.
In this case, through 
denote 
and apply the integration by parts formula twice, returning as a result to the original integral, after which the original integral is expressed from equality.
Comment: In some cases, to find a given integral, the integration by parts formula must be applied several times. Also, the method of integration by parts is combined with other methods.
Example 26.
Find integrals using the method by parts: a) 
; b) 
.
Solution.
b) 
3.1.4. Integration of Fractional-Rational Functions
Fractional rational function(rational fraction) is a function equal to the ratio of two polynomials: 
, Where 
– polynomial of degree 
,
– polynomial of degree
.
The rational fraction is called correct, if the degree of the polynomial in the numerator is less than the degree of the polynomial in the denominator, i.e. 
, otherwise (if 
) rational fraction is called wrong.
Any improper rational fraction can be represented as the sum of a polynomial 
and a proper rational fraction by dividing the numerator by the denominator according to the rule for dividing polynomials:
,
Where 
– whole part from division, 
– proper rational fraction, 
- remainder of the division.
Proper rational fractions of the form:
I. 
;
II. 
;
III. 
;
IV. 
,
Where 
,
,
,
,
,
,
– real numbers and 
(i.e. the square trinomial in the denominator of III and IV fractions has no roots - the discriminant is negative) are called simple rational fractions I, II, III and IV types.
Integrating simple fractions
Integrals of the simplest fractions of four types are calculated as follows.
I) 
.
II) , 
.
III) To integrate the simplest fraction of type III, select a complete square in the denominator and replace 
. After substitution, the integral is divided into two integrals. The first integral is calculated by isolating the derivative of the denominator in the numerator, which gives a tabular integral, and the second integral is converted to the form 
, because 
, which also gives the tabular integral.
;
IV) To integrate the simplest fraction of type IV, select a complete square in the denominator and replace 
. After substitution, the integral is divided into two integrals. The first integral is calculated by substitution 
, and the second using recurrence relations.
Example 27.
Find integrals of simple fractions:
A) 
; 
b) 
.
Solution.
; 
.
V)
A) 
;
Any proper rational fraction whose denominator can be factorized can be represented as a sum of simple fractions. Decomposition into the sum of simple fractions is carried out using the method of indefinite coefficients. It is as follows: 
corresponds to one fraction of the form 
– each factor of the denominator
corresponds to the amount 
;
fractions of the form 
corresponds to one fraction of the form 
corresponds to a fraction of the form
– each square factor of the denominator
fractions of the form
where are the undetermined coefficients.
To find indefinite coefficients, the right side in the form of a sum of simple fractions is brought to a common denominator and transformed. The result is a fraction with the same denominator as on the left side of the equation. Then the denominators are discarded and the numerators are equalized. The result is an identical equality in which the left side is a polynomial with known coefficients, and the right side is a polynomial with unknown coefficients.
There are two ways to determine unknown coefficients: the method of unknown coefficients and the method of partial values. 
Method of undetermined coefficients. 
Because polynomials are identically equal, then the coefficients at the same powers are equal
. Equating coefficients at the same degrees
in the polynomials of the left and right sides, we obtain a system of linear equations. When solving the system, we determine the uncertain coefficients. 
Method of private values. 
Because polynomials are identically equal, then, substituting 
to the left and right sides of any number, we obtain a true equality, linear with respect to the unknown coefficients. Substituting so many values
, how many unknown coefficients there are, we obtain a system of linear equations. Instead of
You can substitute any numbers into the left and right sides, but it is more convenient to substitute the roots of the denominators of fractions. After finding the values of the unknown coefficients, the original fraction is written as a sum of simple fractions in the integrand and the previously discussed integration is carried out over each simple fraction.
1. If the integrand is improper, then it is necessary to present it as the sum of a polynomial and a proper rational fraction (i.e., divide the numerator polynomial by the denominator polynomial with a remainder). If the integrand is correct, we immediately move on to the second point of the diagram.
2. Factor the denominator of a proper rational fraction, if possible.
3. Decompose a proper rational fraction into the sum of simple rational fractions using the method of indefinite coefficients.
4. Integrate the resulting sum of the polynomial and simple fractions.
Example 28.
Find integrals of rational fractions:
; 
; 
b) 
.
Solution.
; 
.
Because the integrand is an improper rational fraction, then we select the whole part, i.e. Let's imagine it as the sum of a polynomial and a proper rational fraction. Divide the polynomial in the numerator by the polynomial in the denominator using a corner.
The original integral will take the form: 
.
Let us decompose a proper rational fraction into a sum of simple fractions using the method of indefinite coefficients:
, we get:
Solving the system of linear equations, we obtain the values of the uncertain coefficients: A = 1; IN = 3.
Then the required expansion has the form: 
.
=
.
b) 
.
.
Let's discard the denominators and equate the left and right sides:
Equating coefficients at the same degrees 
, we get the system:
By solving a system of five linear equations, we find the undetermined coefficients:
.
Let's find the original integral, taking into account the resulting expansion:
.
V) 
.
Let us expand the integrand (proper rational fraction) into a sum of simple fractions using the method of indefinite coefficients. We look for decomposition in the form:
.
Reducing to a common denominator, we get:
Let's discard the denominators and equate the left and right sides:
To find uncertain coefficients, we apply the partial value method. Let's add 
partial values , at which the factors vanish, i.e., we substitute these values into the last expression and get three equations:
; 
;
; 
;
; 
.
Then the required expansion has the form:
Let's find the original integral, taking into account the resulting expansion:
By a definite integral from a continuous function f(x) on the final segment [ a, b] (where ) is the increment of some of its antiderivatives on this segment. (In general, understanding will be noticeably easier if you repeat the topic of the indefinite integral) In this case, the notation is used
As can be seen in the graphs below (the increment of the antiderivative function is indicated by ), a definite integral can be either a positive or a negative number(It is calculated as the difference between the value of the antiderivative in the upper limit and its value in the lower limit, i.e. as F(b) - F(a)).
Numbers a And b are called the lower and upper limits of integration, respectively, and the segment [ a, b] – segment of integration.
Thus, if F(x) – some antiderivative function for f(x), then, according to the definition,
(38)
Equality (38) is called Newton-Leibniz formula . Difference F(b) – F(a) is briefly written as follows:
Therefore, we will write the Newton-Leibniz formula like this:
(39)
Let us prove that the definite integral does not depend on which antiderivative of the integrand is taken when calculating it. Let F(x) and F( X) are arbitrary antiderivatives of the integrand. Since these are antiderivatives of the same function, they differ by a constant term: Ф( X) = F(x) + C. That's why
This establishes that on the segment [ a, b] increments of all antiderivatives of the function f(x) match up.
Thus, to calculate a definite integral, it is necessary to find any antiderivative of the integrand, i.e. First you need to find the indefinite integral. Constant WITH excluded from subsequent calculations. Then the Newton-Leibniz formula is applied: the value of the upper limit is substituted into the antiderivative function b , further - the value of the lower limit a and the difference is calculated F(b) - F(a) . The resulting number will be a definite integral..
At a = b by definition accepted
Example 1.
Solution. First, let's find the indefinite integral:
Applying the Newton-Leibniz formula to the antiderivative
(at WITH= 0), we get
However, when calculating a definite integral, it is better not to find the antiderivative separately, but to immediately write the integral in the form (39).
Example 2. Calculate definite integral
Solution. Using the formula
Properties of a definite integral
Theorem 2.The value of the definite integral does not depend on the designation of the integration variable, i.e.
(40)
Let F(x) – antiderivative for f(x). For f(t) the antiderivative is the same function F(t), in which the independent variable is only designated differently. Hence,
Based on formula (39), the last equality means the equality of the integrals
Theorem 3.The constant factor can be taken out of the sign of the definite integral, i.e.
(41)
Theorem 4.The definite integral of an algebraic sum of a finite number of functions is equal to the algebraic sum of definite integrals of these functions, i.e.
(42)
Theorem 5.If a segment of integration is divided into parts, then the definite integral over the entire segment is equal to the sum of definite integrals over its parts, i.e. If
(43)
Theorem 6.When rearranging the limits of integration, the absolute value of the definite integral does not change, but only its sign changes, i.e.
(44)
Theorem 7(mean value theorem). A definite integral is equal to the product of the length of the integration segment and the value of the integrand at some point inside it, i.e.
(45)
Theorem 8.If the upper limit of integration is greater than the lower one and the integrand is non-negative (positive), then the definite integral is also non-negative (positive), i.e. If
Theorem 9.If the upper limit of integration is greater than the lower one and the functions and are continuous, then the inequality
can be integrated term by term, i.e.
(46)
The properties of the definite integral make it possible to simplify the direct calculation of integrals.
Example 5. Calculate definite integral
Using Theorems 4 and 3, and when finding antiderivatives - table integrals (7) and (6), we obtain
Definite integral with variable upper limit
Let f(x) – continuous on the segment [ a, b] function, and F(x) is its antiderivative. Consider the definite integral
(47)
and through t the integration variable is designated so as not to confuse it with the upper bound. When it changes X the definite integral (47) also changes, i.e. it is a function of the upper limit of integration X, which we denote by F(X), i.e.
(48)
Let us prove that the function F(X) is an antiderivative for f(x) = f(t). Indeed, differentiating F(X), we get
because F(x) – antiderivative for f(x), A F(a) is a constant value.
Function F(X) – one of the infinite number of antiderivatives for f(x), namely the one that x = a goes to zero. This statement is obtained if in equality (48) we put x = a and use Theorem 1 of the previous paragraph.
Calculation of definite integrals by the method of integration by parts and the method of change of variable
where, by definition, F(x) – antiderivative for f(x). If we change the variable in the integrand
then, in accordance with formula (16), we can write
In this expression
antiderivative function for
In fact, its derivative, according to rule of differentiation of complex functions, is equal
Let α and β be the values of the variable t, for which the function
takes values accordingly a And b, i.e.
But, according to the Newton-Leibniz formula, the difference F(b) – F(a) There is
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 for 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.
For a definite integral the following properties are valid:
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.
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
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
