EntranceMethod of integration by parts - examples of solutions. Integration by parts
Let U(x) and V(x) be differentiable functions. Then d(U(x)V(x)) = U(x)dV(x) + V(x)dU(x) . Therefore U(x)dV(x) = d(U(x)V(x)) – V(x)dU(x) . Calculating the integral of both sides of the last equality, taking into account the fact that ∫ d(U(x)V(x))=U(x)V(x)+C, we obtain the relationCalled the integration by parts formula. It is understood in the sense that the set of antiderivatives on the left side coincides with the set of antiderivatives obtained from the right side.
Application of the method of integration by parts
Due to the peculiarities of finding certain quantities, the integration by parts formula is very often used in the following problems:- Mathematical expectation of a continuous random variable. The formula for finding the mathematical expectation and variance of a continuous random variable includes two factors: the polynomial function of x and the distribution density f(x).
- Fourier series expansion. When decomposing, it is necessary to determine the coefficients, which are found by integrating the product of the function f(x) and the trigonometric function cos(x) or sin(x).
Typical decompositions by parts
When using the integration by parts formula, you need to successfully choose U and dV so that the integral obtained on the right side of the formula is easier to find. Let's put U=e x , dV=xdx in the first example. Then dU=e x dx , and 
It is unlikely that the integral ∫ x 2 e x dx can be considered simpler than the original one.
Sometimes it is necessary to apply the integration by parts formula several times, for example, when calculating the integral ∫ x 2 sin(x)dx.
The integrals ∫ e ax cos(bx)dx and ∫ e ax sin(bx)dx are called cyclical and are calculated using the integration by parts formula twice.
Example No. 1. Calculate ∫ xe x dx .
Let's put U=x, dV=e x dx. Then dU=dx , V=e x . Therefore ∫ xe x dx=xe x -∫ e x dx=xe x -e x +C .
Example No. 2. Calculate ∫ xcos(x)dx .
We assume U=x, dV=cos(x)dx. Then dU=dx , V=sin(x) and ∫ xcos(x)dx=xsin(x) - ∫ sin(x)dx = xsin(x)+cos(x)+C
Example No. 3. ∫ (3x+4)cos(x)dx
Solution:
Answer: (3x+4)sin(x)+3cos(x)+C
Let's consider the functions $u=u(x)$ and $v=v(x)$, which have continuous derivatives. According to the properties of differentials, the following equality holds:
$d(u v)=u d v+v d u$
Integrating the left and right sides of the last equality, we get:
$\int d(u v)=\int(u d v+v d u) \Rightarrow u v=\int u d v+\int v d u$
We rewrite the resulting equality as:
$\int u d v=u v-\int v d u$
This formula is called integration by parts formula. With its help, the integral $\int u d v$ can be reduced to finding the integral $\int v d u$, which can be simpler.
Comment
In some cases, the integration formula by parts must be applied repeatedly.
It is advisable to apply the integration by parts formula to integrals of the following form:
1) $\int P_(n)(x) e^(k x) d x$ ; $\int P_(n)(x) \sin (k x) d x$ ; $\int P_(n)(x) \cos (k x) d x$
Here $P_(n)(x)$ is a polynomial of degree $n$, $k$ is some constant. In this case, the polynomial is taken as the function $u$, and the remaining factors are taken as $d v$. For integrals of this type, the integration by parts formula is applied $n$ times.
Examples of solving integrals using this method
Example
Exercise. Find the integral $\int(x+1) e^(2 x) d x$
Solution.
$=\frac((x+1) e^(2 x))(2)-\frac(1)(2) \int e^(2 x) d x=\frac((x+1) e^( 2 x))(2)-\frac(1)(2) \cdot \frac(1)(2) e^(2 x)+C=$
$=\frac((x+1) e^(2 x))(2)-\frac(e^(2 x))(4)+C$
Answer.$\int(x+1) e^(2 x) d x=\frac((x+1) e^(2 x))(2)-\frac(e^(2 x))(4)+C $
Example
Exercise. Find the integral $\int x^(2) \cos x d x$
Solution.
$=x^(2) \sin x-2\left(x \cdot(-\cos) x-\int(-\cos x) d x\right)=$
$=x^(2) \sin x+2 x \cos x-2 \int \cos x d x=$
$=x^(2) \sin x+2 x \cos x-2 \sin x+C=\left(x^(2)-1\right) \sin x+2 x \cos x+C$
Answer.$\int x^(2) \cos x d x=\left(x^(2)-1\right) \sin x+2 x \cos x+C$
2) $\int P_(n)(x) \arcsin x d x$ ; $\int P_(n)(x) \arccos x d x$ ; $\int P_(n)(x)\ln x d x$
Here we assume that $d v=P_(n)(x) d x$, and $u$ are the remaining factors.
Example
Exercise. Find the integral $\int \ln x d x$
Solution. In the original integral, we isolate the functions $u$ and $v$, then perform the integration by parts.
$=x \ln x-\int d x=x \ln x-x+C=x(\ln x-1)+C$
Answer.$\int \ln x d x=x(\ln x-1)+C$
Example
Exercise. Find the integral $\int \arcsin x d x$
Solution. In the original integral, we isolate the functions $u$ and $v$, then perform the integration by parts. To solve this integral, this operation must be repeated 2 times.
$=x \arcsin x-\int \frac(-t d t)(\sqrt(t^(2)))=x \arcsin x+\int \frac(t d t)(t)=x \arcsin x+\int d t= $
$=x \arcsin x+t+C=x \arcsin x+\sqrt(1-x^(2))+C$
Answer.$\int \arcsin x d x=x \arcsin x+\sqrt(1-x^(2))+C$
3) $\int e^(k x+b) \sin (c x+f) d x$ ; $\int e^(k x+b) \cos (c x+f) d x$
In this case, either an exponential or a trigonometric function is taken as $u$. The only condition is that when further applying the integration by parts formula, the same function is taken as the function $u$, that is, either an exponential function or a trigonometric function, respectively.
Example
Exercise. Find the integral $\int e^(2 x+1) \sin x d x$
Solution. In the original integral, we isolate the functions $u$ and $v$, then perform the integration by parts.
$=-e^(2 x+1) \cos x-\int(-\cos x) \cdot \frac(e^(2 x+1))(2) d x=$
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
The method of integration by parts is used when it is necessary to simplify the existing indefinite integral or reduce it to a tabular value. Most often it is used in the presence of exponential, logarithmic, direct and inverse trigonometric formulas and their combinations in the integrand.
The basic formula needed to use this method is:
∫ f (x) d x = ∫ u (x) d (v (x)) = u (x) v (x) - ∫ v (x) d (u (x))
It means that we first need to represent the expression under the integral as the product of the function u (x) and the differential of the function v (x). After this, we calculate the value of the function v (x) by some method (the direct integration method is most often used), and substitute the resulting expressions into the indicated formula, reducing the original integral to the difference u (x) v (x) - ∫ v (x) d(u(x)) . The resulting integral can also be taken using any integration method.
Let's consider a problem in which we need to find the set of antiderivatives of the logarithm function.
Example 1
Evaluate the indefinite integral ∫ ln (x) d x .
Solution
We use the method of integration by parts. To do this, we take ln (x) as a function of u (x) and the remainder of the integrand as d (v (x)). As a result, we obtain that ln (x) d x = u (x) d (v (x)), where u (x) = ln (x), d (v (x)) = d x.
The differential of the function u(x) is d(u(x)) - u"(x)d x = d x x, and the function v(x) can be written as v(x) = ∫ d(v(x)) = ∫ d x = x
Important: the constant C will be considered equal to 0 when calculating the function v (x).
Let’s substitute what we got into the integration by parts formula:
∫ ln (x) d x = u (x) v (x) - ∫ v (x) d (u (x)) = = ln (x) x - ∫ x d x x = ln (x) x - ∫ d x = ln (x) x - x + C 1 = = x (ln (x) - 1) + C
where C = - C 1
Answer:∫ ln (x) d x = x (ln (x) - 1) + C .
The most difficult thing in applying this method is choosing which part of the original expression under the integral to take as u (x) and which as d (v (x)).
Let's look at a few standard cases.
If our condition contains integrals of the form ∫ P n (x) · e a x d x , ∫ P n (x) · sin (a x) d x or ∫ P n (x) · cos (a x) d x , where a is a coefficient and P n (x) is a polynomial of degree n, then P n (x) must be taken as the function u (x).
Example 2
Find the set of antiderivatives of the function f (x) = (x + 1) · sin (2 x) .
Solution
We can take the indefinite integral ∫ (x + 1) · sin (2 x) d x by parts. We take x + 1 as u (x) and sin (2 x) d x as d (v (x)), that is, d (u (x)) = d (x + 1) = d x.
Using direct integration, we obtain:
v (x) = ∫ sin (2 x) d x = - 1 2 cos (2 x)
Substitute into the integration by parts formula:
∫ (x + 1) sin (2 x) d x = u (x) v (x) - ∫ v (x) d (u (x)) = = (x + 1) - 1 2 cos (2 x ) - ∫ - 1 2 cos (2 x) d x = = - 1 2 (x + 1) cos (2 x) + 1 2 ∫ cos (2 x) d (x) = = - 1 2 (x + 1) cos (2 x) + 1 4 sin (2 x) + C
Answer:∫ (x + 1) · sin (2 x) d x = - 1 2 (x + 1) · cos (2 x) + 1 4 sin (2 x) + C .
Example 3
Evaluate the indefinite integral ∫ (x 2 + 2 x) e x d x .
Solution
We take the second order polynomial x 2 + 2 x as u (x) and d (v (x)) - e x d x .
∫ x 2 + 2 x e x d x = u (x) = x 2 + 2 x , d (v (x)) = e x d x d (u (x)) = (2 x + 2) d x , v (x) = ∫ e x d x = e x = = u (x) v (x) - ∫ v (x) d (u (x)) = (x 2 + 2 x) e x - ∫ (2 x + 2) e x d x
To what we have obtained, we must again apply the method of integration by parts:
∫ (2 x + 2) e x d x = (x 2 + 2 x) e x - ∫ 2 x + 2 e x d x = = u (x) = (2 x + 2) , d (v (x)) = e x d x d (u ( x)) = 2 d x , v (x) = ∫ e x d x = e x = = (x 2 + 2 x) e x - (2 x + 2) e x - ∫ v (x) d (u (x)) = = ( x 2 + 2 x) e x - (2 x + 2) e x - ∫ 2 e x d x = = (x 2 + 2 x - 2 x - 2) e x + 2 ∫ e x d x = (x 2 - 2) e x + 2 e x + C = x 2 e x + C
Answer:∫ (x 2 + 2 x) e x d x = x 2 e x + C .
Example 4
Evaluate the integral ∫ x 3 cos 1 3 x d x .
Solution
According to the method of integration by parts, we take u (x) = x 3 and d (v (x)) = cos 1 3 x d x.
In this case, d (u (x)) = 3 x 2 d x and v (x) = ∫ cos 1 3 x d x = 3 sin 1 3 x .
Now let’s substitute the resulting expressions into the formula:
∫ x 3 cos 1 3 x d x = u (x) v (x) - ∫ v (x) d (u)) = = x 3 3 sin 1 3 x - ∫ 3 x 2 3 sin 1 3 x d x = = 3 x 3 sin 1 3 x - 9 ∫ x 2 sin 1 3 x d x
We have an indefinite integral, which again needs to be taken in parts:
∫ x 3 cos 1 3 x d x = 3 x 3 sin 1 3 x - 9 ∫ x 2 sin 1 3 x d x = = u (x) = x 2 , d (v (x)) = sin 1 3 x d x d (u (x )) = 2 x d x , v (x) = ∫ sin 1 3 x d x = - 3 cos 1 3 x = = 3 x 3 sin 1 3 x - 9 - 3 x 2 cos 1 3 x - ∫ - 3 cos 1 3 x · 2 x d x = = 3 x 3 sin 1 3 x + 27 x 2 · cos 1 3 x - 54 ∫ x cos 1 3 x d x
We perform partial integration again:
∫ x 3 cos 1 3 x d x = 3 x 3 sin 1 3 x + 27 x 2 · cos 1 3 x - 54 ∫ x cos 1 3 x d x = = u (x) = x , d (v (x)) = cos 1 3 x d x d (u (x)) = d x , v (x) = ∫ cos 1 3 x d x = 3 sin 1 3 x = = 3 x 3 sin 1 3 x + 27 x 2 cos 1 3 x - 54 3 x sin 1 3 x - ∫ 3 sin 1 3 x d x = = 3 x 3 - 162 x sin 1 3 x + 27 x 2 cos 1 3 x + 162 ∫ sin 1 3 x d x = = (3 x 3 - 162 x) sin 1 3 x + 27 x 2 cos 1 3 x - 486 cos 1 3 x + C = = (3 x 3 - 162 x) sin 1 3 x + (27 x 2 - 486) cos 1 3 x + C
Answer:∫ x 3 cos 1 3 x d x = (3 x 3 - 162 x) sin 1 3 x + (27 x 2 - 486) cos 1 3 x + C .
If our condition contains integrals of the form ∫ P n (x) · ln (a x) d x , ∫ P n (x) · a r c sin (a x) d x , ∫ P n (x) · a r c cos (a x) d x , ∫ P n (x) · a r c t g (a x) d x , ∫ P n (x) · a r c c t g (a x) d x
then we should take as u (x) the functions a r c t g (a x) , a r c c t g (x) , ln (a x) , a r c sin (a x) , a r cos (a x) .
Example 5
Calculate the set of antiderivatives of the function (x + 1) ln (2 x) .
Solution
We take ln (2 x) as u (x) and (x + 1) d x as d (v (x)). We get:
d (u (x)) = (ln (2 x)) " d x = 1 2 x (2 x) " d x = d x x v (x) = ∫ (x + 1) d x = x 2 2 + x
Let's substitute these expressions into the formula:
∫ (x + 1) ln (2 x) d x = u (x) v (x) - ∫ v (x) d (u (x)) = = x 2 2 + x ln 2 x - ∫ x 2 2 + x d x x = = x 2 2 + x ln (2 x) - ∫ x 2 + 1 d x = x 2 2 + x ln 2 x - 1 2 ∫ x d x - ∫ d x = = x 2 2 + x ln (2 x) - x 2 4 - x + C
Answer:∫ (x + 1) ln (2 x) d x = x 2 2 + x ln (2 x) - x 2 4 - x + C .
Example 6
Evaluate the indefinite integral ∫ x · a r c sin (2 x) d x .
Solution
We decide which part to take as u (x) and which part to take as d (v (x)). According to the rule given above, you need to take a r c sin (2 x) as the first function, and d (v (x)) = x d x. We get:
d (u (x)) = (a r c sin (2 x) " d x = 2 x " d x 1 - (2 x) 2 = 2 d x 1 - (2 x) 2 , v (x) = ∫ x d x = x 2 2
Substitute the values into the formula:
∫ x · a r c sin (2 x) d x = u (x) v (x) - ∫ v (x) d (u (x)) = = x 2 2 a r c sin (2 x) - ∫ x 2 2 - 2 d x 1 - (2 x) 2 = x 2 2 a r c sin (2 x) - ∫ x 2 d x 1 - 4 x 2
As a result, we came to the following equality:
∫ x · a r c sin (2 x) d x = x 2 2 a r c sin (2 x) - ∫ x 2 d x 1 - 4 x 2
Now let's calculate the resulting integral ∫ x 2 d x 1 - 4 x 2:
∫ x 2 d x 1 - 4 x 2 = ∫ x 2 d x 4 1 4 - x 2 = 1 2 ∫ x 2 d x 1 4 - x 2 = - 1 2 ∫ - x 2 d x 1 4 - x 2 = = - 1 2 ∫ 1 4 - x 2 - 1 4 1 4 - x 2 d x = - 1 2 1 4 - x 2 d x + 1 8 ∫ d x 1 4 - x 2 = = - 1 2 ∫ 1 4 - x 2 d x + 1 8 a r c sin (2 x)
Here you can apply the method of integration by parts and get:
∫ x 2 d x 1 - 4 x 2 = - 1 2 ∫ 1 4 - x 2 d x + 1 8 a r c sin (2 x) = = u (x) = 1 4 - x 2 , d (v (x)) = d x d (u (x)) = 1 4 - x 2 " d x 2 1 4 - x 2 = - x d x 1 4 - x 2 , v (x) = ∫ d x = x = = - 1 2 u (x) v ( x) - ∫ v (x) d (u (x)) + 1 8 a r c sin (2 x) = = - 1 2 x 1 4 - x 2 - ∫ - x 2 d x 1 4 - x 2 + 1 8 a r c sin (2 x) = = - 1 2 x 1 4 - x 2 - 1 2 ∫ x 2 d x 1 4 - x 2 + 1 8 a r c sin (2 x) = = - 1 2 x 1 4 - x 2 - ∫ x 2 d x 1 - 4 x 2 + 1 8 a r c sin (2 x)
Now our equality looks like this:
∫ x 2 d x 1 - 4 x 2 = - 1 2 x 1 4 - x 2 - ∫ x 2 d x 1 - 4 x 2 + 1 8 a r c sin (2 x)
We see that the integral on the right is similar to that on the left. We move it to another part and get:
2 ∫ x 2 d x 1 - 4 x 2 = - 1 2 x 1 4 - x 2 + 1 8 a r c sin (2 x) + C 1 ⇒ x 2 d x 1 - 4 x 2 = - 1 4 x 1 4 - x 2 + 1 16 a r c sin (2 x) + C 2 x 2 d x 1 - 4 x 2 = - 1 8 x 1 4 - x 2 + 1 16 a r c sin (2 x) + C 2
where C 2 = C 1 2
Let's return to the original variables:
∫ x · a r c sin (2 x) d x = x 2 2 a r c sin (2 x) - ∫ x 2 d x 1 - 4 x 2 = = x 2 2 a r c sin (2 x) - - 1 8 x 1 - 4 x 2 + 1 16 a r c sin (2 x) + C 2 = = 1 2 x 2 - 1 8 a r c sin (2 x) + 1 8 x 1 - 4 x 2 + C
where C = - C 2
Answer:∫ x · a r c sin (2 x) d x = 1 2 x 2 - 1 8 a r c sin (2 x) + 1 8 x 1 - 4 x 2 + C .
If our problem contains an integral of the form ∫ e a · x · sin (b x) d x or ∫ e a · x · cos (b x) d x , then any function can be chosen as u (x).
Example 7
Evaluate the indefinite integral ∫ e x · sin (2 x) d x .
Solution
∫ e x sin (2 x) d x = u (x) = sin (2 x) , d (v (x)) = e x d x d (u (x)) = 2 cos (2 x) d x , v (x) = ∫ e x d x = e x = = u (x) v (x) - ∫ v (x) d (u (x)) = sin (2 x) e x - ∫ e x 2 cos 2 x d x = = sin (2 x) e x - 2 ∫ e x cos (2 x) d x = u (x) = cos (2 x) , d (v (x)) = e x d x d (u (x)) = - 2 sin (2 x) d x , v (x) = ∫ e x d x = e x = = sin (2 x) e x - 2 cos (2 x) e x - ∫ (e x (- 2 sin (2 x) d x)) = = sin (2 x) e x = 2 cos (2 x ) e x - 4 ∫ e x sin (2 x) d x
As a result, we will get:
∫ e x sin (2 x) d x = sin (2 x) e x - 2 cos (2 x) e x - 4 ∫ e x sin (2 x) d x
We see the same integrals on the left and right, which means we can present similar terms:
5 ∫ e x sin (2 x) d x = sin (2 x) e x - 2 cos (2 x) e x ⇒ ∫ e x sin (2 x) d x = 1 5 sin (2 x) e x - 2 5 cos (2 x) e x + C
Answer: ∫ e x sin (2 x) d x = 1 5 sin (2 x) e x - 2 5 cos (2 x) e x + C
This method of solution is standard, and on the right you often get an integral that is identical to the original one.
We looked at the most typical problems in which you can accurately determine which part of the expression to take as d (v (x)) and which part as u (x). In other cases, you have to determine this yourself.
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What is integration by parts? To master this type of integration, let's first remember the derivative of a product:
$((\left(f\cdot g \right))^(\prime ))=(f)"\cdot g+f\cdot (g)"$
The question arises: what do integrals have to do with it? Now let's integrate both sides of this equation. So let's write it down:
$\int(((\left(f\cdot g \right))^(\prime ))\text(d)x=)\int((f)"\cdot g\,\text(d)x+\ int(f\cdot (g)"\,\text(d)x))$
But what is an antiderivative of a stroke? It's just the function itself, which is inside the stroke. So let's write it down:
$f\cdot g=\int((f)"\cdot g\,\text(d)x+\int(f\cdot (g)"\,\text(d)x))$
In this equation, I propose to express the term. We have:
$\int((f)"\cdot g\,\text(d)x=f\cdot g-\int(f\cdot (g)"\,\text(d)x))$
That's what it is integration by parts formula. Thus, we are essentially interchanging the derivative and the function. If initially we had an integral of a stroke multiplied by something, then we get an integral of a new something multiplied by a stroke. That's all the rule. At first glance, this formula may seem complicated and meaningless, but in fact, it can greatly simplify calculations. Let's see.
Examples of integral calculations
Problem 1. Calculate:
\[\int(\ln x\,\text(d)x)\]\[\]
Let's rewrite the expression by adding 1 before the logarithm:
\[\int(\ln x\,\text(d)x)=\int(1\cdot \ln x\,\text(d)x)\]
We have the right to do this because neither the number nor the function will change. Now let's compare this expression with what is written in our formula. The role of $(f)"$ is 1, so we write:
$\begin(align)& (f)"=1\Rightarrow f=x \\& g=\ln x\Rightarrow (g)"=\frac(1)(x) \\\end(align)$
All these functions are in the tables. Now that we have described all the elements that are included in our expression, we will rewrite this integral using the formula for integration by parts:
\[\begin(align)& \int(1\cdot \ln x\,\text(d)x)=x\ln x-\int(x\cdot \frac(1)(x)\text(d )x)=x\ln x-\int(\text(d)x)= \\& =x\ln x-x+C=x\left(\ln x-1 \right)+C \\\ end(align)\]
That's it, the integral has been found.
Problem 2. Calculate:
$\int(x((\text(e))^(-x))\,\text(d)x=\int(x\cdot ((e)^(-x))\,\text(d )x))$
If we take $x$ as the derivative, from which we now need to find the antiderivative, we will get $((x)^(2))$, and the final expression will contain $((x)^(2))( (\text(e))^(-x))$.
Obviously, the problem is not simplified, so we swap the factors under the integral sign:
$\int(x\cdot ((\text(e))^(-x))\,\text(d)x)=\int(((\text(e))^(-x))\cdot x\,\text(d)x)$
Now let’s introduce the notation:
$(f)"=((\text(e))^(-x))\Rightarrow f=\int(((\text(e))^(-x))\,\text(d)x) =-((\text(e))^(-x))$
Let's differentiate $((\text(e))^(-x))$:
$((\left(((\text(e))^(-x)) \right))^(\prime ))=((\text(e))^(-x))\cdot ((\ left(-x \right))^(\prime ))=-((\text(e))^(-x))$
In other words, the minus is added first and then both sides are integrated:
\[\begin(align)& ((\left(((\text(e))^(-x)) \right))^(\prime ))=-((\text(e))^(- x))\Rightarrow ((\text(e))^(-x))=-((\left(((\text(e))^(-x)) \right))^(\prime )) \\& \int(((\text(e))^(-x))\,\text(d)x)=-\int(((\left(((\text(e))^(- x)) \right))^(\prime ))\text(d)x)=-((\text(e))^(-x))+C \\\end(align)\]
Now let's look at the $g$ function:
$g=x\Rightarrow (g)"=1$
We calculate the integral:
$\begin(align)& \int(((\text(e))^(-x))\cdot x\,\text(d)x)=x\cdot \left(-((\text(e ))^(-x)) \right)-\int(\left(-((\text(e))^(-x)) \right)\cdot 1\cdot \text(d)x)= \ \& =-x((\text(e))^(-x))+\int(((\text(e))^(-x))\,\text(d)x)=-x( (\text(e))^(-x))-((\text(e))^(-x))+C=-((\text(e))^(-x))\left(x +1 \right)+C \\\end(align)$
So, we have performed the second integration by parts.
Problem 3. Calculate:
$\int(x\cos 3x\,\text(d)x)$
In this case, what should we take for $(f)"$ and what for $g$? If $x$ acts as a derivative, then during integration we will get $\frac(((x)^(2)))(2 )$, and the first factor will not disappear anywhere - it will be $\frac(((x)^(2)))(2)\cdot \cos 3x$. Therefore, let’s swap the factors again:
$\begin(align)& \int(x\cos 3x\,\text(d)x)=\int(\cos 3x\cdot x\,\text(d)x) \\& (f)"= \cos 3x\Rightarrow f=\int(\cos 3x\,\text(d)x)=\frac(\sin 3x)(3) \\& g=x\Rightarrow (g)"=1 \\\ end(align)$
We rewrite our original expression and expand it according to the integration formula by parts:
\[\begin(align)& \int(\cos 3x\cdot x\ \text(d)x)=\frac(\sin 3x)(3)\cdot x-\int(\frac(\sin 3x) (3)\text(d)x)= \\& =\frac(x\sin 3x)(3)-\frac(1)(3)\int(\sin 3x\,\text(d)x) =\frac(x\sin 3x)(3)+\frac(\cos 3x)(9)+C \\\end(align)\]
That's it, the third problem is solved.
In conclusion, let's take another look at integration by parts formula. How do we choose which factor will be the derivative and which will be the real function? There is only one criterion here: the element that we will differentiate must either give a “beautiful” expression, which will then be reduced, or disappear altogether during differentiation. This concludes the lesson.