Integration - MT1205: Mathematical Analysis for Economists - Business Informatics. Integrals of irrational functions

Definition 1

The set of all antiderivatives of a given function $y=f(x)$, defined on a certain segment, is called the indefinite integral of a given function $y=f(x)$. The indefinite integral is denoted by the symbol $\int f(x)dx $.

Comment

Definition 2 can be written as follows:

\[\int f(x)dx =F(x)+C.\]

Not from everyone irrational function the integral can be expressed in terms of elementary functions. However, most of these integrals can be reduced using substitutions to integrals of rational functions, which can be expressed in terms of elementary functions.

$\int R\left(x,x^(m/n) ,...,x^(r/s) \right)dx $;

$\int R\left(x,\left(\frac(ax+b)(cx+d) \right)^(m/n) ,...,\left(\frac(ax+b)(cx +d) \right)^(r/s) \right)dx $;

$\int R\left(x,\sqrt(ax^(2) +bx+c) \right)dx $.

I

When finding an integral of the form $\int R\left(x,x^(m/n) ,...,x^(r/s) \right)dx $ it is necessary to perform the following substitution:

With this substitution, each fractional power of the variable $x$ is expressed through an integer power of the variable $t$. As a result, the integrand function is transformed into a rational function of the variable $t$.

Example 1

Perform integration:

\[\int \frac(x^(1/2) dx)(x^(3/4) +1) .\]

Solution:

$k=4$ is the common denominator of the fractions $\frac(1)(2) ,\, \, \frac(3)(4) $.

\ \[\begin(array)(l) (\int \frac(x^(1/2) dx)(x^(3/4) +1) =4\int \frac(t^(2) ) (t^(3) +1) \cdot t^(3) dt =4\int \frac(t^(5) )(t^(3) +1) dt =4\int \left(t^( 2) -\frac(t^(2) )(t^(3) +1) \right)dt =4\int t^(2) dt -4\int \frac(t^(2) )(t ^(3) +1) dt =\frac(4)(3) \cdot t^(3) -) \\ (-\frac(4)(3) \cdot \ln |t^(3) +1 |+C)\end(array)\]

\[\int \frac(x^(1/2) dx)(x^(3/4) +1) =\frac(4)(3) \cdot \left+C\]

II

When finding an integral of the form $\int R\left(x,\left(\frac(ax+b)(cx+d) \right)^(m/n) ,...,\left(\frac(ax+ b)(cx+d) \right)^(r/s) \right)dx $ it is necessary to perform the following substitution:

where $k$ is the common denominator of the fractions $\frac(m)(n) ,...,\frac(r)(s) $.

As a result of this substitution, the integrand function is transformed into a rational function of the variable $t$.

Example 2

Perform integration:

\[\int \frac(\sqrt(x+4) )(x) dx .\]

Solution:

Let's make the following substitution:

\ \[\int \frac(\sqrt(x+4) )(x) dx =\int \frac(t^(2) )(t^(2) -4) dt =2\int \left(1 +\frac(4)(t^(2) -4) \right)dt =2\int dt +8\int \frac(dt)(t^(2) -4) =2t+2\ln \left |\frac(t-2)(t+2) \right|+C\]

After making the reverse substitution, we get the final result:

\[\int \frac(\sqrt(x+4) )(x) dx =2\sqrt(x+4) +2\ln \left|\frac(\sqrt(x+4) -2)(\ sqrt(x+4) +2) \right|+C.\]

III

When finding an integral of the form $\int R\left(x,\sqrt(ax^(2) +bx+c) \right)dx $, the so-called Euler substitution is performed (one of three possible substitutions is used).

Euler's first substitution

For the case $a>

Taking the “+” sign in front of $\sqrt(a) $, we get

Example 3

Perform integration:

\[\int \frac(dx)(\sqrt(x^(2) +c) ) .\]

Solution:

Let's make the following substitution (case $a=1>0$):

\[\sqrt(x^(2) +c) =-x+t,\, \, x=\frac(t^(2) -c)(2t) ,\, \, dx=\frac(t ^(2) +c)(2t^(2) ) dt,\, \, \sqrt(x^(2) +c) =-\frac(t^(2) -c)(2t) +t= \frac(t^(2) +c)(2t) .\] \[\int \frac(dx)(\sqrt(x^(2) +c) ) =\int \frac(\frac(t^ (2) +c)(2t^(2) ) dt)(\frac(t^(2) +c)(2t) ) =\int \frac(dt)(t) =\ln |t|+C \]

After making the reverse substitution, we get the final result:

\[\int \frac(dx)(\sqrt(x^(2) +c) ) =\ln |\sqrt(x^(2) +c) +x|+C.\]

Euler's second substitution

For the case $c>0$ it is necessary to perform the following substitution:

Taking the “+” sign in front of $\sqrt(c) $, we get

Example 4

Perform integration:

\[\int \frac((1-\sqrt(1+x+x^(2) ))^(2) )(x^(2) \sqrt(1+x+x^(2) ) ) dx .\]

Solution:

Let's make the following substitution:

\[\sqrt(1+x+x^(2) ) =xt+1.\]

\ \[\sqrt(1+x+x^(2) ) =xt+1=\frac(t^(2) -t+1)(1-t^(2) ) \] \

$\int \frac((1-\sqrt(1+x+x^(2) ))^(2) )(x^(2) \sqrt(1+x+x^(2) ) ) dx = \int \frac((-2t^(2) +t)^(2) (1-t)^(2) (1-t^(2))(2t^(2) -2t+2))( (1-t^(2))^(2) (2t-1)^(2) (t^(2) -t+1)(1-t^(2))^(2) ) dt =\ int \frac(t^(2) )(1-t^(2) ) dt =-2t+\ln \left|\frac(1+t)(1-t) \right|+C$ Having made the reverse substitution, we get the final result:

\[\begin(array)(l) (\int \frac((1-\sqrt(1+x+x^(2) ))^(2) )(x^(2) \sqrt(1+x +x^(2) ) dx =-2\cdot \frac(\sqrt(1+x+x^(2) ) -1)(x) +\ln \left|\frac(x+\sqrt(1 +x+x^(2) ) -1)(x-\sqrt(1+x+x^(2) ) +1) \right|+C=-2\cdot \frac(\sqrt(1+x +x^(2) ) -1)(x) +) \\ (+\ln \left|2x+2\sqrt(1+x+x^(2) ) +1\right|+C) \end (array)\]

Euler's third substitution

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Under irrational understand an expression in which the independent variable %%x%% or the polynomial %%P_n(x)%% of degree %%n \in \mathbb(N)%% is included under the sign radical(from Latin radix- root), i.e. are being built in fractional power. By replacing a variable, some classes of integrands that are irrational with respect to %%x%% can be reduced to rational expressions with respect to a new variable.

The concept of a rational function of one variable can be extended to multiple arguments. If for each argument %%u, v, \dotsc, w%% when calculating the value of a function, only arithmetic operations and raising to an integer power are provided, then we speak of a rational function of these arguments, which is usually denoted %%R(u, v, \ dotsc, w)%%. The arguments of such a function can themselves be functions of the independent variable %%x%%, including radicals of the form %%\sqrt[n](x), n \in \mathbb(N)%%. For example, the rational function $$ R(u,v,w) = \frac(u + v^2)(w) $$ with %%u = x, v = \sqrt(x)%% and %%w = \sqrt(x^2 + 1)%% is a rational function of $$ R\left(x, \sqrt(x), \sqrt(x^2+1)\right) = \frac(x + \sqrt(x ^2))(\sqrt(x^2 + 1)) = f(x) $$ from %%x%% and radicals %%\sqrt(x)%% and %%\sqrt(x^2 + 1 )%%, while the function %%f(x)%% will be an irrational (algebraic) function of one independent variable %%x%%.

Let's consider integrals of the form %%\int R(x, \sqrt[n](x)) \mathrm(d)x%%. Such integrals are rationalized by replacing the variable %%t = \sqrt[n](x)%%, then %%x = t^n, \mathrm(d)x = nt^(n-1)%%.

Example 1

Find %%\displaystyle\int \frac(\mathrm(d)x)(\sqrt(x) + \sqrt(x))%%.

The integrand of the desired argument is written as a function of radicals of degree %%2%% and %%3%%. Since the least common multiple of %%2%% and %%3%% is %%6%%, this integral is an integral of type %%\int R(x, \sqrt(x)) \mathrm(d)x %% and can be rationalized by replacing %%\sqrt(x) = t%%. Then %%x = t^6, \mathrm(d)x = 6t \mathrm(d)t, \sqrt(x) = t^3, \sqrt(x) =t^2%%. Therefore, $$ \int \frac(\mathrm(d)x)(\sqrt(x) + \sqrt(x)) = \int \frac(6t^5 \mathrm(d)t)(t^3 + t^2) = 6\int\frac(t^3)(t+1)\mathrm(d)t. $$ Let's take %%t + 1 = z, \mathrm(d)t = \mathrm(d)z, z = t + 1 = \sqrt(x) + 1%% and $$ \begin(array)(ll ) \int \frac(\mathrm(d)x)(\sqrt(x) + \sqrt(x)) &= 6\int\frac((z-1)^3)(z) \mathrm(d) t = \\ &= 6\int z^2 dz -18 \int z \mathrm(d)z + 18\int \mathrm(d)z -6\int\frac(\mathrm(d)z)(z ) = \\ &= 2z^3 - 9 z^2 + 18z -6\ln|z| + C = \\ &= 2 \left(\sqrt(x) + 1\right)^3 - 9 \left(\sqrt(x) + 1\right)^2 + \\ &+~ 18 \left( \sqrt(x) + 1\right) - 6 \ln\left|\sqrt(x) + 1\right| + C \end(array) $$

Integrals of the form %%\int R(x, \sqrt[n](x)) \mathrm(d)x%% are a special case of fractional linear irrationalities, i.e. integrals of the form %%\displaystyle\int R\left(x, \sqrt[n](\dfrac(ax+b)(cd+d))\right) \mathrm(d)x%%, where %%ad - bc \neq 0%%, which can be rationalized by replacing the variable %%t = \sqrt[n](\dfrac(ax+b)(cd+d))%%, then %%x = \dfrac(dt^n - b)(a - ct^n)%%. Then $$ \mathrm(d)x = \frac(n t^(n-1)(ad - bc))(\left(a - ct^n\right)^2)\mathrm(d)t. $$

Example 2

Find %%\displaystyle\int \sqrt(\dfrac(1 -x)(1 + x))\dfrac(\mathrm(d)x)(x + 1)%%.

Let's take %%t = \sqrt(\dfrac(1 -x)(1 + x))%%, then %%x = \dfrac(1 - t^2)(1 + t^2)%%, $$ \begin(array)(l) \mathrm(d)x = -\frac(4t\mathrm(d)t)(\left(1 + t^2\right)^2), \\ 1 + x = \ frac(2)(1 + t^2), \\ \frac(1)(x + 1) = \frac(1 + t^2)(2). \end(array) $$ Therefore, $$ \begin(array)(l) \int \sqrt(\dfrac(1 -x)(1 + x))\frac(\mathrm(d)x)(x + 1) = \\ = \frac(t(1 + t^2))(2)\left(-\frac(4t \mathrm(d)t)(\left(1 + t^2\right)^2 )\right) = \\ = -2\int \frac(t^2\mathrm(d)t)(1 + t^2) = \\ = -2\int \mathrm(d)t + 2\int \frac(\mathrm(d)t)(1 + t^2) = \\ = -2t + \text(arctg)~t + C = \\ = -2\sqrt(\dfrac(1 -x)( 1 + x)) + \text(arctg)~\sqrt(\dfrac(1 -x)(1 + x)) + C. \end(array) $$

Let's consider integrals of the form %%\int R\left(x, \sqrt(ax^2 + bx + c)\right) \mathrm(d)x%%. In the simplest cases, such integrals are reduced to tabular ones if, after isolating the complete square, a change of variables is made.

Example 3

Find the integral %%\displaystyle\int \dfrac(\mathrm(d)x)(\sqrt(x^2 + 4x + 5))%%.

Considering that %%x^2 + 4x + 5 = (x+2)^2 + 1%%, we take %%t = x + 2, \mathrm(d)x = \mathrm(d)t%%, then $$ \begin(array)(ll) \int \frac(\mathrm(d)x)(\sqrt(x^2 + 4x + 5)) &= \int \frac(\mathrm(d)t) (\sqrt(t^2 + 1)) = \\ &= \ln\left|t + \sqrt(t^2 + 1)\right| + C = \\ &= \ln\left|x + 2 + \sqrt(x^2 + 4x + 5)\right| + C. \end(array) $$

In more difficult cases to find integrals of the form %%\int R\left(x, \sqrt(ax^2 + bx + c)\right) \mathrm(d)x%% are used

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 from complex fractions, which flew past the cash desk in previous articles.

Fourthly, additional integrals from trigonometric functions will be analyzed. In particular, there are methods that avoid time-consuming universal trigonometric substitution.

(2) In the integrand function, we divide the numerator by the denominator term by term.

(3) We use the linearity property of the indefinite integral. In the last integral immediately put the function under the differential sign.

(4) We take the remaining integrals. Note that in a logarithm you can use parentheses rather than a modulus, since .

(5) We carry out a reverse replacement, expressing “te” from the direct replacement:

Masochistic students can differentiate the answer and get the original integrand, as I just did. No, no, I did the check in the right sense =)

As you can see, during the solution we had to use even more than two solution methods, so to deal with such integrals you need confident integration skills and quite a bit of experience.

In practice, of course, the square root is more common, here are three examples for independent decision:

Example 2

Find 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

A witty and beautiful method. Let's take a look at the classics of the genre:

Example 5

Find the indefinite integral

Under the root is a quadratic binomial, and when trying to integrate this example the kettle can suffer for hours. Such an integral is taken in parts and reduced to itself. In principle, it’s not difficult. If you know how.

Let us denote the integral under consideration Latin letter and let's start solving:

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 take the deuce to 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 If a quadratic trinomial is found, then the solution in any case comes down to two 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 necessarily. 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 profitable in this case to denote by , an exponential function 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 into 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 go back to the example with lucky number 13 (honestly, I didn’t guess correctly). This integral is also one of those that can be quite frustrating if you don’t know how to solve.

The solution starts with an artificial transformation:

I think everyone already understands how to divide the numerator by the denominator term by term.

The resulting integral is taken in parts:

For an integral of the form ( – natural number) we derive recurrent reduction formula:
, 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 is expanded into a sum of fractions. But in my practice there is such an example never met so I missed it 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 “complicated” 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:
etc.

What is the idea of ​​the method? The idea is to use transformations and trigonometric formulas to organize only tangents and the tangent derivative into the integrand. That is, we are talking about replacing: . 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.

A couple of creative examples for your own solution:

Example 23

Find the indefinite integral

Example 24

Find the indefinite integral

Yes, in them, of course, you can lower the powers of sine and cosine, and use a universal trigonometric substitution, but the solution will be much more efficient and shorter if it is carried out through tangents. Full solution and answers at the end of the lesson

The class of irrational functions is very wide, so there simply cannot be a universal way to integrate them. In this article we will try to identify the most characteristic types of irrational integrand functions and associate the integration method with them.

There are cases when it is appropriate to use the method of subscribing to the differential sign. For example, when finding indefinite integrals of the form, where p– rational fraction.

Example.

Find the indefinite integral .

Solution.

It is not difficult to notice that . Therefore, we put it under the differential sign and use the table of antiderivatives:

Answer:

.

13. Fractional linear substitution

Integrals of the type where a, b, c, d are real numbers, a, b,..., d, g are natural numbers, are reduced to integrals of a rational function by substitution, where K is the least common multiple of the denominators of the fractions

Indeed, from the substitution it follows that

i.e. x and dx are expressed through rational functions of t. Moreover, each degree of the fraction is expressed through a rational function of t.

Example 33.4. Find the integral

Solution: The least common multiple of the denominators of the fractions 2/3 and 1/2 is 6.

Therefore, we put x+2=t 6, x=t 6 -2, dx=6t 5 dt, Therefore,

Example 33.5. Specify the substitution for finding integrals:

Solution: For I 1 substitution x=t 2, for I 2 substitution

14. Trigonometric substitution

Integrals of type are reduced to integrals of functions that rationally depend on trigonometric functions using the following trigonometric substitutions: x = a sint for the first integral; x=a tgt for the second integral; for the third integral.

Example 33.6. Find the integral

Solution: Let's put x=2 sin t, dx=2 cos tdt, t=arcsin x/2. Then

Here the integrand is a rational function with respect to x and By selecting a complete square under the radical and making a substitution, integrals of the indicated type are reduced to integrals of the type already considered, i.e., to integrals of the type These integrals can be calculated using appropriate trigonometric substitutions.

Example 33.7. Find the integral

Solution: Since x 2 +2x-4=(x+1) 2 -5, then x+1=t, x=t-1, dx=dt. That's why Let's put

Note: Integral type It is expedient to find using the substitution x=1/t.

15. Definite integral

Let a function be defined on a segment and have an antiderivative on it. The difference is called definite integral functions along the segment and denote. So,

The difference is written in the form, then . Numbers are called limits of integration .

For example, one of the antiderivatives for a function. That's why

16 . If c is a constant number and the function ƒ(x) is integrable on , then

that is, the constant factor c can be taken out of the sign of the definite integral.

▼Let’s compose the integral sum for the function with ƒ(x). We have:

Then it follows that the function c ƒ(x) is integrable on [a; b] and formula (38.1) is valid.▲

2. If the functions ƒ 1 (x) and ƒ 2 (x) are integrable on [a;b], then integrable on [a; b] their sum u

that is, the integral of the sum is equal to the sum of the integrals.

▼
▲

Property 2 applies to the sum of any finite number of terms.

3.

This property can be accepted by definition. This property is also confirmed by the Newton-Leibniz formula.

4. If the function ƒ(x) is integrable on [a; b] and a< с < b, то

that is, the integral over the entire segment is equal to the sum of the integrals over the parts of this segment. This property is called the additivity of a definite integral (or the additivity property).

When dividing the segment [a;b] into parts, we include point c in the number of division points (this can be done due to the independence of the limit of the integral sum from the method of dividing the segment [a;b] into parts). If c = x m, then the integral sum can be divided into two sums:

Each of the written sums is integral, respectively, for the segments [a; b], [a; s] and [s; b]. Passing to the limit in the last equality as n → ∞ (λ → 0), we obtain equality (38.3).

Property 4 is valid for any location of points a, b, c (we assume that the function ƒ (x) is integrable on the larger of the resulting segments).

So, for example, if a< b < с, то

(properties 4 and 3 were used).

5. “Theorem on mean values.” If the function ƒ(x) is continuous on the interval [a; b], then there is a tonka with є [a; b] such that

▼By the Newton-Leibniz formula we have

where F"(x) = ƒ(x). Applying the Lagrange theorem (the theorem on the finite increment of a function) to the difference F(b)-F(a), we obtain

F(b)-F(a) = F"(c) (b-a) = ƒ(c) (b-a).▲

Property 5 (“the mean value theorem”) for ƒ (x) ≥ 0 has a simple geometric meaning: the value of the definite integral is equal, for some c є (a; b), to the area of ​​a rectangle with height ƒ (c) and base b-a ( see fig. 170). Number

is called the average value of the function ƒ(x) on the interval [a; b].

6. If the function ƒ (x) maintains its sign on the segment [a; b], where a< b, то интегралимеет тот же знак, что и функция. Так, если ƒ(х)≥0 на отрезке [а; b], то

▼By the “mean value theorem” (property 5)

where c є [a; b]. And since ƒ(x) ≥ 0 for all x О [a; b], then

ƒ(с)≥0, b-а>0.

Therefore ƒ(с) (b-а) ≥ 0, i.e. ▲

7. Inequality between continuous functions on the interval [a; b], (a

▼Since ƒ 2 (x)-ƒ 1 (x)≥0, then when a< b, согласно свойству 6, имеем

Or, according to property 2,

Note that it is impossible to differentiate inequalities.

8. Estimation of the integral. If m and M are, respectively, the smallest and largest values ​​of the function y = ƒ (x) on the segment [a; b], (a< b), то

▼Since for any x є [a;b] we have m≤ƒ(x)≤M, then, according to property 7, we have

Applying Property 5 to the extreme integrals, we obtain

▲

If ƒ(x)≥0, then property 8 is illustrated geometrically: the area of ​​a curvilinear trapezoid is enclosed between the areas of rectangles whose base is , and whose heights are m and M (see Fig. 171).

9. The modulus of a definite integral does not exceed the integral of the modulus of the integrand:

▼Applying property 7 to the obvious inequalities -|ƒ(x)|≤ƒ(x)≤|ƒ(x)|, we obtain

It follows that

▲

10. The derivative of a definite integral with respect to a variable upper limit is equal to the integrand in which the integration variable is replaced by this limit, i.e.

Calculating the area of ​​a figure is one of the most difficult problems in area theory. In the school geometry course, we learned to find the areas of basic geometric shapes, for example, a circle, triangle, rhombus, etc. However, much more often you have to deal with calculating the areas of more complex figures. When solving such problems, one has to resort to integral calculus.

In this article we will consider the problem of calculating the area of ​​a curvilinear trapezoid, and we will approach it in a geometric sense. This will allow us to find out the direct connection between the definite integral and the area of ​​a curvilinear trapezoid.