Find the graph of the figure with bounded lines. Examples

We begin to consider the actual process of calculating the double integral and get acquainted with its geometric meaning.

The double integral is numerically equal to the area of ​​the plane figure (the region of integration). This is the simplest form of double integral, when the function of two variables is equal to one: .

First, let's look at the problem in general form. Now you will be quite surprised how simple everything really is! Let's calculate the area of ​​a flat figure bounded by lines. For definiteness, we assume that on the segment . The area of ​​this figure is numerically equal to:

Let's depict the area in the drawing:

Let's choose the first way to traverse the area:

Thus:

And immediately an important technical technique: iterated integrals can be calculated separately. First the inner integral, then the outer integral. I highly recommend this method to beginners in the subject.

1) Let's calculate the internal integral, and the integration is carried out over the variable “y”:

The indefinite integral here is the simplest, and then the banal Newton-Leibniz formula is used, with the only difference that the limits of integration are not numbers, but functions. First, we substituted the upper limit into the “y” (antiderivative function), then the lower limit

2) The result obtained in the first paragraph must be substituted into the external integral:

A more compact representation of the entire solution looks like this:

The resulting formula is exactly the working formula for calculating the area of ​​a plane figure using the “ordinary” definite integral! Watch the lesson Calculating Area Using a Definite Integral, there she is at every step!

That is, problem of calculating area using double integral not much different from the problem of finding the area using a definite integral! In fact, it's the same thing!

Accordingly, no difficulties should arise! I won’t look at very many examples, since you, in fact, have repeatedly encountered this task.

Example 9

Solution: Let's depict the area in the drawing:

Let us choose the following order of traversal of the area:

Here and further I will not dwell on how to traverse the area, since very detailed explanations were given in the first paragraph.

Thus:

As I already noted, it is better for beginners to calculate iterated integrals separately, and I will stick to the same method:

1) First, using the Newton-Leibniz formula, we deal with the internal integral:

2) The result obtained in the first step is substituted into the external integral:

Point 2 is actually finding the area of ​​a plane figure using a definite integral.

Answer:

This is such a stupid and naive task.

An interesting example for an independent solution:

Example 10

Using a double integral, calculate the area of ​​a plane figure bounded by the lines , ,

An approximate example of a final solution at the end of the lesson.

In Examples 9-10, it is much more profitable to use the first method of traversing the area; curious readers, by the way, can change the order of traversal and calculate the areas using the second method. If you do not make a mistake, then, naturally, you will get the same area values.

But in some cases, the second method of traversing the area is more effective, and at the end of the young nerd’s course, let’s look at a couple more examples on this topic:

Example 11

Using a double integral, calculate the area of ​​a plane figure bounded by lines,

Solution: We are looking forward to two parabolas with a quirk that lie on their sides. There is no need to smile; similar things occur quite often in multiple integrals.

What is the easiest way to make a drawing?

Let's imagine a parabola in the form of two functions:
– the upper branch and – the lower branch.

Similarly, imagine a parabola in the form of upper and lower branches.

Next, point-wise plotting of graphs rules, resulting in such a bizarre figure:

We calculate the area of ​​the figure using the double integral according to the formula:

What happens if we choose the first method of traversing the area? Firstly, this area will have to be divided into two parts. And secondly, we will observe this sad picture: . Integrals, of course, are not of a super-complicated level, but... there is an old mathematical saying: those who are close to their roots do not need a test.

Therefore, from the misunderstanding given in the condition, we express the inverse functions:

Inverse functions in this example have the advantage that they specify the entire parabola at once without any leaves, acorns, branches and roots.

According to the second method, the area traversal will be as follows:

Thus:

As they say, feel the difference.

1) We deal with the internal integral:

We substitute the result into the outer integral:

Integration over the variable “y” should not be confusing; if there were a letter “zy”, it would be great to integrate over it. Although who read the second paragraph of the lesson How to calculate the volume of a body of revolution, he no longer experiences the slightest awkwardness with integration according to the “Y” method.

Also pay attention to the first step: the integrand is even, and the interval of integration is symmetrical about zero. Therefore, the segment can be halved, and the result can be doubled. This technique is commented in detail in the lesson. Efficient methods for calculating the definite integral.

What to add…. All!

Answer:

To test your integration technique, you can try to calculate . The answer should be exactly the same.

Example 12

Using a double integral, calculate the area of ​​a plane figure bounded by lines

This is an example for you to solve on your own. It is interesting to note that if you try to use the first method of traversing the area, the figure will no longer have to be divided into two, but into three parts! And, accordingly, we get three pairs of repeated integrals. Sometimes it happens.

The master class has come to an end, and it’s time to move on to the grandmaster level - How to calculate double integral? Examples of solutions. I’ll try not to be so maniacal in the second article =)

I wish you success!

Solutions and answers:

Example 2:Solution: Let's depict the area on the drawing:

Let us choose the following order of traversal of the area:

Thus:
Let's move on to inverse functions:


Thus:
Answer:

Example 4:Solution: Let's move on to direct functions:


Let's make the drawing:

Let's change the order of traversing the area:

Answer:

Back forward

Attention! Slide previews are for informational purposes only and may not represent all the features of the presentation. If you are interested in this work, please download the full version.

Keywords: integral, curvilinear trapezoid, area of ​​figures bounded by lilies

Equipment: marker board, computer, multimedia projector

Lesson type: lesson-lecture

Lesson Objectives:

• educational: to create a culture of mental work, create a situation of success for each student, and create positive motivation for learning; develop the ability to speak and listen to others.
• developing: formation of independent thinking of the student in applying knowledge in various situations, the ability to analyze and draw conclusions, development of logic, development of the ability to correctly pose questions and find answers to them. Improving the formation of computational skills, developing students’ thinking in the course of completing proposed tasks, developing an algorithmic culture.
• educational: to form concepts about a curvilinear trapezoid, about an integral, to master the skills of calculating the areas of plane figures

Teaching Method: explanatory and illustrative.

During the classes

In previous classes we learned to calculate the areas of figures whose boundaries are broken lines. In mathematics, there are methods that allow you to calculate the areas of figures bounded by curves. Such figures are called curvilinear trapezoids, and their area is calculated using antiderivatives.

Curvilinear trapezoid ( slide 1)

A curved trapezoid is a figure bounded by the graph of a function, ( sh.m.), straight x = a And x = b and x-axis

Various types of curved trapezoids ( slide 2)

We consider various types of curvilinear trapezoids and notice: one of the straight lines is degenerate to a point, the role of the limiting function is played by the straight line

Area of ​​a curved trapezoid (slide 3)

Fix the left end of the interval A, and the right one X we will change, i.e., we move the right wall of the curvilinear trapezoid and get a changing figure. The area of ​​a variable curvilinear trapezoid bounded by the graph of the function is an antiderivative F for function f

And on the segment [ a; b] area of ​​a curvilinear trapezoid formed by the function f, is equal to the increment of the antiderivative of this function:

Exercise 1:

Find the area of ​​a curvilinear trapezoid bounded by the graph of the function: f(x) = x 2 and straight y = 0, x = 1, x = 2.

Solution: ( according to the algorithm slide 3)

Let's draw a graph of the function and lines

Let's find one of the antiderivatives of the function f(x) = x 2 :

Self-test on slide

Integral

Consider a curvilinear trapezoid defined by the function f on the segment [ a; b]. Let's break this segment into several parts. The area of ​​the entire trapezoid will be divided into the sum of the areas of smaller curved trapezoids. ( slide 5). Each such trapezoid can be approximately considered a rectangle. The sum of the areas of these rectangles gives an approximate idea of ​​the entire area of ​​the curved trapezoid. The smaller we divide the segment [ a; b], the more accurately we calculate the area.

Let us write these arguments in the form of formulas.

Divide the segment [ a; b] into n parts by dots x 0 =a, x1,...,xn = b. Length k- th denote by xk = xk – xk-1. Let's make a sum

Geometrically, this sum represents the area of ​​the figure shaded in the figure ( sh.m.)

Sums of the form are called integral sums for the function f. (sh.m.)

Integral sums give an approximate value of the area. The exact value is obtained by passing to the limit. Let's imagine that we are refining the partition of the segment [ a; b] so that the lengths of all small segments tend to zero. Then the area of ​​the composed figure will approach the area of ​​the curved trapezoid. We can say that the area of ​​a curved trapezoid is equal to the limit of integral sums, Sc.t. (sh.m.) or integral, i.e.,

Definition:

Integral of a function f(x) from a before b called the limit of integral sums

= (sh.m.)

Newton-Leibniz formula.

We remember that the limit of integral sums is equal to the area of ​​a curvilinear trapezoid, which means we can write:

Sc.t. = (sh.m.)

On the other hand, the area of ​​a curved trapezoid is calculated using the formula

S k.t. (sh.m.)

Comparing these formulas, we get:

= (sh.m.)

This equality is called the Newton-Leibniz formula.

For ease of calculation, the formula is written as:

= = (sh.m.)

Tasks: (sh.m.)

1. Calculate the integral using the Newton-Leibniz formula: ( check on slide 5)

2. Compose integrals according to the drawing ( check on slide 6)

3. Find the area of ​​the figure bounded by the lines: y = x 3, y = 0, x = 1, x = 2. ( Slide 7)

Finding the areas of plane figures ( slide 8)

How to find the area of ​​figures that are not curved trapezoids?

Let two functions be given, the graphs of which you see on the slide . (sh.m.) Find the area of ​​the shaded figure . (sh.m.). Is the figure in question a curved trapezoid? How can you find its area using the property of additivity of area? Consider two curved trapezoids and subtract the area of ​​the other from the area of ​​one of them ( sh.m.)

Let's create an algorithm for finding the area using animation on a slide:

1. Graph functions
2. Project the intersection points of the graphs onto the x-axis
3. Shade the figure obtained when the graphs intersect
4. Find curvilinear trapezoids whose intersection or union is the given figure.
5. Calculate the area of ​​each of them
6. Find the difference or sum of areas

Oral task: How to obtain the area of ​​a shaded figure (tell using animation, slide 8 and 9)

Homework: Work through the notes, No. 353 (a), No. 364 (a).

Bibliography

1. Algebra and the beginnings of analysis: a textbook for grades 9-11 of evening (shift) school / ed. G.D. Glaser. - M: Enlightenment, 1983.
2. Bashmakov M.I. Algebra and the beginnings of analysis: a textbook for 10-11 grades of secondary school / Bashmakov M.I. - M: Enlightenment, 1991.
3. Bashmakov M.I. Mathematics: textbook for institutions beginning. and Wednesday prof. education / M.I. Bashmakov. - M: Academy, 2010.
4. Kolmogorov A.N. Algebra and beginnings of analysis: textbook for grades 10-11. educational institutions / A.N. Kolmogorov. - M: Education, 2010.
5. Ostrovsky S.L. How to make a presentation for a lesson?/ S.L. Ostrovsky. – M.: September 1st, 2010.

Example1 . Calculate the area of ​​the figure bounded by the lines: x + 2y – 4 = 0, y = 0, x = -3, and x = 2

Let's construct a figure (see figure) We construct a straight line x + 2y – 4 = 0 using two points A(4;0) and B(0;2). Expressing y through x, we get y = -0.5x + 2. Using formula (1), where f(x) = -0.5x + 2, a = -3, b = 2, we find

S = = [-0.25=11.25 sq. units

Example 2. Calculate the area of ​​the figure bounded by the lines: x – 2y + 4 = 0, x + y – 5 = 0 and y = 0.

Solution. Let's construct the figure.

Let's construct a straight line x – 2y + 4 = 0: y = 0, x = - 4, A(-4; 0); x = 0, y = 2, B(0; 2).

Let's construct a straight line x + y – 5 = 0: y = 0, x = 5, C(5; 0), x = 0, y = 5, D(0; 5).

Let's find the point of intersection of the lines by solving the system of equations:

x = 2, y = 3; M(2; 3).

To calculate the required area, we divide the triangle AMC into two triangles AMN and NMC, since when x changes from A to N, the area is limited by a straight line, and when x changes from N to C - by a straight line

For triangle AMN we have: ; y = 0.5x + 2, i.e. f(x) = 0.5x + 2, a = - 4, b = 2.

For triangle NMC we have: y = - x + 5, i.e. f(x) = - x + 5, a = 2, b = 5.

By calculating the area of ​​each triangle and adding the results, we find:

sq. units

sq. units

9 + 4, 5 = 13.5 sq. units Check: = 0.5AC = 0.5 sq. units

Example 3. Calculate the area of ​​a figure bounded by lines: y = x 2 , y = 0, x = 2, x = 3.

In this case, you need to calculate the area of ​​a curved trapezoid bounded by the parabola y = x 2 , straight lines x = 2 and x = 3 and the Ox axis (see figure) Using formula (1) we find the area of ​​the curvilinear trapezoid

= = 6 sq. units

Example 4. Calculate the area of ​​the figure bounded by the lines: y = - x 2 + 4 and y = 0

Let's construct the figure. The required area is enclosed between the parabola y = - x 2 + 4 and the Ox axis.

Let's find the intersection points of the parabola with the Ox axis. Assuming y = 0, we find x = Since this figure is symmetrical about the Oy axis, we calculate the area of ​​the figure located to the right of the Oy axis, and double the result obtained: = +4x]sq. units 2 = 2 sq. units

Example 5. Calculate the area of ​​a figure bounded by lines: y 2 = x, yx = 1, x = 4

Here you need to calculate the area of ​​a curvilinear trapezoid bounded by the upper branch of the parabola 2 = x, Ox axis and straight lines x = 1 and x = 4 (see figure)

According to formula (1), where f(x) = a = 1 and b = 4, we have = (= sq. units.

Example 6 . Calculate the area of ​​the figure bounded by the lines: y = sinx, y = 0, x = 0, x= .

The required area is limited by the half-wave of the sinusoid and the Ox axis (see figure).

We have - cosx = - cos = 1 + 1 = 2 sq. units

Example 7. Calculate the area of ​​the figure bounded by the lines: y = - 6x, y = 0 and x = 4.

The figure is located under the Ox axis (see figure).

Therefore, we find its area using formula (3)

= =

Example 8. Calculate the area of ​​the figure bounded by the lines: y = and x = 2. Construct the y = curve from the points (see figure). Thus, we find the area of ​​the figure using formula (4)

Example 9 .

X 2 + y 2 = r 2 .

Here you need to calculate the area enclosed by the circle x 2 + y 2 = r 2 , i.e. the area of ​​a circle of radius r with the center at the origin. Let's find the fourth part of this area by taking the limits of integration from 0

before; we have: 1 = = [

Hence, 1 =

Example 10. Calculate the area of ​​a figure bounded by lines: y= x 2 and y = 2x

This figure is limited by the parabola y = x 2 and the straight line y = 2x (see figure) To determine the intersection points of the given lines, we solve the system of equations: x 2 – 2x = 0 x = 0 and x = 2

Using formula (5) to find the area, we obtain

= \- -fl -- Г -1-±Л_ 1V1 -l-l-Ii-^ 3) |_ 2 3V 2 / J 3 24 24* Example 2. Let's calculate the area limited by the sinusoid y = sinXy, the Ox axis and the straight line (Fig. 87). Applying formula (I), we obtain A 2 S= J sinxdx= [-cos x]Q =0 -(-1) = lf Example 3. Calculate the area limited by the arc of the sinusoid ^у = sin jc, enclosed between two adjacent intersection points with the Ox axis (for example, between the origin and the point with the abscissa i). Note that from geometric considerations it is clear that this area will be twice the area of ​​​​the previous example. However, let's do the calculations: I 5= | s\nxdx= [ - cosх)* - - cos i-(-cos 0)= 1 + 1 = 2. o Indeed, our assumption turned out to be correct. Example 4. Calculate the area bounded by the sinusoid and the Ox axis at one period (Fig. 88). Preliminary calculations suggest that the area will be four times larger than in Example 2. However, after making calculations, we obtain “i Г,*i S - \ sin x dx = [ - cos x]0 = = - cos 2l -(-cos 0) = - 1 + 1 = 0. This result requires clarification. To clarify the essence of the matter, we also calculate the area limited by the same sinusoid y = sin l: and the Ox axis in the range from l to 2i. Applying formula (I), we obtain 2l $2l sin xdx=[ - cosх]l = -cos 2i~)-c05i=- 1-1 =-2. Thus, we see that this area turned out to be negative. Comparing it with the area calculated in exercise 3, we find that their absolute values ​​are the same, but the signs are different. If we apply property V (see Chapter XI, § 4), we get 2l I 2l J sin xdx= J sin * dx [ sin x dx = 2 + (- 2) = 0What happened in this example is not an accident. Always the area located below the Ox axis, provided that the independent variable changes from left to right, is obtained when calculated using integrals. In this course we will always consider areas without signs. Therefore, the answer in the example just discussed will be: the required area is 2 + |-2| = 4. Example 5. Let's calculate the area of ​​the BAB shown in Fig. 89. This area is limited by the Ox axis, the parabola y = - xr and the straight line y - = -x+\. Area of ​​a curvilinear trapezoid The required area OAB consists of two parts: OAM and MAV. Since point A is the intersection point of a parabola and a straight line, we will find its coordinates by solving the system of equations 3 2 Y = mx. (we only need to find the abscissa of point A). Solving the system, we find l; = ~. Therefore, the area has to be calculated in parts, first square. OAM and then pl. MAV: .... G 3 2, 3 G xP 3 1/2 U 2. QAM-^x (base of a curved trapezoid) into n equal parts; this partition is carried out using points x 1, x 2, . .. x k , ... x n-1 . Let us draw straight lines through these points parallel to the y-axis. Then the given curvilinear trapezoid will be divided into n parts, into n narrow columns. The area of ​​the entire trapezoid is equal to the sum of the areas of the columns.

Let us consider the k-th column separately, i.e. a curved trapezoid whose base is a segment. Let's replace it with a rectangle with the same base and height equal to f(x k) (see figure). The area of ​​the rectangle is equal to \(f(x_k) \cdot \Delta x_k \), where \(\Delta x_k \) is the length of the segment; It is natural to consider the resulting product as an approximate value of the area of ​​the kth column.

If we now do the same with all the other columns, we will come to the following result: the area S of a given curvilinear trapezoid is approximately equal to the area S n of a stepped figure made up of n rectangles (see figure):
\(S_n = f(x_0)\Delta x_0 + \dots + f(x_k)\Delta x_k + \dots + f(x_(n-1))\Delta x_(n-1) \)
Here, for the sake of uniformity of notation, we assume that a = x 0, b = x n; \(\Delta x_0 \) - length of the segment, \(\Delta x_1 \) - length of the segment, etc.; in this case, as we agreed above, \(\Delta x_0 = \dots = \Delta x_(n-1) \)

So, \(S \approx S_n \), and this approximate equality is more accurate, the larger n.
By definition, it is believed that the required area of ​​a curvilinear trapezoid is equal to the limit of the sequence (S n):
$$ S = \lim_(n \to \infty) S_n $$

Problem 2(about moving a point)
A material point moves in a straight line. The dependence of speed on time is expressed by the formula v = v(t). Find the movement of a point over a period of time [a; b].
Solution. If the movement were uniform, then the problem would be solved very simply: s = vt, i.e. s = v(b-a). For uneven movement, you have to use the same ideas on which the solution to the previous problem was based.
1) Divide the time interval [a; b] into n equal parts.
2) Consider a period of time and assume that during this period of time the speed was constant, the same as at time t k. So we assume that v = v(t k).
3) Let’s find the approximate value of the point’s movement over a period of time; we’ll denote this approximate value as s k
\(s_k = v(t_k) \Delta t_k \)
4) Find the approximate value of displacement s:
\(s \approx S_n \) where
\(S_n = s_0 + \dots + s_(n-1) = v(t_0)\Delta t_0 + \dots + v(t_(n-1)) \Delta t_(n-1) \)
5) The required displacement is equal to the limit of the sequence (S n):
$$ s = \lim_(n \to \infty) S_n $$

Let's summarize. Solutions to various problems were reduced to the same mathematical model. Many problems from various fields of science and technology lead to the same model in the process of solution. This means that this mathematical model must be specially studied.

The concept of a definite integral

Let us give a mathematical description of the model that was built in the three considered problems for the function y = f(x), continuous (but not necessarily non-negative, as was assumed in the considered problems) on the interval [a; b]:
1) split the segment [a; b] into n equal parts;
2) make up the sum $$ S_n = f(x_0)\Delta x_0 + f(x_1)\Delta x_1 + \dots + f(x_(n-1))\Delta x_(n-1) $$
3) calculate $$ \lim_(n \to \infty) S_n $$

In the course of mathematical analysis it was proven that this limit exists in the case of a continuous (or piecewise continuous) function. He is called a certain integral of the function y = f(x) over the segment [a; b] and denoted as follows:
\(\int\limits_a^b f(x) dx \)
The numbers a and b are called the limits of integration (lower and upper, respectively).

Let's return to the tasks discussed above. The definition of area given in Problem 1 can now be rewritten as follows:
\(S = \int\limits_a^b f(x) dx \)
here S is the area of ​​the curved trapezoid shown in the figure above. This is geometric meaning of a definite integral.

The definition of the displacement s of a point moving in a straight line with a speed v = v(t) over the period of time from t = a to t = b, given in Problem 2, can be rewritten as follows:

Newton-Leibniz formula

First, let's answer the question: what is the connection between the definite integral and the antiderivative?

The answer can be found in Problem 2. On the one hand, the displacement s of a point moving in a straight line with a speed v = v(t) over the period of time from t = a to t = b is calculated by the formula
\(S = \int\limits_a^b v(t) dt \)

On the other hand, the coordinate of a moving point is an antiderivative for speed - let's denote it s(t); this means that the displacement s is expressed by the formula s = s(b) - s(a). As a result we get:
\(S = \int\limits_a^b v(t) dt = s(b)-s(a) \)
where s(t) is the antiderivative of v(t).

The following theorem was proven in the course of mathematical analysis.
Theorem. If the function y = f(x) is continuous on the interval [a; b], then the formula is valid
\(S = \int\limits_a^b f(x) dx = F(b)-F(a) \)
where F(x) is the antiderivative of f(x).

The given formula is usually called Newton-Leibniz formula in honor of the English physicist Isaac Newton (1643-1727) and the German philosopher Gottfried Leibniz (1646-1716), who received it independently of each other and almost simultaneously.

In practice, instead of writing F(b) - F(a), they use the notation \(\left. F(x)\right|_a^b \) (it is sometimes called double substitution) and, accordingly, rewrite the Newton-Leibniz formula in this form:
\(S = \int\limits_a^b f(x) dx = \left. F(x)\right|_a^b \)

When calculating a definite integral, first find the antiderivative, and then carry out a double substitution.

Based on the Newton-Leibniz formula, we can obtain two properties of the definite integral.

Property 1. The integral of the sum of functions is equal to the sum of the integrals:
\(\int\limits_a^b (f(x) + g(x))dx = \int\limits_a^b f(x)dx + \int\limits_a^b g(x)dx \)

Property 2. The constant factor can be taken out of the integral sign:
\(\int\limits_a^b kf(x)dx = k \int\limits_a^b f(x)dx \)

Calculating the areas of plane figures using a definite integral

Using the integral, you can calculate the areas not only of curved trapezoids, but also of plane figures of a more complex type, for example, the one shown in the figure. The figure P is limited by straight lines x = a, x = b and graphs of continuous functions y = f(x), y = g(x), and on the segment [a; b] the inequality \(g(x) \leq f(x) \) holds. To calculate the area S of such a figure, we will proceed as follows:
\(S = S_(ABCD) = S_(aDCb) - S_(aABb) = \int\limits_a^b f(x) dx - \int\limits_a^b g(x) dx = \)
\(= \int\limits_a^b (f(x)-g(x))dx \)

So, the area S of a figure bounded by straight lines x = a, x = b and graphs of functions y = f(x), y = g(x), continuous on the segment and such that for any x from the segment [a; b] the inequality \(g(x) \leq f(x) \) is satisfied, calculated by the formula
\(S = \int\limits_a^b (f(x)-g(x))dx \)

Table of indefinite integrals (antiderivatives) of some functions

$$ \int 0 \cdot dx = C $$ $$ \int 1 \cdot dx = x+C $$ $$ \int x^n dx = \frac(x^(n+1))(n+1 ) +C \;\; (n \neq -1) $$ $$ \int \frac(1)(x) dx = \ln |x| +C $$ $$ \int e^x dx = e^x +C $$ $$ \int a^x dx = \frac(a^x)(\ln a) +C \;\; (a>0, \;\; a \neq 1) $$ $$ \int \cos x dx = \sin x +C $$ $$ \int \sin x dx = -\cos x +C $$ $ $ \int \frac(dx)(\cos^2 x) = \text(tg) x +C $$ $$ \int \frac(dx)(\sin^2 x) = -\text(ctg) x +C $$ $$ \int \frac(dx)(\sqrt(1-x^2)) = \text(arcsin) x +C $$ $$ \int \frac(dx)(1+x^2 ) = \text(arctg) x +C $$ $$ \int \text(ch) x dx = \text(sh) x +C $$ $$ \int \text(sh) x dx = \text(ch ) x +C $$