A)Direct integration.
Finding integrals of functions based on the direct application of the properties of indefinite integrals and a table of basic integration formulas. Let's consider an example of finding the integral of a function by direct integration.
Example:
∫(X–3) 2 d X= ∫(X 2 –6X+9)d X= ∫X 2d X- 6∫X d X+9∫d X=X 3 ∕3 -3X 2 +9X+S.
In the vast majority of cases, we are dealing with integrals of functions that cannot be found by direct integration. In this case, it is necessary to make a substitution (replace the variable).
b)Integration by substitution (variable replacement).
Integration by substitution, or as it is often called, the variable substitution method, is one of the more effective and common methods of integration. The substitution method is to move from a given integration variable to another variable in order to simplify the integrand expression and reduce it to one of the tabular types of integrals. In this case, the choice of substitution is decided by the performer individually, because there are no general rules indicating which substitution in in this case take.
Example: Find the integral ∫ e 2х+3 d X.
Let us introduce a new variable t associated with X following dependency 2 X+ 3 =t.
Let's take the differentials of the left and right sides of this equality: 2d X=dt;d X=dt/2.
Now instead of 2 X+ 3 иd X Let us substitute their values into the integrand. Then we get: ∫ e 2х+3 d X=∫e t dt= e t + C. Returning to the previous variable, we finally obtain the expression:
∫e 2х+3 d X=e 2x+3 + C.
To make sure that the integral is taken correctly, you need an antiderivative function e 2x+ 3 differentiate and check whether there will be Is its derivative equal to the integrand function:
(e 2x+ 3)" =e 2x+ 3 (2 X+3)" =e 2x+ 3 .
3. Definite integral and its properties.
The concept of a definite integral is widely used in many fields of science and technology. With its help, areas bounded by curves, volumes of arbitrary shape, power and work of a variable force, the path of a moving body, moments of inertia and many other quantities are calculated.
IN 
In the vast majority of cases, the concept of a definite integral is introduced when solving problems of determining the area of a curvilinear trapezoid. Let there be a continuous function y =f( X) on the segment [ a,c]. A figure bounded by the curve y=f( X) ordinates A Oh, V A P and the segment [ a,c] the x-axis is called a curvilinear trapezoid (Fig. 1).
Let us set ourselves the task: determine the area S of a curved trapezoid A A o A P V. To do this, we divide the segment [ a,c] on P not necessary equal parts and designate the division points as follows: A=X O < X 1 < X 2 ‹ … ‹ X P = in.
From the division points we restore perpendiculars to the intersection with the curve y = f( X). Thus, we divided the entire area bounded by the curve into P elementary curvilinear trapezoids. Let's restore from arbitrary points each segment ∆ X i ordinatef(C i) until it intersects with the curve y =f( X). Next, we will construct a stepped figure consisting of rectangles with a base ∆ X i and height f(C i). Elementary Square ith the rectangle will be S i =f(C i)(X i -X i -1 ), and the entire area S P the resulting stepped figure will be equal to the sum of the areas of the rectangles:
S P=f(C o)( X 1 -X o) +f(C 1)( X 2 -X 1 ) + … +f(C P- 1)(X P -X P- 1).
To abbreviate the entry of this amount, enter the symbol 
(sigma) – a sign meaning the summation of quantities. Then
S P
=
.
This amount S P, which is called the integral sum, can be either greater or less than the true value of a given area. The closest value to the true value of the area will be the limit of the sum, provided that the elementary segments will be crushed ( p→
), and the length itself large segment ∆X max will tend to zero, i.e.:
S=
(4)
This cumulative sum limit (if it exists) is called definite integral from functionf( X) on the segment [ A,V] and denote: 
=
(5)
(reads “definite integral of A before V ef from x de x”).
Numbers A And V are called the lower and upper limits of integration, respectively, f( X) – subintegral function; X– integration variable. Using formulas (4) and (5) we can write. That the area of a curvilinear trapezoid is numerically equal to the integral of the function limiting the trapezoid, taken over the integration interval [A,V]:
.
This fact expresses the geometric meaning of a definite integral.
Let us consider the properties of the definite integral.
1. The definite integral does not depend on the designation of the variable, i.e.: 
=
.
2. The definite integral of an algebraic sum is equal to the algebraic sum of definite integrals of each term:
=
f 1 ( X)d x +
f 2 ( X)d X+ ….
We have seen that the derivative has numerous uses: the derivative is the speed of movement (or, more generally, the speed of any process); derivative is slope tangent to the graph of a function; using the derivative, you can examine a function for monotonicity and extrema; the derivative helps solve optimization problems.
But in real life have to decide and inverse problems: for example, along with the problem of finding speed according to a known law of motion, there is also the problem of restoring the law of motion according to a known speed. Let's consider one of these problems.
Example 1. Moves in a straight line material point, the speed of its movement at time t is given by the formula u = tg. Find the law of motion.
Solution. Let s = s(t) be the desired law of motion. It is known that s"(t) = u"(t). This means that to solve the problem you need to choose function s = s(t), whose derivative is equal to tg. It's not hard to guess that
Let us immediately note that the example is solved correctly, but incompletely. We found that, in fact, the problem has infinitely many solutions: any function of the form 
an arbitrary constant can serve as a law of motion, since
To make the task more specific, we needed to fix the initial situation: indicate the coordinate of a moving point at some point in time, for example, at t=0. If, say, s(0) = s 0, then from the equality we obtain s(0) = 0 + C, i.e. S 0 = C. Now the law of motion is uniquely defined:
In mathematics, reciprocal operations are assigned different names, come up with special notations: for example, squaring (x 2) and extracting square root sine(sinх) and arcsine(arcsin x), etc. The process of finding the derivative of a given function is called differentiation, and the inverse operation, i.e. the process of finding a function from a given derivative - integration.
The term “derivative” itself can be justified “in everyday terms”: the function y - f(x) “produces into existence” new feature y"= f"(x) The function y = f(x) acts as a “parent”, but mathematicians, naturally, do not call it a “parent” or “producer”, they say that it is, in relation to the function y"=f"(x), the primary image, or, in short, the antiderivative.
Definition 1. The function y = F(x) is called antiderivative for the function y = f(x) on a given interval X if for all x from X the equality F"(x)=f(x) holds.
In practice, the interval X is usually not specified, but is implied (as the natural domain of definition of the function).
Here are some examples:
1) The function y = x 2 is antiderivative for the function y = 2x, since for all x the equality (x 2)" = 2x is true.
2) the function y - x 3 is antiderivative for the function y-3x 2, since for all x the equality (x 3)" = 3x 2 is true.
3) The function y-sinх is antiderivative for the function y = cosx, since for all x the equality (sinx)" = cosx is true.
4) The function is antiderivative for a function on the interval since for all x > 0 the equality is true
In general, knowing the formulas for finding derivatives, it is not difficult to compile a table of formulas for finding antiderivatives.
We hope you understand how this table is compiled: the derivative of the function, which is written in the second column, is equal to the function that is written in the corresponding row of the first column (check it, don’t be lazy, it’s very useful). For example, for the function y = x 5 the antiderivative, as you will establish, is the function (see the fourth row of the table).
Notes: 1. Below we will prove the theorem that if y = F(x) is an antiderivative for the function y = f(x), then the function y = f(x) has infinitely many antiderivatives and they all have the form y = F(x ) + C. Therefore, it would be more correct to add the term C everywhere in the second column of the table, where C is an arbitrary real number.
2. For the sake of brevity, sometimes instead of the phrase “the function y = F(x) is an antiderivative of the function y = f(x),” they say F(x) is an antiderivative of f(x).”
2. Rules for finding antiderivatives
When finding antiderivatives, as well as when finding derivatives, not only formulas are used (they are listed in the table on p. 196), but also some rules. They are directly related to the corresponding rules for calculating derivatives.
We know that the derivative of a sum is equal to the sum of its derivatives. This rule generates the corresponding rule for finding antiderivatives.
Rule 1. The antiderivative of a sum is equal to the sum of the antiderivatives.
We draw your attention to the somewhat “lightness” of this formulation. In fact, one should formulate the theorem: if the functions y = f(x) and y = g(x) have antiderivatives on the interval X, respectively y-F(x) and y-G(x), then the sum of the functions y = f(x)+g(x) has an antiderivative on the interval X, and this antiderivative is the function y = F(x)+G(x). But usually, when formulating rules (and not theorems), they leave only keywords- this makes it more convenient to apply the rule in practice
Example 2. Find the antiderivative for the function y = 2x + cos x.
Solution. The antiderivative for 2x is x"; the antiderivative for cox is sin x. This means that the antiderivative for the function y = 2x + cos x will be the function y = x 2 + sin x (and in general any function of the form Y = x 1 + sinx + C) .
We know that the constant factor can be taken out of the sign of the derivative. This rule generates the corresponding rule for finding antiderivatives.
Rule 2. Constant multiplier can be taken out as a sign of an antiderivative.
Example 3.
Solution. a) The antiderivative for sin x is -soz x; This means that for the function y = 5 sin x the antiderivative function will be the function y = -5 cos x.
b) The antiderivative for cos x is sin x; This means that the antiderivative of a function is the function
c) The antiderivative for x 3 is the antiderivative for x, the antiderivative for the function y = 1 is the function y = x. Using the first and second rules for finding antiderivatives, we find that the antiderivative for the function y = 12x 3 + 8x-1 is the function
Comment. As is known, the derivative of a product is not equal to the product of derivatives (the rule for differentiating a product is more complex) and the derivative of a quotient is not equal to the quotient of derivatives. Therefore, there are no rules for finding the antiderivative of the product or the antiderivative of the quotient of two functions. Be careful!
Let us obtain another rule for finding antiderivatives. We know that the derivative of the function y = f(kx+m) is calculated by the formula
This rule generates the corresponding rule for finding antiderivatives.
Rule 3. If y = F(x) is an antiderivative for the function y = f(x), then the antiderivative for the function y=f(kx+m) is the function
Indeed,
This means that it is an antiderivative for the function y = f(kx+m).
The meaning of the third rule is as follows. If you know that the antiderivative of the function y = f(x) is the function y = F(x), and you need to find the antiderivative of the function y = f(kx+m), then proceed like this: take the same function F, but instead of the argument x, substitute the expression kx+m; in addition, do not forget to write “correction factor” before the function sign
Example 4. Find antiderivatives for given functions:
Solution, a) The antiderivative for sin x is -soz x; This means that for the function y = sin2x the antiderivative will be the function
b) The antiderivative for cos x is sin x; This means that the antiderivative of a function is the function
c) The antiderivative for x 7 means that for the function y = (4-5x) 7 the antiderivative will be the function
3. Indefinite integral
We have already noted above that the problem of finding an antiderivative for a given function y = f(x) has more than one solution. Let's discuss this issue in more detail.
Proof. 1. Let y = F(x) be the antiderivative for the function y = f(x) on the interval X. This means that for all x from X the equality x"(x) = f(x) holds. Let us find the derivative of any function of the form y = F(x)+C:
(F(x) +C) = F"(x) +C = f(x) +0 = f(x).
So, (F(x)+C) = f(x). This means that y = F(x) + C is an antiderivative for the function y = f(x).
Thus, we have proven that if the function y = f(x) has an antiderivative y=F(x), then the function (f = f(x) has infinitely many antiderivatives, for example, any function of the form y = F(x) +C is an antiderivative.
2. Let us now prove that specified type functions, the entire set of antiderivatives is exhausted.
Let y=F 1 (x) and y=F(x) be two antiderivatives for the function Y = f(x) on the interval X. This means that for all x from the interval X the following relations hold: F^ (x) = f (X); F"(x) = f(x).
Let's consider the function y = F 1 (x) -.F(x) and find its derivative: (F, (x) -F(x))" = F[(x)-F(x) = f(x) - f(x) = 0.
It is known that if the derivative of a function on an interval X is identically equal to zero, then the function is constant on the interval X (see Theorem 3 from § 35). This means that F 1 (x) - F (x) = C, i.e. Fx) = F(x)+C.
The theorem has been proven.
Example 5. The law of change of speed with time is given: v = -5sin2t. Find the law of motion s = s(t), if it is known that at time t=0 the coordinate of the point was equal to the number 1.5 (i.e. s(t) = 1.5).
Solution. Since speed is a derivative of the coordinate as a function of time, we first need to find the antiderivative of the speed, i.e. antiderivative for the function v = -5sin2t. One of such antiderivatives is the function , and the set of all antiderivatives has the form:
To find specific meaning constant C, let's use initial conditions, according to which, s(0) = 1.5. Substituting the values t=0, S = 1.5 into formula (1), we get:
Substituting the found value of C into formula (1), we obtain the law of motion that interests us:
Definition 2. If a function y = f(x) has an antiderivative y = F(x) on an interval X, then the set of all antiderivatives, i.e. the set of functions of the form y = F(x) + C is called the indefinite integral of the function y = f(x) and is denoted by:
(read: “indefinite integral ef from x de x”).
In the next paragraph we will find out what is hidden meaning the indicated designation.
Based on the table of antiderivatives available in this section, we will compile a table of the main indefinite integrals:
Based on the above three rules for finding antiderivatives, we can formulate the corresponding integration rules.
Rule 1. Integral of the sum of functions equal to the sum integrals of these functions:
Rule 2. The constant factor can be taken out of the integral sign:
Rule 3. If
Example 6. Find indefinite integrals:
Solution, a) Using the first and second rules of integration, we obtain:
Now let’s use the 3rd and 4th integration formulas:
As a result we get:
b) Using the third rule of integration and formula 8, we obtain:
c) For immediate location For a given integral, we have neither the corresponding formula nor the corresponding rule. In such cases, pre-executed identity transformations expression contained under the integral sign.
Let's take advantage trigonometric formula Degree reduction:
Then we find sequentially:
A.G. Mordkovich Algebra 10th grade
Calendar-thematic planning in mathematics, video in mathematics online, Mathematics at school
This lesson is the first in a series of videos on integration. In it we will analyze what an antiderivative of a function is, and also study the elementary methods of calculating these very antiderivatives.
In fact, there is nothing complicated here: essentially it all comes down to the concept of derivative, which you should already be familiar with. :)
I’ll note right away that since this is the very first lesson in our new topic, there won't be any today complex calculations and formulas, but what we will study today will form the basis for much more complex calculations and constructions when calculating complex integrals and squares.
In addition, when starting to study integration and integrals in particular, we implicitly assume that the student is already at least familiar with the concepts of derivatives and has at least basic skills in calculating them. Without a clear understanding of this, there is absolutely nothing to do in integration.
However, here lies one of the most common and insidious problems. The fact is that, when starting to calculate their first antiderivatives, many students confuse them with derivatives. As a result, in exams and independent work stupid and offensive mistakes are made.
Therefore, now I will not give a clear definition of an antiderivative. In return, I suggest you see how it is calculated using a simple specific example.
What is an antiderivative and how is it calculated?
We know this formula:
\[((\left(((x)^(n)) \right))^(\prime ))=n\cdot ((x)^(n-1))\]
This derivative is calculated simply:
\[(f)"\left(x \right)=((\left(((x)^(3)) \right))^(\prime ))=3((x)^(2))\ ]
Let's look carefully at the resulting expression and express $((x)^(2))$:
\[((x)^(2))=\frac(((\left(((x)^(3)) \right))^(\prime )))(3)\]
But we can write it this way, according to the definition of a derivative:
\[((x)^(2))=((\left(\frac(((x)^(3)))(3) \right))^(\prime ))\]
And now attention: what we just wrote down is the definition of an antiderivative. But to write it correctly, you need to write the following:
Let us write the following expression in the same way:
If we generalize this rule, we can derive the following formula:
\[((x)^(n))\to \frac(((x)^(n+1)))(n+1)\]
Now we can formulate a clear definition.
An antiderivative of a function is a function whose derivative is equal to the original function.
Questions about the antiderivative function
It would seem a fairly simple and understandable definition. However, upon hearing it, the attentive student will immediately have several questions:
- Let's say, okay, this formula is correct. However, in this case, with $n=1$, we have problems: “zero” appears in the denominator, and we cannot divide by “zero”.
- The formula is limited to degrees only. How to calculate the antiderivative, for example, of sine, cosine and any other trigonometry, as well as constants.
- Existential question: is it always possible to find an antiderivative? If yes, then what about the antiderivative of the sum, difference, product, etc.?
On last question I'll answer right away. Unfortunately, the antiderivative, unlike the derivative, is not always considered. No such thing universal formula, by which from any initial construction we will obtain a function that will be equal to this similar construction. As for powers and constants, we’ll talk about that now.
Solving problems with power functions
\[((x)^(-1))\to \frac(((x)^(-1+1)))(-1+1)=\frac(1)(0)\]
As we see, this formula for $((x)^(-1))$ does not work. The question arises: what works then? Can't we count $((x)^(-1))$? Of course we can. Let's just remember this first:
\[((x)^(-1))=\frac(1)(x)\]
Now let's think: the derivative of which function is equal to $\frac(1)(x)$. Obviously, any student who has studied this topic at least a little will remember that this expression is equal to the derivative of the natural logarithm:
\[((\left(\ln x \right))^(\prime ))=\frac(1)(x)\]
Therefore, we can confidently write the following:
\[\frac(1)(x)=((x)^(-1))\to \ln x\]
You need to know this formula, just like the derivative of a power function.
So what we know so far:
- For a power function - $((x)^(n))\to \frac(((x)^(n+1)))(n+1)$
- For a constant - $=const\to \cdot x$
- A special case of a power function is $\frac(1)(x)\to \ln x$
And if we start multiplying and dividing the simplest functions, how then can we calculate the antiderivative of a product or quotient. Unfortunately, analogies with the derivative of a product or quotient do not work here. Any standard formula does not exist. For some cases, there are tricky special formulas - we will get acquainted with them in future video lessons.
However, remember: general formula, a similar formula for calculating the derivative of a quotient and a product does not exist.
Solving real problems
Task No. 1
Let's each power functions Let's calculate separately:
\[((x)^(2))\to \frac(((x)^(3)))(3)\]
Returning to our expression, we write the general construction:
Problem No. 2
As I already said, prototypes of works and private “right through” are not considered. However, here you can do the following:
We broke down the fraction into the sum of two fractions.
Let's do the math:
The good news is that knowing the formulas for calculating antiderivatives, you are already able to calculate more complex designs. However, let's go further and expand our knowledge a little more. The fact is that many constructions and expressions, which, at first glance, have nothing to do with $((x)^(n))$, can be represented as a power with rational indicator, namely:
\[\sqrt(x)=((x)^(\frac(1)(2)))\]
\[\sqrt[n](x)=((x)^(\frac(1)(n)))\]
\[\frac(1)(((x)^(n)))=((x)^(-n))\]
All these techniques can and should be combined. Power expressions Can
- multiply (degrees add);
- divide (degrees are subtracted);
- multiply by a constant;
- etc.
Solving power expressions with rational exponent
Example #1
Let's calculate each root separately:
\[\sqrt(x)=((x)^(\frac(1)(2)))\to \frac(((x)^(\frac(1)(2)+1)))(\ frac(1)(2)+1)=\frac(((x)^(\frac(3)(2))))(\frac(3)(2))=\frac(2\cdot (( x)^(\frac(3)(2))))(3)\]
\[\sqrt(x)=((x)^(\frac(1)(4)))\to \frac(((x)^(\frac(1)(4))))(\frac( 1)(4)+1)=\frac(((x)^(\frac(5)(4))))(\frac(5)(4))=\frac(4\cdot ((x) ^(\frac(5)(4))))(5)\]
In total, our entire construction can be written as follows:
Example No. 2
\[\frac(1)(\sqrt(x))=((\left(\sqrt(x) \right))^(-1))=((\left(((x)^(\frac( 1)(2))) \right))^(-1))=((x)^(-\frac(1)(2)))\]
Therefore we get:
\[\frac(1)(((x)^(3)))=((x)^(-3))\to \frac(((x)^(-3+1)))(-3 +1)=\frac(((x)^(-2)))(-2)=-\frac(1)(2((x)^(2)))\]
In total, collecting everything into one expression, we can write:
Example No. 3
To begin with, we note that we have already calculated $\sqrt(x)$:
\[\sqrt(x)\to \frac(4((x)^(\frac(5)(4))))(5)\]
\[((x)^(\frac(3)(2)))\to \frac(((x)^(\frac(3)(2)+1)))(\frac(3)(2 )+1)=\frac(2\cdot ((x)^(\frac(5)(2))))(5)\]
Let's rewrite:
I hope I won't surprise anyone if I say that what we've just studied is just the most simple calculations primitive, the most elementary structures. Let's now look a little more complex examples, in which, in addition to the tabular antiderivatives, you will also need to remember school curriculum, namely, abbreviated multiplication formulas.
Solving more complex examples
Task No. 1
Let us recall the formula for the squared difference:
\[((\left(a-b \right))^(2))=((a)^(2))-ab+((b)^(2))\]
Let's rewrite our function:
We now have to find the prototype of such a function:
\[((x)^(\frac(2)(3)))\to \frac(3\cdot ((x)^(\frac(5)(3))))(5)\]
\[((x)^(\frac(1)(3)))\to \frac(3\cdot ((x)^(\frac(4)(3))))(4)\]
Let's put everything together into a common design:
Problem No. 2
In this case, we need to expand the difference cube. Let's remember:
\[((\left(a-b \right))^(3))=((a)^(3))-3((a)^(2))\cdot b+3a\cdot ((b)^ (2))-((b)^(3))\]
Taking this fact into account, we can write it like this:
Let's transform our function a little:
We count as always - for each term separately:
\[((x)^(-3))\to \frac(((x)^(-2)))(-2)\]
\[((x)^(-2))\to \frac(((x)^(-1)))(-1)\]
\[((x)^(-1))\to \ln x\]
Let us write the resulting construction:
Problem No. 3
At the top we have the square of the sum, let's expand it:
\[\frac(((\left(x+\sqrt(x) \right))^(2)))(x)=\frac(((x)^(2))+2x\cdot \sqrt(x )+((\left(\sqrt(x) \right))^(2)))(x)=\]
\[=\frac(((x)^(2)))(x)+\frac(2x\sqrt(x))(x)+\frac(x)(x)=x+2((x) ^(\frac(1)(2)))+1\]
\[((x)^(\frac(1)(2)))\to \frac(2\cdot ((x)^(\frac(3)(2))))(3)\]
Let's write the final solution:
Now attention! Very important thing, with which it is connected lion's share errors and misunderstandings. The fact is that until now, counting antiderivatives using derivatives and bringing transformations, we did not think about what the derivative of a constant is equal to. But the derivative of a constant is equal to “zero”. This means that you can write the following options:
- $((x)^(2))\to \frac(((x)^(3)))(3)$
- $((x)^(2))\to \frac(((x)^(3)))(3)+1$
- $((x)^(2))\to \frac(((x)^(3)))(3)+C$
This is very important to understand: if the derivative of a function is always the same, then the same function has an infinite number of antiderivatives. We can simply add any constant numbers to our antiderivatives and get new ones.
It is no coincidence that in the explanation of the problems that we just solved it was written “Write down general form primitives." Those. It is already assumed in advance that there is not one of them, but a whole multitude. But, in fact, they differ only in the constant $C$ at the end. Therefore, in our tasks we will correct what we did not complete.
Once again we rewrite our constructions:
In such cases, you should add that $C$ is a constant - $C=const$.
In our second function we get the following construction:
And the last one:
And now we really got what was required of us in the original condition of the problem.
Solving problems of finding antiderivatives with a given point
Now that we know about constants and the peculiarities of writing antiderivatives, it is quite logical that next type problems when, from the set of all antiderivatives, it is required to find one single one that would pass through given point. What is this task?
The fact is that all antiderivatives of a given function differ only in that they are shifted vertically by a certain number. And this means that no matter what point on coordinate plane we didn’t take it, one antiderivative will definitely pass, and, moreover, only one.
So, the tasks that we will now solve are formulated as follows: it’s not easy to find an antiderivative, knowing the formula of the original function, but to choose exactly one of them that passes through a given point, the coordinates of which will be given in the problem statement.
Example #1
First, let’s simply count each term:
\[((x)^(4))\to \frac(((x)^(5)))(5)\]
\[((x)^(3))\to \frac(((x)^(4)))(4)\]
Now we substitute these expressions into our construction:
This function must pass through the point $M\left(-1;4 \right)$. What does it mean that it passes through a point? This means that if instead of $x$ we put $-1$ everywhere, and instead of $F\left(x \right)$ we put $-4$, then we should get the correct numerical equality. Let's do this:
We see that we have an equation for $C$, so let's try to solve it:
Let's write down the very solution we were looking for:
Example No. 2
First of all, it is necessary to reveal the square of the difference using the abbreviated multiplication formula:
\[((x)^(2))\to \frac(((x)^(3)))(3)\]
The original construction will be written as follows:
Now let's find $C$: substitute the coordinates of point $M$:
\[-1=\frac(8)(3)-12+18+C\]
We express $C$:
It remains to display the final expression:
Solving trigonometric problems
As final chord In addition to what we have just discussed, I propose to consider two more complex tasks, which contain trigonometry. In them, in the same way, you will need to find antiderivatives for all functions, then select from this set the only one that passes through the point $M$ on the coordinate plane.
Looking ahead, I would like to note that the technique that we will now use to find antiderivatives of trigonometric functions, in fact, is a universal technique for self-testing.
Task No. 1
Let's remember the following formula:
\[((\left(\text(tg)x \right))^(\prime ))=\frac(1)(((\cos )^(2))x)\]
Based on this, we can write:
Let's substitute the coordinates of point $M$ into our expression:
\[-1=\text(tg)\frac(\text( )\!\!\pi\!\!\text( ))(\text(4))+C\]
Let's rewrite the expression taking this fact into account:
Problem No. 2
This will be a little more difficult. Now you'll see why.
Let's remember this formula:
\[((\left(\text(ctg)x \right))^(\prime ))=-\frac(1)(((\sin )^(2))x)\]
To get rid of the “minus”, you need to do the following:
\[((\left(-\text(ctg)x \right))^(\prime ))=\frac(1)(((\sin )^(2))x)\]
Here is our design
Let's substitute the coordinates of point $M$:
In total, we write down the final construction:
That's all I wanted to tell you about today. We studied the term antiderivatives itself, how to count them from elementary functions, as well as how to find an antiderivative passing through a specific point on the coordinate plane.
I hope this lesson will help you at least a little to understand this complex topic. In any case, it is on antiderivatives that indefinite and indefinite integrals are constructed, so it is absolutely necessary to calculate them. That's all for me. See you again!