Definition 1
The antiderivative $F(x)$ for the function $y=f(x)$ on the segment $$ is a function that is differentiable at each point of this segment and the following equality holds for its derivative:
Definition 2
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 $.
From the table of derivatives and Definition 2 we obtain the table of basic integrals.
Example 1
Check the validity of formula 7 from the table of integrals:
\[\int tgxdx =-\ln |\cos x|+C,\, \, C=const.\]
Let's differentiate the right-hand side: $-\ln |\cos x|+C$.
\[\left(-\ln |\cos x|+C\right)"=-\frac(1)(\cos x) \cdot (-\sin x)=\frac(\sin x)(\cos x) =tgx\]
Example 2
Check the validity of formula 8 from the table of integrals:
\[\int ctgxdx =\ln |\sin x|+C,\, \, C=const.\]
Let's differentiate the right-hand side: $\ln |\sin x|+C$.
\[\left(\ln |\sin x|\right)"=\frac(1)(\sin x) \cdot \cos x=ctgx\]
The derivative turned out to be equal to the integrand. Therefore, the formula is correct.
Example 3
Check the validity of formula 11" from the table of integrals:
\[\int \frac(dx)(a^(2) +x^(2) ) =\frac(1)(a) arctg\frac(x)(a) +C,\, \, C=const .\]
Let's differentiate the right-hand side: $\frac(1)(a) arctg\frac(x)(a) +C$.
\[\left(\frac(1)(a) arctg\frac(x)(a) +C\right)"=\frac(1)(a) \cdot \frac(1)(1+\left( \frac(x)(a) \right)^(2) ) \cdot \frac(1)(a) =\frac(1)(a^(2) ) \cdot \frac(a^(2) ) (a^(2) +x^(2) ) \]
The derivative turned out to be equal to the integrand. Therefore, the formula is correct.
Example 4
Check the validity of formula 12 from the table of integrals:
\[\int \frac(dx)(a^(2) -x^(2) ) =\frac(1)(2a) \ln \left|\frac(a+x)(a-x) \right|+ C,\, \, C=const.\]
Let's differentiate the right-hand side: $\frac(1)(2a) \ln \left|\frac(a+x)(a-x) \right|+C$.
$\left(\frac(1)(2a) \ln \left|\frac(a+x)(a-x) \right|+C\right)"=\frac(1)(2a) \cdot \frac( 1)(\frac(a+x)(a-x) ) \cdot \left(\frac(a+x)(a-x) \right)"=\frac(1)(2a) \cdot \frac(a-x)( a+x) \cdot \frac(a-x+a+x)((a-x)^(2) ) =\frac(1)(2a) \cdot \frac(a-x)(a+x) \cdot \ frac(2a)((a-x)^(2) ) =\frac(1)(a^(2) -x^(2) ) $The derivative turned out to be equal to the integrand. Therefore, the formula is correct.
Example 5
Check the validity of formula 13" from the table of integrals:
\[\int \frac(dx)(\sqrt(a^(2) -x^(2) ) ) =\arcsin \frac(x)(a) +C,\, \, C=const.\]
Let's differentiate the right-hand side: $\arcsin \frac(x)(a) +C$.
\[\left(\arcsin \frac(x)(a) +C\right)"=\frac(1)(\sqrt(1-\left(\frac(x)(a) \right)^(2 ) ) ) \cdot \frac(1)(a) =\frac(a)(\sqrt(a^(2) -x^(2) ) ) \cdot \frac(1)(a) =\frac( 1)(\sqrt(a^(2) -x^(2) ) ) \]
The derivative turned out to be equal to the integrand. Therefore, the formula is correct.
Example 6
Check the validity of formula 14 from the table of integrals:
\[\int \frac(dx)(\sqrt(x^(2) \pm a^(2) ) =\ln |x+\sqrt(x^(2) \pm a^(2) ) |+ C,\, \, C=const.\]
Let's differentiate the right-hand side: $\ln |x+\sqrt(x^(2) \pm a^(2) ) |+C$.
\[\left(\ln |x+\sqrt(x^(2) \pm a^(2) ) |+C\right)"=\frac(1)(x+\sqrt(x^(2) \pm a^(2) ) \cdot \left(x+\sqrt(x^(2) \pm a^(2) ) \right)"=\frac(1)(x+\sqrt(x^(2) \ pm a^(2) ) \cdot \left(1+\frac(1)(2\sqrt(x^(2) \pm a^(2) ) ) \cdot 2x\right)=\] \[ =\frac(1)(x+\sqrt(x^(2) \pm a^(2) ) \cdot \frac(\sqrt(x^(2) \pm a^(2) ) +x)( \sqrt(x^(2) \pm a^(2) ) =\frac(1)(\sqrt(x^(2) \pm a^(2) ) ) \]
The derivative turned out to be equal to the integrand. Therefore, the formula is correct.
Example 7
Find the integral:
\[\int \left(\cos (3x+2)+5x\right) dx.\]
Let's use the sum integral theorem:
\[\int \left(\cos (3x+2)+5x\right) dx=\int \cos (3x+2)dx +\int 5xdx .\]
Let us use the theorem about placing a constant factor outside the integral sign:
\[\int \cos (3x+2)dx +\int 5xdx =\int \cos (3x+2)dx +5\int xdx .\]
According to the table of integrals:
\[\int \cos x dx=\sin x+C;\] \[\int xdx =\frac(x^(2) )(2) +C.\]
When calculating the first integral, we use rule 3:
\[\int \cos (3x+2) dx=\frac(1)(3) \sin (3x+2)+C_(1) .\]
Hence,
\[\int \left(\cos (3x+2)+5x\right) dx=\frac(1)(3) \sin (3x+2)+C_(1) +\frac(5x^(2) )(2) +C_(2) =\frac(1)(3) \sin (3x+2)+\frac(5x^(2) )(2) +C,\, \, C=C_(1 ) +C_(2) \]
Principal integrals that every student should know
The listed integrals are the basis, the basis of the fundamentals. These formulas should definitely be remembered. When calculating more complex integrals, you will have to use them constantly.
Pay special attention to formulas (5), (7), (9), (12), (13), (17) and (19). Don't forget to add an arbitrary constant C to your answer when integrating!
Integral of a constant
∫ A d x = A x + C (1)Integrating a Power Function
In fact, it was possible to limit ourselves to only formulas (5) and (7), but the rest of the integrals from this group occur so often that it is worth paying a little attention to them.
∫ x d x = x 2 2 + C (2)
∫ x 2 d x = x 3 3 + C (3)
∫ 1 x d x = 2 x + C (4)
∫ 1 x d x = ln | x | +C (5)
∫ 1 x 2 d x = − 1 x + C (6)
∫ x n d x = x n + 1 n + 1 + C (n ≠ − 1) (7)
Integrals of exponential functions and hyperbolic functions
Of course, formula (8) (perhaps the most convenient for memorization) can be considered as special case formulas (9). Formulas (10) and (11) for the integrals of the hyperbolic sine and hyperbolic cosine are easily derived from formula (8), but it is better to simply remember these relations.
∫ e x d x = e x + C (8)
∫ a x d x = a x ln a + C (a > 0, a ≠ 1) (9)
∫ s h x d x = c h x + C (10)
∫ c h x d x = s h x + C (11)
Basic integrals of trigonometric functions
A mistake that students often make is that they confuse the signs in formulas (12) and (13). Remembering that the derivative of the sine is equal to the cosine, for some reason many people believe that the integral of the function sinx is equal to cosx. This is not true! The integral of sine is equal to “minus cosine”, but the integral of cosx is equal to “just sine”:
∫ sin x d x = − cos x + C (12)
∫ cos x d x = sin x + C (13)
∫ 1 cos 2 x d x = t g x + C (14)
∫ 1 sin 2 x d x = − c t g x + C (15)
Integrals that reduce to inverse trigonometric functions
Formula (16), leading to the arctangent, is naturally a special case of formula (17) for a=1. Similarly, (18) is a special case of (19).
∫ 1 1 + x 2 d x = a r c t g x + C = − a r c c t g x + C (16)
∫ 1 x 2 + a 2 = 1 a a r c t g x a + C (a ≠ 0) (17)
∫ 1 1 − x 2 d x = arcsin x + C = − arccos x + C (18)
∫ 1 a 2 − x 2 d x = arcsin x a + C = − arccos x a + C (a > 0) (19)
More complex integrals
It is also advisable to remember these formulas. They are also used quite often, and their output is quite tedious.
∫ 1 x 2 + a 2 d x = ln | x + x 2 + a 2 | +C (20)
∫ 1 x 2 − a 2 d x = ln | x + x 2 − a 2 | +C (21)
∫ a 2 − x 2 d x = x 2 a 2 − x 2 + a 2 2 arcsin x a + C (a > 0) (22)
∫ x 2 + a 2 d x = x 2 x 2 + a 2 + a 2 2 ln | x + x 2 + a 2 | + C (a > 0) (23)
∫ x 2 − a 2 d x = x 2 x 2 − a 2 − a 2 2 ln | x + x 2 − a 2 | + C (a > 0) (24)
General rules of integration
1) Integral of the sum of two functions equal to the sum corresponding integrals: ∫ (f (x) + g (x)) d x = ∫ f (x) d x + ∫ g (x) d x (25)
2) The integral of the difference of two functions is equal to the difference of the corresponding integrals: ∫ (f (x) − g (x)) d x = ∫ f (x) d x − ∫ g (x) d x (26)
3) The constant can be taken out of the integral sign: ∫ C f (x) d x = C ∫ f (x) d x (27)
It is easy to see that property (26) is simply a combination of properties (25) and (27).
4) Integral of complex function, If internal function is linear: ∫ f (A x + B) d x = 1 A F (A x + B) + C (A ≠ 0) (28)
Here F(x) is an antiderivative for the function f(x). Please note: this formula only works when the inner function is Ax + B.
Important: there is no universal formula for the integral of the product of two functions, as well as for the integral of a fraction:
∫ f (x) g (x) d x = ? ∫ f (x) g (x) d x = ? (thirty)
This does not mean, of course, that a fraction or product cannot be integrated. It’s just that every time you see an integral like (30), you will have to invent a way to “fight” it. In some cases, integration by parts will help you, in others you will have to make a change of variable, and sometimes even “school” algebra or trigonometry formulas can help.
A simple example of calculating the indefinite integral
Example 1. Find the integral: ∫ (3 x 2 + 2 sin x − 7 e x + 12) d xLet us use formulas (25) and (26) (the integral of the sum or difference of functions is equal to the sum or difference of the corresponding integrals. We obtain: ∫ 3 x 2 d x + ∫ 2 sin x d x − ∫ 7 e x d x + ∫ 12 d x
Let us remember that the constant can be taken out of the integral sign (formula (27)). The expression is converted to the form
3 ∫ x 2 d x + 2 ∫ sin x d x − 7 ∫ e x d x + 12 ∫ 1 d x
Now let's just use the table of basic integrals. We will need to apply formulas (3), (12), (8) and (1). Let's integrate power function, sine, exponential and constant 1. Let’s not forget to add an arbitrary constant C at the end:
3 x 3 3 − 2 cos x − 7 e x + 12 x + C
After elementary transformations we get the final answer:
X 3 − 2 cos x − 7 e x + 12 x + C
Test yourself by differentiation: take the derivative of the resulting function and make sure that it is equal to the original integrand.
Summary table of integrals
| ∫ A d x = A x + C |
| ∫ x d x = x 2 2 + C |
| ∫ x 2 d x = x 3 3 + C |
| ∫ 1 x d x = 2 x + C |
| ∫ 1 x d x = ln | x | +C |
| ∫ 1 x 2 d x = − 1 x + C |
| ∫ x n d x = x n + 1 n + 1 + C (n ≠ − 1) |
| ∫ e x d x = e x + C |
| ∫ a x d x = a x ln a + C (a > 0, a ≠ 1) |
| ∫ s h x d x = c h x + C |
| ∫ c h x d x = s h x + C |
| ∫ sin x d x = − cos x + C |
| ∫ cos x d x = sin x + C |
| ∫ 1 cos 2 x d x = t g x + C |
| ∫ 1 sin 2 x d x = − c t g x + C |
| ∫ 1 1 + x 2 d x = a r c t g x + C = − a r c c t g x + C |
| ∫ 1 x 2 + a 2 = 1 a a r c t g x a + C (a ≠ 0) |
| ∫ 1 1 − x 2 d x = arcsin x + C = − arccos x + C |
| ∫ 1 a 2 − x 2 d x = arcsin x a + C = − arccos x a + C (a > 0) |
| ∫ 1 x 2 + a 2 d x = ln | x + x 2 + a 2 | +C |
| ∫ 1 x 2 − a 2 d x = ln | x + x 2 − a 2 | +C |
| ∫ a 2 − x 2 d x = x 2 a 2 − x 2 + a 2 2 arcsin x a + C (a > 0) |
| ∫ x 2 + a 2 d x = x 2 x 2 + a 2 + a 2 2 ln | x + x 2 + a 2 | + C (a > 0) |
| ∫ x 2 − a 2 d x = x 2 x 2 − a 2 − a 2 2 ln | x + x 2 − a 2 | + C (a > 0) |
Download the table of integrals (part II) from this link
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We study the concept « integral »
Integration was known back in Ancient Egypt. Of course, not in its modern form, but still. Since then, mathematicians have written many books on this topic. Especially distinguished themselves Newton And Leibniz , but the essence of things has not changed.
How to understand integrals from scratch? No way! To understand this topic you will still need a basic understanding of the basics. mathematical analysis. We already have information about , necessary for understanding integrals, on our blog.
Indefinite integral
Let us have some function f(x) .
Indefinite integral function f(x) this function is called F(x) , whose derivative is equal to the function f(x) .
In other words, an integral is a derivative in reverse or an antiderivative. By the way, read about how in our article.
The antiderivative exists for everyone continuous functions. Also, a constant sign is often added to the antiderivative, since the derivatives of functions that differ by a constant coincide. The process of finding the integral is called integration.
Simple example:
In order not to constantly calculate antiderivatives elementary functions, it is convenient to summarize them in a table and use ready-made values.
Complete table of integrals for students
Definite integral
When dealing with the concept of an integral, we are dealing with infinitesimal quantities. The integral will help to calculate the area of the figure, the mass of the inhomogeneous body, the distance traveled at uneven movement path and much more. It should be remembered that an integral is an infinite sum large quantity infinitesimal terms.
As an example, imagine a graph of some function.
How to find the area of a figure bounded by the graph of a function? Using an integral! Let us divide the curvilinear trapezoid, limited by the coordinate axes and the graph of the function, into infinitesimal segments. This way the figure will be divided into thin columns. The sum of the areas of the columns will be the area of the trapezoid. But remember that such a calculation will give an approximate result. However, the smaller and narrower the segments, the more accurate the calculation will be. If we reduce them to such an extent that the length tends to zero, then the sum of the areas of the segments will tend to the area of the figure. This is a definite integral, which is written like this:
Points a and b are called limits of integration.
« Integral »
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Rules for calculating integrals for dummies
Properties of the indefinite integral
How to solve an indefinite integral? Here we will look at the properties of the indefinite integral, which will be useful when solving examples.
- The derivative of the integral is equal to the integrand:
- The constant can be taken out from under the integral sign:
- The integral of the sum is equal to the sum of the integrals. This is also true for the difference:
Properties of a definite integral
- Linearity:
- The sign of the integral changes if the limits of integration are swapped:
- At any points a, b And With:
We have already found out that a definite integral is the limit of a sum. But how to get a specific value when solving an example? For this there is the Newton-Leibniz formula:
Examples of solving integrals
Below we will consider the indefinite integral and examples with solutions. We suggest you figure out the intricacies of the solution yourself, and if something is unclear, ask questions in the comments.
To reinforce the material, watch a video about how integrals are solved in practice. Don't despair if the integral is not given right away. Contact a professional service for students, and any triple or curved integral over a closed surface will be within your power.
