The integral of the sine sin (x) is equal to the cosine, and with a minus sign. Many people often make mistakes because they cannot remember that the derivative of a sine is equal to a minus cosine, and that of a cosine is equal to a sine with a plus sign.
Those who study the original should remember that right side a constant should be added
This constant is determined with an additional condition.
The sine graph looks like
Sine is odd, and cosine is even function, so when integrating a minus sign appears. At the beginning, everything seems simple and clear to everyone. But sooner or later the time comes to complicate the integral, that is, to integrate the sine of a double angle, triple argument, etc. And many have difficulties with integration. To derive the integral formula for sin (k*x), we will carry out all the calculations from the beginning. In order to reduce the integral to a tabular formula, you need to enter a coefficient under the differential, but this will change the integral itself. Therefore, we simultaneously divide by the coefficient
Knowing this formula, we write the integral of the sine of a double angle in one line 
Next we can integrate the sine of the triple angle 
etc.
int(sin(k*x)=-1/k*cos(k*x).
Using the same formula, the integral of the sine of half an angle is derived, which is equal to minus 2 cosine of half an angle. 
The integral of the sine of one third x is equal to 
Common examples of sine integration
Example 1. Find the integral of sin(4*x).
Solution: Using the integration formula we find
Example 2. Calculate the integral of sin(5*x).
Solution: Perform integrations
Example 3. Integrate the expression sin(7*x).
Example 4. Find the integral of the function y=sin(x/5).
Solution: Find indefinite integral
Once you learn how to calculate simple integrals of sine, you can move on to definite integral
Example 5. Find the initial value of sin(x) which equals 2 at zero.
Solution: Calculate the initial
From the condition on the initial one we find the constant
-cos(0)+C=2;
C=2+cos(0)=3.
We return to the original one and substitute the found constant 
This is the answer to the problem.
Example 7. Integrate the sine of the double angle y=sin(2*x) from 0 to 45 degrees.
Solution: Write the integral of the sine and substitute the limits of integration 
According to the physical content, a definite integral equal to area figures bounded by the sin(x) function and the x-axis.
But the definite integral and area are not the same thing. The integral can be negative, but the area cannot. If a function has a large area under the x-axis, then its definite integral is negative. 
Square curved trapezoid equal to the integral of the difference between the equations of the upper curve and the lower one. 
In this case, the upper curve is the x-axis or y = 0. The lower one is the sine graph. Therefore, the formula for the area of the sine function is equal to 1, or a definite integral modulo. 
If a function is antisymmetric with respect to the abscissa axis, then its integral is equal to zero, and the area is equal to the double integral of the graph above the abscissa axis. For example, the integral of the sine of a double angle from -45 to 45 degrees is zero
At the same time, the area of the shaded figure is equal to one. 
Integrals of trigonometric functions.
Examples of solutions
On this lesson we will consider integrals of trigonometric functions, that is, the filling of the integrals will be sines, cosines, tangents and cotangents in various combinations. All examples will be analyzed in detail, accessible and understandable even for a teapot.
To successfully study integrals of trigonometric functions, you must have a good understanding of the simplest integrals, as well as master some integration techniques. You can familiarize yourself with these materials in lectures Indefinite integral. Examples of solutions And .
And now we need: Table of integrals, Derivatives table And Directory of trigonometric formulas. All methodological manuals can be found on the page Mathematical formulas and tables. I recommend printing everything out. I especially focus on trigonometric formulas, they should be in front of your eyes– without this, work efficiency will noticeably decrease.
But first, about what integrals are in this article No. There are no integrals of the form , 
- cosine, sine, multiplied by some polynomial (less often something with a tangent or cotangent). Such integrals are integrated by parts, and to learn the method, visit the lesson Integration by parts. Examples of solutions. Also here there are no integrals with “arches” - arctangent, arcsine, etc., they are also most often integrated by parts.
When finding integrals of trigonometric functions, a number of methods are used:
(4) We use the tabular formula 
, the only difference is that instead of “X” we have a complex expression.
Example 2
Example 3
Find the indefinite integral.
A classic of the genre for those who are drowning in the competition. As you probably noticed, in the table of integrals there is no integral of tangent and cotangent, but, nevertheless, such integrals can be found.
(1) We use the trigonometric formula
(2) We bring the function under the differential sign.
(3) We use the table integral 
.
Example 4
Find the indefinite integral.
This is an example for an independent solution, the full solution and answer are at the end of the lesson.
Example 5
Find the indefinite integral.
Our degrees will gradually increase =).
First the solution:
(1) We use the formula
(2) We use the main trigonometric identity 
, from which it follows that 
.
(3) Divide the numerator by the denominator term by term.
(4) We use the linearity property of the indefinite integral.
(5) We integrate using the table.
Example 6
Find the indefinite integral.
This is an example for an independent solution, the full solution and answer are at the end of the lesson.
There are also integrals of tangents and cotangents, which are in higher powers. The integral of the tangent cubed is discussed in the lesson How to calculate the area of a flat figure? Integrals of tangent (cotangent) to the fourth and fifth powers can be obtained on the page Complex integrals.
Reducing the degree of the integrand
This technique works when the integrand functions are stuffed with sines and cosines in even degrees. To reduce the degree, use trigonometric formulas 
, 
and , and the last formula is often used in the opposite direction: 
.
Example 7
Find the indefinite integral.
Solution:
In principle, there is nothing new here, except that we applied the formula 
(lowering the degree of the integrand). Please note that I have shortened the solution. As you gain experience, the integral of can be found orally; this saves time and is quite acceptable when finishing assignments. In this case, it is advisable not to describe the rule 
, first we verbally take the integral of 1, then of .
Example 8
Find the indefinite integral.
This is an example for an independent solution, the full solution and answer are at the end of the lesson.
This is the promised degree increase:
Example 9
Find the indefinite integral.
First the solution, then the comments:
(1) Prepare the integrand to apply the formula 
.
(2) We actually apply the formula.
(3) We square the denominator and take the constant out of the integral sign. It could have been done a little differently, but, in my opinion, it was more convenient.
(4) We use the formula
(5) In the third term we again reduce the degree, but using the formula 
.
(6) We present similar terms (here I divided term by term 
and did the addition).
(7) Actually, we take the integral, the linearity rule 
and the method of subsuming a function under the differential sign is performed orally.
(8) Combing the answer.
! In an indefinite integral, the answer can often be written in several ways
In the example just considered, the final answer could have been written differently - opening the brackets and even doing this before integrating the expression, that is, the following ending to the example is quite acceptable:
It’s quite possible that this option is even more convenient, I just explained it the way I’m used to solving it myself). Here is another typical example for an independent solution:
Example 10
Find the indefinite integral. 
This example can be solved in two ways, and you may succeed two completely different answers(more precisely, they will look completely different, but from a mathematical point of view they will be equivalent). Most likely, you will not see the most rational method and will suffer with opening brackets and using other trigonometric formulas. Most effective solution given at the end of the lesson.
To summarize the paragraph, we conclude: any integral of the form 
, where and – even numbers, is solved by the method of reducing the degree of the integrand.
In practice, I came across integrals with 8 and 10 degrees, and I had to solve their terrible mess by lowering the degree several times, resulting in long, long answers.
Variable Replacement Method
As mentioned in the article Variable change method in indefinite integral, the main prerequisite for using the replacement method is the fact that in the integrand there is a certain function and its derivative:
(functions are not necessarily in the product)
Example 11
Find the indefinite integral.
We look at the table of derivatives and notice the formulas, 
, that is, in our integrand there is a function and its derivative. However, we see that during differentiation, cosine and sine mutually transform into each other, and the question arises: how to perform a change of variable and what do we mean by sine or cosine?! The question can be solved by scientific poking: if we carry out the replacement incorrectly, then nothing good will come of it.
A general guideline: in similar cases, you need to designate the function that is in the denominator.
We interrupt the solution and make a replacement
Everything is fine in the denominator, everything depends only on , now it remains to find out what it will turn into.
To do this, we find the differential:
Or, in short:
From the resulting equality, using the rule of proportion, we express the expression we need:
So: 
Now our entire integrand depends only on and we can continue solving
Ready. Let me remind you that the purpose of the replacement is to simplify the integrand; in this case, everything came down to integration power function according to the table.
It is no coincidence that I described this example in such detail; this was done for the purpose of repetition and reinforcement of the lesson materials Variable change method in indefinite integral.
And now two examples for your own solution:
Example 12
Find the indefinite integral.
Example 13
Find the indefinite integral.
Complete solutions and answers at the end of the lesson.
Example 14
Find the indefinite integral.
Here again, in the integrand, there are sine and cosine (a function with a derivative), but in a product, and a dilemma arises - what do we mean by sine or cosine?
You can try to carry out a replacement using scientific poking, and if nothing works, then designate it as another function, but there is:
General guideline: you need to designate the function that, figuratively speaking, is in an “uncomfortable position”.
We see that in in this example the student cosine “suffers” from the degree, but the sine sits freely, on its own.
Therefore, let's make a replacement: 
If anyone still has difficulties with the algorithm for replacing a variable and finding the differential, then you should return to the lesson Variable change method in indefinite integral.
Example 15
Find the indefinite integral.
Let's analyze the integrand, what should be denoted by ?
Let's remember our guidelines:
1) The function is most likely in the denominator;
2) The function is in an “inconvenient position”.
By the way, these guidelines are valid not only for trigonometric functions.
The sine fits both criteria (especially the second), so a replacement suggests itself. In principle, the replacement can already be carried out, but first it would be nice to figure out what to do with? First, we “pinch off” one cosine:
We reserve for our “future” differential
And we express it through sine using the main trigonometric identity:
Now here's the replacement:
General rule: If in the integrand one of the trigonometric functions (sine or cosine) is in odd degree, then you need to “bite off” one function from the odd degree, and designate another function behind it. We are talking only about integrals where there are cosines and sines.
In the example considered, we had a cosine at an odd power, so we plucked one cosine from the power, and designated it as a sine.
Example 16
Find the indefinite integral.
Degrees are taking off =).
This is an example for you to solve on your own. Complete solution and the answer at the end of the lesson.
Universal trigonometric substitution
Universal trigonometric substitution is a common case of the variable replacement method. You can try to use it when you “don’t know what to do.” But in fact there are some guidelines for its application. Typical integrals where the universal trigonometric substitution needs to be applied are the following integrals: 
, , , 
etc.
Example 17
Find the indefinite integral.
The universal trigonometric substitution in this case is implemented in the following way. Let's make a replacement: . I don’t use the letter , but the letter , this is not some kind of rule, it’s just that, again, I’m used to solving things this way.
Here it is more convenient to find the differential; for this, from equality, I express:
I attach an arctangent to both parts: 
Arctangent and tangent cancel each other out:
Thus: 
In practice, you don’t have to describe it in such detail, but simply use finished result:
!
The expression is valid only if under the sines and cosines we simply have “X’s”, for the integral 
(which we'll talk about later) everything will be a little different!
When replacing, sines and cosines turn into the following fractions:
, , these equalities are based on well-known trigonometric formulas: 
, 
So, the final design could look like this:
Let's carry out a universal trigonometric substitution: 
For integration rational functions of the form R(sin x, cos x), a substitution is used, which is called the universal trigonometric substitution. Then . Universal trigonometric substitution often results in large calculations. Therefore, whenever possible, use the following substitutions. 
Integration of functions rationally dependent on trigonometric functions
1. Integrals of the form ∫ sin n xdx , ∫ cos n xdx , n>0a) If n is odd, then one power of sinx (or cosx) should be entered under the sign of the differential, and from the remaining even power should be passed to the opposite function.
b) If n is even, then we use formulas for reducing the degree
2. Integrals of the form ∫ tg n xdx , ∫ ctg n xdx , where n is an integer.
Formulas must be used
3. Integrals of the form ∫ sin n x cos m x dx
a) Let m and n be of different parities. We use the substitution t=sin x if n is odd or t=cos x if m is odd.
b) If m and n are even, then we use formulas for reducing the degree
2sin 2 x=1-cos2x , 2cos 2 x=1+cos2x .
4. Integrals of the form
If the numbers m and n are of the same parity, then we use the substitution t=tg x. It is often convenient to use the trigonometric unit technique.
5. ∫ sin(nx) cos(mx)dx , ∫ cos(mx) cos(nx)dx , ∫ sin(mx) sin(nx)dx
Let us use the formulas for converting the product of trigonometric functions into their sum
Examples
1. Calculate the integral ∫ cos 4 x·sin 3 xdx .
We make the replacement cos(x)=t. Then ∫ cos 4 x sin 3 xdx = 
2. Calculate the integral.
Making the replacement sin x=t , we get
3. Find the integral.
We make the replacement tg(x)=t. Substituting, we get
Note that the replacement ctg(x)=t is more convenient here, since then , and therefore
Integrating expressions of the form R(sinx, cosx)
Example No. 1. Calculate integrals: 
Solution.
a) Integration of expressions of the form R(sinx, cosx), where R is a rational function of sin x and cos x, are converted into integrals of rational functions using the universal trigonometric substitution tg(x/2) = t.
Then we have
A universal trigonometric substitution makes it possible to go from an integral of the form ∫ R(sinx, cosx) dx to an integral of a fractional rational function, but often such a substitution leads to cumbersome expressions. Under certain conditions, simpler substitutions are effective:
- If the equality R(-sin x, cos x) = -R(sin x, cos x)dx is satisfied, then the substitution cos x = t is applied.
- If the equality R(sin x, -cos x) = -R(sin x, cos x)dx holds, then the substitution sin x = t.
- If the equality R(-sin x, -cos x) = R(sin x, cos x)dx holds, then the substitution tgx = t or ctg x = t.
let us apply the universal trigonometric substitution tg(x/2) = t.
Then
Since the fraction is improper, then, isolating the whole part, we get
Returning to the original variable we will have
b) In the second example, consider an important special case, when the general expression ∫ R(sinx, cosx) dx is ∫ sin m x cos n xdx . In this particular case, if m is odd, the substitution cos x = t should be applied. If n is odd, the substitution sin x = t should be applied. If both type exponents are even non-negative numbers (in particular, one of them can be equal to zero), then perform the replacement using the known trigonometric formulas:
In this case 
Answer: 
Previously, given a given function, guided by various formulas and rules, we found its derivative. The derivative has numerous uses: it is the speed of movement (or, more generally, the speed of any process); slope tangent to the graph of a function; using the derivative, you can examine the function for monotonicity and extrema; it helps solve optimization problems.
But along with the problem of finding the speed according to a known law of motion, there is also an inverse problem - the problem of restoring the law of motion according to a known speed. Let's consider one of these problems.
Example 1. A material point moves in a straight line, the speed of its movement at time t is given by the formula v=gt. Find the law of motion.
Solution. Let s = s(t) be the desired law of motion. It is known that s"(t) = v(t). This means that to solve the problem you need to select a function s = s(t), the derivative of which is equal to gt. It is not difficult to guess that \(s(t) = \frac(gt^ 2)(2)\).
\(s"(t) = \left(\frac(gt^2)(2) \right)" = \frac(g)(2)(t^2)" = \frac(g)(2) \ cdot 2t = gt\)
Answer: \(s(t) = \frac(gt^2)(2) \)
Let us immediately note that the example is solved correctly, but incompletely. We got \(s(t) = \frac(gt^2)(2) \). In fact, the problem has infinitely many solutions: any function of the form \(s(t) = \frac(gt^2)(2) + C\), where C is an arbitrary constant, can serve as a law of motion, since \(\left (\frac(gt^2)(2) +C \right)" = gt \)
To make the problem 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 s(t) = (gt 2)/2 + C we get: s(0) = 0 + C, i.e. C = s 0. Now the law of motion is uniquely defined: s(t) = (gt 2)/2 + s 0.
In mathematics, mutually inverse operations are given different names and special notations are invented, for example: squaring (x 2) and extracting square root(\(\sqrt(x) \)), sine (sin x) 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, is integration.
The term “derivative” itself can be justified “in everyday terms”: the function y = f(x) “gives birth” to a new function y" = f"(x). The function y = f(x) acts as if it were 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) , primary image, or primitive.
Definition. The function y = F(x) is called antiderivative for the function y = f(x) on the interval X if the equality F"(x) = f(x) holds for \(x \in X\)
In practice, the interval X is usually not specified, but is implied (as the natural domain of definition of the function).
Let's give examples.
1) The function y = x 2 is antiderivative for the function y = 2x, since for any x the equality (x 2)" = 2x is true
2) The function y = x 3 is antiderivative for the function y = 3x 2, since for any x the equality (x 3)" = 3x 2 is true
3) The function y = sin(x) is antiderivative for the function y = cos(x), since for any x the equality (sin(x))" = cos(x) is true
When finding antiderivatives, as well as derivatives, not only formulas are used, 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 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. If F(x) is an antiderivative for f(x), then kF(x) is an antiderivative for kf(x).
Theorem 1. 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 \(y=\frac(1)(k)F(kx+m) \)
Theorem 2. If y = F(x) is an antiderivative for the function y = f(x) on the interval X, then the function y = f(x) has infinitely many antiderivatives, and they all have the form y = F(x) + C.
Integration methods
Variable replacement method (substitution method)
The method of integration by substitution involves introducing a new integration variable (that is, substitution). In this case, the given integral is reduced to a new integral, which is tabular or reducible to it. Common methods there is no selection of substitutions. The ability to correctly determine substitution is acquired through practice.
Let it be necessary to calculate the integral \(\textstyle \int F(x)dx \). Let's make the substitution \(x= \varphi(t) \) where \(\varphi(t) \) is a function that has a continuous derivative.
Then \(dx = \varphi " (t) \cdot dt \) and based on the invariance property of the integration formula for the indefinite integral, we obtain the integration formula by substitution:
\(\int F(x) dx = \int F(\varphi(t)) \cdot \varphi " (t) dt \)
Integration of expressions of the form \(\textstyle \int \sin^n x \cos^m x dx \)
If m is odd, m > 0, then it is more convenient to make the substitution sin x = t.
If n is odd, n > 0, then it is more convenient to make the substitution cos x = t.
If n and m are even, then it is more convenient to make the substitution tg x = t.
Integration by parts
Integration by parts - applying the following formula for integration:
\(\textstyle \int u \cdot dv = u \cdot v - \int v \cdot du \)
or:
\(\textstyle \int u \cdot v" \cdot dx = u \cdot v - \int v \cdot u" \cdot dx \)
