Integrals from trigonometric functions.
Examples of solutions

In this lesson we will look at 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 get acquainted 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 independent decision, the complete solution and answer is 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 more high degrees. 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's another one typical example for 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. The most effective solution is 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 perform 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 integrating the 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 the scientific method, 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 this example, the student cosine “suffers” from the degree, and 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 basic 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 replace: . 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 the 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:

Examples of solutions of integrals by parts are considered in detail, the integrand of which is the product of a polynomial by an exponential (e to the x power) or by a sine (sin x) or a cosine (cos x).

Content

See also: Method of integration by parts
Table of indefinite integrals
Methods for calculating indefinite integrals
Basic elementary functions and their properties

Formula for integration by parts

When solving examples in this section, the integration by parts formula is used:
;
.

Examples of integrals containing the product of a polynomial and sin x, cos x or e x

Here are examples of such integrals:
, , .

To integrate such integrals, the polynomial is denoted by u, and the remaining part by v dx.

Next, apply the integration by parts formula. Below is given detailed solution

these examples.

Examples of solving integrals

Example with exponent, e to the power of x
.

Determine the integral:
Let us introduce the exponent under the differential sign:.

e - x dx = - e - x d(-x) = - d(e - x)

Let's integrate by parts.
.
Here
.
.
.
We also integrate the remaining integral by parts.
.

Finally we have:

An example of defining an integral with sine
.

Calculate the integral:

e - x dx = - e - x d(-x) = - d(e - x)

Let's introduce sine under the differential sign: here u = x 2 , v = cos(2 x+3) ( , du = )′ x 2

dx


We also integrate the remaining integral by parts. To do this, introduce the cosine under the differential sign. here u = x, v = sin(2 x+3)

We also integrate the remaining integral by parts.

, du = dx

An example of defining an integral with sine
.

Example of the product of a polynomial and cosine

e - x dx = - e - x d(-x) = - d(e - x)

Let's introduce the cosine under the differential sign: here u = x 2 + 3 x + 5 , v = cos(2 x+3) ( sin 2 x )′ x 2

x 2 + 3 x + 5 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 ,
a) 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's use the formulas for converting the product of trigonometric functions into their sum:

• sin α cos β = ½(sin(α+β)+sin(α-β))
• cos α cos β = ½(cos(α+β)+cos(α-β))
• sin α sin β = ½(cos(α-β)-cos(α+β))

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

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.

In this case, to find the integral
let us apply the universal trigonometric substitution tg(x/2) = t.
Then Answer:

Table of antiderivatives ("integrals"). Table of integrals. Tabular not definite integrals. (The simplest integrals and integrals with a parameter). Formulas for integration by parts. Newton-Leibniz formula.

Table of antiderivatives ("integrals"). indefinite integrals.
(The simplest integrals and integrals with a parameter).	(The simplest integrals and integrals with a parameter).
Integral of a power function.
	An integral that reduces to the integral of a power function if x is driven under the differential sign.
Integral of an exponential, where a is a constant number.	Integral of a complex exponential function.
Integral of an exponential function.	An integral equal to the natural logarithm.
	An integral equal to the natural logarithm.
Integral: "Long logarithm".	Integral: "High logarithm".
Integral: "Long logarithm".
An integral, where x in the numerator is placed under the differential sign (the constant under the sign can be either added or subtracted), is ultimately similar to an integral equal to the natural logarithm.	Cosine integral.
Sine integral.	Integral equal to tangent.
Integral equal to cotangent.
Integral equal to both arcsine and arccosine	An integral equal to both arcsine and arccosine.
An integral equal to both arctangent and arccotangent.	Integral equal to cosecant.
Integral equal to secant.	Integral equal to arcsecant.
Integral equal to secant.	Integral equal to secant.
Integral equal to arccosecant.	Integral equal to the hyperbolic sine.
Integral equal to hyperbolic cosine.	Integral equal to the hyperbolic sine, where sinhx is the hyperbolic sine in the English version.
Integral equal to the hyperbolic cosine, where sinhx is the hyperbolic sine in the English version.	Integral equal to the hyperbolic tangent.
Integral equal to the hyperbolic cotangent.	Integral equal to the hyperbolic secant.

Integral equal to the hyperbolic cosecant.

Formulas for integration by parts. Integration rules.
Formulas for integration by parts. Newton-Leibniz formula. Rules of integration.
Integrating a product (function) by a constant:
Integrating the sum of functions:
indefinite integrals: Formula for integration by parts
definite integrals: Formula for integration by parts	Newton-Leibniz formula

Where F(a),F(b) are the values ​​of the antiderivatives at points b and a, respectively.

Table of derivatives. Tabular derivatives. Derivative of the product. Derivative of the quotient. Derivative of a complex function.

If x is an independent variable, then:
	Table of derivatives. Tabular derivatives."table derivative" - ​​yes, unfortunately, this is exactly how they are searched for on the Internet
	Derivative of a power function
Derivative of the exponent	Derivative of a complex exponential function
Derivative of exponential function	Derivative of a logarithmic function
	Derivative of the natural logarithm
Derivative of the natural logarithm of a function	Derivative of sine
Derivative of cosine	Derivative of cosecant
Derivative of arcsine	Derivative of arc cosine
Derivative of arcsine	Derivative of arc cosine
Tangent derivative	Derivative of cotangent
Derivative of arctangent	Derivative of arc cotangent
Derivative of arctangent	Derivative of arc cotangent
Derivative of arcsecant	Derivative of arccosecant
Derivative of arcsecant	Derivative of arccosecant
Derivative of the hyperbolic sine Derivative of the hyperbolic sine in the English version	Derivative of hyperbolic cosine Derivative of hyperbolic cosine in English version
Derivative of hyperbolic tangent	Derivative of hyperbolic cotangent
Derivative of the hyperbolic secant	Derivative of the hyperbolic cosecant

Rules of differentiation. Derivative of the product. Derivative of the quotient.
Derivative of a complex function.
Derivative of a product (function) by a constant:
Derivative of sum (functions):
Derivative of the product (functions):
Derivative of the quotient (of functions):

Derivative of a complex function:

Properties of logarithms. Basic formulas for logarithms. Decimal (lg) and natural logarithms (ln).
Basic logarithmic identity
Let's show how any function of the form a b can be made exponential. Since a function of the form e x is called exponential, then

Any function of the form a b can be represented as a power of ten

Natural logarithm ln (logarithm to base e = 2.718281828459045...) ln(e)=1; log(1)=0

Taylor series. Taylor series expansion of a function. It turns out that the majority practically encountered

mathematical functions can be represented with any accuracy in the vicinity of a certain point in the form of power series containing powers of a variable in increasing order. For example, in the vicinity of the point x=1: When using series called Taylor's rows,

mixed functions containing, say, algebraic, trigonometric and exponential functions can be expressed as purely algebraic functions. Using series, you can often quickly perform differentiation and integration.

1) The Taylor series in the neighborhood of point a has the form: , where f(x) is a function that has derivatives of all orders at x=a. Rn- remainder term

2)

in the Taylor series is determined by the expression

3) The k-th coefficient (at x k) of the series is determined by the formula A special case of the Taylor series is the Maclaurin (=McLaren) series

(the expansion occurs around the point a=0)

at a=0

members of the series are determined by the formula

1. In order for the function f(x) to be expanded into a Taylor series on the interval (-R;R), it is necessary and sufficient that the remainder term in the Taylor (Maclaurin (=McLaren)) formula for this function tends to zero as k →∞ on the specified interval (-R;R).

2. It is necessary that there are derivatives for a given function at the point in the vicinity of which we are going to construct the Taylor series.

Properties of Taylor series.

If f is an analytic function, then its Taylor series at any point a in the domain of definition of f converges to f in some neighborhood of a.

There are infinitely differentiable functions whose Taylor series converges, but at the same time differs from the function in any neighborhood of a. For example:

Taylor series are used in approximation (approximation - scientific method, which consists in replacing some objects with others, in one sense or another close to the original ones, but simpler) functions by polynomials. In particular, linearization ((from linearis - linear), one of the methods of approximate representation of closed nonlinear systems, in which the study of a nonlinear system is replaced by the analysis of a linear system, in some sense equivalent to the original one.) equations occurs by expanding into a Taylor series and cutting off all terms above first order.

Thus, almost any function can be represented as a polynomial with a given accuracy.

Examples of some common decompositions power functions into the Maclaurin series (=McLaren, Taylor in the vicinity of point 0) and Taylor in the vicinity of point 1. The first terms of the expansions of the main functions in the Taylor and McLaren series.

Examples of some common expansions of power functions in Maclaurin series (=McLaren, Taylor in the vicinity of point 0)

Examples of some common Taylor series expansions in the vicinity of point 1

Basic trigonometric formulas and basic substitutions are presented. Methods for integrating trigonometric functions are outlined - integration of rational functions, product of power functions of sin x and cos x, product of a polynomial, exponential and sine or cosine, integration of inverse trigonometric functions. Non-standard methods are affected.

Content

Standard methods for integrating trigonometric functions

General approach

First, if necessary, the integrand must be transformed so that the trigonometric functions depend on a single argument, which is the same as the integration variable.

For example, if the integrand depends on sin(x+a) And cos(x+b), then you should perform the conversion:
cos (x+b) = cos (x+a - (a-b)) = cos (x+a) cos (b-a) + sin ( x+a ) sin (b-a).
Then make the replacement z = x+a.

As a result, trigonometric functions will depend only on the integration variable z. When trigonometric functions depend on one argument that coincides with the integration variable (let's say it's z), that is, the integrand consists only of functions like, sin z, cos z, tg z ctg z
.
, then you need to make a substitution Such a substitution leads to the integration of rational or irrational functions (if there are roots) and allows one to calculate the integral if it is integrated into.

elementary functions

However, you can often find other methods that allow you to evaluate the integral in a shorter way, based on the specifics of the integrand. Below is a summary of the main such methods.

Methods for integrating rational functions of sin x and cos x Rational functions from And sin x cos x Rational functions from, sin x are functions formed from and any constants using the operations of addition, subtraction, multiplication, division and raising to an integer power. They are designated as follows: R(sin x, cos x)
.
.

This may also include tangents and cotangents, since they are formed by dividing sine by cosine and vice versa.
Integrals of rational functions have the form:
Methods for integrating rational trigonometric functions are as follows. and any constants using the operations of addition, subtraction, multiplication, division and raising to an integer power. They are designated as follows: R 1) Substitution always leads to the integral of a rational fraction. However, in some cases, there are substitutions (these are presented below) that lead to shorter calculations. Rational functions from.
2) If R and any constants using the operations of addition, subtraction, multiplication, division and raising to an integer power. They are designated as follows: R cos x → - cos x 3) If R multiplied by -1 when replacing sin x.
sin x → - sin x and any constants using the operations of addition, subtraction, multiplication, division and raising to an integer power. They are designated as follows: R, then the substitution t = 1) Substitution always leads to the integral of a rational fraction. However, in some cases, there are substitutions (these are presented below) that lead to shorter calculations. 4) If R 3) If R does not change as with simultaneous replacement , And, then the substitution t = tg x.

or t =
, , .

ctg x

Examples:

Product of power functions of cos x and sin x

Integrals of the form are integrals of rational trigonometric functions. Therefore, the methods outlined in the previous section can be applied to them. Methods based on the specifics of such integrals are discussed below. If m and n - Rational functions from, then the substitution t = sin x rational numbers

, then one of the substitutions t =

;
;
;
.

the integral is reduced to the integral of the differential binomial.
.

If m and n are integers, then integration is performed using reduction formulas:

Example:
, ,
Integrals of the product of a polynomial and sine or cosine

;
.

or t =
, .

Integrals of the product of a polynomial, exponential and sine or cosine

Example:
, ,
where P(x) is a polynomial in x, integrated using Euler’s formula
e iax = cos ax + isin ax(where i 2 = - 1 ).
To do this, using the method outlined in the previous paragraph, calculate the integral
.
By separating the real and imaginary parts from the result, the original integrals are obtained.

the integral is reduced to the integral of the differential binomial.
.

Non-standard methods for integrating trigonometric functions

Below are a number of non-standard methods that allow you to perform or simplify the integration of trigonometric functions.

Dependence on (a sin x + b cos x)

If the integrand depends only on a sin x + b cos x, then it is useful to apply the formula:
,
Where .

For example

Resolving fractions from sines and cosines into simpler fractions

Consider the integral
.
The simplest method of integration is to decompose the fraction into simpler ones using the transformation:
sin(a - b) = sin(x + a - (x + b)) = sin(x+a) cos(x+b) - cos(x+a) sin(x+b)

Integrating fractions of the first degree

When calculating the integral
,
it is convenient to isolate the integer part of the fraction and the derivative of the denominator
a 1 sin x + b 1 cos x = A (a sin x + b cos x) + B (a sin x + b cos x)′ .
The constants A and B are found by comparing the left and right sides.

References:
N.M. Gunter, R.O. Kuzmin, Collection of problems on higher mathematics, "Lan", 2003.

See also: