Practical lesson 2. Integration of trigonometric functions
The topic of the lesson is quite interesting. Integration. In general the integration process is very creative. Some people prefer to do crosswords, but we, mathematicians, always have a pleasant activity: to think and solve our tasks.
Integration of trigonometric functions. What should we do if under the integral sign we have a sine, cosine, tangent and cotangent connected by some signs of operations, and so on. There is a universal trigonometric substitution. Pay attention to the word «universal». Why? It turns out that all these functions are easily expressed in terms of the tangent of the half angle t=tg(x/2). It turns out that sin (x) is expressed through tg (x/2): sin (x)=2tg(x/2)/(tg(x/2))^2+1=2t/(1+t^2); cos (x) is expressed as cos(x)=(1-(tg(x/2)^2)/((tg(x/2))^2+1)= (1-t^2)/(1+t^2). It is clear that both tg(x)=sin(x)/cos(x) and ctg(x)=cos(x)/sin(x) are expressed. In addition, x is easily expressed using this universal substitution and the differential dx=2dt/(1+t^2) is found. thus, all functions are expressed in terms of the tangent of the half angle tg (x/2). We could say that we have nothing else to consider, but the trouble is that this substitution sometimes leads to very cumbersome calculations.
Example 1. There is dx/(sin(x)-2cos(x)) under the integral sign. When do we apply the universal trigonometric substitution? When we do not see any other techniques. When there are no other formulas that would allow us to transform this. Although, there are some ways here, of course. That's when nothing else comes to mind. Besides, it's better when sin(x), cos(x) and other functions that occur under the integral have the first power.
So, let's apply the universal trigonometric substitution. This is t=tg(x/2). Let's once again write out what we are going to replace all the expressions that occur here with. We need to know what we are going to replace sin(x)=2t/(1+t^2), cos(x)=(1-t^2)/ (1+t^2) and dx = 2dt/(1+t^2) with. These are the substitutions. Let us return to the integral. We write what we have in all places. The numerator of the fraction becomes quite complex, but we are not in a hurry (see the video). We can do the following: reduce the fractions to a common denominator, especially here where the denominators are the same. However, for me, it is always more convenient to multiply the numerator and denominator by 1+t^2. Let's see what the result is (see the video). There is 2dt/(2t-2+2t^2) under the integral sign. Obviously, we reduce it by 2. In the denominator, write the powers of t in the descending order. So (see the video), there is dt/(t^2+t-1) under the sign of the integral.
In general, the purpose of substitutions is to lead to a fractional-rational function, which we have. This is the simplest fraction. We can factor it out, but it is still easier for us to find out the full square of the denominator. Thus, this is under the integral sign: (dt+1/2)/(t +½)^2 -5/4. We have a tabular integral, only we need to add ½ in the place of t under the sign of the differential in the numerator: dt+1/2. The tabular integral – (see video).
Let's simplify these expressions a little (see the video) not to have such a cumbersome expression: we multiply the numerator and denominator of the fraction by 2 again, and get rid of the four floors. (See video)
We are almost at the end. tg(x/2)=t, so then we replace the variable t with tg(x/2) wherever it occurs (see the video). We have the answer. It is great.
Let me tell you again: the universal trigonometric substitution, despite its universality, often leads to really cumbersome expressions that the desire arises to calculate the integral in some other way. What other ways exist?
A little bit of theory. If you have a function under the integral sign that connects sin(x) and cos(x), we introduce the following terms: the integrand is odd with respect to sin(x), with respect to cos(x), and even with respect to both sin(x) and cos(x). What does it mean? What does odd with respect to sine mean? It means that if the integrand contains sin (x), we replace it with -sin(x), then the minus sign can be placed before the entire function, the same goes for cos(x). What is even with respect to sin(x) and cos(x) at the same time? This means that if we substitute sin(x) and cos(x) for -sin(x) and-cos(x), then after the transformation we come to the original function. Note what substitutions are used in each of these cases (see the video). The easiest way is to consider this in practice, to analyze the situation.
Example 2. There is (1-sin(x)) dx/cos (x) under the integral sign.
If we substitute cos(x) for -cos(x), then we can take out the minus. That is, the function is odd with respect to cos (x). This is the substitution t=sin (x) according to the table. The function is odd with respect to cos (x) of the integrand, then we replace the variable t=sin(x). We should get the integral where there is dt. What are we going to circle as dt? This is cos(x)dx (see the video). However, we do not see cos(x) anywhere here. What does this mean? We need to perform an additional transformation: multiply the numerator and denominator by cos(x). Let's do this. (See the video) with the sign of the integral (1-sin (x)) cos(x) dx / (cos(x))^2. cos(x)dx=dt. (cos(x))^2 is perfectly expressed in sin(x): (cos(x))^2 =1-(sin(x))^2=1-t^2 (see video). That's it, we're ready to move on to the new t variable. We have the numerator 1-t, the denominator 1-t^2 and here we have dt (see the video) under the integral sign: (1-t)/ (1-t^2). After reducing by the formula of the difference of squares 1-t^2 =(1-t)*(1+t), we have dt/(t+1) under the integral sign. Then we get the answer. We need to return to the variable x: t=sin(x). We have: ln|sin(x)+1|+C (see the video). We have the answer. By the way, the modulus can be removed here, because the expression under the logarithm sign is greater than or equal to 0. The problem is solved.
Example 3. There is tg(x) dx under the integral sign. I have already mentioned that calculating integrals is a very creative task, and the same integral allows you to apply many different methods. Let's take example 3. First, tg (x) is not in the table of integrals, and all substitutions are reduced to sine and cosine. So first we will express the tangent: tg(x)=sin(x)/cos (x) (see the video). Let's see what methods work here. In fact, I wouldn't even turn to trigonometry. The easiest way is to add the function under the differential sign. I write - (1/cos(x))*(- sin(x)) dx under the integral sign (see video). There is the function cos(x) and there is its derivative, which turns out to be a multiplier before dx, taking into account the minus sign (see the video). The simplest solution is under the integral sign: - (dcos(x))/(cos(x)) (see video). We immediately get a tabular integral. Thus, we do not analyze whether this is a function of trigonometric functions. In general, we do not even need the theory of this lesson with this approach. We only need to know the table of derivatives, the table of integrals, and the method of adding functions under the differential sign. We get -ln|cos(x)|+C.
Now let's consider other methods. This was the first method. We are not going to consider the second method: the universal trigonometric substitution is t=tg(x/2). It is not too difficult, but it's still not the optimal way. There is the third way.
We write this integral in terms of sin(x) and cos (x). There is sin(x)/cos (x) under the integral sign. Then let's analyze it using this table (see the video). We have to check that the condition is odd with respect to sine, odd with respect to sine, or even with respect to sine and cosine at the same time.
If we substitute minus sine x for the sine, we can take minus out the integral sign, the same applies to the cosine, and third, if we change the sign of both the sine and the cosine, we get the original integral. This means that all substitutions work (see the video).
We have considered the first method; in addition, there are five other ways to solve this problem.
I want to use this variable replacement t= sin(x) as an example. We are going to use t=sin(x). Let's write t= sin(x), and then we should have dt=cos(x)dx. There is no cosine multiplier here, so we take and multiply under the integral sign: (sin(x)/cos(x)^2)*cos(x)dx (see the video). There is sin(x)=t, cos(x)dx=dt, and cos (x)^2 left. By the basic trigonometric identity cos(x)^2=1-sin(x)^2, i.e. 1-t^2. We get: (see video). This is an improper fraction, a fractional-rational function, we must divide with the remainder (see the video). What do we have? (see the video). All the integrals turned out to be tabular; and we have t –ln|(t-1)/(t+1)|+C.
We have this integral and make substitutions. We have t in the numerator and 1 - t^2 in the denominator and cos(x)dx=dt. We take out -1/2. There is -1/2 * under the integral sign: 2tdt/(t^2-1) (See the video). We do it to add 2t under the differential sign. (see the video). -1/2ln|t^2-1|+C. Then we replace t with sin(x): -1/2ln|sin (x)^2-1|+C. Obviously, the modulus is equal to sin(x)^2-1= cos (x)^2. We get -1/2ln| cos(x)^2 |+C (although some further transformations are possible). The problem is solved. We should always use trigonometric formulas (see the video) where possible. Here is the simplest example, when you have an even degree of sine or cosine under the integral sign. There are reduction formulas: cos (a)^2=(1+cos(2a))/2. We write 1/2 and there is (1+cos(6x))dx under the integral sign. We almost get the answer 1/2x. This is the integral of the first summand. 1/12sin(6x)+C is the integral of the second summand.
We have considered all the examples. I wish you success!