 Hi, I'm Zor, welcome to a new Zor education. Today I would like to solve the same problem I was solving in the previous lecture, integrate a certain trigonometric expression, and I would like to do it in two different ways, different from whatever I have suggested in the previous lecture. And we will get some different results probably, right? Alright, so this lecture is part of the course of advanced mathematics for teenagers in Unizor.com. I do suggest you to watch this lecture from the website, because the website not only has a video, but also detailed notes, especially for this lecture it will be important because I'm probably not going to go through all the calculations, I'll just refer you to the notes, I'll just show the idea and how to approach the problem, and then I will probably not spend so much time going into the detailed calculations. So, what's the problem we are talking about? We are talking about this problem, integrate this expression. Now, the previous lecture was actually addressing this particular integral, and the approach which I suggested was, if I will multiply by square root of 2 over 2, numerator and denominator. Now, in denominator, considering square root of 2 over 2 is sine of pi over 4 and cosine pi over 4. So, if I will multiply it, then I can say that this is the cosine x times cosine pi over 4 minus sine x and sine pi over 4, which is a formula for a cosine of a sum of x plus pi over 4. That simplifies the whole expression, and it's much easier to later on to integrate it. Now, there was a result which I have received in this previous lecture, which I'm going to write down here. So, the result of this integration is square root of 2 over 4, natural logarithm of 1 plus sine of x plus pi over 4 minus natural logarithm of 1 minus sine of x plus pi over 4 plus c, some constant. So, this is the answer to this integral using this particular methodology which I have suggested. Now, I would like to suggest another one, actually another two, but let's start with the first one. So, basically, my purpose to integrate this is, my first step is to simplify it somehow. So, in the previous lecture, I simplified by multiplying by square root of 2 over 2 and converting into cosine of x plus pi over 4. Now, I will do it differently. Now, I will recall that every trigonometric function can be rationalized through tangent of the half angle, right? Remember, sine x is equal to 2 tangent x divided by 1 plus tangent square of x, and the cosine of x is equal to 1 minus tangent square of x divided by 1 plus tangent square. So, x over 2 everywhere. Okay? Now, what does it give me? Well, it gives me the new way how this integral looks. So, they will have common denominator in the bottom, so it will go to the top. So, in the top, I will have 1 plus tangent of x over 2. And here, I will have cosine, which is 1 minus tangent square of x over 2 minus sine, which is minus 2 tangent of x over 2 dx. Now, why is it simpler? Well, there is a very important observation which you can make. Remember this, what is derivative of tangent? It's 1 plus tangent square, okay? And here, oh, this is supposed to be square, right? Square. Now, here, I have this 1 plus tangent square. Well, this is x over 2, but it doesn't really matter, because I can always substitute this and multiply it by 2, right? So, now, we have x over 2 everywhere. So, I mean, we can substitute y is equal to x over 2. It doesn't really matter. But what's important is that 1 plus tangent square of this times dx2 is exactly differential of tangent of x over 2. And this, now, with obvious substitution, looks like 2 integral dy over 1 minus y square minus 2y, which is a rational function, which we know basically how to do it. Now, this is a quadratic polynomial, which means I can always express it as 1 minus y minus something times y minus something and have it presented in, like, sum of 2 integrals, where I have only the first degree of polynomial on the bottom. So, let me just do that. That's a simple operation. So, we need roots of this particular equation. So, if I have y square plus 2y minus 1 is equal to 0, we need the roots of this, which is 2 minus 2 plus minus square root of 4 plus 4, which is minus 2 plus minus square root of 8, which is 2 square root of 2 divided by 2, which is minus 1 plus minus square root of 2, right? Which means my polynomial is equal to y minus, which means plus 1 minus square root of 2y plus 1 plus square root of 2, right? Let me just check. y square is y square, y would be 1 and square root of 2 and another one and minus square root of 2. So, it's 2y and here I have 1 minus square root of 2 times 1 plus square root of 2, which is 1 minus 2, which is minus 1, so that's what it is. Which means that I can say that this is equal to 2 integral, well, let me put minus here, that would be easy, and here I will have y plus 1 minus square root of 2, this is supposed to be 2 times 1 y plus 1 plus square root of 2, which is obviously equals to minus 2 integral dy divided by y plus 1 minus square root of 2 and then dy and then I will have some integral of this, but I need some multipliers. So, let me just think about multipliers. So, I need minus here, right? So, y would be cancelling out and so I will have this plus this, minus this, minus this, which is minus 2, minus minus 2, this plus, sorry, this is plus, that's why. So, I have plus 2 minus minus 2, so it's 2 square root of 2. So, now I have to divide by 2 square root of 2, right? So, again, the common denominator is obviously this. Now, if I will bring it to common denominator, I will have this minus this, which is 2 square root of 2 and that's why I have to divide by 2 square root of 2. Alright, and this, as we know is, obviously, we can do this. So, it's minus, well, instead of 1 over square root of 2, I can put square root of 2 over 2. That's kind of easier for me. And then I will have, now, what is this? This is integral of 1 over something. Well, this constant, I can obviously put it into differential because it's a constant. So, it's a logarithm of y plus 1 minus square root of 2 where y is equal to tangent of x over 2 plus 1 minus square root of 2 minus logarithm of tangent of x over 2 plus 1 plus square root of 2. Something like this. Do I have the same answer? Plus and minus. Yeah, plus c. Now, look at this and look at this. Well, they look quite differently. And, well, here is the challenge for you. If somebody can bring this into this, I will publish it on that website, which contains this particular problem and do the proper attribution, obviously, to the author of this. Now, for myself, I really didn't really know how to do it and didn't spend much time. So, instead, I've decided to graph this and graph that. Well, there are web-based services. You can put the formula in and they can draw you a graph of this. So, I did this and the graphs were relatively similar as far as I can judge. But shifted by a constant from one to another, which means that I cannot directly convert this into this. I need this constant. So, there is some kind of a constant which I should add to this to convert it into this without a constant. And I don't know what this constant actually is. So, I did another thing. I did a difference between this function and this function as one big function and again graphed it. And lo and behold, my graph was a straight line parallel to x-actors, which means it's a constant. So, the difference between this formula and this formula is a constant unknown to me. And again, I didn't spend much time to convert one into another. But what's interesting is that a different approach gave you a completely different formula for an integral, right? Okay. So, that's my first approach, additional approach to whatever I did in the previous lecture. And now I have yet another one for the same integral. I just want to demonstrate how many different things can be done and how many different ways you can use to approach the same problem. And if you get more or less the same well, not more or less the same results, well, difference by a constant actually, then you are right. But anyway, I wanted to present you yet another way which I think is very interesting because it's related to a famous Wheeler's formula and complex numbers. Again, as you see, we are doing something completely different, completely different approach. We are using complex numbers and Wheeler's formula, which is basically e to the power of ix is equal to cosine x plus i sin x. I hope you remember it. If no, again, my lectures on complex numbers and trigonometry contain this particular formula, it's really amazing thing. They don't seem to be related exponential function and trigonometric functions. But when you have an imaginary number i in the exponent, then that's exactly how it looks. Now, why did I need this? Well, for a very simple reason because now I can express cosine and sine in terms of e to the power of ix and e to the power of anything is very easy to integrate. Now, obviously, whenever I will get involved in integration of trigonometric functions, I have to really explain what it is. We never spoke about derivative, for instance, of the functions with complex argument and never talked about integrals. So this is definitely a deficiency of this approach. But let me just tell you that basically the whole theory, whatever I was explaining about real arguments, functions with real arguments is definitely expandable to the functions with complex argument. All these limits and stuff like this, it's all basically transported without any problems. So the only thing is whenever I was talking in case of the limits, for instance, when I was talking about approaching one number approaching as a sequence, for instance, to a limit, I was using the difference between these numbers. Well, in case of complex numbers, I just cannot use just the difference. I have to use the norm, which is basically this. So when this goes to zero, it means both of them are going to zero. So with this particular, well, I should say a lame excuse for not really presenting the whole theory of differentiating and integrating of the functions with complex arguments, I will just use exactly the same methodology which I really approached, which I really addressed before, with real functions. I will just use it without any modification because, again, I'm just saying that, trust me, that the theory is basically built in a very, very similar fashion. Now, from this, what can we say? Well, e to the power minus ix is equal to e to the power of i minus x, right? So it's cosine of minus x plus i sine of minus x. Now cosine of minus x is the same as cosine of x. Now sine is an odd function, which means it's minus i sine x. Now these two, this and this, allow me to express cosine of x as, so if I will add this and this, my i sine will cancel out and I will have one half e to the power of ix plus e to the power minus ix. Now if I will subtract from this, I subtract this, I will have 2i, so sine is equal to 1 over 2i e to the power of ix minus e to the power minus ix. Now what is 1 over i? Well, 1 over i is obviously equal to minus i because i square is equal to 1, so minus i square is equal to, I mean i square is equal to minus 1, so minus i square is equal to 1. This is equal to minus 1 half e to the power of ix minus i to the power. Okay, so why did they do it? For very simple reason. I can substitute it and I will have everything related to, this will be 1 half e to the power ix plus e to the power minus ix minus and minus, so it's plus 1 half e to the power of ix minus e to the power minus ix, right? So what do we have now? 1 over i is equal to minus i, I made a mistake. Alright, so now what I can say is that this is equal to 2 integral, okay, this is dx, right? dx and here, what do we have here? We have 1 plus i e to the power of ix and 1 minus i e to the power of minus ix. Now, what do we do next? Well, we know that e to the power of x derivative is equal to e to the power of x, right? We know that derivative of e to the power of iax is equal to a e to the power of ax, right? Now, again, I'm making just a leap of face and I'm saying that e to the power of ix is equal to i e to the power of ix. The proof of that would be exactly the same as the proof of that. But again, I didn't do it. All I did was with real numbers and this is a complex number. So, again, leap of face, just trust me, this is exactly the same. Now, what does it give it to me? Well, very simple. Let's just multiply by e to the power of ix. Here I will have square and here I will have nothing, right? If I will multiply it by i and divide it by i, I can replace this with differential of e to the power of ix. Now, everything is expressed in terms of e to the power of ix, right? Which obviously means that I can just substitute and I will have 2 over i, well, it's minus 2i integral dy divided by 1 plus i y square plus 1 minus i. Now, this is slightly different from the case when I was able to represent this as the product of two linear functions, although in complex analysis you can do that. But in any case, what I'm saying right now that all the further details of how to integrate this are relatively technical, there is no idea about the whole thing. The idea was here to substitute with trigonometric functions with exponential, with a complex argument. So, this integral seems to be relatively simple and I can refer you to notes for this lecture to basically continue or I do suggest actually to do it yourself. And again, don't hesitate to deal with complex numbers in exactly the same way as you dealt with rational numbers. Just remember something, some simple things. Like, for instance, in one particular case, if you will divide it by 1 minus i, for instance, so you will have 1 plus something, you will have 1 minus i plus i divided by minus i, which is actually simplified to i, right? Because i times 1 is i, i times minus i is minus minus 1, which is 1. So, that can be simplified. And the square root of this, the square root of i will be necessary to substitute y square, also addressed in one of the lectures. The square root of i is actually square root of 2 over 2, 1 plus i. That's one of the lectures on complex numbers actually. So, using these two things, you basically reduce this integral to some multiplier, which I don't want to do whatever is multiplier, dz divided by 1 plus z square. So, if you divide by 1 minus i here, and then you have a square root of i like this, and that would be your final integral, which is, happened to be arc tangent of z, right? Plus z. And you have some multiplier here. So, the whole thing is, my answer to this, by the way, is, my answer is minus square root of 2 i arc tangent of i plus 1 e to the power of i x divided by square root of 2. Well, kind of complicated and absolutely not the same as this one. And quite frankly, I don't really know what to do with this. I mean, it works. I try to differentiate this, and it does come up with this. All i's cancel out, everything is fine. But, you know, I'm not really sure how they're related because it's supposed to be, you know, some kind of a function of complex argument, and they never actually defined even what arc tangent of complex argument is. So, nevertheless, you can, well, you can consider this as a curiosity of mathematics, which requires a lot of maybe additional research and definitions and theory and theorems to basically say that this is exactly the function derivative of which is equal to this. Right? Because we never defined what is, for instance, arc tangent of some complex number. All right. Anyway, as I was saying, it's just for a curiosity purpose. It's really a very nice exercise. And I actually love that formula, the Euler's formula, which express e to the power of ix as cosine x plus i sin x. It's really a remarkable thing. Anyway, I do suggest you to go through the notes very detail and very attentively. You probably would be, it would probably be very beneficial if you will try to do everything yourself. And then you can compare it with whatever I have. In any case, thanks very much. That's it for today. Good luck.