 So let's see. When I study, I'm going to say, if I see, and then I'm going to say something like A. And then the usual form of this guy, this guy, for example, in the past, I can't remember. But I'm going to say S squared, Lape S squared, S squared. Then what is that the volumetric change that I have to do? The volumetric change is going to be X equals 2. I need to get rid of this guy because I'm going to factor out here, right? So I'm going to say, if I'm going to be given by P over A in a formula that gives me after I factor out this guy here, I'm going to get all this last. Okay, so what happens then? Let me put it in here. So, no reading. So, if I have the extra Lape S squared, then I have to put more extra here. But I want you to tell me what I need here. So, if I plug in this position, P over A, instead of my S, Lape S squared, Lape S squared, times P over A, and then I'm going to put something here, S squared, correct? All this is inside this formula. So, when I put these things, then I'm going to have P squared, Lape S squared, P squared, over A squared, and something here, S squared. Now, these days, I'm going to be able to cancel out. Yes? I'm going to be able to factor out here. I'm going to have P squared 1 plus, I don't know, something here, S squared. And then these people are going to have P squared over 1 plus something here, S squared. And I want an identity here. But let me get rid of these miserable S squared. Right? So, what is that? I know that 1 plus 100 squared is equal to? Equal to? Second squared. Second squared. So, when I see the S squared, Lape S squared, times S squared, my change of coordinate should be P over A, second squared. Because then, I want an identity, right? Because when I put it over there, if I put my channel over here, until it's going to cost 2, every place, all these ones last, is in the square root of that, it's just going to be, this time, this one of theta. You know this because you said it covered this part. Okay? Now, this is all the only possibility. This is 1 because she needs to be. But instead of a plus here, I have a minus. B squared minus A squared S. What is my change of coordinate? So, I do understand from this computation. I have to have P over A. X has to be equal to? I'm not thinking that they will be equal, right? Because it makes no sense. They will be equal than what I have here. S squared will be, and I see. So, what should be the thing that I have to put here instead of standard? Because standard is not going to work. What will work? Stay in that. Because now, this change of minus here. Now, he told me I have to use sine of theta. It comes a minus. I have a sine of theta right here. Theta right here. And what do I know? Now, what minus sine of theta is equal to? Sine of theta. Sine of theta. So, I'm going to have P over sine of theta. And, it's going to be cosine of theta. Which is something that I think will work. Right? Yes? So, this is very easy to use. This is very easy to use. So, I see that here is that I know what I'm using here is that I know that one class change in square is theta. That is cosine. And I'm going to use here, what do I need to do here? Now, what do I need? These things in here. Now, what I have is the other way around. So, I know that we're going to make identity like this. But that gives me something. So, what I have now is this. And I need to put something there to see a problem. And then, my bigger question is going to be, what is that to make? It's very clear because you have a beautiful minus to 3. You're very simple as to 1. What do you think is 5? So, I'm going to get 3 times 3 times 3 is 9, divided by 9 is 1. So, this is cosine of theta. That's looking at this. On the other side, which has 1 minus sine square of theta. And then, I'm going to divide sine square of theta. And you see that 1 over sine square is cos theta. Right? So, you know the way around. According to my law, we are not going to be told we have an identity like this for this cos theta is 4. And this is 15 times theta. So, this is just theta. 15 times theta is 4. That's the answer. Ah, okay, okay. And no, don't. Because I think I have to solve for this, right? My problem is that here, so let's be honest, it's much easier if I change it to that. This here. And look at me. And here. We're going to have these two problems in my chair. I kind of want to be 5 times 3 is 1. So, as you can see from here, I'm just going to put integral of square root of theta squared. And my DSP, what am I doing? Of theta plus 1. And over here, I have theta squared theta. This is my work, and then I compare it and put it. Is this right? Yes. Is this right? Yes. Okay. And then the square plus 1. So how does this begin? It looks like tiny cube, tiny square denominators. What is the derivative of this function? For this thing, how do we need to have a square here? And I don't have to just integrate this tangent as I want. What can I do there? I have to use my formula, tangent square equals to theta root root is there. So it says, you root for the tangent equals to theta squared of theta minus 1. It can be a times theta, in many different cases. Because now what we're going to do over here, and that's a trivial integral. Integrate that, and then we have to go back. Is that I want to integrate, and this is what I need to do. You, cube, or X3, and you, plan a constant. But I'm not done, baby. I'm not done because you is theta not theta. These guys are not theta. So over here is everything. This is equals to theta, cube, theta, theta not theta. From this one, what is the outflow? Outflow, of course. We know that tangent is 1,100, so this is clear for problems. But this is before you get it to be too short, and you've got an idea of what is what you have to do when you have this kind of, and I wrote them in the general, this is the easiest way. And it has something, a polynomial whose degree has to be smaller than this. So what you have to do is that you have to know what you need to do. That you need to know what you need to do. Right? So what do we do? We want to undo this kind of undo it. So what we do is we try let's together you'll be about this integral, and let's go to the inside. So this is 2x plus 5 over I get in 2 integrals. So I want to say this is equal to 8 over, 8, I don't know what it is, but I'm going to give it out. 8 over x minus 3, 3 over the other factor that is exactly. It has to be irreducible, meaning you cannot reduce this polynomial so this has to be irreducible solenoid. You cannot be factorial anymore. It has to be irreducible. Now what I want to do here is I want to take home a denominator. I'm going to take home a denominator. As you know, that's what I'm saying. I'm going to do it in this way. So, home a denominator is a product of this group. So this is going to be 3 minus 5. So this is class 7. And then I check this home denominator and that first factor. And I said, what is missing? Oh, this guy is missing. So this would be 8 times x plus 7 plus, if I compare this with this, what is missing? x minus 3, so this would be plus 3 times x minus 3. Then I'm going to point the top and see what I get. If it's an unoblique, then whatever has said so I can only compare as I told you in elementary school of kindergarten, apples with apples. If it's 2 or equal then that means that 2 has to be equal to 8 plus 3. So this implies that 8 plus 3 has to be equal to 2. These values are here 7, 8 minus 3d has to be equal to what? In your system that you were using I mean, you were using 5 times 3, counter this factor and solve for 8. What do you want me to do to be good on tl, solve for b, solving it here, what do you want? From this one, I'm going to say that this equation over here implies that b is equal to minus 10. Don't be like this. The mean is equal to 5. Okay, so that is just algebra. So what do I get? So b is 10 over 10. So what is this? 20 minus 11, so 11 times b which is what?