 Hi, I'm Zor. Welcome to Unisor Education. Today we will talk about another method of integration. The last lecture was about integration by parts, and today I will talk about the method which can be called substitution of variables, variable substitution. Actually it's a little simpler than integration by parts, but it's a very powerful method. Between these two, integration by parts and substitution, you basically have as much tools as possible to have to integrate whatever is possible to integrate. And don't forget that there are things which are just impossible to integrate simply, and we'll just be content with this. Anyway, so this lecture is part of the course of Advanced Mathematics, which is presented on Unisor.com. I do suggest you to watch this lecture from the website because it has notes and for registered students it has the ability to participate in the educational process like enroll in some course or topic or take exams, for instance, any number of times you want. And the site is completely free, no advertisement. So let's talk about integration using the variable substitution. Alright, let's just consider something which is very simple thing. You have certain integral, and let's say you know the answer to this. There is some kind of a function which you know which is an integral of f of x, which basically means that the derivative of function g of x is equal to f of x. So this is given. So let's say we know this. Now, let's consider a composite function which is g of w of x. And let's take its derivative. Okay. So its derivative is equal to. Now you remember that if you have a compound function, compound, composite, whatever, if you have a compound function, then to derive, to take derivative, you have to first differentiate the function itself, the outer function, so to speak, and we know its derivative. So it's f. And put as a parameter this wx. And then we have to multiply it by derivative of the inner function. So this is the rule which we know from differentiation. Alright, fine. Now, these are two different representations of the same thing, derivative of this function. Alright, so now let's integrate it. Well, if the functions are equal, then indefinite integrals from these functions must also be equal. Well, there is a constant involved, but let's just skip it for a while because if we will use the symbol of integral, it basically implies that there is some kind of a constant in it. Alright. Now, let's just slightly modify this particular expression. First of all, this is integral of derivative, right? So integral of a derivative, we know it's equal to the original function. Well, let's say plus c. And this we will slightly modify using the following thing. Now, derivative of w times dx is basically a differential of wx, right? And this is the formula of, oops, my small mistake here. The point of the whole thing is it's not g of x, it's g of w of x, right? Okay, so this is basically an expression of substitution of variable. So if we know this, then this is also true for some function w. See, instead of x here and here and here, we use any other function, w of x. Well, obviously, differentiable, etc. So instead of x, we use this, instead of dx, we use this, and instead of the answer. So what it actually means that if we know this, we can very easily find out all these types of integrals. So if we have some complicated function under our integral, if we are able to represent it in this particular way, that's the guarantee that we can very easily derive, we can very easily find out what the indefinite integral is. Well, that's it for the theory, so let's go to practice. I'll put this formula again on the top. So integral of f of w of x d w x equals to g of w of x plus c. Provided, obviously, the integral of f of x dx equals g of x plus c. So this is given, then this is true. Alright, my first example is integral x e to the power of x square dx. Okay, so what can we do? You see, this x square, that's really very unpleasant kind of power. So we have to somehow simplify it. Fortunately for us, we notice that there is this x, and we remember that x square derivative is equal to 2x, right? So if I will combine this and this together, I will have almost this, right? So I will have 1f, 1 half of d of x square, right? Differential of x square is derivative of x square, which is 2x, and 2 will be cancelling out times dx, which is this one, right? So this integral is the same as integral of e to the power of x square. Now, I put these parenthesis, so this square is interpreted as the square of the x, not square of e to the power of x, right? That's two different things. So e to the power of x square is basically e to the power of x times e to the power of x, which is e to the power of 2x. e to the power of x square is a completely different thing, right? So inner function here is x square, and this I will put 1 half in front of it dx square. Now, let's recall that integral of e to the power of x dx is equal to e to the power of x plus c, because derivative of e to the power of x is e to the power of x, right? So I will use this basically as this, but instead of x, I have a function x square, which means that the answer to this is e to the power of x square plus c. Well, 1 half, I'm sorry. Let's check it out. Let's get the derivative of this and see if we will get this. Okay, 1 half e to the power of x square derivative equals to 1. 1 half goes out. Now we have to differentiate e to the power of x square. Now derivative of e to the power of anything is e to the power of anything times derivative of the inner function. Now inner function is x square, derivative is 2x. 2 is cancelling out. So I have x times this, which is exactly what we need. So we were using this and substituted instead of x here, here and here in all three places, x square. Now, well, are we so smart we just realized it? Well, basically yes. Based on whatever experience you might accumulate, you just see this is x square and definitely you don't like this x square, but you have this x and now you just realize that x times dx is, well, with some coefficient, it's basically differential of x square, with coefficient of 1 half. That's basically a guess, if you wish. Again, that's why integration is an art more than skill. So you have to basically accumulate enough experience and knowledge to guess these things if it's possible to guess, obviously. Sometimes it's impossible. But obviously my examples are only those kinds where it is possible to do something like this. The next example is integral sine x times cosine square x dx. Okay, here again we immediately notice that this cosine square is some kind of a function which looks complicated, but this sine combined with dx, sine is, if you remember, its derivative of the cosine with a minus sign, right? So instead of sine x, we can put minus cosine x derivative, right? And if I will multiply it by dx, that's the same thing as minus derivative of cosine x. So the whole thing is equal to integral of cosine square x minus dx cosine of x. Now this is simpler because we know that integral of x square dx, this is just a power function and it's equal to 1 third x to the power of 3 plus constant, right? Because the derivative of x to the power of 3 is 3x square. 3 will cancel out, I will have only x square. So this looks exactly like this, instead of x I have a cosine x. So I will just basically use the same thing, so it's minus 1 third. And instead of x I will substitute cosine of x to the power of 3 plus c. Again, what prompted me this? Sine, cosine, I remember that sine is a derivative of the cosine with a minus sign, so I combine them together and I have a uniform dependency only on the cosine. So everything depends on the cosine right now and I can apply this substitution, basically substituting cosine x for another variable if you wish. Or in this formula substitute x for cosine, doesn't matter. Alright, and my last example, integral logarithms sine x cosine x dx. Well, I took everything kind of about logarithms and trigonometric functions, power function will be here, etc., so all together. Well, obviously the first guess, the very intelligent guess is to combine this, cosine, remember it's a derivative of sine, so derivative of sine times dx is d sine x, right? Okay, so now, what I'm saying is if I knew this integral, if I knew this integral, I would just take this solution, g of x and substitute sine of x instead of x, right? So let's concentrate on this one. Integral logarithm x dx equals 2. Well, here we can do it by parts, right? So this is like u, this is dv, so it's by multiplication, x times logarithm x minus integral x d logarithm x, right? Now, d logarithm x is dx divided by x, so this is obviously cancelling out, so it's 1 and integral is equal to x plus c. So I know this integral. Well, let's just substitute. We will get the result of this would be sine x logarithm of sine x minus sine x plus c, right? How does it look? Let's just check it out just in case. All right, let's take the derivative of this one. Now, this is a product, so the derivative of this product is derivative of the first function, which is cosine of x times the second plus first function without modification and then derivative of the second one. Derivative of this one, derivative of logarithm of sine, it's a compound function, so first derivative of logarithm is 1 over whatever is under logarithm times derivative of the inner function cosine x, derivative of sine minus derivative of sine minus cosine x. So what do I have? This reduces, this is 1, now minus cosine plus cosine, so I have only this one, which is exactly the same as this. Well, works fine. So these are my three examples of how substitution of variables actually looks like. I do suggest you to basically do everything whatever I just did by yourself. Go to theunisor.com, all these examples are there with answers. So just check yourself out and good luck. Thank you.