 Hi, I'm Zor. Welcome to Unisor Education. Today I would like to consider a few examples of integration by parts. Now this lecture is part of the Advanced Mathematics course for high school students and teenagers. It's presented on Unisor.com. The course is free, so I recommend you to go to this website to watch it directly because every lecture has very detailed comments, and if you are registered then you will be able to take exams, for instance, anytime you want, etc. So, why I'm considering all these examples of integration? There is a very important difference between differentiation and integration. Differentiation actually has certain rules. Every elementary function is known to have certain derivative, and if you remember it, basically any however complex formula which contains the combination of elementary functions can be differentiated without any problems. Just follow the rules and you'll get there. With integration it's not as easy. Well, there are certain functions which you can construct which are not actually very easy to integrate, because to integrate you it's a little bit of a guessing game, obviously based on your experience, but in any way if you are presented the function and you have to find its integral which is another function derivative of which will be the first, the original one, that's not always easy. I mean there are certain rules you can try, but no guarantee that you will succeed, and you will reduce your integration into something relatively simple which you know about. So that's why integration is a little bit more of an art rather than a skill, if you wish. And that goes to the whole philosophical concept of this course. I'm always trying to emphasize that the purpose of this course is not as much to give you certain skills to approach certain practical problems. It's definitely not. Most likely will not use in your practical profession all these different things which you learn from this course. However, I view this course as a development tool, development of your creativity, your analytical abilities, your logic, etc. So you can consider integration as one of the very important parts of this development process because it's really very creative. You really cannot just follow the rules and guaranteed to succeed integrating the function. You really have to think about this. Well, obviously based on your experience, maybe you will find something similar to something else which you did before. So all these circumstances make integration more of an art and a perfect development tool for your creativity. Okay, with this opening statement, let's just go down to business. So today we'll talk about integration by parts. I have just examples which I will consider just straight integration and I will show you what I'm thinking about how to do this. And all these examples are dedicated to integration by parts. So before going to that, I will remind you what is integration by parts. So, well, obviously you can go to a previous lecture where I'm doing this and I'll just repeat very, very briefly. So the concept of integration by parts is based on derivative of the product of two functions. So if you have two functions and you would like to have derivative of this function, then the rule is that's the rule. So it's the first function times derivative of the second plus the first derivative times the second. From this, we can integrate, which means we can take the function, the derivative of which is this. So if I will integrate this and this, well, integral of sum is the sum of integrals, I will get this one, right? From this, I can notice that what is this? This is differentiation and integration. These are two opposite operations. I mean, integration is defined actually as an opposite to differentiation. To integrate, you have to find the function. Derivative of this is this. But this is a derivative of uv, right? So this is uv. From which follows the following rule, integral uv? Well, I deliberately admit the dx, obviously, u and v are all functions of some argument x and you always have to put dx everywhere. So I will put it here. So integral of u and derivative of v is equal to uv minus integral of v and derivative of u. So this is basically your integration by parts. In its most, you know, concise form, if you wish. Sometimes it's even written like this. Now, another very brief comment, whenever you're talking about integral of this, you're not talking about a particular function derivative of which is equal to this. Or in this case, you're not talking about a particular function derivative of this is equal to whatever is under integral. You're talking about familiar functions. All of them are different from each other by a constant, right? So we know that because if you add a constant to any function, the derivative will be exactly the same. That's why plus c is implied here and applied here. That's why I don't really write it down unless in a very last expression about integration of certain function, which is equal to certain other function, then I will use the plus c. If I have integrals here and here, this constant is implied in both cases, so I don't really write it down. So let's remember this and let's approach all our problems. Example number one, integral x, natural logarithm x dx. Now, let me just again remind you that integral of u dv, the derivative times dx is actually derivative of v is equal to uv minus integral of v dv, right? So that's my integration by parts. So this is, okay, that's the same thing. Now, look at this. What's the complication here? I mean, if it was something like a polynomial, then I know how to do the integration, right? Now, this is a combination of a polynomial and logarithm. Now, and here is the creative part. What's interesting about logarithm? That's what's interesting. The derivative of logarithm is much simpler than logarithm. It's kind of a polynomial, if you wish, with a negative x to the power of minus one. So, what's important is I would like to change this in such a way that I will have to differentiate the logarithm. So how can I do it? Well, very simply, let's consider this as x and then, sorry, as logarithm x multiplied by v star dx. That's what I would like to have. So this would be my u, this would be my v. Then, if I will apply my integration by parts, in the next step, I will have to differentiate u, which is logarithm. And that would simplify the picture. So my question is what is a v derivative of which is equal to x? Well, I know what it is. That's x squared divided by 2 derivative, right? What's the derivative of x squared? 2x divided by 2 is x, and that's exactly what I need. So I have replaced x with a derivative of some function, and now I can use my, so this is v. And now I can use my rule, which is u times v, which is x squared divided by 2 times logarithm x minus integral of v and derivative of u. Now v is this times derivative of logarithm dx. So now this differentiation will simplify my expression, and I will get this, which is equal to, now this is... Now, by the way, since I'm using the function logarithm, so that implies that I'm always considering only positive x, right? So if it's positive, then I can just divide by x numerator and denominator, and I have this, and the answer is what's integral of x divided by 2? Which function, if I take a derivative, gives me x divided by 2? Well, to get x, I need x squared, but now I need to multiply by something. x squared gives me 2x. I need x divided by 2, so I have to divide it by 4. And now I can put plus c, because I have no integrals anymore, so I cannot imply. So this is the answer, x squared divided by 2, logarithm x minus x squared divided by 4 plus c. Well, obviously I can simplify it, factor out x squared, that's a simple thing. Alright, that's it, that's my first thing. Now, what I do recommend in cases of integration, always check if you are right by differentiating your results. So let's just very quickly differentiate this and see if we will get x logarithm x. So we are talking about integral of x logarithm x dx, right? So if I will differentiate this, I have to have this. Well, let's try. Derivative of product of two functions is the first function times derivative of the second one plus the second one times derivative of the first, 2x divided by 2. Minus derivative of this is 2x divided by 4 and the constant we skip. Now, look at this. This is x divided by 2 and this is x divided by 2 if I will reduce by 2 with a minus sign. So what's left? This. And obviously 2 is reducing and I have x logarithm x exactly what needed to be this. Okay, so that's a mandatory check which I myself might actually not always do during this lecture, but I certainly encourage you to do always whenever you are doing some integration. Number two, integral e to the power of x cosine x dx. Okay, now let's think about other things. Again, how can I simplify this in some way? Well, I know that e to the power of x, if you take derivative, it will be e to the power of x. So that might not actually help. Now with a cosine, can I simplify? From cosine, I will get minus sign. From minus sign, I get minus cosine. So I'm not sure I will get the simplification in that, but I may get some kind of equation because after a certain number of integration by parts, from cosine, I will get to sine and from sine, I will get the cosine and I might get actually an equation where this integral would be actually part of this equation. Look at what I'm doing. So first, I'm changing this to sine x derivative. Since derivative of sine is cosine, this is equal to this, right? Now I will apply the rule of integration by part. So this is my u and this is my derivative of v. So it would be e to the power of x times sine x minus integral sine x and derivative of e to the power of x, which is e to the power of x dx. So you see what I'm talking about? If I will do it again, I might actually get again integral of e to the power of x and the cosine. And then I will get some kind of equation, okay? Okay, this retains minus. Again, sine will actually minus sine. This minus and sine, this is cosine derivative, right? Derivative of cosine is minus sine. So it's minus sine. So again, this is u and this is derivative of v. So I will get plus their product into the power of x cosine x minus integral of this times derivative of this, which is the same. Now you see where I can get my equation? This is integral and this is exactly the same as this one, right? So if I will solve this equation for this, how can I do it? Let's say this is equal to a. So a equals to e to the power of x sine x plus e to the power of x cosine x minus a. From which a, our integral is equal to a goes here, so it's one half e to the power of x times sine x plus cosine x. I'm not going to do the checking, but I do encourage you always to do the checking. Plus c obviously. Alright, so that's the answer. So what is some kind of a lesson to learn here? Sometimes you cannot explicitly get the result of your integration, but you get to something which is exactly the same and you can include it into equation like this. So that's just another approach to this. So what I'm saying is that sometimes you have this approach, sometimes you have that approach. The number of problems which you solve is really enriching your repertoire of different approaches which you can use. And integration by parts is one thing. Now if you have something like cosine or sine, you can consider that maybe integration by parts twice will give you a certain equation. This is just another way of doing this. Next. Okay, integral x x plus 1 dx. Well, the way how I can approach it is basically there are different ways to approach it. Let me tell you the easy way. Easy way is this. So plus 1 minus 1 equals x plus 1 times this which is x plus 1 to the power of 3 halves, right? 1 and 1 half, so it's 1 and a half dx minus integral of x plus 1 to the power of 1 half dx, right? Now whether it's dx or d of x plus 1 doesn't really matter because x is equal to 1 and x plus 1 is equal to 1. So dx is equal to d of x plus 1. And you can actually say that this the whole thing is equal to integral of x plus 1 to the power of 3 second g of x plus 1 minus integral of x plus 1 1 half dx plus 1. Now integral of the power we know what it is. It's the power plus 1, right? So this will be x plus 1 to the power of 5 second and I have to multiply by 2 fifths. So the derivative would be 5 second which would be reduced by this times x plus 1 to the power of 3 second. Same thing here. 1 half, so I need x plus 1 to the power of 3 second but I have to multiply by 2 thirds to neutralize this and plus c. So this is one of the answers which is perfectly legitimate way to approach this problem. Now I would still like to do it differently, maybe at least, and use the integration by parts. I would like to retain this answer. x plus 1 dx. So what else can I do here? Well, I might actually consider the square root of x plus 1 to be the derivative. What does it make the function v? It's x plus 1 to the power of 3 second multiplied by 2 thirds, right? Derivative of this is 3 seconds times x plus 1 to the power of 3 second minus 1 which is 1 half which is this, right? So this is my v. Okay, therefore this is equal to and this is my u and this is my v. Now their product which is 2 third x x plus 1 to the power of 3 second minus integral of v and u derivative. Now v is this one which is 2 third x plus 1 to the power of 3 second dx because the derivative of u derivative of x is 1. So that's my answer here. Now again, here I have basically just a power function. So it's equal to minus. So what is the function derivative of which is equal to this? Well, first of all let me retain this. Now I have to have x plus 1 to the 3 second so that would be x plus 1 to the power of 5 seconds times 2 fifths plus c, right? The derivative of this one is equal to x plus 1 to the power of 3 seconds. 5 second minus 1 is 3 seconds and the multiplier would reduce each other. Okay, so that's slightly different answer than this one as you see, right? Now it's probably a little bit of a challenge to say and to prove that these are exactly the same thing. They are exactly the same thing. So let's take the factor out, let's say x plus 1 to the power of 3 seconds here and x plus 1 to the power of 3 seconds there. You would see exactly the same expressions in parenthesis. So these are two different approaches with basically the same, which I do encourage actually you to prove that this is the same, but this is the same, I checked. So two different approaches and they basically give the same results although it looks a little different. By the way, I remember a joke. No matter how you calculate you get the same results, that's mathematics. If you get different results, it's a creative accounting. Alright, so we are in mathematics, we are not in accounting and that's why these answers are exactly the same. Okay, next. Now a little bit more x-lagorism square x dx. Well, a little bit more difficult than the one before and because this is the logarithm square, right? Before our first problem was x-lagorism x dx and we did it. And our purpose was to differentiate logarithm, right? So somehow we have to make logarithm the function which you will differentiate here, right? So how can we do it? Well, very simply, we'll do relatively similar to whatever it is. We will write it like this. So the derivative of x-square divided by 2 is x. So this is our v derivative and this is our u. And now this is equal to their product x-square divided by 2 logarithm square x minus integral of the function x-square divided by 2 times derivative of logarithm. Now what is derivative of logarithm square? Derivative of logarithm square is 2 logarithm x, that's the derivative of the function square, the power, times derivative of the inner function logarithm which is 1x, 1 over x. Now, why is it better? This is better for the same reason before. We are actually reducing, you see the power of logarithm now is 1, used to be 2. So if we will do it again, we will reduce it to a polynomial dx equals, let's do it again. So let me just replace this, x-square divided by 2 would be x and 2 divided by 2 would be 2. Right? That's where it is. So, well, actually, you know what? The simplest thing for me right now is just to apply my answer which I remember here because this is exactly what we were doing as the problem number one. So I'll just do it right now. I will write this and that's my answer. Obviously, if I did not solve my problem number one, I would have to do exactly the same integration by parts with this thing. So I will separate it logarithm x times derivative of x-square divided by 2 and that's why I would have x-square divided by 2 logarithm x divided by 2 logarithm right. That's what it is. Minus another integral and everything would be like in the problem number one. Okay, next. Okay, next. Integral of arc tangents x dx. Okay, obviously, I would like to differentiate arc tangents which means I will take this one as u and this one as dv. First, we have their product. So dx is dv or these derivative times, times dx, right? So this is one, basically. So I will have x times arc tangents x minus x times derivative of arc tangents x dx, right? What's the problem right now? Well, the problem is I have no idea what derivative of arc tangents is. I mean, I can obviously take a look at this but what if I don't remember? Well, I do remember that tangent of arc tangent x equals x. That's the definition of arc tangents, right? Our tangent is equal to, tangent of, our tangent is equal to original. So if I will take a derivative from both sides, oops, I forgot what's derivative of tangent. Well, all I remember is sine is a cosine and cosine is minus sine and also I remember the formula for a product. So that's a problem, right? Okay, so what is the tangent? So let me just make it first my research. What is this? Well, this is sine divided by cosine x. Okay, I will consider it again. I don't remember the derivative of the ratio. I do remember derivative of the product. So I will represent it as sine x times 1 over cosine x. That I might actually remember. So that's sine x times derivative of 1 over cosine. Well, that's cosine to the power of minus 1. So it's minus 1 divided by cosine squared times derivative of the cosine is minus sine. So that would be plus sine x. Okay, plus derivative of sine, which is cosine times the second one. So what is this? It's sine squared divided by cosine squared, which is tangent squared, plus 1. Okay, so I've got the derivative of the tangent. That's good. It's 1 plus tangent squared. Now I will use it here. So derivative of this is derivative of tangent, which is 1 plus tangent squared of whatever argument is, times derivative of the inner function, right? This is the composite, which I don't know. Now, tangent of arc tangent is x, so I can change this to x squared. So that's the derivative of the left part. Derivative of the right part is 1 derivative of x. From which our tangent x derivative is equal to 1 over 1 plus x squared. Okay, now, here all these efforts were only because I just don't remember what is the derivative. And, well, quite frankly, I did remember it. I just wanted to show you that if you don't remember something, there is always a way to derive it somehow. If you know the basic principles and some fundamental things like the derivative of the sine is a cosine and the derivative of the cosine is minus sine, that's actually enough to derive everything else. You don't have to load into your memory all these huge tables of all different functions, etc. Just remember a couple of functions. Everything else will just go directly from there by derivation. Okay, so, and that's what this is. So now what I can say is that this is equal to x arc tangent x minus integral x divided by 1 plus x squared. What can I do with this? That's another challenge. Well, I know that 1 over something is usually the result of differentiation of logarithm, right? Now I'm just making some intelligent guess. So if I have logarithm of 1 plus x squared, the derivative of this is equal to 1 over 1 plus x squared. But now I don't have just logarithm of x, I have logarithm of some inner function which is the composition and I have to multiply it by the derivative of the inner function which is 1 plus x squared derivative would be 2x, right? Oops, look at this. So my intelligent guess was actually correct because if I will do this instead of this I will put here, I will get exactly what I need. So that's why I can say that this is equal to x arc tangent x minus 1 half logarithm of 1 plus x squared plus c. So a little guess about this thing gave us this expression because the derivative of this is equal to this. Now what actually kind of prompted this? Well, the fact that 1 plus x squared is in the denominator and in the numerator I have x to the first degree, which is, well, with a factor of 2 is derivative of the inner function which is 1 plus x squared. Okay, that's good. Now my last problem is x times arc tangent x. Well, I actually would like to do exactly the same thing. I want somehow to include arc tangent into differentiation. So I would like arc tangent to be u. In which case my v should be 1 half of x squared because the derivative of this is exactly x. It's equal to 1, sorry, before the integral. I have the product u and v which is 1 half x squared arc tangent x minus integral of v which is 1 half x squared times derivative of arc tangents. Now I remember what it is. Now this is easy because what you can do here is the following. You can put, instead of x squared, you can put x squared plus 1 minus 1. So that's x squared plus 1 divided by 1 plus x squared dx and minus 1. And this minus will be plus 1 divided by 1 plus x squared dx, right? 1 half. Now basically I know everything because this is 1 arc tangent x minus 1 half x. Integral of 1 is x, right? Plus 1 half. Now this is something which we have already calculated a couple of times. This is 1 over 1 over 1 over plus x squared is a derivative from arc tangent. So I will have arc tangent x plus c. And that's the answer. I mean obviously you can combine this and this to make it a little bit more compact but that's the answer. Okay, so these were examples of how you can apply the rules of integration by parts to different kind of scenarios. I would recommend you to go through the notes for this lecture on unison.com and try to solve just by yourself all these integrals and see if you get the result. It's just very useful exercise. And well other than that I can also add that integration by parts is very important tool for you. And you obviously have to remember it. This is the main formulas. And again, do not think that whatever we're doing right now here is directly applicable to practical life. Yes, maybe but rarely. It's not really such a common place thing. What it is really applicable is directly to development of your creativity and analytical abilities and logic. I would recommend you to view all these exercises as exercises for your brain in as much as gym is exercise for your muscles. On this very bright note, this is the end of this lecture and I'll see you next. Thank you. Good luck.