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About this Lesson
 Type: Video Tutorial
 Length: 12:25
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 133 MB
 Posted: 07/01/2009
This lesson is part of the following series:
Calculus (279 lessons, $198.00)
Calculus: Techniques of Integration (28 lessons, $40.59)
Calculus: Integration by Parts (5 lessons, $8.91)
Taught by Professor Edward Burger, this lesson comes from a comprehensive Calculus course. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/calculus. The full course covers limits, derivatives, implicit differentiation, integration or antidifferentiation, L'Hopital's Rule, functions and their inverses, improper integrals, integral calculus, differential calculus, sequences, series, differential equations, parametric equations, polar coordinates, vector calculus and a variety of other AP Calculus, College Calculus and Calculus II topics.
Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.
He has also taught at UTAustin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".
Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, "Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas" and of the textbook "The Heart of Mathematics: An Invitation to Effective Thinking". He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The "Journal of Number Theory" and "American Mathematical Monthly". His areas of specialty include number theory, Diophantine approximation, padic analysis, the geometry of numbers, and the theory of continued fractions.
Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.
About this Author
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 04/23/2010

He is ALWAYS GREAT!!!! MUST WATCH!! THANK YOU!!!! Hopefully people can truly appreciate how rare it is to have the ability to watch one of the best professors utilize all skills of teaching to produce such power packed lessons that stick in the brain. Every detail is thought of in the video presentation. THANKS AGAIN!
Techniques of Integration
Integration by Parts
Introduction to Integration by Parts Page [1 of 2]
Okay, now this is really something that I personally just think is great and I want to try to lead us into this by looking at a very simple integral, just to warm us up. So here's a very simple integral.
The simple integral is dx. Not a big deal, this is one that we learned early on in calculus. This is just the . S, not a big deal. But here's the question: did you ever wonder what the integral of the natural log function is? You see, it's not . That doesn't make sense, because if you take the derivative of , you don't get this, you get . So what is the function, whose derivative is natural log? Well, that's sort of a huge question mark. So how would you make sense out of this? Now, will any technique work that we might have seen earlier, for example, a usubstitution? I don't think so. It's not a fraction stuff, so any type of technique, like a partial fraction thing, I don't know exactly how that's going to work out. So, in fact, I don't know how to crack this open. And that is the challenge and that is the thing that I want us to consider. How would you crack this thing open? This is a very famous function, the natural log, and certainly it deserves to be integrated. So with that as the inspiration, what I want to do is now introduce a new method for integrating that, in many instances, will allow us to crack integrals wide open.
Okay, now let's just think about this new method, first of all, in context. Remember that integration is just the undoing of differentiation. So, in fact, any technique that we have in differentiation somehow, at least philosophically, must have reflections in the integration theory. And, in fact, we've seen things like this, for example, with the u dusubstitution. So with u dusubstitution, in fact, that's just employing the idea of the chain rule, but backwards. With a usubstitution, you let u = some clump, and then you realize that the derivative of the lump is somewhere sitting outside here. And, in that case, you can make a substitution and, if you want to check your answer, you would have to take the derivative of your answer, and taking the derivative will always involve the chain rule, because the usubstitution integration technique is sort of the chain rule in reverse.
So, well what other techniques have we seen for differentiating functions? Maybe we can actually find reflections of them to integrating techniques if we run them backwards. Well, that's exactly what I want to propose right now. And so let me remind you, first of all, of the product rule. So the product rule is great. The product rule rules. So here we go, the product rule. So this is now a fact about differentiations, so let's now think back to the old differentiation going forward idea. Suppose you want to take the derivative with respect to x of the product of two things. So let's say you have two things, maybe f(x)g(x). Well, the product rule says you take the first and multiply it by the derivative of the second plus the second times the derivative of the first. And that's the product rule. Great! Now, let's just think about this for a second and realize that if I were to take one of these terms, let's say I take this term, and I bring it over to this side here, just bring this term over to this side here. If I do that, I would see the following: I would see (f(x)g(x)), and then I'll bring this over to this side, so I subtract off f(x)g'(x). That would equal g(x)f'(x). Now, what does that do for us? It looks sort of complicated now. This the good oldfashioned, familiar product rule, and now I'm messing it up a little bit. Well, what I want to do now is actually integrate everything through. So let's now just take the antiderivative of everything. Let's integrate. If we integrate everything, what do we get? Well, we'll have to integrate this side with respect to x and then we'll have to integrate this side with respect to x. Now, integrating this term is actually pretty easy, because remember the integral is just the antiderivative. So if I integrate everything here, this side, and then integrate this side with respect to x, then this piece here, the integral and the derivative, just basically cancel each other out and I'm just left with that answer. So what I'm just left with is f(x)g(x) minus, and then I have the integral here, so the integral of f(x)g'(x)dx. And what's that going to equal on this side? On this side that's going to equal the integral of g(x)f'(x)dx. Now, that looks a little peculiar, but, in some sense, it is an integration formula. It's always true. And we can actually employ this fact as an integration technique. Now let's take a look at what it's saying.
Suppose I had to figure out this integral and I see that that integral is made up of a product of two people, one person and another person. And one person I see, let's say this one, can be easily differentiated. I can take the derivative of that person. And the other person, f'(x), I realize is a derivative of some other function. So let me say it again: I see an integral that's actually formed by the product of two things. One thing I can differentiate and one thing I can integrate, the integral of f' is just f. Then it turns out this formula tells me that that integral equals the product of the two functions themselves, f and g, minus this integral. And notice this integral is actually markedly different than this one, because here, instead of having a g(f'), I have an f(g'). So I switch the roles of one what thing I'm taking the derivative of and what thing I'm not taking the derivative of.
So let me recap. If I see an integral that's formed by the product of two things, one of which I can integrate very easily and one of which I can differentiate very easily, then that integral is equal to just the product of the original functions that I have, the f after I integrate, and then the g, which is sitting right in front of me, minus a new integral, where the roles of the derivative have swapped. So instead of being g(f'), I now have f(g'). This integral might be easier to integrate than this one. And if that's the case, then this is actually a useful technique, a useful fact. I can take this hard integral and convert it into an integral that might be easily rectified. This technique is called integration by parts, where I have two parts to an integral, one part, which I can easily integrate, and the other factor I can easily differentiate. And then I see that that integral is identical to the product of the two functions minus the integral where I swap the roles of what we're differentiating and what we're integrating. And this integral might be easier to resolve. This is a technique called integration by parts and all you can see is it just comes from the product rule. It's just the product rule now, in some sense, read backwards, or the integral form.
All right, let me just state this in a slightly different way, although it's absolutely the same, but some people like to think of it this way and, quite frankly, I like to think of it this way, too. Some people write the integration by parts technique, or that formula I just showed you, using u's and v's, because if you think of usubstitution, we write u a lot. This is sort of a substitution where you have two characters in our play, not just a u, but now we have a companion v. So it gets a little intriguing, a little bit of romance. The u is no longer alone. Suppose you have the integral of u dv. So let me actually say what this is implying. It's saying that I'm integrating two things and one thing I can easily integrate and the other thing I can easily differentiate. The integral of dv is just v and the derivative of u is just du. So I'm being a little informal here. Well then, the point is that this integral will be just the product of the two people themselves, so u v, minus a new integral, where I switch the roles of these guys. So this would be v du. This is a common way of expressing, a least in shorthand, integration by parts. But I want to try to convince you that, in fact, it's nothing more than what I just wrote here. Of course, you can see that, but if you could see that, you could see that it's nothing more than what I just wrote here, except now it's this piece here integral of u, and this is the dv. So, in fact, if I try to convert the dictionary here, this would be the u and this would be the dv, the thing that's easily integrated, the thing that's easily differentiated. So there's the integral of u dv, and that equals, well, u v, so that's going to be u v minus, there's the minus sign, and a new integral, where I swap the roles. So now this is the v function and this is the derivative of the u function, so du. So, in fact, this is a shorthand way of expressing this idea. The integral of u dv = uv minus the integral of v du. Using the product rule backwards produces a new integration technique called integration by parts.
Now, how do you actually apply integration by parts to an integral question? The answer is we'll have to figure out a very clever choice for the two characters, the two substitutions. It's sort of like a two substitutions sort of issue. We're going to have a usubstitution and a vsubstitution. And we're going to try to pick the usubstitution to be some function that's in our picture that can be easily differentiated, and then the v function should be a function that can be easily integrated. So we're going to look at a product of two things and ask, "Which one can I easily integrate? Which one can I easily differentiate?" and then produce an equivalent, but different integral, which will allow us, potentially to crack the original one. And I leave you just with the remark that we still have not cracked the integral of . Maybe integration by parts will allow us to crack the natural log wide open. We'll see. I'll see you at the next lecture.
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He is ALWAYS GREAT!!!! MUST WATCH!! THANK YOU!!!! Hopefully people can truly appreciate how rare it is to have the ability to watch one of the best professors utilize all skills of teaching to produce such power packed lessons that stick in the brain. Every detail is thought of in the video presentation. THANKS AGAIN!