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About this Lesson
 Type: Video Tutorial
 Length: 9:19
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 100 MB
 Posted: 06/26/2009
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.
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 Thinkwell
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Techniques of Integration
Integration by Parts
Inspirational Examples of Integration by Parts Page [1 of 2]
All right, let's take a look at some more inspirational examples of how to use the integration by parts technique to actually crack open some really nastylooking integrals. The theme here, what we're looking for, is a product of two things. So when could we possibly use integration by parts? When we see a product of two things inside the integral. One thing can be easily differentiated and one thing can be easily integrated. If we see that, that means that we might have a good candidate for using the technique of integration by parts. So let's take a look at a couple of examples here together.
The first one is the integral xe^x dx. Now, we could try the different techniques here, like a usubstitution, but that won't work, because there's no way to get rid of the x or the e^x term with this one usubstitution. But is this a candidate potentially for an integration by parts? Well, I see I do, in fact, have a product and notice that this term can be easily integrated or differentiated, and similarly this term alone can be easily integrated or differentiated. In fact, the integral of this, or the derivative of this is the same, it's just e^x. So that remains unchanged. So, in fact, this looks like a really good candidate for potentially using integration by parts.
So now, how do you set this up? We're going to have a u and its derivative, we're going to have a dv and v. And the real challenge in these questions is to figure out what character will play what role. Now, let's think about it. What I want to do is convert this integral into an easier integral. So what I'm hoping is that whatever I elect to call u and whatever I elect to call dv, that's what I'm going to place in here using these things. When I take the dv and integrate it to get v and take the u and differentiate it to get du, I want that new integral to be an easier integral than the original one. That's the integral that I'll have to face later. So let's think about this. Well, if I take e and differentiate it or integrate it, I'm going to get itself. So if I put e^x here, then I get an e^x here. Or if I put the e^x here, it's integral would be e^x here. So no matter where I put e^x, I'm going to get e^x in the other spot. But what about x? If I put x here, when I take the derivative I just get 1, whereas if I put x here, when I take the integral I get . And that sounds a little more complicated to me. I'm trying to make things easier, so what I'm going to try to do is my first attempt is to put the x as the person I want to take the derivative of. Since taking the derivative of x actually will reduce the complexity of the question, I'll just have 1 instead of x, but that will then force me to put the other thing, what's left, the e^x dx, in the dv slot. So that's how I'm going to try to fill up my table. Now let's go through and actually fill up everything else.
Well, if u = x, what's the derivative? It's just 1 dx. And if dv = e^x dx, if you integrate to get back to what v is, it's just e^x, just itself. So now I've filled up the table. And now, using integration by parts, let's apply the formula and see what we get. If we apply the formula, what I see is the integral of xe^x dx equals  well, let's see. This original integral was this times this. So my new integral is going to be what? It's going to be this axis. But I first have to put in just the u times the v. So I always start out with u times v, which is xe^x, and then subtract the new integral. So remember the first integral was this axis, and so I replace it now by this axis. So that's the integral of e^x dx. And look, that integral is really easy. So, in fact, this just equals xe^x minus  and the integral of e^x = e^x + c.
So look at the power of integration by parts. It allows us to take this thing and rip it apart into may pieces, and then knit it back together in sort of a different way, to produce a much easier integral. That's the power of integration by parts. The tricky thing is to figure out who should play what character. And you cannot be discouraged at all if, when you try these things, you end up putting in the wrong thing for u and v and dv. So, you might try to make a guess for u and a guess for dv, and it might turn out not to be a good one. You know what you do? You just switch the roles and try again. This is definitely something that takes a lot of practice and, at the beginning, is just trial and error. I'm trying to talk through with you how I think about them, so you can sort of see how I'm thinking about it, but in reality it's just tons of practice. I have been taking these kind of integrals for about 106 years, or at least it seems that way. It's only been 102.
Now, let's try another example. How about this one, the integral of x ln x dx? Again, the same principle. I see the product of two things, and let's see if I can easily differentiate one and easily integrate the other. Well, integrating the natural log is actually a pretty tough thing to do. In fact, you can use the technique of integration by parts to do it. But differentiating it is a walk in the park. It's easy to take the derivative, it's just . And what about this? Could I now integrate x? I can certainly integrate x, it's just . So it seems like this is a good choice for the dv thing, and this is a good choice for the u thing.
So we're going to set this up by putting in our cast of characters. I always start off this way, u du, dv and v. And I'm going to make sure that everyone gets covered, so I'll take the derivative of this term. So I'll make that the u term. I can differentiate that real easily, and I'll take what's left over and put it here in the dv, so x dx. Notice that the first slots always to be filled up will be the u and the dv, because my formula for integration by parts has an integral of u dv.
Okay, now let's fill up the rest of the table by taking a derivative and one integral. The derivative of ln x is just dx, and the integral of x is just . So that's pretty easy, there's my table. Now, let's apply this rule and see what happens. If we apply the rule, I see that the integral x ln x dx equals  okay, well let's see. The first integral is this one, so that equals, first of all, the u times the v. So that's going to be ln x. That's the product of those two things. Always the product of the top line, and then minus  and now what's the new integral? Well notice that the original integral was this one. So the new integral will be this on. So this is the first integral and this will be the new integral. And that new integral is what? Well, it's going to be dx. So there's the new integral. Now is that integral easier to look at than the original thing? Well, let's see. If not, then maybe we didn't make a very clever substitution. But if so, then we might be good shape. So this is ln x minus  and notice that the x down here cancels with one of those, so I'm left with this going away then just having one power of x left. And that comes out in front, so I'm just integrating now the integral of x dx. Well, look how easy that is. So I've reduced this to a really easy question. I have the ln x, and then I'm left with times the integral of x, which is , which makes this + c. So, in fact, you can see that this integration by parts is a fantastic way, a powerful way of cracking open things that can be split apart into an easily differentiated thing and an easily integrated thing, and you see a quite attractive answer. And you can always check your answer, by the way. You can always check it. That's the great thing about integrals. All you've got to do is take the derivative. And to take the derivative after using integration by parts, will always require the product rule, because integration by parts is the integral's answer to the derivative's product rule.
Okay, well have some fun with these things and don't forget that it's okay to make a mistake in who gets cast in what role, but then try to think through it ahead as I did, and say, "If I differentiate that or integrate that, will it make the associated new integral easier or more difficult to crack open?" 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!