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Calculus: Application of Integration by Parts


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About this Lesson

  • Type: Video Tutorial
  • Length: 9:32
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 103 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 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 UT-Austin 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, p-adic 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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Techniques of Integration
Integration by Parts
Repeated Application of Integration by Parts Page [1 of 2]
All right, let's take a look at some more inspirational examples of the technique of integration by parts.
So the next one I want to consider is the following: look at the integral x^2 sin x dx. Okay, now again, we notice that we see a product of two functions and, in this case, either of them can be easily differentiated or integrated. So, in fact, the technique of integration by parts seems like it might be a powerful thing here, because I do, in fact, have a product of two functions and we can potentially figure out an appropriate one to integrate and an appropriate one to differentiate.
Now, so how would you do this? Well again, a great technique is trial and error. And that's one that I really hope that you'll employ and not get frustrated with, when, in fact, you hit those errors. In fact, making mistakes is the most important and the most effective way of learning anything, especially mathematics. You've got to make the requisite mistakes, in order to really understand something. So when you see someone do a worked out example and they do it without mistakes, that means they're really old. But when they were young, like in case, they made mistakes all the time. In fact, that's what made me, I think, a math teacher, was all the mistakes I made. So if you start to realize that you're making more mistakes than maybe other people, you might be destined to be a math teacher. And then maybe you'll be standing here under these hot lights and talking to yourself, because you'd be there, too. Wait, if you're here, but you're there - I don't understand. Anyway, the point is to make a lot of mistakes and don't be frustrated by that. But let's try to think through and reason and see if we can figure out what a good choice might be.
So, remember the idea, the theme here is to convert this integral into an integral that actually might be easier. So I want to try to reduce the complexity of the integral, and that's going to determine what will play the role of the u and what will play the role of the dv.
Now, let's think about the sine for a second. If you take the derivative of sine, you get cosine. If you take the integral of sine, you get minus cosine. So basically there's no change in complexity if I integrate it or differentiate it. So from sine's point of view, it doesn't make a difference if I put it here or here. It doesn't reduce the complexity all. No matter what, I'm going to get a cosine or a minus cosine.
Let's take a look at the x^2 now. Does it make a difference whether I integrate it or differentiate it? Which would give me a more easy kind of object? Well, if I differentiate it, I'll get 2x, whereas if I integrate I'll get . Well, I'd rather deal with 2x than , so let me actually differentiate this person, and then that will force my hand and make me integrate that person. So that's how I'm going to pick my choices. But, in fact, like I said, if you pick something else, that's great, too. It might not work, but it's still great. Great and working, by the way, are not synonymous.
Okay, so let's put in now what remains, sin x - notice I keep the dx there, just to remind me that if I really brought that over to this side, I'd see and that's the derivative, but I'm writing them this way. So if I take the derivative of this, fill in the rest of the chart - I always fill these two in first, I see 2x dx. And if I integrate this, then I see minus cosine. A great mistake is to forget the negative sign, but remember the derivative of cosine is negative sin, so to make it positive I need a minus out in front there.
Okay, now what does this integral become, the integral x^2 sin x dx? I'm going to use this formula, so here, playing the role of u tonight, is x^2, and dv is being played by sin x dx. So what does this integral equal? Well, it always equals u times v, so I have to multiply these two things together, so I see -(-x^2) cos x, and then minus, and then a new integral. And what's the new integral? Well, the original integral, of course, was this integral, and so the new integral will be this other one. And what is that? That's going to be -2x cos x dx. That negative sign I can pull out in front and it would change this sign from a negative, with that negative sign will make this a positive, and I would see 2x cos x dx.
Okay, now is this integral an integral that we can actually integrate? Well, actually, I don't know how to do that exactly right now. But I notice that one of these people can be easily integrated and one of these people can be easily differentiated. So maybe I should take this new integral and actually apply integration by parts on it, so do sort of an integration by parts within an integration by parts. That's right! You can actually do integration by parts, and then what's in it an integration by parts, and then what's in it an integration by parts. So let's see if we can now apply integration by parts again. It's always good to stop and ask yourself, "Is this really progress or not?" Well, it actually is, because the original thing, I had a trig function and x^2. Now I've got an integral, where I've got a trig function and just an x thing. In fact, that 2 can be pulled right outside here. In fact, I'll do that right now. I'm going to pull the 2 right out, and now it's no longer there. So it really it progress. I removed the 2. But if I do this process again, arguing by analogy, if I remove the 2 here, I should remove the x here and I'll be left with something very simple.
So let's now do an aside problem, a little sub-problem, and the sub-problem is can we find this integral? And so let me write that integral here. That's the integral x cos x dx. It is simpler than this integral, so I seem to be moving in the right direction. And let's see if another application of the chain rule will drive us home. So this is a multi-step question. So again, we have our characters u, du, dv, v. And so what should I pick with these things? Well again, I need to think about this. I want to reduce the complexity. So if I let u = x, the derivative of that = 1. So let me put in the x here and then put everything else here. If I take the derivative, I just get 1 times dx, and if I integrate this, I just get sin x dx. So filling in the chart is not too bad, as long as we're careful.
Now, I want to come back to the integral at hand. So the integral at hand would look like what? It would look like the integral x cos x dx. So what do I do first? I first take u times v, so that's these two things - oh, wait. There should be no dx there, of course. If I integrate, then the dx should go away, so that's sort of a little typo. I apologize for that. Don't look at that. And you've got to make it seem as though it's supposed to be there, so let's make it a special box, to point out that this is a good point. It's important to sort of fit everything into the lecture, which I did quite poorly. Anyway, here we go. So it's going to be u times v, so that's going to be x sin x. And then I subtract a new integral. Well, this was the original integral, you can see it. And so this will be the new integral. And the new integral, notice, is just sin x times 1 times dx. So that's sin x dx. And that actually is an easy integral to do. So, in fact, that progression we saw with an x^2 in front, we applied integration by parts and just got an x in front. And then we applied the integration by parts technique again and we got just a 1 in front. So, in fact, we did reduce the complexity, and this equals x sin x - and what's the integral of sine? It's negative cosine, which will make this thing a positive, and I have cos x. So this is the integral of that piece.
Now you're saying, "Well, is that the answer?" Well, no, because remember that's just the subliminal problem. The original question was here and we've just got it down to this. So putting all this together, I see this equals -x^2 cos x plus - and then 2 times the integral we just found. And 2 times that integral, which is right here. Don't forget to multiply that by that orange 2, so I see 2x sin x + 2cos x + c. Never forget the c.
So, the original question was very complicated, but we split it up. We got it to an easier integral, but that integral required integration by parts yet again and we did that. And we actually were able to resolve that last integral. Putting everything together, we finally produced the answer. So integration by parts sometimes needs to be applied more than once. All right, congratulations and I'll see you at the next lecture.

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