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Calculus: Trig Substitutions on Rational Powers

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  • Type: Video Tutorial
  • Length: 9:21
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 100 MB
  • Posted: 06/27/2009

This lesson is part of the following series:

Calculus (279 lessons, $198.00)
Calculus: Techniques of Integration (28 lessons, $40.59)
Calculus: Intro to Trigonometric Substitution (3 lessons, $4.95)

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 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
Introduction to Trigonometric Substitution
Trigonometric Substitutions on Rational Powers Page [1 of 2]
All right, let's try another trigonometric substitution type integral. So here's the one I want to take a look at, the integral of.
Now, we can try some methods and see if we can take the integral using a long laundry list of techniques. For example, the easiest thing is just to see if you can just do it. I can't. A u du-substitution is also something that's worth thinking about before you go any further. If I let u equal the inside, that would require me to have a 2x somewhere dangling somewhere else. There is no 2x, that's missing. So, in fact, that technique won't work. But what I begin to notice is that when I look at this, I see something squared plus something squared, and then this in the exponent means I'm taking the square root. Oh, I admit there's also a 3, which means I'm also cubing, but I see still the , and then all that stuff cubed. When I see that kind of thing without any x's to make a substitution, it sort of makes me wonder if really there's a right triangle lurking in the background, and if I untangle that right triangle, it might provide a substitution of a trigonometric nature. So let's see if we can now construct a right triangle that might correspond to some of the things we're seeing in this integral.
Now, since I see the sum of two numbers squared, namely, 1^2 + x^2, it makes me think that, in fact, this will be the hypotenuse of some right triangle and these two things individually without the squares might represent some legs. So let me just - I can put the legs anywhere I want. I could put an x here and a 1 here, or an x here or a 1 here. It doesn't matter. I'll put it like this. And if we use the Pythagorean theorem, what's this? Well, it'll be 1^2 + x^2, but I've got to take the square root. So that'll be , and that's exactly the quantity that I see here. In fact, let me just remind you that, in fact, I could write this, report what I just said, as . So there it is.
All right, now if I do that, where's the trigonometry? Well, the trigonometry can come in right now by taking a look at what the trig functions would be for a particular angle. And again, you can pick any angle you want. It doesn't make a difference. There's not always one right one, but let's just call this angle right there theta. So if I call this angle theta, we can report some things. You might actually want to list all the trig functions. That sometimes is handy. I'll just list a few of them here, so, for example, let's list the cosine. Cosine is . So that's . And another natural one, just looking at this picture so it seems easy, is this ratio. And what's ? Well, is tangent, so let's write down tangent. Tangent is just . You could write down the other one's too. For example, you could write down sine. Sine would be . You could write down cosecant, you can write down secant, you can write cotangent, you can write them all down. In fact, it's sometimes not a bad idea to write them all down and look at all of them. But here are two and let's take a look at these two.
Why do I like these two? Well, notice what this integral is. This integral is just . Well, that's , that actually is cos , given this picture. So this thing right here, this bubble right here, that 1 over that stuff cubed is just cos^3 . So that 1 over stuff can be replaced by cos^3 . So there's sort of a neat substitution. I could write this now as the integral of cos^3 dx. Do you see that? So the invisible 1 over the square root of that stuff cubed is just this thing cubed, so it's cos^3.
All right, now, of course, the annoying problem again remains that I've got this dx and I have 's here. So I want to figure out what dx equals in terms of d. How do I do that? Take any relationship at all that links 's and x's together, and then differentiate that. Now, what would that look like? Well, you could pick any one of these or any thing else, but always try to pick, if you can, the simplest one. This is the simplest one that I see, so I'm going to actually take that object and differentiate it with respect to x. So I take a simple relationship, tan = x, and differentiate it with respect to x. Now, to take the derivative of the first piece, we'll always have to use a little chain rule, because I have 's, not x's. So there's an inside and an outside. What's the derivative of tan blop? It's sec^2 blop. So this would be sec^2 blop. And then I multiply that by the derivative of the inside with respect to x, but that's just d dx. And that will equal the derivative of x. It's just 1. And now, if I multiply everything through by dx, the differential dx, it will cancel here and appear here. And so what I see is sec^2 d = dx. And there is a relationship that links the ex to stuff with 's.
Now, if you picked a different identity, you would get an equivalent answer, but it might not be as simple. So always try to search for an identity, looking at the right triangle. Try to figure out a trig identity that's actually very easy and malleable and actually designed nicely to be differentiated so you can find this relationship. Anyway, there's the relationship and so back here, in place of this dx, I can write this as the integral cos^3 , and in place of dx I'm going to put sec^2 d. So now I got rid of the dx thing and happily I now have d's. It looks like I have more stuff. In fact, this looks like a pretty complicated integral, except what exactly is secant? Secant is . So, in fact, this equals the integral of cos^3 · d. Well, that's great, because now I see on the bottom and a cos^3 on the top. All the cosines on the bottom cancel with all but one cosine of the top. So, in fact, this integral actually becomes the integral of cos d. Well, that's real easy, that's just sin + c.
So, it seems like we're home free. Wow, the answer is sin . But remember, no, the original question was couched in terms of x's and our answer is in terms of 's. So how do we get around that? Well, we return back to our dictionary. This is the dictionary that tells us how to relate 's to x. And I want sin . Well, what's sin ? Well, it turns out, as we know, it's . So this would be + c. And so that conversion is the last part of the trig substitution and it allows us to start here, create a right triangle, produce a substitution in terms of angles, and then convert our integral into an integral that just has thetas. We integrate that and then we bring back the triangle, in order to figure out how to get rid of the theta and replacing it by x's.
So this is a really powerful technique. Whenever you see things of the form a square plus a square square rooted or a square minus a square square rooted. Even if it's on the top or the bottom or even raised to greater powers. Trigonometric substitution - gotta love it!

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