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Calculus: Converting Radicals to Trig Expressions


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About this Lesson

  • Type: Video Tutorial
  • Length: 10:01
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 107 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: 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 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
Converting Radicals into Trigonometric Expressions Page [1 of 2]
You know, a lot of times when you take a look at an integral, you'll see things that make it really complicated. And you'll see things sometimes either in the top or the bottom of some sort of complicated-looking expression, and you'll see things like . You might see something like , and all sorts of really weird things. In particular, when you see a root of two things and they're each either the difference or the sum of two perfect squares. So, for example, this is 12 - x2, this is x2 + 12, and so forth. You can imagine putting a 4 here and so forth.
Now, those are complicated things, and if you imagine an integral sign written right in front of it, that's sort of pretty threatening. How would you deal with that? Now imagine that on the denominator of some place. How would you deal with these things? Well, in fact, there's a way of dealing with this and the technique that I'm going to describe is called trig substitution, which actually is just shorthand for saying trigonometric substitution. And so, it's a technique, which will allow us to convert these square roots and square into potentially more basic fundamental, easier to handle objects. That's always the strategy with any kind of integration technique. You take an integral that's really complex and you try to convert it into an equivalent integral, for which the new integral version is now easier to look at and solve.
So what's the theme of this? Well, I really want to sort of outline the sort of strategy and the theme of this, so you can really see it coming and see why it's even called trig substitution. So whenever you see roots and two numbers that are being either added or subtracted together and those numbers themselves can be thought of as perfect squares, then you know what you should thing? The same thing that I think, which is this is information about some right triangle hidden far away from us. In fact, if you think about it, these are just sides of a right triangle that follow from the Pythagorean theorem. Remember the Pythagorean theorem says leg2 + leg2 = hypotenuse2. So in some sense this must be some legs and hypotenuses and I'm trying to figure out what they are. So let me try to show you sort of the general way of thinking about these people, so that when you see them coming, you can figure out how to untangle it and how to make a trigonometric substitution. Okay, so let's not worry about any kind of technical details about integration now, let's just try to get a sense of how to decode this and think about this as a right triangle.
Let's look at the first example here. I'm going to convert those values somehow to the lengths of sides of a right triangle. Now how could I possibly do that? Well, you see, since I see the difference here, what that must mean that this thing must be the side of a right triangle that must represent a leg. And this person right here, think of it as 12, if you will, that must represent the hypotenuse. Now, I'm going to write this in and then we'll go over it and you'll see why I say must. And this x must represent one side. If I were to do that - so notice what I did. I saw a difference here, and when I saw a difference, I took this person, but got rid of the square, and made that the hypotenuse, took the thing here that I'm subtracting, took off the square and made it one of the sides. What's this third side? Pythagorean theorem. So the Pythagorean theorem says if I call this ???, ???2 + x2 = 12, which is 1. And so if I solve for ???, I see ??? equals, well, I bring this over and I see 1 - x2, and then I take the square root and technically a +, but it is a length so it's just the positive, square root. And look, that's exactly what I had up here. So this intuition was pretty good. This thing represents this length.
Now, how did I see that coming? I saw that coming because since I saw the difference, that difference in the squares meant, in my mind, that I had to do this step of bringing this over to the other side. If I'm bringing this over to the dark side, that means that the person that was in front must have been the hypotenuse. So that negative sign in front tips me off that that must have been a hypotenuse, this must be one of the sides and the other side must be this length. It doesn't make a difference where I put the x. If I were to put the x over here, then the length of this side would have worked out to this. It doesn't make a difference how you place the x down, but it's important that the 1 becomes the hypotenuse. Since I see 1 minus, that's the step of taking this term and bringing it over. So this must be a leg and that must be the hypotenuse. And so this ??? mark length turns out to be precisely what we saw, 1 - x2. So notice how I took this expression and I was able to convert that into a picture that involves a right triangle. Let's try with the second example, just to sort of really solidify these ideas.
So let's take another right triangle and try to do the same kind of thing. Well, now here I see the sum. So it makes me thing that maybe this, in fact, is going to represent the hypotenuse itself and these are the two sides. When I square them and then add them up, I get the square of the hypotenuse. So since I see them being added, it makes me wonder if, in fact, it makes sense to think about this as really the two sides. So if I call this, for example, x, and I call this, for example, 2, then the question is what is this length? So that's the thing I want to look at now. So what is the ??? length? Well, I'll use the Pythagorean theorem. The Pythagorean theorem says I take x2 + 22 = ???2. And so if I solve this, I see that ??? = . And that's exactly what we have here. So, in fact, that ??? length is precisely that value, .
So again, my thinking is that since I see a plus sign, these must be the two sides of the two legs of a right triangle that have been squared, so that when I take the square root, this thing produces the hypotenuse. If I see a difference, that means that this must have been the hypotenuse, this must have been a leg, and then if I solve for the other leg, it requires me, in the Pythagorean theorem, to subtract and take the square root.
So that `s how I can convert a thing like this into a picture that looks like a right triangle. Now, why is this at all relevant or useful in terms of integration? Well, because now we can use trigonometry to help us out. For example, look at this piece right here. If take this angle and call it theta, then how can I describe this thing here? Well, that thing actually is nothing more than with respect to theta, because adjacent is divided by the hypotenuse, which is 1. So, in fact, this number is just of this angle, which has a name; it's called cosine. So, in fact, this whole thing is actually cos . So instead of all that square root and the squares and all that stuff that's really awful, I can replace this by cosine. That's a really nice trigonometric substitution for this value. So if I had an integral that had that in it, I could easily throw that away and replace it by cos , which is a simpler object potentially to work with.
What about here? Well, if I call this angle phi, what would I have? Well, I'd see that this thing is what? Well, now it's a little bit tricky. Let's see if we can think about this. It's actually going to be this thing over 2, so let's think about cosine would be here. cos would be , which would be 2 divided by that square root. It's not exactly what we want, I have to admit. But if I take the reciprocal of both sides, if I flip both sides, is actually secant. So, in fact, this is identical to sec . I flip this and then I have to flip this. And that would equal . And if I thereby multiply through both sides by 2, I see that this would equal 2sec . So this square root, sort of ugly-looking thing can be converted to a different variable. It's a substitution, because there's no x's here, now it's . But now it has a trigonometric flavor. This thing equals, with respect to this triangle, 2sec . This thing, with respect to this triangle, equals cos . And so what I'm doing here is actually taking these x's and I'm making a substitution, , , whatever, of some angle and what happens is that substitution leads me to introduce trigonometric functions.
So whenever you see things like this, making a substitution and drawing a triangle like this is a potentially powerful way of converting the complicated integral with square roots and squares to a different integral with trigonometric functions and maybe that integral can be resolved. So we'll take a look at actually how to incorporate this general theme into integration. I'll see you there. Bye.

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