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About this Lesson
 Type: Video Tutorial
 Length: 7:50
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 84 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: Integrals  Other Trig Function Powers (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 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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Techniques of Integration
Integrals Involving Powers of Other Trig Functions
Integrals with Even Powers of Secant and Any Power of Tangent Page [1 of 1]
Okay, so let's just think a little bit more about integrals involving secants and tangents. Now if you see a product of a whole bunch of secants and a whole bunch of tangents and it turns out that the power on the secant is even, then there's another great technique that will allow you to crack that integral right open. So if you see a power of tangents together with a power of secants and the power on the secant is even, great technique. So let me show you what the technique is with a particular example, and then the example will illustrate the method, and the method always works.
So let's integrate together tan^4 x sec sec^6 x dx. Now remember, if the exponent on the tangent is odd, then there's a technique I can do. I can peel off one of the tangents and then make a substitution. But now the tangent is even, but happily the power on the secant in this particular example is also even, so this is a technique that will always work when you have an even power on the secant. And now, here's the idea: the idea is to peel off two copies of the secant. If you do that, what are you left with? So we'll do that here. tan^4 x sec^4 x sec^2 x dx. So I just peeled off two copies. That leaves me with four extra ones here. And now, what I do is I want to get rid of the secant. So the next step is to get rid of these secants here. And the way you do that is by using the Pythagoreantype identity. In this case, instead of solving for tan^2 x, we solve for sec^2 x. And you can see how that would look. sec^2 x = tan^2 x + 1. So let's make that substitution right now. This is always the next step, to keep all the tangents there, so I have tan^4. And then, well, this sec^4 is really . And so what I see here is . Notice that = sec^4. So I make that substitution and then I keep this thing just as is. Don't touch that. Put that off to the side for a rainy day. And now, I've made this substitution, and I'm going to do a u dusubstitution, and the u dusubstitution I'm that I'm going to make is going to be to let u = tan, and let me just say why in words. Because if let u = tan, the derivative of tangent is sec^2, and that's right here. So that's the power of this. You see, if I have an even exponent on secant, if I pull off two of them, what I'm going to be left with is still an even number. So if I have an even number left, I can always write it as some square to some power. And that allows me to make this substitution, and so I can always convert everything to tangents. And then I get that sec^2 dangling here, which will be my u dusubstitution. So if I let u = tan x, then that implies that du = sec^2 x dx. And so this now really elaborate integral converts to a quite modest one. I see tan^4 x  that's just . So the integral becomes a really simple integral, especially compared to this.
So what do you do? Well, what I would do in this case is actually square that out. So that's going to be a little teeny task. So u^4, I'll keep that out in front, and then I have . So if you square that out by FOIL, I see u^4, the first times the first, the inside and the outside terms contribute a 2u^2, and then 1 1 = 1. And if I now distribute through, I'd see u^8 + 2u^6 + u^4 du. And that's what I have to integrate. But integrating that, that's a breeze. That's just . You know, when you get to this part of the problem, you can just do that almost mindlessly. So there it is. Now, of course, this is the shorthand way. Let's now insert the long version, the tangent. If I insert the long version, the tangent, then what I see here, if I put tangent in for u, I see . And so it turns out that's what this integral equals, and so, in fact, this integral, the integral of tan^4 x sec^6 x actually equals this. Sort of amazing. And so the method is actually very clear in the following context. It's very simple what always to do. What you do is if you see an even power on the secant, pull off two of them. That's going to by your du at the end. What you're left with, what remains if you have an even number of secants to begin with and you pull out two, you'll still have an even number left. Now make your Pythagoreantype identity substitution using this and convert all the secant stuff, except for that last one, to tangents. And then once you do that, those have a whole bunch of tangents and a sec^2 at the very end. That's going to be your du. Let u = tan, so convert all those things to u's and then all that junk there becomes a du. You've got an integral that even I can do. And then when you plug back in, there's your answer.
Okay, well, I hope that you have had enough of these special cases of trig functions. We were talking about integrating powers of secants and tangents. In fact, this is part three. And I don't know about you, but I've really had enough of this, so I think it's really time to leave these things behind and really move on. By the way, if you don't use flammable paper, then very little flame. But anyway, you've heard of the burning of Atlanta? This is the burning of the tangent. I'll see you at the next lecture, unless we get quarantined by the Fire Department. Byebye.
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