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About this Lesson
 Type: Video Tutorial
 Length: 10:27
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 111 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.
About this Author
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Techniques of Integration
Integrals Involving Powers of Other Trig Functions
Integrals with Odd Powers of Tangent and Any Power of Secant Page [1 of 2]
Okay, well now I want to take a look at what happens if you have integrals that have products of tangents and secants and the power on the tangent is an odd power. Now, as I was saying, of course, these seem like a hodge podge of possibilities, but these are the possibilities for which there are methods to actually crack the integrals. So now I want to take a look at the case where we have powers of tangents, powers of secants multiplied together, and the power on tangent is odd. What do you do?
Well, to warm up to this  so the real goal, by the way, is to just integrate tan^3. We already know what tangent is, that's an odd power, of course, it's to the first power. But we already know that's . But what about tan^3? So let's see if we can work toward that.
Now, the way to get to this is to actually come up with a Pythagoreantype identity that's going to be appropriate for tangent. Well, we know sin^2 x + cos^2 x = 1. Now, how can we come up with an identity that has tangent squares in it? Well, let's just take this identity, in fact, and divide through by cos^2. So if we take this identity and divide through by cos^2, what do I see? I'll see . It's just the Pythagorean identity. I just divided everything through by cos^2. But = tan^2 x. So this is tan^2 x. That's just a 1, and = sec^2 x. And so, if I bring this 1 over, I discover a really neat formula for tan^2 x. It equals sec^2 x  1. And I have that written down right here. So this is sort of the analog identity to this sine and cosine identity, and it turns out it will be used in this particular case when you have an odd power on the tangent to reduce things. So let's take a look at this in action and see how we can use this new Pythagoreantype identity to actually conquer the question at hand.
So what do I do? Well, the first step is to actually rip off one of the tangents. So if we rip off one of the tangents, what I see is tan^2 x tan x dx. And now what do I do? Well now what I'll do is I'll make this substitution for tan^2 and I'll put this thing in right here. Now, why is that a good idea? Well it's going to be a good idea because then I'm allowed to do some usubstitutions. So if I now insert that in what I see is the integral of (sec^2 x  1) tan x dx. Now, do you like that? Answer, of course not, of course you don't like that. But it actually is progress. Let's take the tan x and distribute it across these two terms. If I do that, here's what I get: I'll see the integral of sec^2 x tan x. So I'll write that as tan x first, and then I'll put in the sec^2 x dx. So that's this piece. And if I distribute to this piece here, I see a minus, a negative sign, the integral of tan x dx. So our original integral actually now can be written in this way. Now, why is that beneficial? Well, it's beneficial because this thing is just a u dusubstitution. If I let u = tan, then du = sec^2 x. So, in fact, this is just the integral of u du. So let me write that down. So for the tangent part, tan x sec^2 x dx, let's let u = tan x. If I do that, then I see that du = sec^2 x dx. So this integral becomes the integral of u and this piece here is du. And what's that? That's just . And what's u? Let's put it back to longhand and we see it's just . So that first integral right there  I would do it this way. I'm going to put a fancy star here, and then I'll put the fancy star here. So that's going to be . So that's that integral. Now, what about this integral right here? Well, that's the integral that we already talked about. It's the . So the bottom line is if we go back to the original question, what's the integral of tan^3, what I see is, after all this substitution, tan^3 x dx =  . That is pretty involved. But I was able to take a very complicated integral and, by using this new type of Pythagorean identity, able to reduce it to an integral that can be done. Many steps, but each step is not that bad.
All right, why don't we try just on last one together? How about tan^3 x sec^5 x dx? Okay, so I see an odd power on tangent, so I'm going to use this method. Remember the method is to peel off one of the tangents, take the even power of tangents left over, use the Pythagoreantype identity, get everything in terms of secants with that one extra tangent hanging on there, and then make a substitution in this case, which is going to be u = sec. Now, let's see what happens.
So, first of all, I peel off this. And what I'm going to do is peel off one of these guys, so I'm going to have a 2 here. And so I owe you one. And then I'm going to peel off one of these guys and I owe you one of those. And so, what do I owe you? Well, now I owe you this. I owe you secant, there's the secant, and here's the tangent. So I've done absolutely nothing. I've peeled off one of these, so there's the tan^3, and I've peeled off one of these, there's sec^5. Now I use this identity and I insert a sec^2 x  1, that's the tan^2 part, and then I copy everything else down. And just to bias the jury, I'm going to put the rest of the stuff in parentheses. You don't have to do this, but I will. So I still haven't done anything. I've just used this little formula in breaking things up. But what's the power of this? Well, when I make the substitution now, letting u = sec x, then look what happens. What's du? Well, the derivative of sec x = sec tan x dx. So it's exactly that piece right there. So when I see an odd power of tangent and some secants lying around, the method is you pluck off a tangent and you pluck off a secant and put them aside and save them for a rainy day, and then take the even tangents left over, convert them to secants, keep the other secants there, and make the substitution. When you do that, the integral now becomes an integral just with a lot of u's in it. This is going to be u^2  1 u^4. And all of that just becomes du. So this whole integral is now looking like the integral of, if I distribute, u^6  u^4 du. Well, this integral is easy. This is just . What could be easier? And now I just plug back in what u is. So I put back the longhand and I see . And so this really complicated integral with tangents and secants and so forth, really can be cracked in a pretty easy way. All you do, if you have an odd power of tangent, is peel off one of them. If you have some secants, peel off a secant, too. It doesn't make a difference whether you have an odd or even here. But then you take that even thing and you now convert that to secants and then just make a substitution u = sec. The derivative will be that stuff that you pushed off to the side. You have an easy integral and your life is great.
Congratulations on conquering what happens when you have tangents and secants with an odd power on the tangent. I'll see you at the next lecture.
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