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Calculus: Trig Substitution - Definite Integral 2


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About this Lesson

  • Type: Video Tutorial
  • Length: 9:04
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 97 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: Trigonometric Substitution Strategy (3 lessons, $5.94)

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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Recent Reviews

Saved my calculus life!
~ Thomas45

my test score went up a WHOPPING 40 points once i started watching Professor burgers videos! i do have one thing i did not get out of this and it is how to change the limits of the integral to radians in the very beginning...

Saved my calculus life!
~ Thomas45

my test score went up a WHOPPING 40 points once i started watching Professor burgers videos! i do have one thing i did not get out of this and it is how to change the limits of the integral to radians in the very beginning...

Techniques of Integration
Trigonometric Substitution Strategy
Trig Substitution Involving a Definite Integral - Part Two Page [1 of 3]
All right, so we've spent a lot of time trying to actually untangle this integral. That required a whole bunch of substitutions and using trigonometric identities. And we got it all the way down, starting with here. And after making our trig substitution, we ended up with an integral that we were able to evaluate and we got this. The question now is, "Well, how do you get from here to the actual definite integral, where we actually have points of integration?" So let's now pick up the action there, and hopefully you gave this a try on your own, and we'll see how our answers compare.
So if we now just remove all of this stuff and just keep the question and the answer thus far, then we see that this thing, which is all in terms of x, can be written in this fashion, but, unfortunately, here we have 's. So I'd like to go back to that dictionary and translate from the theta language back to the x language. And how do I do that? Well, the dictionary is given just by this right triangle that we created earlier. So if we look at this dictionary, what can we figure out here? Well, first of all, what is cot ? Well, cot is . That's how I think about it. And so' what's tan ? tan = . And so what's cotangent? It would be , so that would equal . So that's what cotangent equals.
Now, what about ? Well, that's sort of a peculiar thing, because is . How you get that in terms of x's seems a little bit weird. And so you can remember some of the other little dictionary facts that we already found. And actually any one of these will do, in order for us to figure out what is, because I'll just use the inverse trig function of whatever particular thing in the dictionary we decide to use. So it behooves us to choose the easiest one, which seems to be this one right here. So if I look at this thing, I see that sin = . So now how can I solve this for ? Well, if I think about the inverse function for sine, which is arcsine, then I could say that equals the angle whose sine is . So what that means is arcsin = . So I can actually solve for , although you might not find it particularly satisfying. I could say that = arcsin . And why is that? Because is the angle whose sin = . So these two statements, in fact, are identical. Now, you can use any trig function you want, but I'll just use one that seems to be the easiest.
Now, if we come back, I've now used my dictionary to translate these things, and so let's see what we get. So what does this equal? By the way, look at all these negative sings. There are way too many negative signs here. So if I actually factor out -1 here, then these will become positive and I'll have an extra -1 out here, which will make this positive. So, in fact, I can just throw away all the negative signs by just factoring out -1. So let's actually do that. That's the first thing I'll do. So this is going to be . And now in replace of cotangent, I put down what we found, which is , and then I add, because remember I've taken care of all those negative signs, . And , well, it's sort of this weird thing, arcsin . Great! So now we've inserted all the data and I now have the actual integral value for this.
Now, what I want to do is make that a definite integral, so instead of putting + c, I'm now going to evaluate this from the two points of integration, namely, all the way up to 2, and let's see what I get. So now the conversion is over and now we're just going to actually plug in some numbers. So let's put in 2 first and see what we get. When we put in 2, first of all, I still have this stuff way out in front, so that's . When I put in 2 here, I se 2^2, which is 4, and 4 - 3 = 1, so I'm taking the , which is just 1. So I see . That seems pretty much okay to me. And then if I put a 2 in here, I see arcsin . So what's the angle whose sine = ? Well, that's 60 degrees or, the way we like to think about it in radiants, . Oh, but I want to put a plus sign there. You've got to put a plus sign, because if I put the 1 here, see the plus? Do you see it here? I don't think so. So I'd better put it in, and we'd better put it in, in a really red color.
All right, so we add those two things there and that's the value of this piece. Now I subtract from that what I get when I plug in . So let's plug in and see what I get.
Well, I still start off with this . And now what do I have? If I put in here, I see , which is 3. And 3 - 3 = 0. So, in fact = 0, so this whole term will actually drop out in this case. And what am I left with? I'm left with the arcsin . = 1, so I'm looking at arcsin 1. So what is that? That's the angle whose sin = 1. So what is that? Well, that's , or 90 degrees. So all I see here is 0 + . And so what's the answer? Well, the answer is this equals - well, you could do all sorts of things here, I guess, if you want. It depends on how much you want to do. Well, let's do all sorts of stuff.
So let's distribute the to each of these terms. Don't forget that red plus sign. So I'll see 3 divided by, so there's the 3, and then = 3. So that actually looks pretty attractive, and then plus, and then here I see , so there's that piece. And then minus, and then here I see , so that's this piece right here. So there are tons of cancellations here. This is just a 1 + - . So that's a fine answer, or you might actually want to get a common denominator here and combine all these 's together. Well, it's a little frivolous, but if you want to do it, that's fine. So I need a common bottom, so that's gonna be here. And so what I see here is 1 minus, and then this would be , 3 - 3, and then a . So there's an answer. In fact, there's another answer. Suppose you say, "Oh, gosh, I hate having square roots in the denominator." You know, there are people out there - this is true - there are people out there that absolutely hate having square roots in the denominator. Those people are wacko, absolutely wacko. But suppose that you said, "Gee, I really don't want a square root in the denominator." It's no problem, just multiply by 1. If you multiply by 1 , it doesn't change the value, but then there is no square root downstairs and we'd see , and now this will give me a net gain of 3 and 3 2 = 6, and so I'd see this answer. So you could say the answer is 1 - . That numerical value turns out to be the value of this definite integral.
So a lot of steps involved here. First, there was the converting of this integral to an integral that involved trigonometric objects and thetas. Evaluating that integral actually required a little bit of work. And then after that, we had to reconvert back into x's and then actually evaluate. So major work, and hopefully you got either this answer or this answer, or this answer, or even this answer. Any one of those answers and all I have to say is congratulations. If you made a mistake, just try to find your mistake and learn from it. Again, a great success. I'll see you at the next lecture.

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