WEBVTT 00:00:00.000 --> 00:00:02.330 The following content is provided under a Creative 00:00:02.330 --> 00:00:03.610 Commons license. 00:00:03.610 --> 00:00:05.750 Your support will help MIT OpenCourseWare 00:00:05.750 --> 00:00:09.455 continue to offer high quality educational resources for free. 00:00:09.455 --> 00:00:12.540 To make a donation or to view additional materials 00:00:12.540 --> 00:00:16.150 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:16.150 --> 00:00:21.780 at ocw.mit.edu. 00:00:21.780 --> 00:00:24.560 PROFESSOR: So we're going to continue 00:00:24.560 --> 00:00:29.420 to talk about trig integrals and trig substitutions. 00:00:29.420 --> 00:00:31.520 This is maybe the most technical part 00:00:31.520 --> 00:00:34.560 of this course, which maybe is why professor Jerison decided 00:00:34.560 --> 00:00:37.520 to just take a leave, go AWOL just now and let me take over 00:00:37.520 --> 00:00:38.610 for him. 00:00:38.610 --> 00:00:41.960 But I'll do my best to help you learn this technique 00:00:41.960 --> 00:00:45.680 and it'll be useful for you. 00:00:45.680 --> 00:00:48.880 So we've talked about trig integrals involving 00:00:48.880 --> 00:00:51.460 sines and cosines yesterday. 00:00:51.460 --> 00:00:53.060 There's another whole world out there 00:00:53.060 --> 00:00:58.850 that involves these other trig polynomials-- trig functions, 00:00:58.850 --> 00:01:00.520 secant and tangent. 00:01:00.520 --> 00:01:03.460 Let me just make a little table to remind you what they are. 00:01:03.460 --> 00:01:05.210 Because I have trouble remembering myself, 00:01:05.210 --> 00:01:07.630 so I enjoy the opportunity to go back 00:01:07.630 --> 00:01:09.321 to remind myself of this stuff. 00:01:09.321 --> 00:01:09.820 Let's see. 00:01:09.820 --> 00:01:15.860 The secant is one over one of those things, which one is it? 00:01:15.860 --> 00:01:18.420 It's weird, it's 1 over the cosine. 00:01:18.420 --> 00:01:24.710 And the cosecant is 1 over the sin. 00:01:24.710 --> 00:01:26.800 Of course the tangent, we know. 00:01:26.800 --> 00:01:31.840 It's the sine over the cosine and the cotangent 00:01:31.840 --> 00:01:36.000 is the other way around. 00:01:36.000 --> 00:01:37.700 So when you put a "co" in front of it, 00:01:37.700 --> 00:01:43.590 it exchanges sine and cosine. 00:01:43.590 --> 00:01:46.840 Well, I have a few identities involving tangent and secant 00:01:46.840 --> 00:01:50.270 up there, in that little prepared blackboard up above. 00:01:50.270 --> 00:01:53.680 Maybe I'll just go through and check that out 00:01:53.680 --> 00:01:57.604 to make sure that we're all on the same page with them. 00:01:57.604 --> 00:01:59.020 So I'm going to claim that there's 00:01:59.020 --> 00:02:00.560 this trig identity at the top. 00:02:00.560 --> 00:02:07.310 Secant squared is 1 plus the tangent. 00:02:07.310 --> 00:02:10.590 So let's just check that out. 00:02:10.590 --> 00:02:12.670 So the secant is 1 over the cosine, 00:02:12.670 --> 00:02:14.890 so secant squared is 1 over cosine squared. 00:02:14.890 --> 00:02:17.100 And then whenever you see a 1 in trigonometry, 00:02:17.100 --> 00:02:23.010 you'll always have the option of writing as cos^2 + sin^2. 00:02:27.890 --> 00:02:32.210 And if I do that, then I can divide the cos^2 into that 00:02:32.210 --> 00:02:34.300 first term. 00:02:34.300 --> 00:02:36.380 And I get 1 + sin^2 / cos^2. 00:02:36.380 --> 00:02:37.940 / ^2. 00:02:37.940 --> 00:02:42.430 Which is the tangent squared. 00:02:42.430 --> 00:02:44.180 So there you go. 00:02:44.180 --> 00:02:47.820 That checks the first one. 00:02:47.820 --> 00:02:49.320 That's the main trig identity that's 00:02:49.320 --> 00:02:51.330 going to be behind what I talk about today. 00:02:51.330 --> 00:02:53.460 That's the trigonometry identity part. 00:02:53.460 --> 00:02:57.040 How about this piece of calculus. 00:02:57.040 --> 00:03:05.490 Can we calculate what the derivative of tan x is? 00:03:05.490 --> 00:03:12.350 Actually, I'm going to do that on this board. 00:03:12.350 --> 00:03:19.110 So tan x = sin x / cos x. 00:03:19.110 --> 00:03:21.250 So I think I was with you when we learned 00:03:21.250 --> 00:03:23.500 about the quotient rule. 00:03:23.500 --> 00:03:26.030 Computing the derivative of a quotient. 00:03:26.030 --> 00:03:30.340 And the rule is, you take the numerator 00:03:30.340 --> 00:03:34.330 and you-- sorry, you take the derivative of the numerator, 00:03:34.330 --> 00:03:36.510 which is cosine. 00:03:36.510 --> 00:03:38.570 And you multiply it by the denominator, 00:03:38.570 --> 00:03:41.590 so that gives you cos^2. 00:03:41.590 --> 00:03:46.480 And then you take the numerator, take minus the numerator, 00:03:46.480 --> 00:03:49.400 and multiply that by the derivative of the denominator, 00:03:49.400 --> 00:03:53.240 which is -sin x. 00:03:53.240 --> 00:03:57.870 And you put all that over the square of the denominator. 00:03:57.870 --> 00:04:02.490 And now I look at that and before my eyes I see the same 00:04:02.490 --> 00:04:07.420 trig identity, cos^2 + sin^2 = 1, appearing there. 00:04:07.420 --> 00:04:13.850 This is 1 / cos^2(x) which is secant squared. 00:04:13.850 --> 00:04:15.022 And, good. 00:04:15.022 --> 00:04:16.230 So that's what the claim was. 00:04:16.230 --> 00:04:20.420 The derivative of the tangent is the secant squared. 00:04:20.420 --> 00:04:22.260 That immediately gives you an integral. 00:04:22.260 --> 00:04:25.240 Namely, the integral of secant squared is the tangent. 00:04:25.240 --> 00:04:28.510 That's the fundamental theorem of calculus. 00:04:28.510 --> 00:04:32.360 So we verified the first integral there. 00:04:32.360 --> 00:04:34.490 Well, let's just do the second one as well. 00:04:34.490 --> 00:04:40.280 So if I want to differentiate the secant, 00:04:40.280 --> 00:04:41.360 derivative of the secant. 00:04:41.360 --> 00:04:46.500 So that's d/dx of 1 over the cosine. 00:04:46.500 --> 00:04:47.817 And again, I have a quotient. 00:04:47.817 --> 00:04:49.900 This one's a little easier because the numerator's 00:04:49.900 --> 00:04:51.010 so simple. 00:04:51.010 --> 00:04:54.310 So I take the derivative of the numerator, which is 0. 00:04:54.310 --> 00:04:56.580 And then I take the numerator, I take 00:04:56.580 --> 00:05:00.830 minus the numerator times the derivative of the denominator. 00:05:00.830 --> 00:05:05.610 Which is -sin x, and put all that 00:05:05.610 --> 00:05:09.900 over the square of the same denominator. 00:05:09.900 --> 00:05:12.809 So one minus sign came from the quotient rule, 00:05:12.809 --> 00:05:14.350 and the other one came because that's 00:05:14.350 --> 00:05:16.580 the derivative of the cosine. 00:05:16.580 --> 00:05:21.010 But they cancel, and so I get sin / cos^2, 00:05:21.010 --> 00:05:25.640 which is sin / cos times 1 / cos and so that's the secant, 00:05:25.640 --> 00:05:32.160 that's 1 / cos, times tan x. 00:05:32.160 --> 00:05:34.190 So, not hard. 00:05:34.190 --> 00:05:36.300 That verifies that the derivative of secant 00:05:36.300 --> 00:05:37.990 is secant tangent. 00:05:37.990 --> 00:05:41.800 And it tells you that the integral of that weird thing 00:05:41.800 --> 00:05:43.830 in case you ever want to know, the integral 00:05:43.830 --> 00:05:47.910 of the secant tangent is the secant. 00:05:47.910 --> 00:05:50.470 Well, there are a couple more integrals 00:05:50.470 --> 00:05:54.000 that I want to do for you. 00:05:54.000 --> 00:05:57.050 Where I can't sort of work backwards like that. 00:05:57.050 --> 00:06:07.060 Let's calculate the integral of the tangent. 00:06:07.060 --> 00:06:09.380 Just do this straight out. 00:06:09.380 --> 00:06:22.610 So the tangent is the sine divided by the cosine. 00:06:22.610 --> 00:06:28.520 And now there's a habit of mind, that I hope you get into. 00:06:28.520 --> 00:06:33.310 When you see the cosine and you're calculating an integral 00:06:33.310 --> 00:06:35.950 like this, it's useful to remember 00:06:35.950 --> 00:06:37.700 what the derivative of the cosine is. 00:06:37.700 --> 00:06:41.420 Because maybe it shows up somewhere else in the integral. 00:06:41.420 --> 00:06:43.540 And that happens here. 00:06:43.540 --> 00:06:50.300 So that suggests we make a substitution. u = cos x. 00:06:50.300 --> 00:06:55.620 Which means du = -sin x dx. 00:06:55.620 --> 00:06:59.020 That's the numerator, except for the minus sign. 00:06:59.020 --> 00:07:08.570 And so I can rewrite this as, under the substitution, 00:07:08.570 --> 00:07:13.120 I can rewrite this as -du, that's the numerator, 00:07:13.120 --> 00:07:19.600 sin x dx is -du, divided by u. 00:07:19.600 --> 00:07:22.150 Well, I know how to do that integral too. 00:07:22.150 --> 00:07:24.210 That gives me the natural log, doesn't it. 00:07:24.210 --> 00:07:31.150 So this is -ln(u) plus a constant. 00:07:31.150 --> 00:07:32.360 I'm not quite done. 00:07:32.360 --> 00:07:34.750 I have to back-substitute and replace 00:07:34.750 --> 00:07:39.080 this new variable that I've made up, called u, with what it is. 00:07:39.080 --> 00:07:46.760 And what you get is -ln(cos x). 00:07:46.760 --> 00:07:50.410 So the integral of the tangent is minus log cosine. 00:07:50.410 --> 00:07:55.590 Now, you find these tables of integrals 00:07:55.590 --> 00:07:56.910 in the back of the book. 00:07:56.910 --> 00:07:58.320 Things like that. 00:07:58.320 --> 00:08:01.060 I'm not sure how much memorization Professor Jerison 00:08:01.060 --> 00:08:04.210 is going to ask of you, but there 00:08:04.210 --> 00:08:05.770 is a certain amount of memorization 00:08:05.770 --> 00:08:06.937 that goes on in calculus. 00:08:06.937 --> 00:08:08.520 And this is one of the kinds of things 00:08:08.520 --> 00:08:12.380 that you probably want to know. 00:08:12.380 --> 00:08:15.450 Let me do one more integral. 00:08:15.450 --> 00:08:18.100 I think I'm making my way through a prepared board here, 00:08:18.100 --> 00:08:20.350 let's see. 00:08:20.350 --> 00:08:23.060 Good. 00:08:23.060 --> 00:08:28.940 So the integral of the tangent is minus log cosine. 00:08:28.940 --> 00:08:36.090 I'd also like to know what the integral of the secant of x is. 00:08:36.090 --> 00:08:41.280 And I don't know a way to kind of go straight at this, 00:08:41.280 --> 00:08:45.580 but let me show you a way to think your way through to it. 00:08:45.580 --> 00:08:52.720 If I take these two facts, tangent prime is what it is, 00:08:52.720 --> 00:08:56.280 and secant prime is what it is, and add them together, 00:08:56.280 --> 00:08:57.420 I get this fact. 00:08:57.420 --> 00:09:05.400 That the derivative of the sec x + tan x is, 00:09:05.400 --> 00:09:08.350 well, it's the sum of these two things. 00:09:08.350 --> 00:09:11.060 Secant squared plus secant tangent. 00:09:11.060 --> 00:09:15.180 And there's a secant that occurs in both of those terms. 00:09:15.180 --> 00:09:17.140 So I'll factor it out. 00:09:17.140 --> 00:09:21.440 And that gives me, I'll put it over here. 00:09:21.440 --> 00:09:24.516 There's the secant of x that occurs in both terms. 00:09:24.516 --> 00:09:26.390 And then in one term, there's another secant. 00:09:26.390 --> 00:09:32.740 And in the other term, there's a tangent. 00:09:32.740 --> 00:09:36.970 So that's interesting somehow, because this same term appears 00:09:36.970 --> 00:09:40.200 on both sides of this equation. 00:09:40.200 --> 00:09:49.160 Let's write u, for that sec x + tan x. 00:09:49.160 --> 00:09:55.120 And so the equation that I get is u' = u sec x. 00:10:00.930 --> 00:10:03.860 I've just made a direct substitution. 00:10:03.860 --> 00:10:05.620 Just decide that I'm going to write u 00:10:05.620 --> 00:10:07.730 for that single thing that occurs 00:10:07.730 --> 00:10:09.600 on both sides of the equation. 00:10:09.600 --> 00:10:16.440 So u' is on the left, and u * sec x is on the right. 00:10:16.440 --> 00:10:18.560 Well, there's my secant. 00:10:18.560 --> 00:10:20.210 That I was trying to integrate. 00:10:20.210 --> 00:10:28.560 And what it tells you is that = u' / u. 00:10:28.560 --> 00:10:35.550 Just divide both sides by u, and I get this equation. u' / u, 00:10:35.550 --> 00:10:36.910 that has a name. 00:10:36.910 --> 00:10:38.880 Not sure that professor Jerison's 00:10:38.880 --> 00:10:41.604 used this in this class, but u' / u, 00:10:41.604 --> 00:10:43.270 we've actually used something like that. 00:10:43.270 --> 00:10:45.560 It's on the board right now. 00:10:45.560 --> 00:10:47.650 It's a logarithmic derivative. 00:10:47.650 --> 00:10:57.750 It is the derivative of the national logarithm of u. 00:10:57.750 --> 00:10:59.940 Maybe it's easier to read this from right to left, 00:10:59.940 --> 00:11:02.780 if I want to calculate the derivative of the logarithm, 00:11:02.780 --> 00:11:06.380 well, the chain rule says I get the derivative of u times 00:11:06.380 --> 00:11:11.100 the derivative of the log function, which is 1 / u. 00:11:11.100 --> 00:11:25.964 So often u' / u is called the logarithmic derivative. 00:11:25.964 --> 00:11:27.130 But it's done what I wanted. 00:11:27.130 --> 00:11:31.740 Because it's expressed the secant as a derivative. 00:11:31.740 --> 00:11:38.630 And I guess I should put in what u is. 00:11:38.630 --> 00:11:46.090 It's the secant plus the tangent. 00:11:46.090 --> 00:11:48.500 And so that implies that the integral-- 00:11:48.500 --> 00:11:49.510 Integrate both sides. 00:11:49.510 --> 00:11:56.840 That says that the integral of sec x dx, is ln(sec x + tan x). 00:12:02.280 --> 00:12:09.500 So that's the last line in this little memo that I created. 00:12:09.500 --> 00:12:14.500 That we can use now for the rest of the class. 00:12:14.500 --> 00:12:19.320 Any questions about that trick? 00:12:19.320 --> 00:12:23.740 It's a trick, I have nothing more to say about it. 00:12:23.740 --> 00:12:24.540 OK. 00:12:24.540 --> 00:12:30.210 So, the next thing I-- oh yes, so now 00:12:30.210 --> 00:12:33.060 I want to make the point that using these rules 00:12:33.060 --> 00:12:38.350 and some thought, you can now integrate 00:12:38.350 --> 00:12:41.650 most trigonometric polynomials. 00:12:41.650 --> 00:12:44.850 Most things that involve powers of sines and cosines 00:12:44.850 --> 00:12:48.080 and tangents and secants and everything else. 00:12:48.080 --> 00:12:56.790 For example, let's try to integrate the integral of sec^4 00:12:56.790 --> 00:12:59.180 x. 00:12:59.180 --> 00:13:01.914 Big power of the secant function. 00:13:01.914 --> 00:13:03.830 Well, there are too many secants there for me. 00:13:03.830 --> 00:13:05.680 So let's take some away. 00:13:05.680 --> 00:13:09.730 And I can take them away by using that trig identity, 00:13:09.730 --> 00:13:12.300 sec^2 = 1 + tan^2. 00:13:12.300 --> 00:13:16.680 So I'm going to replace two of those secants by 1 + tan^2. 00:13:21.790 --> 00:13:25.190 That leaves me with two left over. 00:13:25.190 --> 00:13:27.920 Now there was method to my madness. 00:13:27.920 --> 00:13:31.020 Because I've got a secant squared left over there. 00:13:31.020 --> 00:13:35.280 And secant squared is the derivative of tangent. 00:13:35.280 --> 00:13:40.970 So that suggests a substitution. 00:13:40.970 --> 00:13:48.440 Namely, let's say, let's let u = tan x, so that du = sec^2 x dx. 00:13:50.980 --> 00:13:55.700 And I have both terms that occur in my integral 00:13:55.700 --> 00:14:01.800 sitting there very nicely. 00:14:01.800 --> 00:14:05.020 So this is the possibility of making this substitution 00:14:05.020 --> 00:14:08.240 and seeing a secant squared up here as part 00:14:08.240 --> 00:14:10.330 of the differential here. 00:14:10.330 --> 00:14:13.910 That's why it was a good idea for me to take two 00:14:13.910 --> 00:14:16.350 of the secants and write them as 1 + tan^2. 00:14:19.170 --> 00:14:22.250 So now I can continue this. 00:14:22.250 --> 00:14:25.780 Under that substitution, I get 1. 00:14:25.780 --> 00:14:32.810 Oh yeah, and I should add the other fact, that-- 00:14:32.810 --> 00:14:38.460 Well I guess it's obvious that tangent squared is u^2. 00:14:38.460 --> 00:14:41.070 So I get 1 + u^2. 00:14:41.070 --> 00:14:50.110 And then du-- sec^2 sec^2 x dx, that is du. 00:14:50.110 --> 00:14:52.430 Well that's pretty easy to integrate. 00:14:52.430 --> 00:14:56.690 So I get u + u^3 / 3. 00:14:56.690 --> 00:14:57.950 Plus a constant. 00:14:57.950 --> 00:15:00.020 And then I just have to back-substitute. 00:15:00.020 --> 00:15:03.600 Put things back in terms of the original variables. 00:15:03.600 --> 00:15:12.220 And that gives me tan x plus tangent cubed over 3. 00:15:12.220 --> 00:15:15.610 And there's the answer. 00:15:15.610 --> 00:15:17.070 So we could spend a lot more time 00:15:17.070 --> 00:15:21.810 doing more examples of this kind of polynomial trig thing. 00:15:21.810 --> 00:15:27.440 It's probably best for you to do some practice on your own. 00:15:27.440 --> 00:15:30.020 Because I want to talk about other things, also. 00:15:30.020 --> 00:15:32.800 And what I want to talk about is the use 00:15:32.800 --> 00:15:39.380 of these trig identities in making really trig substitution 00:15:39.380 --> 00:15:51.590 integration. 00:15:51.590 --> 00:15:54.320 So we did a little bit of this yesterday, 00:15:54.320 --> 00:15:56.990 and I'll show you some more examples today. 00:15:56.990 --> 00:15:59.520 Let's start with a pretty hard example right off the bat. 00:15:59.520 --> 00:16:07.700 So this is going to be the integral of dx over x^2 times 00:16:07.700 --> 00:16:10.200 the square root of 1+x^2. 00:16:14.690 --> 00:16:18.120 It's a pretty bad looking integral. 00:16:18.120 --> 00:16:20.190 So how can we approach this? 00:16:20.190 --> 00:16:25.390 Well, the square root is the ugliest part of the integral, 00:16:25.390 --> 00:16:27.240 I think. 00:16:27.240 --> 00:16:30.160 What we should try to do is write this square root 00:16:30.160 --> 00:16:31.600 in some nicer way. 00:16:31.600 --> 00:16:37.180 That is, figure out a way to write 1+x^2 as a square. 00:16:37.180 --> 00:16:40.140 That'll get rid of the square root. 00:16:40.140 --> 00:16:44.540 So there is an example of a way to write 1 plus something 00:16:44.540 --> 00:16:46.140 squared in a different way. 00:16:46.140 --> 00:16:47.760 And it's right up there. 00:16:47.760 --> 00:16:50.790 sec^2 = 1 + tan^2. 00:16:50.790 --> 00:16:53.450 So I want to use that idea. 00:16:53.450 --> 00:16:56.650 And when I see this form, that suggests 00:16:56.650 --> 00:16:59.520 that we make a trig substitution and write 00:16:59.520 --> 00:17:04.682 x as the tangent of some new variable. 00:17:04.682 --> 00:17:06.140 Which you might as well call theta, 00:17:06.140 --> 00:17:09.060 to because it's like an angle. 00:17:09.060 --> 00:17:15.960 Then 1+x^2 is the secant squared. 00:17:15.960 --> 00:17:18.620 According to that trig identity. 00:17:18.620 --> 00:17:23.970 And so the square root of 1+x^2 is sec(theta). 00:17:29.780 --> 00:17:31.070 Right? 00:17:31.070 --> 00:17:36.220 So this identity is the reason that the substitution 00:17:36.220 --> 00:17:37.120 is going to help us. 00:17:37.120 --> 00:17:39.430 Because it gets rid of the square root 00:17:39.430 --> 00:17:43.550 and replaces it by some other trig function. 00:17:43.550 --> 00:17:46.080 I'd better be able to get rid of the dx, too. 00:17:46.080 --> 00:17:48.520 That's part of the substitution process. 00:17:48.520 --> 00:17:50.520 But we can do that, because I know 00:17:50.520 --> 00:17:52.440 what the derivative of the tangent is. 00:17:52.440 --> 00:17:56.070 It's secant squared. 00:17:56.070 --> 00:17:58.880 So dx / d theta is sec^2(theta). 00:17:58.880 --> 00:18:01.150 So dx is sec^2(theta) d theta. 00:18:04.460 --> 00:18:07.350 So let's just substitute all of that stuff 00:18:07.350 --> 00:18:10.280 in, and rewrite the entire integral in terms 00:18:10.280 --> 00:18:11.750 of our new variable, theta. 00:18:11.750 --> 00:18:14.570 So dx is in the numerator. 00:18:14.570 --> 00:18:18.210 That's sec^2(theta) d theta. 00:18:18.210 --> 00:18:21.197 And then the denominator, well, it has an x^2. 00:18:21.197 --> 00:18:22.030 That's tan^2(theta). 00:18:24.740 --> 00:18:28.660 And then there's this square root. 00:18:28.660 --> 00:18:31.690 And we know what that is in terms of theta. 00:18:31.690 --> 00:18:32.730 It's sec(theta). 00:18:36.470 --> 00:18:41.920 OK, now. we've done the trig substitution. 00:18:41.920 --> 00:18:43.940 I've gotten rid of the square root, 00:18:43.940 --> 00:18:46.050 I've got everything in terms of trig functions 00:18:46.050 --> 00:18:48.020 of the new variable. 00:18:48.020 --> 00:18:49.860 Pretty complicated trig function. 00:18:49.860 --> 00:18:52.550 This often happens, you wind up with a complete scattering 00:18:52.550 --> 00:18:55.050 of different trig functions in the numerator and denominator 00:18:55.050 --> 00:18:56.250 and everything. 00:18:56.250 --> 00:18:58.950 A systematic thing to do here is to put everything 00:18:58.950 --> 00:19:09.110 in terms of sines and cosines. 00:19:09.110 --> 00:19:12.600 Unless you can see right away, how it's going to simplify, 00:19:12.600 --> 00:19:15.010 the systematic thing to do is to rewrite 00:19:15.010 --> 00:19:17.110 in terms of sines and cosines. 00:19:17.110 --> 00:19:19.300 So let's do that. 00:19:19.300 --> 00:19:20.260 So let's see. 00:19:20.260 --> 00:19:23.520 The secant squared, secant is 1 over cosine. 00:19:23.520 --> 00:19:28.920 So I'm going to put a cosine squared in the denominator. 00:19:28.920 --> 00:19:32.070 Oh, I guess the first thing I can do is cancel. 00:19:32.070 --> 00:19:33.270 Let's do that. 00:19:33.270 --> 00:19:34.200 That's clever. 00:19:34.200 --> 00:19:35.700 You were all thinking that too. 00:19:35.700 --> 00:19:37.220 Cancel those. 00:19:37.220 --> 00:19:39.860 So now I just get one cosine denominator 00:19:39.860 --> 00:19:44.382 from the secant there in the numerator. 00:19:44.382 --> 00:19:46.840 It's still pretty complicated, secant over tangent squared, 00:19:46.840 --> 00:19:48.110 who knows. 00:19:48.110 --> 00:19:49.410 Well, we'll find out. 00:19:49.410 --> 00:19:52.450 Because the tangent is sine over cosine. 00:19:52.450 --> 00:19:55.940 So I should put a sine squared where the tangent was, 00:19:55.940 --> 00:19:59.330 and a cosine squared up there. 00:19:59.330 --> 00:20:02.180 And I still have d theta. 00:20:02.180 --> 00:20:04.300 And now you see some more cancellation occurs. 00:20:04.300 --> 00:20:07.950 That's the virtue of writing things out in this way. 00:20:07.950 --> 00:20:14.940 So now, the square here cancels with this cosine. 00:20:14.940 --> 00:20:20.580 And I'm left with cos(theta) d theta / sin^2(theta). 00:20:25.340 --> 00:20:26.860 That's a little simpler. 00:20:26.860 --> 00:20:33.300 And it puts me in a position to use the same idea I just used. 00:20:33.300 --> 00:20:35.100 I see the sine here. 00:20:35.100 --> 00:20:37.090 I might look around in this integral 00:20:37.090 --> 00:20:39.940 to see if its derivative occurs anywhere. 00:20:39.940 --> 00:20:43.890 The differential of the sine is the cosine. 00:20:43.890 --> 00:20:49.720 And so I'm very much inclined to make another substitution. 00:20:49.720 --> 00:20:54.250 Say, u, direct substitution this time. 00:20:54.250 --> 00:20:56.150 And say u is the cosine of theta. 00:20:56.150 --> 00:20:59.110 Because then du-- Oh, I'm sorry. 00:20:59.110 --> 00:21:01.850 Say, u is the sine of theta. 00:21:01.850 --> 00:21:05.980 Because then du is cos(theta) d theta. 00:21:13.540 --> 00:21:20.140 And then this integral becomes, well, the numerator just is du. 00:21:20.140 --> 00:21:22.660 The denominator is u^2. 00:21:22.660 --> 00:21:25.550 And I think we can break out the champagne, 00:21:25.550 --> 00:21:28.169 because we can integrate that one. 00:21:28.169 --> 00:21:29.710 Finally get rid of the integral sign. 00:21:29.710 --> 00:21:30.210 Yes sir. 00:21:30.210 --> 00:21:37.842 STUDENT: [INAUDIBLE] 00:21:37.842 --> 00:21:40.300 PROFESSOR: OK, how do I know to make u equal to sine rather 00:21:40.300 --> 00:21:41.640 than cosine. 00:21:41.640 --> 00:21:45.540 Because I want to see du appear up here. 00:21:45.540 --> 00:21:49.360 If I'd had a sine up here, that would be a signal to me 00:21:49.360 --> 00:21:53.510 that maybe I should say let u be the cosine. 00:21:53.510 --> 00:21:54.280 OK? 00:21:54.280 --> 00:21:56.980 Also, because this thing in the denominator is something 00:21:56.980 --> 00:21:57.870 I want to get rid of. 00:21:57.870 --> 00:21:59.050 It's in the denominator. 00:21:59.050 --> 00:22:01.030 So I'll get rid of it by wishful thinking 00:22:01.030 --> 00:22:04.940 and just call it something else. 00:22:04.940 --> 00:22:07.010 It works pretty well in this case. 00:22:07.010 --> 00:22:10.400 Wishful thinking doesn't always work so well. 00:22:10.400 --> 00:22:18.510 So I integrate u^(-2) du, and I get - -1/u plus a constant, 00:22:18.510 --> 00:22:21.340 and I'm done with the calculus part of this problem. 00:22:21.340 --> 00:22:22.580 I've done the integral now. 00:22:22.580 --> 00:22:25.736 Gotten rid of the integral sign. 00:22:25.736 --> 00:22:27.360 But I'm not quite done with the problem 00:22:27.360 --> 00:22:29.640 yet, because I have to work my way back 00:22:29.640 --> 00:22:31.320 through two substitutions. 00:22:31.320 --> 00:22:33.120 First, this one. 00:22:33.120 --> 00:22:34.620 And then this one. 00:22:34.620 --> 00:22:38.800 So this first substitution isn't so bad to get rid of, to undo, 00:22:38.800 --> 00:22:40.200 to back-substitute. 00:22:40.200 --> 00:22:43.630 Because u is just sin(theta). 00:22:43.630 --> 00:22:46.740 And so 1/u is, I guess a fancy way 00:22:46.740 --> 00:22:52.540 to write it is the cosecant of theta. 00:22:52.540 --> 00:22:54.890 1 over the sine is the cosecant. 00:22:54.890 --> 00:23:00.257 So I get -csc(theta) plus a constant. 00:23:00.257 --> 00:23:01.590 Is there a question in the back? 00:23:01.590 --> 00:23:02.090 Yes sir? 00:23:02.090 --> 00:23:07.010 STUDENT: [INAUDIBLE] 00:23:07.010 --> 00:23:09.000 PROFESSOR: I'm sorry, my hearing is so bad. 00:23:09.000 --> 00:23:12.500 STUDENT: [INAUDIBLE] 00:23:12.500 --> 00:23:16.070 PROFESSOR: How did I know this substitution 00:23:16.070 --> 00:23:19.950 in the first place. 00:23:19.950 --> 00:23:22.510 It's because of the 1 + x^2. 00:23:22.510 --> 00:23:24.510 And I want to make use of the trig identity 00:23:24.510 --> 00:23:26.380 in the upper left-hand corner. 00:23:26.380 --> 00:23:29.430 I'll make you a table in a few minutes that will put 00:23:29.430 --> 00:23:30.870 all this in a bigger context. 00:23:30.870 --> 00:23:32.350 And I think it'll help you then. 00:23:32.350 --> 00:23:35.880 OK, I'll promise. 00:23:35.880 --> 00:23:39.570 So, what I want to try to talk about right 00:23:39.570 --> 00:23:43.710 now is how to rewrite a term like this. 00:23:43.710 --> 00:23:47.240 A trig term like this, back in terms of x. 00:23:47.240 --> 00:23:51.020 So I want to undo this trick substitution. 00:23:51.020 --> 00:23:55.190 This is a trig sub. 00:23:55.190 --> 00:23:59.350 And what I want to do now is try to undo that trig sub. 00:23:59.350 --> 00:24:00.970 And I'll show you a general method 00:24:00.970 --> 00:24:03.210 for undoing trig substitutions. 00:24:03.210 --> 00:24:05.100 This happens quite often. 00:24:05.100 --> 00:24:07.150 I don't know what the cosecant of theta is. 00:24:07.150 --> 00:24:11.290 But I do know what the tangent of theta is. 00:24:11.290 --> 00:24:14.060 So I want to make a relation between them. 00:24:14.060 --> 00:24:20.920 OK, so undoing. 00:24:20.920 --> 00:24:24.420 Trig subs. 00:24:24.420 --> 00:24:29.420 So let's go back to where trigonometry always comes from, 00:24:29.420 --> 00:24:32.130 this right angled triangle. 00:24:32.130 --> 00:24:35.540 The theta in the corner, and then these three sides. 00:24:35.540 --> 00:24:37.590 This one's called the hypotenuse. 00:24:37.590 --> 00:24:41.430 This one is called the adjacent side, 00:24:41.430 --> 00:24:45.690 and that one's called the opposite side. 00:24:45.690 --> 00:24:52.040 And now, let's find out where x lies in this triangle. 00:24:52.040 --> 00:24:55.500 Let's try to write the sides of this triangle in terms of x. 00:24:55.500 --> 00:24:58.670 And what I know is, x is the tangent of theta. 00:24:58.670 --> 00:25:01.310 So the tangent of theta, tangent of this angle, 00:25:01.310 --> 00:25:03.830 is opposite divided by adjacent. 00:25:03.830 --> 00:25:05.890 Did you learn SOH CAH TOA? 00:25:05.890 --> 00:25:09.520 OK, so it's opposite divided by adjacent. 00:25:09.520 --> 00:25:10.580 Is the tangent. 00:25:10.580 --> 00:25:13.650 So there are different ways to do that, 00:25:13.650 --> 00:25:15.530 but why not just do it in the simplest way 00:25:15.530 --> 00:25:21.950 and suppose that the adjacent is 1, and the opposite is x. 00:25:21.950 --> 00:25:24.530 This is correct now, isn't it? 00:25:24.530 --> 00:25:27.890 I get the correct value for the tangent of theta 00:25:27.890 --> 00:25:32.910 by saying that the lengths of those are 1 and x. 00:25:32.910 --> 00:25:39.160 And that means that the hypotenuse has length 1 + x^2. 00:25:39.160 --> 00:25:40.840 Well, here's a triangle. 00:25:40.840 --> 00:25:44.060 I'm interested in computing the cosecant of theta. 00:25:44.060 --> 00:25:50.840 Where's that appear in the triangle? 00:25:50.840 --> 00:25:51.510 Well, let's see. 00:25:51.510 --> 00:25:55.570 The cosecant of theta is 1 over the sine. 00:25:55.570 --> 00:26:02.830 And the sine is opposite over hypotenuse. 00:26:02.830 --> 00:26:15.240 So the cosecant is hypotenuse over opposite. 00:26:15.240 --> 00:26:20.680 And the hypotenuse is the square root of 1 + x^2, 00:26:20.680 --> 00:26:24.680 and the opposite is x. 00:26:24.680 --> 00:26:26.200 And so I've done it. 00:26:26.200 --> 00:26:29.650 I've undone the trig substitution. 00:26:29.650 --> 00:26:32.100 I've figured out what this cosecant of theta 00:26:32.100 --> 00:26:35.360 is, in terms of x. 00:26:35.360 --> 00:26:39.490 And so the final answer is minus the square root of 1+x^2, 00:26:39.490 --> 00:26:43.470 over x, plus a constant, and there's an answer 00:26:43.470 --> 00:26:53.030 to the original problem. 00:26:53.030 --> 00:26:57.160 This took two boards to go through this. 00:26:57.160 --> 00:27:00.800 I illustrated several things. 00:27:00.800 --> 00:27:02.870 Actually, this three half boards. 00:27:02.870 --> 00:27:05.190 I illustrated this use of trig substitution, 00:27:05.190 --> 00:27:07.910 and I'll come back to that in a second. 00:27:07.910 --> 00:27:10.870 I illustrated patience. 00:27:10.870 --> 00:27:15.320 I illustrated rewriting things in terms of sines and cosines, 00:27:15.320 --> 00:27:17.080 and then making a direct substitution 00:27:17.080 --> 00:27:19.550 to evaluate an integral like this. 00:27:19.550 --> 00:27:23.520 And then there's this undoing all of those substitutions. 00:27:23.520 --> 00:27:25.670 And it culminated with undoing the trig sub. 00:27:25.670 --> 00:27:31.100 So let's play a game here. 00:27:31.100 --> 00:27:35.070 Why don't we play the game where you 00:27:35.070 --> 00:27:46.970 give me-- So, there's a step in here that I should have done. 00:27:46.970 --> 00:27:56.390 I should've said this is -cos(arctan(theta)) plus 00:27:56.390 --> 00:27:58.812 a constant. 00:27:58.812 --> 00:28:00.520 The most straightforward thing you can do 00:28:00.520 --> 00:28:02.760 is to say since x is the tangent of theta, 00:28:02.760 --> 00:28:06.950 that means that-- sorry, if x, that means that theta 00:28:06.950 --> 00:28:08.970 is the arctangent of x. 00:28:08.970 --> 00:28:11.761 And so let's just put in theta as the arctangent of x, 00:28:11.761 --> 00:28:12.760 and that's what you get. 00:28:12.760 --> 00:28:15.020 So really, what I just did for you 00:28:15.020 --> 00:28:19.720 was to show you a way to compute some trig function applied 00:28:19.720 --> 00:28:23.900 to the inverse of another trig function. 00:28:23.900 --> 00:28:29.860 I computed cosecant of the arctangent by this trick. 00:28:29.860 --> 00:28:32.260 So now, let's play the game where you give me 00:28:32.260 --> 00:28:35.150 a trig function and an inverse trig function, 00:28:35.150 --> 00:28:46.950 and I try to compute what the composite is. 00:28:46.950 --> 00:28:47.510 OK. 00:28:47.510 --> 00:28:58.530 So who can give me a trig function. 00:28:58.530 --> 00:29:03.310 Has to be one of these standard ones. 00:29:03.310 --> 00:29:03.900 STUDENT: Tan. 00:29:03.900 --> 00:29:04.870 PROFESSOR: Tangent. 00:29:04.870 --> 00:29:07.034 Alright. 00:29:07.034 --> 00:29:07.950 How about another one? 00:29:07.950 --> 00:29:11.120 STUDENT: Sine. 00:29:11.120 --> 00:29:12.350 PROFESSOR: Sine. 00:29:12.350 --> 00:29:15.180 Do we have agreement on sine. 00:29:15.180 --> 00:29:16.910 STUDENT: [INAUDIBLE] 00:29:16.910 --> 00:29:17.760 PROFESSOR: Secant? 00:29:17.760 --> 00:29:27.802 STUDENT: [INAUDIBLE] 00:29:27.802 --> 00:29:29.510 PROFESSOR: Right, csc has the best cheer. 00:29:29.510 --> 00:29:30.301 So that's the game. 00:29:30.301 --> 00:29:34.867 We have to compute, try to compute, that composite. 00:29:34.867 --> 00:29:35.950 Something wrong with this? 00:29:35.950 --> 00:29:45.459 STUDENT: [INAUDIBLE] 00:29:45.459 --> 00:29:47.000 PROFESSOR: What does acceptable mean? 00:29:47.000 --> 00:29:48.810 Don't you think-- so the question is, 00:29:48.810 --> 00:29:51.900 isn't this a perfectly acceptable final answer. 00:29:51.900 --> 00:29:53.760 It's a correct final answer. 00:29:53.760 --> 00:29:56.850 But this is much more insightful. 00:29:56.850 --> 00:30:00.919 And after all the original thing was involving square roots 00:30:00.919 --> 00:30:02.460 and things, this is the kind of thing 00:30:02.460 --> 00:30:03.920 you might hope for is an answer. 00:30:03.920 --> 00:30:07.720 This is just a nicer answer for sure. 00:30:07.720 --> 00:30:10.210 And likely to be more useful to you when you go on 00:30:10.210 --> 00:30:13.830 and use that answer for something else. 00:30:13.830 --> 00:30:18.610 OK, so let's try to do this this. 00:30:18.610 --> 00:30:22.260 Undo a trig substitution that involved a cosecant. 00:30:22.260 --> 00:30:23.910 And I manipulate around, and I find 00:30:23.910 --> 00:30:27.540 myself trying to find out what's the tangent of theta. 00:30:27.540 --> 00:30:29.240 So here's how we go about it. 00:30:29.240 --> 00:30:33.970 I draw this triangle. 00:30:33.970 --> 00:30:37.590 Theta is the angle here. 00:30:37.590 --> 00:30:43.570 This is the adjacent, opposite, hypotenuse. 00:30:43.570 --> 00:30:49.030 So, the first thing is how can I make the cosecant appear here, 00:30:49.030 --> 00:30:49.950 csc x. 00:30:49.950 --> 00:30:52.640 What dimensions should I give to the sides in order 00:30:52.640 --> 00:31:01.400 for the cosecant of x, sorry, in order for theta 00:31:01.400 --> 00:31:02.880 to be the cosecant of x. 00:31:02.880 --> 00:31:05.280 This thing is theta. 00:31:05.280 --> 00:31:17.810 So, that means that the cosecant of x-- 00:31:17.810 --> 00:31:24.830 that means the cosecant of theta should be x. 00:31:24.830 --> 00:31:28.970 Theta is the arccosecant, so x is the cosecant of theta. 00:31:28.970 --> 00:31:31.980 So, what'll I take the sides to be, to get the cosecant? 00:31:31.980 --> 00:31:37.050 The cosecant is 1 over the sine. 00:31:37.050 --> 00:31:47.960 And the sine is the opposite over the hypotenuse. 00:31:47.960 --> 00:31:49.910 So I get hypotenuse over opposite. 00:31:49.910 --> 00:31:53.750 And that's supposed to be what x is. 00:31:53.750 --> 00:31:56.540 So I could make the opposite anything 00:31:56.540 --> 00:31:59.190 I want, but the simplest thing is to make it 1. 00:31:59.190 --> 00:32:00.080 Let's do that. 00:32:00.080 --> 00:32:04.300 And then what does that mean about the rest of the sides? 00:32:04.300 --> 00:32:06.142 Hypotenuse had better be x. 00:32:06.142 --> 00:32:07.350 And then I've recovered this. 00:32:07.350 --> 00:32:12.350 So here's a triangle that exhibits the correct angle. 00:32:12.350 --> 00:32:14.680 This remaining side is going to be useful to us. 00:32:14.680 --> 00:32:20.550 And it is the square root of x^2 - 1. 00:32:20.550 --> 00:32:24.060 So I've got a triangle of the correct angle theta, 00:32:24.060 --> 00:32:27.400 and now I want to compute the tangent of that angle. 00:32:27.400 --> 00:32:28.350 Well, that's easy. 00:32:28.350 --> 00:32:31.020 That's opposite divided by adjacent. 00:32:31.020 --> 00:32:34.690 So I get 1 over the square root of x^2 - 1. 00:32:38.270 --> 00:32:40.590 Very flexible tool that'll be useful to you 00:32:40.590 --> 00:32:42.200 in many different times. 00:32:42.200 --> 00:32:44.170 Whenever you have to undo a trig substitution, 00:32:44.170 --> 00:32:49.100 this is likely to be useful. 00:32:49.100 --> 00:32:51.740 OK, that was a good game. 00:32:51.740 --> 00:32:52.740 No winners in this game. 00:32:52.740 --> 00:32:53.700 We're all winners. 00:32:53.700 --> 00:33:01.970 No losers, we're all winners. 00:33:01.970 --> 00:33:02.740 OK. 00:33:02.740 --> 00:33:06.340 So, good. 00:33:06.340 --> 00:33:09.810 So let me make this table of the different trig substitutions, 00:33:09.810 --> 00:33:11.010 and how they can be useful. 00:33:11.010 --> 00:33:20.520 Summary of trig substitutions. 00:33:20.520 --> 00:33:28.540 So over here, we have, if you see, 00:33:28.540 --> 00:33:58.090 so if your integrand contains, make a substitution to get. 00:33:58.090 --> 00:34:01.086 So if your integrand contains, I'll 00:34:01.086 --> 00:34:02.710 write these things out as square roots. 00:34:02.710 --> 00:34:07.670 If it contains the square root a^2 - x^2, 00:34:07.670 --> 00:34:10.560 this is what we talked about on Thursday. 00:34:10.560 --> 00:34:14.470 When I was trying to find the area of that piece of a circle. 00:34:14.470 --> 00:34:18.130 There, I suggested that we should make the substitution 00:34:18.130 --> 00:34:20.260 x = a cos(theta). 00:34:22.890 --> 00:34:28.920 Or, x = a sin(theta). 00:34:28.920 --> 00:34:31.400 Either one works just as well. 00:34:31.400 --> 00:34:36.430 And there's no way to prefer one over the other. 00:34:36.430 --> 00:34:41.110 And when you make the substitution, x = a cos(theta), 00:34:41.110 --> 00:34:44.030 you get a^2 - a^2 cos^2(theta). 00:34:44.030 --> 00:34:45.340 theta. 00:34:45.340 --> 00:34:48.280 1 - cos^2 is sin^2. 00:34:48.280 --> 00:34:50.620 So you get a sin(theta). 00:34:55.270 --> 00:34:59.880 So this expression becomes equal to this expression 00:34:59.880 --> 00:35:03.950 under that substitution. 00:35:03.950 --> 00:35:04.800 And then you go on. 00:35:04.800 --> 00:35:06.626 Then you've gotten rid of the square root, 00:35:06.626 --> 00:35:08.250 and you've got a trigonometric integral 00:35:08.250 --> 00:35:10.610 that you have to try to do. 00:35:10.610 --> 00:35:14.680 If you made the substitution a sin(theta), you'd get a a^2 - 00:35:14.680 --> 00:35:21.580 a^2 sin^2, which is a cos(theta). 00:35:21.580 --> 00:35:24.530 And then you can go ahead as well. 00:35:24.530 --> 00:35:26.430 We just saw another example. 00:35:26.430 --> 00:35:29.510 Namely, if you have a^2 + x^2. 00:35:29.510 --> 00:35:35.530 That's like the example we had up here. a = 1 in this example. 00:35:35.530 --> 00:35:36.510 What did we do? 00:35:36.510 --> 00:35:42.150 We tried the substitution x = a tan(theta). 00:35:42.150 --> 00:35:44.910 And the reason is that I can plug into the trig identity 00:35:44.910 --> 00:35:46.950 up here in the upper left. 00:35:46.950 --> 00:35:51.210 And replace a^2 + x^2 by a sec(theta). 00:35:55.020 --> 00:35:59.790 Square root of the secant squared. 00:35:59.790 --> 00:36:03.410 There's one more thing in this table. 00:36:03.410 --> 00:36:06.520 Sort of, the only remaining sum or difference 00:36:06.520 --> 00:36:07.544 of terms like this. 00:36:07.544 --> 00:36:09.460 And that's what happens if you have x^2 - a^2. 00:36:14.050 --> 00:36:17.790 So there, I think we can make a substitution a sec(theta). 00:36:21.300 --> 00:36:27.031 Because, after all, sec^2(theta)-- so x^2 - a^2-- 00:36:27.031 --> 00:36:27.530 Sorry. 00:36:32.310 --> 00:36:34.790 Let's see what happens when I make that substitution. 00:36:34.790 --> 00:36:42.850 x^2 - a^2 = a^2 sec^2(theta) - a^2. 00:36:42.850 --> 00:36:45.460 Under this substitution. 00:36:45.460 --> 00:36:48.690 That's sec^2 - 1. 00:36:48.690 --> 00:36:51.390 Well, put the 1 on the other side. 00:36:51.390 --> 00:36:53.720 And you find tan^2, coming out. 00:36:53.720 --> 00:36:56.710 So this is a a^2 tan^2(theta). 00:36:59.440 --> 00:37:02.680 And so that's what you get, a times tan(theta). 00:37:02.680 --> 00:37:07.850 After I take the square root, I get a tan(theta). 00:37:07.850 --> 00:37:13.080 So these are the three basic trig substitution forms. 00:37:13.080 --> 00:37:15.580 Where trig substitutions are useful to get 00:37:15.580 --> 00:37:17.920 rid of expressions like this, and replace them 00:37:17.920 --> 00:37:22.670 by trigonometric expressions. 00:37:22.670 --> 00:37:25.140 And then you use this trick, you do the integral if you can 00:37:25.140 --> 00:37:27.265 and then you use this trick to get rid of the theta 00:37:27.265 --> 00:37:37.460 at the end. 00:37:37.460 --> 00:37:40.630 So now, the last thing I want to talk about today 00:37:40.630 --> 00:37:59.180 is called completing the square. 00:37:59.180 --> 00:38:03.440 And that comes in because unfortunately, 00:38:03.440 --> 00:38:08.350 not every square root of a quadratic 00:38:08.350 --> 00:38:10.390 has such a simple form. 00:38:10.390 --> 00:38:16.200 You will often encounter things that are not just 00:38:16.200 --> 00:38:18.080 the square root of something simple. 00:38:18.080 --> 00:38:19.140 Like one of these forms. 00:38:19.140 --> 00:38:27.537 Like there might be a middle term in there. 00:38:27.537 --> 00:38:29.120 I don't actually have time to show you 00:38:29.120 --> 00:38:33.200 an example of how this comes out in a sort of practical example. 00:38:33.200 --> 00:38:35.760 But it does happen quite frequently. 00:38:35.760 --> 00:38:38.580 And so I want to show you how to deal with things 00:38:38.580 --> 00:38:40.550 like the following example. 00:38:40.550 --> 00:38:47.380 Let's try to integrate dx over x^2 + 4x, 00:38:47.380 --> 00:38:55.090 the square root of x^2 + 4x. 00:38:55.090 --> 00:38:59.010 So there's a square root of some square, some quadratic. 00:38:59.010 --> 00:39:02.010 It's very much like this business. 00:39:02.010 --> 00:39:04.830 But it isn't of any of these forms. 00:39:04.830 --> 00:39:06.360 And so what I want to do is show you 00:39:06.360 --> 00:39:10.540 how to rewrite it in one of those forms using substitution, 00:39:10.540 --> 00:39:11.120 again. 00:39:11.120 --> 00:39:14.520 All this is about substitution. 00:39:14.520 --> 00:39:26.100 So the game is to rewrite quadratic as something 00:39:26.100 --> 00:39:30.950 like x plus something or other. 00:39:30.950 --> 00:39:32.470 Plus some other constant. 00:39:32.470 --> 00:39:35.690 So write it, try to write it, in the form of a square 00:39:35.690 --> 00:39:42.340 plus or minus another constant. 00:39:42.340 --> 00:39:45.170 And then we'll go on from there. 00:39:45.170 --> 00:39:51.110 So let's do that in this case. x^2, x^2 + 4x. 00:39:51.110 --> 00:39:53.390 Well, if you square this form out, 00:39:53.390 --> 00:39:57.280 then the middle term is going to be 2ax. 00:39:57.280 --> 00:40:00.920 So that, since I have a middle term here, 00:40:00.920 --> 00:40:03.640 I pretty much know what a has to be. 00:40:03.640 --> 00:40:08.220 The only choice in order to get something like x^2 + 4x out 00:40:08.220 --> 00:40:11.440 of this, is to take a to be 2. 00:40:11.440 --> 00:40:16.950 Because then, this is what you get. 00:40:16.950 --> 00:40:19.790 This isn't quite right yet, but let's compute what I have here. 00:40:19.790 --> 00:40:24.490 x^2 + 4x, so far so good, plus 4, 00:40:24.490 --> 00:40:26.190 and I don't have a plus 4 here. 00:40:26.190 --> 00:40:31.394 So I have to fix that by subtracting 4. 00:40:31.394 --> 00:40:32.310 So that's what I mean. 00:40:32.310 --> 00:40:33.393 I've completed the square. 00:40:33.393 --> 00:40:38.370 The word for this process of eliminating the middle term 00:40:38.370 --> 00:40:41.850 by using the square of an expression like that. 00:40:41.850 --> 00:40:44.830 That's called completing the square. 00:40:44.830 --> 00:40:48.010 And we can use that process to compute this integral. 00:40:48.010 --> 00:40:51.040 So let's do that. 00:40:51.040 --> 00:40:54.170 So I can rewrite this integral, rewrite this denominator 00:40:54.170 --> 00:41:00.100 like this. 00:41:00.100 --> 00:41:03.860 And then I'm going to try to use one of these forms over here. 00:41:03.860 --> 00:41:07.900 So in order to get a single variable there, 00:41:07.900 --> 00:41:12.500 instead of something complicated like x + 2, 00:41:12.500 --> 00:41:15.090 I'm inclined to come up with another variable name 00:41:15.090 --> 00:41:20.330 and write it equal, write x + 2 as that other variable name. 00:41:20.330 --> 00:41:29.950 So here's another little direct substitution. u = x + 2. 00:41:29.950 --> 00:41:31.260 Figure out what du is. 00:41:31.260 --> 00:41:36.610 That's pretty easy. 00:41:36.610 --> 00:41:41.580 And then rewrite the integral in those terms. 00:41:41.580 --> 00:41:44.880 So dx = du. 00:41:44.880 --> 00:41:47.270 And then in the denominator I have, 00:41:47.270 --> 00:41:49.100 well, I have the square root of that. 00:41:49.100 --> 00:41:55.100 Oh yeah, so I think as part of this I'll write out what x^2 + 00:41:55.100 --> 00:41:56.550 4x is. 00:41:56.550 --> 00:41:59.520 The point is, it's equal to u^2 - 4. 00:42:04.170 --> 00:42:06.540 x^2 + 4x = u^2 - 4. 00:42:12.950 --> 00:42:20.260 There's the data box containing the substitution data. 00:42:20.260 --> 00:42:22.640 And so now I can put that in. 00:42:22.640 --> 00:42:24.960 I have x^2 + 4x there. 00:42:24.960 --> 00:42:31.330 In terms of u, that's u^2 - 4. 00:42:31.330 --> 00:42:33.690 Well, now I'm in a happier position because I can look 00:42:33.690 --> 00:42:37.060 for u^2 - 4 for something like that in my table here. 00:42:37.060 --> 00:42:40.430 And it actually sits down here. 00:42:40.430 --> 00:42:43.365 So except for the use of the letter x 00:42:43.365 --> 00:42:45.470 here instead of u over there. 00:42:45.470 --> 00:42:47.510 That tells me what I want. 00:42:47.510 --> 00:42:55.100 So to handle this, what I should use is a trig substitution. 00:42:55.100 --> 00:42:58.020 And the trig substitution that's suggested is, 00:42:58.020 --> 00:43:03.870 according to the bottom line with a = 2, so a^2 = 4. 00:43:03.870 --> 00:43:07.600 The suggestion is, I should take x-- 00:43:07.600 --> 00:43:13.909 But I'd better not use the letter x any more. 00:43:13.909 --> 00:43:15.950 But I don't have a letter x, I have the letter u. 00:43:15.950 --> 00:43:21.310 I should take u equal to 2 secant. 00:43:21.310 --> 00:43:23.110 And then some letter I haven't used before. 00:43:23.110 --> 00:43:28.820 And theta is available. 00:43:28.820 --> 00:43:30.560 This is a look-up table process. 00:43:30.560 --> 00:43:33.240 I see the square root of u^2 - 4, 00:43:33.240 --> 00:43:35.250 I see that that's of this form. 00:43:35.250 --> 00:43:38.230 I'm instructed to make this substitution. 00:43:38.230 --> 00:43:40.540 And that's what I just did. 00:43:40.540 --> 00:43:43.600 Let's see how it works out. 00:43:43.600 --> 00:43:47.690 So that means the du is 2, OK. 00:43:47.690 --> 00:43:50.480 What's the derivative of the secant? 00:43:50.480 --> 00:43:52.660 Secant tangent. 00:43:52.660 --> 00:43:59.050 So du = 2 sec(theta) tan(theta). 00:43:59.050 --> 00:44:05.690 And u^2 - 4 is, here's the payoff. 00:44:05.690 --> 00:44:07.170 I'm supposed to be able to rewrite 00:44:07.170 --> 00:44:09.350 that in terms of the tangent. 00:44:09.350 --> 00:44:19.650 According to this. u u^2 - 4 is 4 secant squared minus 4. 00:44:19.650 --> 00:44:22.450 And secant squared minus 1 is tangent squared. 00:44:22.450 --> 00:44:24.400 So this is 4 tan^2(theta). 00:44:30.890 --> 00:44:36.807 Right, yeah? 00:44:36.807 --> 00:44:37.640 STUDENT: [INAUDIBLE] 00:44:37.640 --> 00:44:42.200 PROFESSOR: But I squared it. 00:44:42.200 --> 00:44:44.180 And now I'll square root it. 00:44:44.180 --> 00:44:50.650 And I'll get a 2 and this tangent will go away. 00:44:50.650 --> 00:44:56.130 So there's my data box for this substitution. 00:44:56.130 --> 00:45:15.560 And let's go on to another board. 00:45:15.560 --> 00:45:20.980 So where I'm at is the integral of du over the square root 00:45:20.980 --> 00:45:22.010 of u^2 - 4. 00:45:26.850 --> 00:45:30.060 And I have all the data I need here 00:45:30.060 --> 00:45:33.090 to rewrite that in terms of theta. 00:45:33.090 --> 00:45:38.170 So du = 2 sec(theta) tan(theta) d theta. 00:45:41.720 --> 00:45:44.140 And the denominator is 2 tan(theta). 00:45:47.760 --> 00:45:48.590 Ha. 00:45:48.590 --> 00:45:53.040 Well, so some very nice simplification happens here. 00:45:53.040 --> 00:45:55.940 The 2's cancel. 00:45:55.940 --> 00:45:58.260 And the tangents cancel. 00:45:58.260 --> 00:46:01.900 And I'm left with trying to work with the integral, sec(theta) d 00:46:01.900 --> 00:46:03.920 theta. 00:46:03.920 --> 00:46:06.610 And luckily enough at the very beginning of the hour, 00:46:06.610 --> 00:46:08.480 I worked out how to compute the integral 00:46:08.480 --> 00:46:09.885 of the secant of theta. 00:46:09.885 --> 00:46:11.670 And there it is. 00:46:11.670 --> 00:46:25.000 So this is ln(sec(theta) + tan(theta)) plus a constant. 00:46:25.000 --> 00:46:27.660 And we're done with the calculus part. 00:46:27.660 --> 00:46:29.390 There's no more integral there. 00:46:29.390 --> 00:46:31.490 But I still am not quite done with the problem, 00:46:31.490 --> 00:46:37.180 because again I have these two substitutions to try to undo. 00:46:37.180 --> 00:46:40.220 So let's undo them one by one. 00:46:40.220 --> 00:46:42.240 Let's see. 00:46:42.240 --> 00:46:44.130 I have this trig substitution here. 00:46:44.130 --> 00:46:47.210 And I could use my triangle trick, if I need to. 00:46:47.210 --> 00:46:49.280 But maybe I don't need to. 00:46:49.280 --> 00:46:51.220 Let's see, do I know what the secant of theta 00:46:51.220 --> 00:46:53.090 is in terms of u? 00:46:53.090 --> 00:46:54.620 Well, I do. 00:46:54.620 --> 00:46:55.730 So I get ln(u/2). 00:46:58.980 --> 00:47:01.740 Do I know what the tangent is in terms of u? 00:47:01.740 --> 00:47:02.730 Well, I do. 00:47:02.730 --> 00:47:04.180 It's here. 00:47:04.180 --> 00:47:06.500 So I lucked out, in this case. 00:47:06.500 --> 00:47:09.930 And I don't have to go through and use that triangle trick. 00:47:09.930 --> 00:47:15.700 So the tangent of theta is the square root of u^2 - 4, over 2. 00:47:19.000 --> 00:47:20.080 Good. 00:47:20.080 --> 00:47:24.220 So I've undone this trig substitution. 00:47:24.220 --> 00:47:28.440 I'm not quite done yet because my answer is involved with u. 00:47:28.440 --> 00:47:30.780 And what I wanted originally was x. 00:47:30.780 --> 00:47:33.780 But this direct substitution that I started with 00:47:33.780 --> 00:47:35.250 is really easy to deal with. 00:47:35.250 --> 00:47:39.780 I can just put x + 2 every time I see a u. 00:47:39.780 --> 00:47:45.590 So this is the natural logarithm of (x+2)/2 plus the square 00:47:45.590 --> 00:47:46.090 root... 00:47:46.090 --> 00:47:51.330 What's going to happen when I put x + 2 in place for u here? 00:47:51.330 --> 00:47:54.490 You know what you get. 00:47:54.490 --> 00:47:59.230 You get exactly what we started with. 00:47:59.230 --> 00:48:00.260 Right? 00:48:00.260 --> 00:48:04.850 I put x + 2 in place of the u here. 00:48:04.850 --> 00:48:13.210 I get x^2 + 4x. 00:48:13.210 --> 00:48:15.330 So I've gotten back to a function purely 00:48:15.330 --> 00:48:21.200 in terms of x OK, that's a good place to quit. 00:48:21.200 --> 00:48:24.840 Have a great little one-day break. 00:48:24.840 --> 00:48:27.320 I guess this class doesn't meet on Monday anyway. 00:48:27.320 --> 00:48:28.477 Bye.