1 00:00:00 --> 00:00:01 2 00:00:01 --> 00:00:03 The following content is provided under a Creative 3 00:00:03 --> 00:00:04 Commons License. 4 00:00:04 --> 00:00:06 Your support will help MIT OpenCourseWare continue to 5 00:00:06 --> 00:00:09 offer high quality educational resources for free. 6 00:00:09 --> 00:00:12 To make a donation or to view additional materials from 7 00:00:12 --> 00:00:16 hundreds of MIT courses, visit MIT OpenCourseWare 8 00:00:16 --> 00:00:21 at ocw.mit.edu. 9 00:00:21 --> 00:00:26 PROFESSOR: So we're going to continue to talk about trig 10 00:00:26 --> 00:00:29 integrals and trig substitutions. 11 00:00:29 --> 00:00:32 This is maybe the most technical part of this course, 12 00:00:32 --> 00:00:35 which maybe is why professor Jerison decided to just take a 13 00:00:35 --> 00:00:38 leave, go AWOL just now and let me take over for him. 14 00:00:38 --> 00:00:43 But I'll do my best to help you learn this technique and 15 00:00:43 --> 00:00:45 it'll be useful for you. 16 00:00:45 --> 00:00:49 So we've talked about trig integrals involving sines 17 00:00:49 --> 00:00:51 and cosines yesterday. 18 00:00:51 --> 00:00:53 There's another whole world out there that involves these 19 00:00:53 --> 00:00:57 other trig polynomials. 20 00:00:57 --> 00:00:58 Trig functions. 21 00:00:58 --> 00:01:00 Secant and tangent. 22 00:01:00 --> 00:01:03 Let me just make a little table to remind you what they are. 23 00:01:03 --> 00:01:05 Because I have trouble remembering myself, so I enjoy 24 00:01:05 --> 00:01:09 the opportunity to go back to remind myself of this stuff. 25 00:01:09 --> 00:01:09 Let's see. 26 00:01:09 --> 00:01:15 The secant is one over one of those things, which one is it? 27 00:01:15 --> 00:01:18 It's weird, it's 1 / cos. 28 00:01:18 --> 00:01:24 And the cosecant = 1 / sin. 29 00:01:24 --> 00:01:26 Of course the tangent, we know. 30 00:01:26 --> 00:01:32 It's the sin / cos and the cotangent is the 31 00:01:32 --> 00:01:36 other way around. 32 00:01:36 --> 00:01:40 So when you put a co in front of it, it exchanges 33 00:01:40 --> 00:01:43 sine and cosine. 34 00:01:43 --> 00:01:46 Well, I have a few identities involving tangent and secant 35 00:01:46 --> 00:01:50 up there, in that little prepared blackboard up above. 36 00:01:50 --> 00:01:54 Maybe I'll just go through and check that out to make sure 37 00:01:54 --> 00:01:57 that we're all on the same page with them. 38 00:01:57 --> 00:01:59 So I'm going to claim that there's this trig 39 00:01:59 --> 00:02:00 identity at the top. 40 00:02:00 --> 00:02:07 Sec^2 = 1 + tangent. 41 00:02:07 --> 00:02:10 So let's just check that out. 42 00:02:10 --> 00:02:14 So the sec = 1 / cos, so sec ^2 = 1 / cos ^2. 43 00:02:14 --> 00:02:17 And then whenever you see a 1 in trigonometry, you'll always 44 00:02:17 --> 00:02:23 have the option of writing as cos ^2 + sin^2. 45 00:02:23 --> 00:02:27 46 00:02:27 --> 00:02:31 And if I do that, then I can divide the cos ^2 47 00:02:31 --> 00:02:34 into that first term. 48 00:02:34 --> 00:02:37 And I get 1 + sin ^2 / cos ^2. 49 00:02:37 --> 00:02:42 Which is the tan ^2. 50 00:02:42 --> 00:02:44 So there you go. 51 00:02:44 --> 00:02:47 That checks the first one. 52 00:02:47 --> 00:02:50 That's the main trig identity that's going to be behind 53 00:02:50 --> 00:02:51 what I talk about today. 54 00:02:51 --> 00:02:53 That's the trigonometry identity part. 55 00:02:53 --> 00:02:57 How about this piece of calculus. 56 00:02:57 --> 00:03:05 Can we calculate what the derivative of the tan x is. 57 00:03:05 --> 00:03:12 Actually, I'm going to do that on this board. 58 00:03:12 --> 00:03:19 So the tangent of x = sin x / cos x. 59 00:03:19 --> 00:03:21 So I think I was with you when we learned about 60 00:03:21 --> 00:03:23 the quotient rule. 61 00:03:23 --> 00:03:26 Computing the derivative of a quotient. 62 00:03:26 --> 00:03:31 And the rule is, you take the numerator and you -- sorry, you 63 00:03:31 --> 00:03:36 take the derivative of the numerator, which is cosine. 64 00:03:36 --> 00:03:38 And you multiply it by the denominator, so 65 00:03:38 --> 00:03:41 that gives you cos ^2. 66 00:03:41 --> 00:03:45 And then you take the numerator, take minus the 67 00:03:45 --> 00:03:48 numerator, and multiply that by the derivative of the 68 00:03:48 --> 00:03:53 denominator, which is - sin x. 69 00:03:53 --> 00:03:57 And you put all that over the square of the denominator. 70 00:03:57 --> 00:04:02 And now I look at that and before my eyes I see the same 71 00:04:02 --> 00:04:07 trig identity, cosine ^2 + sine ^2 = 1, appearing there. 72 00:04:07 --> 00:04:13 This is 1 / cosine ^2 x, which is secant ^2. 73 00:04:13 --> 00:04:15 And, good. 74 00:04:15 --> 00:04:16 So that's what the claim was. 75 00:04:16 --> 00:04:20 The derivative of the tangent is the secant ^2. 76 00:04:20 --> 00:04:22 That immediately gives you an integral. 77 00:04:22 --> 00:04:25 Namely, the integral of secant ^2 is the tangent. 78 00:04:25 --> 00:04:28 That's the fundamental theorem of calculus. 79 00:04:28 --> 00:04:32 So we verified the first integral there. 80 00:04:32 --> 00:04:34 Well, let's just do the second one as well. 81 00:04:34 --> 00:04:40 So if I want to differentiate the secant, derivative 82 00:04:40 --> 00:04:41 of the secant. 83 00:04:41 --> 00:04:46 So that's d/dx of 1 / cosine. 84 00:04:46 --> 00:04:47 And again, I have a quotient. 85 00:04:47 --> 00:04:49 This one's a little easier because the 86 00:04:49 --> 00:04:51 numerator's so simple. 87 00:04:51 --> 00:04:54 So I take the derivative of the numerator, which is 0. 88 00:04:54 --> 00:04:58 And then I take the numerator, I take minus the numerator your 89 00:04:58 --> 00:05:00 times the derivative of the denominator. 90 00:05:00 --> 00:05:07 Which is - sin x, and put all that over the square 91 00:05:07 --> 00:05:09 of the same denominator. 92 00:05:09 --> 00:05:13 So 1 - sin x came from the quotient rule, and the other 93 00:05:13 --> 00:05:16 one came because that's the derivative of the cosine. 94 00:05:16 --> 00:05:22 But they cancel, and so I get sin / cos ^2, which is (sin / 95 00:05:22 --> 00:05:25 cos) ( 1 / cos) and so that's the secant, that's (1 96 00:05:25 --> 00:05:32 / cos), times tan x. 97 00:05:32 --> 00:05:34 So, not hard. 98 00:05:34 --> 00:05:36 That verifies that the c derivative of secant 99 00:05:36 --> 00:05:37 is secant tangent. 100 00:05:37 --> 00:05:41 And it tells you that the integral of that weird thing in 101 00:05:41 --> 00:05:44 case you ever want to know, the integral of the secant 102 00:05:44 --> 00:05:47 tangent is the secant. 103 00:05:47 --> 00:05:50 Well, there are a couple more integrals that I 104 00:05:50 --> 00:05:54 want to do for you. 105 00:05:54 --> 00:05:57 Where I can't sort of work backwards like that. 106 00:05:57 --> 00:06:07 Let's calculate the integral of the tangent. 107 00:06:07 --> 00:06:09 Just do this straight out. 108 00:06:09 --> 00:06:22 So the tangent is the sine / cosine. 109 00:06:22 --> 00:06:28 And now there's a habit of mine, that I hope you get into. 110 00:06:28 --> 00:06:33 When you see the cosine and you're calculating an integral 111 00:06:33 --> 00:06:36 like this, it's useful to remember what the derivative 112 00:06:36 --> 00:06:37 of the cosine is. 113 00:06:37 --> 00:06:41 Because maybe it shows up somewhere else in the integral. 114 00:06:41 --> 00:06:43 And that happens here. 115 00:06:43 --> 00:06:50 So that suggests we make a substitution. u = cos x. 116 00:06:50 --> 00:06:55 Which means du = - sin x dx. 117 00:06:55 --> 00:06:59 That's the numerator, except for the minus sign. 118 00:06:59 --> 00:07:08 And so I can rewrite this as, under the substitution, I can 119 00:07:08 --> 00:07:13 rewrite this as - du, that's the numerator, sin x 120 00:07:13 --> 00:07:19 dx is - du / by u. 121 00:07:19 --> 00:07:22 Well, I know how to do that integral too. 122 00:07:22 --> 00:07:24 That gives me the natural log, doesn't it. 123 00:07:24 --> 00:07:31 So this is - ln u + a constant. 124 00:07:31 --> 00:07:32 I'm not quite done. 125 00:07:32 --> 00:07:35 I have to back-substitute and replace this new variable that 126 00:07:35 --> 00:07:39 I've made up, called u, with what it is. 127 00:07:39 --> 00:07:46 And what you get is - ln cos x. 128 00:07:46 --> 00:07:50 So the integral of the tangent is - ln cos. 129 00:07:50 --> 00:07:55 Now, you find these tables of integrals in the 130 00:07:55 --> 00:07:56 back of the book. 131 00:07:56 --> 00:07:58 Things like that. 132 00:07:58 --> 00:08:01 I'm not sure how much memorization Professor Jerison 133 00:08:01 --> 00:08:05 is going to ask of you, but there is a certain amount 134 00:08:05 --> 00:08:07 of memorization that goes on in calculus. 135 00:08:07 --> 00:08:08 And this is one of the kinds of things that you 136 00:08:08 --> 00:08:12 probably want to know. 137 00:08:12 --> 00:08:15 Let me do one more integral. 138 00:08:15 --> 00:08:17 I think I'm making my way through a prepared 139 00:08:17 --> 00:08:20 board here, let's see. 140 00:08:20 --> 00:08:23 Good. 141 00:08:23 --> 00:08:28 So the integral of the tangent is - ln cos. 142 00:08:28 --> 00:08:36 I'd also like to know what the integral of the secant of x is. 143 00:08:36 --> 00:08:41 And I don't know a way to kind of go straight at this, but let 144 00:08:41 --> 00:08:45 me show you a way to think your way through to it. 145 00:08:45 --> 00:08:53 If I take these two facts, tan' is what it is, and sec' is 146 00:08:53 --> 00:08:57 what it is, and add them together, I get this fact. 147 00:08:57 --> 00:09:07 That the derivative of the sec x + tan x is, well, it's the 148 00:09:07 --> 00:09:08 sum of these two things. 149 00:09:08 --> 00:09:11 Secant ^2 + secant tangent. 150 00:09:11 --> 00:09:15 And there's a secant that occurs in both of those terms. 151 00:09:15 --> 00:09:17 So I'll factor it out. 152 00:09:17 --> 00:09:21 And that gives me, I'll put it over here. 153 00:09:21 --> 00:09:24 There's the secant of x that occurs in both terms. 154 00:09:24 --> 00:09:26 And then in one term, there's another secant. 155 00:09:26 --> 00:09:32 And in the other term, there's a tangent. 156 00:09:32 --> 00:09:36 So that's interesting somehow, because this same term appears 157 00:09:36 --> 00:09:40 on both sides of this equation. 158 00:09:40 --> 00:09:49 Let's write u, for that secant x + tangent of x. 159 00:09:49 --> 00:10:00 And so the equation that I get is u' = u (sec x). 160 00:10:00 --> 00:10:03 I've just made a direct substitution. 161 00:10:03 --> 00:10:06 Just decide that I'm going to write u for that single 162 00:10:06 --> 00:10:09 thing that occurs on both sides of the equation. 163 00:10:09 --> 00:10:16 So u' is on the left, and u * sec x is on the right. 164 00:10:16 --> 00:10:18 Well, there's my secant. 165 00:10:18 --> 00:10:20 That I was trying to integrate. 166 00:10:20 --> 00:10:28 And what it tells you is that the secant of x = u' / by u. 167 00:10:28 --> 00:10:33 Just divide both sides by u, and I get this equation. 168 00:10:33 --> 00:10:36 u' / u, that has a name. 169 00:10:36 --> 00:10:39 Not sure that professor Jerison's used this in 170 00:10:39 --> 00:10:42 this class, but u' / u, we've actually used 171 00:10:42 --> 00:10:43 something like that. 172 00:10:43 --> 00:10:45 It's on the board right now. 173 00:10:45 --> 00:10:47 It's a logarithmic derivative. 174 00:10:47 --> 00:10:57 It is the derivative of the national logarithm of u. 175 00:10:57 --> 00:11:00 Maybe it's easier to read this from right to left, if I want 176 00:11:00 --> 00:11:03 to calculate the derivative of the logarithm, well, the chain 177 00:11:03 --> 00:11:07 rule says I get the derivative of u times the derivative of 178 00:11:07 --> 00:11:11 the ln function, which is 1 / u. 179 00:11:11 --> 00:11:26 So often u' / u is called the logarithmic derivative. 180 00:11:26 --> 00:11:27 But it's done what I wanted. 181 00:11:27 --> 00:11:31 Because it's expressed the secant as a derivative. 182 00:11:31 --> 00:11:38 And I guess I should put in what u is. 183 00:11:38 --> 00:11:46 It's the secant + the tangent. 184 00:11:46 --> 00:11:49 And so that implies that the integral, integrate both sides. 185 00:11:49 --> 00:11:53 That says that the integral of the sec x dx, is 186 00:11:53 --> 00:12:02 ln( sec x + tan x). 187 00:12:02 --> 00:12:09 So that's the last line in this little memo that I created. 188 00:12:09 --> 00:12:14 That we can use now for the rest of the class. 189 00:12:14 --> 00:12:19 Any questions about that trick? 190 00:12:19 --> 00:12:23 It's a trick, I have nothing more to say about it. 191 00:12:23 --> 00:12:24 OK. 192 00:12:24 --> 00:12:31 So, the next thing I -- oh yes, so now I want to make the point 193 00:12:31 --> 00:12:38 that using these rules and some thought, you can now integrate 194 00:12:38 --> 00:12:41 most trigonometric polynomials. 195 00:12:41 --> 00:12:45 Most things that involve powers of sines and cosines and 196 00:12:45 --> 00:12:48 tangents and secants and everything else. 197 00:12:48 --> 00:12:52 For example, let's try to integrate the integral 198 00:12:52 --> 00:12:59 of sec^ 4 (x). 199 00:12:59 --> 00:13:02 Big power of the secant function. 200 00:13:02 --> 00:13:03 Well, there are too many secants there for me. 201 00:13:03 --> 00:13:05 So let's take some away. 202 00:13:05 --> 00:13:09 And I can take them away by using that trig identity, 203 00:13:09 --> 00:13:11 secant ^2 = 1 + tan^2. 204 00:13:12 --> 00:13:15 So I'm going to replace two of those secants 205 00:13:15 --> 00:13:21 by 1 + tangent ^2. 206 00:13:21 --> 00:13:25 That leaves me with two left over. 207 00:13:25 --> 00:13:27 Now there was method to my madness. 208 00:13:27 --> 00:13:31 Because I've got a secant ^2 left over there. 209 00:13:31 --> 00:13:35 And secant ^2 is the derivative of tangent. 210 00:13:35 --> 00:13:40 So that suggests a substitution. 211 00:13:40 --> 00:13:46 Namely, let's say, let's let u = tangent of x, so that 212 00:13:46 --> 00:13:50 du = secant ^2 x dx. 213 00:13:50 --> 00:13:55 And I have both terms that occur in my integral 214 00:13:55 --> 00:14:01 sitting there very nicely. 215 00:14:01 --> 00:14:05 So this is the possibility of making this substitution and 216 00:14:05 --> 00:14:08 seeing a secant squared up here as part of the 217 00:14:08 --> 00:14:10 differential here. 218 00:14:10 --> 00:14:14 That's why it was a good idea for me to take two of the 219 00:14:14 --> 00:14:19 secants and write them as 1 + tangent ^2. 220 00:14:19 --> 00:14:22 So now I can continue this. 221 00:14:22 --> 00:14:25 Under that substitution, I get 1. 222 00:14:25 --> 00:14:34 Oh yeah, and I should add the other fact, that, well I 223 00:14:34 --> 00:14:38 guess it's obvious that tangent ^2 = u ^2. 224 00:14:38 --> 00:14:41 So I get 1 + u ^2. 225 00:14:41 --> 00:14:50 And then du sec^2 (x) dx, that is du. 226 00:14:50 --> 00:14:52 Well that's pretty easy to integrate. 227 00:14:52 --> 00:14:56 So I get u + u ^3 / 3. 228 00:14:56 --> 00:14:57 Plus a constant. 229 00:14:57 --> 00:15:00 And then I just have to back-substitute. 230 00:15:00 --> 00:15:03 Put things back in terms of the original variables. 231 00:15:03 --> 00:15:12 And that gives me tan x + tan ^3 / 3. 232 00:15:12 --> 00:15:15 And there's the answer. 233 00:15:15 --> 00:15:18 So we could spend a lot more time doing more examples of 234 00:15:18 --> 00:15:21 this kind of polynomial trig thing. 235 00:15:21 --> 00:15:27 It's probably best for you to do some practice on your own. 236 00:15:27 --> 00:15:30 Because I want to talk about other things, also. 237 00:15:30 --> 00:15:34 And what I want to talk about is the use of these trig 238 00:15:34 --> 00:15:51 identities in making really trig substitution integration. 239 00:15:51 --> 00:15:55 So we did a little bit of this yesterday, and I'll show you 240 00:15:55 --> 00:15:56 some more examples today. 241 00:15:56 --> 00:15:59 Let's start with a pretty hard example right off the bat. 242 00:15:59 --> 00:16:07 So this is going to be the integral of dx / x ^2 ( the 243 00:16:07 --> 00:16:14 square root of 1 + x ^2). 244 00:16:14 --> 00:16:18 It's a pretty bad looking integral. 245 00:16:18 --> 00:16:20 So how can we approach this? 246 00:16:20 --> 00:16:24 Well, the square root is the ugliest part of 247 00:16:24 --> 00:16:27 the integral, I think. 248 00:16:27 --> 00:16:29 What we should try to do is write this square 249 00:16:29 --> 00:16:31 root in some nicer way. 250 00:16:31 --> 00:16:37 That is, figure out a way to write 1 + x ^2 as a square. 251 00:16:37 --> 00:16:40 That'll get rid of the square root. 252 00:16:40 --> 00:16:44 So there is an example of a way to write 1 plus something 253 00:16:44 --> 00:16:46 squared in a different way. 254 00:16:46 --> 00:16:47 And it's right up there. 255 00:16:47 --> 00:16:50 Secant ^2 = 1 + tangent ^2. 256 00:16:50 --> 00:16:53 So I want to use that idea. 257 00:16:53 --> 00:16:57 And when I see this form, that suggests that we make a trig 258 00:16:57 --> 00:17:04 substitution and write x as the tangent of some new variable. 259 00:17:04 --> 00:17:06 Which you might as well call theta, to because 260 00:17:06 --> 00:17:09 it's like an angle. 261 00:17:09 --> 00:17:15 Then 1 + x ^2 is the secant ^2. 262 00:17:15 --> 00:17:18 According to that trig identity. 263 00:17:18 --> 00:17:31 And so the square root of 1 + x ^2 = sec theta. right? 264 00:17:31 --> 00:17:36 So this identity is the reason that the substitution 265 00:17:36 --> 00:17:37 is going to help us. 266 00:17:37 --> 00:17:40 Because it gets rid of the square root and replaces it 267 00:17:40 --> 00:17:43 by some other trig function. 268 00:17:43 --> 00:17:46 I'd better be able to get rid of the dx, too. 269 00:17:46 --> 00:17:48 That's part of the substitution process. 270 00:17:48 --> 00:17:51 But we can do that, because I know what the derivative 271 00:17:51 --> 00:17:52 of the tangent is. 272 00:17:52 --> 00:17:56 It's secant ^2. 273 00:17:56 --> 00:17:58 So dx d theta is secant ^2 theta. 274 00:17:58 --> 00:18:04 So dx is the secant ^2 theta ( d theta). 275 00:18:04 --> 00:18:08 So let's just substitute all of that stuff in, and rewrite the 276 00:18:08 --> 00:18:11 entire integral in terms of our new variable, theta. 277 00:18:11 --> 00:18:14 So dx is in the numerator. 278 00:18:14 --> 00:18:18 That's secant ^2 theta d theta. 279 00:18:18 --> 00:18:21 And then the denominator, well, it has an x ^2. 280 00:18:21 --> 00:18:24 That's tangent ^2 theta. 281 00:18:24 --> 00:18:28 And then there's this square root. 282 00:18:28 --> 00:18:31 And we you know what that is in terms of theta. 283 00:18:31 --> 00:18:36 It's secant of theta. 284 00:18:36 --> 00:18:41 OK, now. we've done the trig substitution. 285 00:18:41 --> 00:18:44 I've gotten rid of the square root, I've got everything in 286 00:18:44 --> 00:18:48 terms of trig functions of the new variable. 287 00:18:48 --> 00:18:49 Pretty complicated trig function. 288 00:18:49 --> 00:18:52 This often happens, you wind up with a complete scattering of 289 00:18:52 --> 00:18:55 different trig functions in the numerator and denominator 290 00:18:55 --> 00:18:56 and everything. 291 00:18:56 --> 00:18:59 A systematic thing to do here is to put everything in 292 00:18:59 --> 00:19:09 terms of sines and cosines. 293 00:19:09 --> 00:19:12 Unless you can see right away, how it's going to simplify, the 294 00:19:12 --> 00:19:15 systematic thing to do is to rewrite in terms of 295 00:19:15 --> 00:19:17 sines and cosines. 296 00:19:17 --> 00:19:19 So let's do that. 297 00:19:19 --> 00:19:20 So let's see. 298 00:19:20 --> 00:19:23 The secant ^2, secant is 1 / cosine. 299 00:19:23 --> 00:19:28 So I'm going to put a cosine ^2 in the denominator. 300 00:19:28 --> 00:19:32 Oh, I guess the first thing I can do is cancel. 301 00:19:32 --> 00:19:33 Let's do that. 302 00:19:33 --> 00:19:34 That's clever. 303 00:19:34 --> 00:19:35 You were all thinking that too. 304 00:19:35 --> 00:19:37 Cancel those. 305 00:19:37 --> 00:19:40 So now I just get one cosine denominator from the secant 306 00:19:40 --> 00:19:44 there in the numerator. 307 00:19:44 --> 00:19:48 It's still pretty complicated, secant / tangent ^2, who knows. 308 00:19:48 --> 00:19:49 Well, we'll find out. 309 00:19:49 --> 00:19:52 Because the tangent is sine / cosine. 310 00:19:52 --> 00:19:55 So I should put a sine ^2 where the tangent was, 311 00:19:55 --> 00:19:59 and a cosine ^2 up there. 312 00:19:59 --> 00:20:02 And I still have d theta. 313 00:20:02 --> 00:20:04 And now you see some more cancellation occurs. 314 00:20:04 --> 00:20:07 That's the virtue of writing things out in this way. 315 00:20:07 --> 00:20:14 So now, the square here cancels with this cosine. 316 00:20:14 --> 00:20:22 And I'm left with cosine theta d theta, / sine 317 00:20:22 --> 00:20:25 square root of theta. 318 00:20:25 --> 00:20:26 That's a little simpler. 319 00:20:26 --> 00:20:33 And it puts me in a position to use the same idea I just used. 320 00:20:33 --> 00:20:35 I see the sine here. 321 00:20:35 --> 00:20:37 I might look around in this integral to see if its 322 00:20:37 --> 00:20:39 derivative occurs anywhere. 323 00:20:39 --> 00:20:43 The differential of the sine is the cosine. 324 00:20:43 --> 00:20:49 And so I'm very much inclined to make another substitution. 325 00:20:49 --> 00:20:54 Say, u, direct substitution this time. 326 00:20:54 --> 00:20:56 And say u as the cosine of theta. 327 00:20:56 --> 00:20:59 Because then du -- oh, I'm sorry. 328 00:20:59 --> 00:21:01 Say, u as the sine of theta. 329 00:21:01 --> 00:21:13 Because then du is cosine of theta d theta. 330 00:21:13 --> 00:21:20 And then this integral becomes, well, the numerator just is du. 331 00:21:20 --> 00:21:22 The denominator is u ^2. 332 00:21:22 --> 00:21:27 And I think we can break out the champagne, because we 333 00:21:27 --> 00:21:28 can integrate that one. 334 00:21:28 --> 00:21:29 Finally get rid of the integral sign. 335 00:21:29 --> 00:21:29 Yes sir. 336 00:21:29 --> 00:21:37 STUDENT: [INAUDIBLE] 337 00:21:37 --> 00:21:40 PROFESSOR: OK, how do I know to make u = sine 338 00:21:40 --> 00:21:41 rather than cosine. 339 00:21:41 --> 00:21:45 Because I want to see du appear up here. 340 00:21:45 --> 00:21:49 If I'd had a sine up here, that would be a signal to 341 00:21:49 --> 00:21:53 me that maybe I should say let u be the cosine. 342 00:21:53 --> 00:21:54 OK? 343 00:21:54 --> 00:21:56 Also, because this thing in the denominator is something 344 00:21:56 --> 00:21:57 I want to get rid of. 345 00:21:57 --> 00:21:59 It's in the denominator. 346 00:21:59 --> 00:22:01 So I'll get rid of it by wishful thinking and just 347 00:22:01 --> 00:22:04 call it something else. 348 00:22:04 --> 00:22:07 It works pretty well in this case. 349 00:22:07 --> 00:22:10 Wishful thinking doesn't always work so well. 350 00:22:10 --> 00:22:18 So I integrate u ^ - 2 du, and I get - 1 / u + a constant, and 351 00:22:18 --> 00:22:21 I'm done with the calculus part of this problem. 352 00:22:21 --> 00:22:22 I've done the integral now. 353 00:22:22 --> 00:22:25 Gotten rid of the integral sign. 354 00:22:25 --> 00:22:28 But I'm not quite done with the problem yet, because I have to 355 00:22:28 --> 00:22:31 work my way back through two substitutions. 356 00:22:31 --> 00:22:33 First, this one. 357 00:22:33 --> 00:22:34 And then this one. 358 00:22:34 --> 00:22:37 So this first substitution isn't so bad to get rid of. 359 00:22:37 --> 00:22:38 To undo. 360 00:22:38 --> 00:22:40 To back-substitute. 361 00:22:40 --> 00:22:43 Because u is just the sine of theta. 362 00:22:43 --> 00:22:47 And so 1 / u is, I guess a fancy way to write it is 363 00:22:47 --> 00:22:52 the cosecant of theta. 364 00:22:52 --> 00:22:54 1 / sin = cosecant. 365 00:22:54 --> 00:23:00 So I get - cosecant of theta + a constant. 366 00:23:00 --> 00:23:01 Is there a question in the back? 367 00:23:01 --> 00:23:01 Yes sir? 368 00:23:01 --> 00:23:07 STUDENT: [INAUDIBLE] 369 00:23:07 --> 00:23:09 PROFESSOR: I'm sorry, my hearing is so bad. 370 00:23:09 --> 00:23:12 STUDENT: [INAUDIBLE] 371 00:23:12 --> 00:23:16 PROFESSOR: How did I know substitution 372 00:23:16 --> 00:23:19 in the first place. 373 00:23:19 --> 00:23:22 It's because of the 1 + x ^2. 374 00:23:22 --> 00:23:24 And I want to make use of the trig identity in the 375 00:23:24 --> 00:23:26 upper left-hand corner. 376 00:23:26 --> 00:23:29 I'll make you a table in a few minutes that will put all 377 00:23:29 --> 00:23:30 this in a bigger context. 378 00:23:30 --> 00:23:32 And I think it'll help you then. 379 00:23:32 --> 00:23:35 OK, I'll promise. 380 00:23:35 --> 00:23:41 So, what I want to try to talk about right now is how to 381 00:23:41 --> 00:23:43 rewrite a term like this. 382 00:23:43 --> 00:23:47 A trig term like this, back in terms of x. 383 00:23:47 --> 00:23:51 So I want to undo this trick substitution. 384 00:23:51 --> 00:23:55 This is a trig sub. 385 00:23:55 --> 00:23:59 And what I want to do now is try to undo that trig sub. 386 00:23:59 --> 00:24:02 And I'll show you a general method for undoing 387 00:24:02 --> 00:24:03 trig substitutions. 388 00:24:03 --> 00:24:05 This happens quite often. 389 00:24:05 --> 00:24:07 I don't know what the cosecant of theta is. 390 00:24:07 --> 00:24:11 But I do know what the tangent of theta is. 391 00:24:11 --> 00:24:14 So I want to make a relation between them. 392 00:24:14 --> 00:24:20 OK, so undoing. 393 00:24:20 --> 00:24:24 Trig subs. 394 00:24:24 --> 00:24:29 So let's go back to where trigonometry always comes from, 395 00:24:29 --> 00:24:32 this right angled triangle. 396 00:24:32 --> 00:24:35 The theta in the corner, and then these three sides. 397 00:24:35 --> 00:24:37 This one's called the hypotenuse. 398 00:24:37 --> 00:24:41 This one is called the adjacent side, and that one's 399 00:24:41 --> 00:24:45 called the opposite side. 400 00:24:45 --> 00:24:52 And now, let's find out where x lies in this triangle. 401 00:24:52 --> 00:24:55 Let's try to write the sides of this triangle in terms of x. 402 00:24:55 --> 00:24:58 And what I know is, x is the tangent of theta. 403 00:24:58 --> 00:25:01 So the tangent of theta, tangent of this angle, 404 00:25:01 --> 00:25:03 is opposite / adjacent. 405 00:25:03 --> 00:25:05 Did you learn SOH CAH TOA. 406 00:25:05 --> 00:25:09 OK, so it's opposite / adjacent. 407 00:25:09 --> 00:25:10 Is the tangent. 408 00:25:10 --> 00:25:14 So there are different ways to do that, but why not just do it 409 00:25:14 --> 00:25:17 in the simplest way and suppose that the adjacent is 1, 410 00:25:17 --> 00:25:21 and the opposite is x. 411 00:25:21 --> 00:25:24 This is correct now, isn't it? 412 00:25:24 --> 00:25:28 I get the correct value for the tangent of theta by saying 413 00:25:28 --> 00:25:32 that the lengths of those are 1 and x. 414 00:25:32 --> 00:25:39 And that means that the hypotenuse has length 1 + x ^2. 415 00:25:39 --> 00:25:40 Well, here's a triangle. 416 00:25:40 --> 00:25:44 I'm interested in computing the cosecant of theta. 417 00:25:44 --> 00:25:50 Where's that appear in the triangle? 418 00:25:50 --> 00:25:51 Well, let's see. 419 00:25:51 --> 00:25:55 The cosecant of theta = 1 / sin. 420 00:25:55 --> 00:26:02 And the sine is opposite / hypotenuse. 421 00:26:02 --> 00:26:15 So the cosecant is hypotenuse / opposite. 422 00:26:15 --> 00:26:20 And the hypotenuse is the square root of 1 + x ^2, 423 00:26:20 --> 00:26:24 and the opposite is x. 424 00:26:24 --> 00:26:26 And so I've done it. 425 00:26:26 --> 00:26:29 I've undone the trig substitution. 426 00:26:29 --> 00:26:32 I've figured out what this cosecant of theta 427 00:26:32 --> 00:26:35 is, in terms of x. 428 00:26:35 --> 00:26:41 And so the final answer is - square root of 1 + x ^2 / x + a 429 00:26:41 --> 00:26:53 constant, and there's an answer to the original problem. 430 00:26:53 --> 00:26:57 This took two boards to go through this. 431 00:26:57 --> 00:27:00 I illustrated several things. 432 00:27:00 --> 00:27:02 Actually, this three half boards. 433 00:27:02 --> 00:27:05 I illustrated this use of trig substitution, and I'll come 434 00:27:05 --> 00:27:07 back to that in a second. 435 00:27:07 --> 00:27:10 I illustrated patience. 436 00:27:10 --> 00:27:15 I illustrated rewriting things in terms of sines and cosines, 437 00:27:15 --> 00:27:18 and then making a direct substitution to evaluate 438 00:27:18 --> 00:27:19 an integral like this. 439 00:27:19 --> 00:27:23 And then there's this undoing all of those substitutions. 440 00:27:23 --> 00:27:25 And it culminated with undoing the trig sub. 441 00:27:25 --> 00:27:31 So let's play a game here. 442 00:27:31 --> 00:27:37 Why don't we play the game where you give me. 443 00:27:37 --> 00:27:46 So, there's a step in here that I should have done. 444 00:27:46 --> 00:27:55 I should've said this is - csc ( arc tan of 445 00:27:55 --> 00:27:58 theta) + a constant. 446 00:27:58 --> 00:28:01 The most straightforward thing you can do is to say since x is 447 00:28:01 --> 00:28:05 the tangent of theta, that means that -- sorry, if x, 448 00:28:05 --> 00:28:08 that means that theta is the arc tangent of x. 449 00:28:08 --> 00:28:11 And so let's just put in theta as the arc tangent of x, 450 00:28:11 --> 00:28:12 and that's what you get. 451 00:28:12 --> 00:28:16 So really, what I just did for you was to show you a way to 452 00:28:16 --> 00:28:21 compute some trig function applied to the inverse of 453 00:28:21 --> 00:28:23 another trig function. 454 00:28:23 --> 00:28:29 I computed cosecant of the arc tangent by this trick. 455 00:28:29 --> 00:28:33 So now, let's play the game where you give me a trig 456 00:28:33 --> 00:28:36 function and an inverse trig function, and I try to compute 457 00:28:36 --> 00:28:46 what the composite is. 458 00:28:46 --> 00:28:47 OK. 459 00:28:47 --> 00:28:58 So who can give me a trig function. 460 00:28:58 --> 00:29:03 Has to be one of these standard ones. 461 00:29:03 --> 00:29:03 STUDENT: Tan. 462 00:29:03 --> 00:29:04 PROFESSOR: Tangent. 463 00:29:04 --> 00:29:07 Alright. 464 00:29:07 --> 00:29:07 How about another one? 465 00:29:07 --> 00:29:11 STUDENT: Sine. 466 00:29:11 --> 00:29:12 PROFESSOR: Sine. 467 00:29:12 --> 00:29:15 Do we have agreement on sine. 468 00:29:15 --> 00:29:16 STUDENT: [INAUDIBLE] 469 00:29:16 --> 00:29:17 PROFESSOR: Secant? 470 00:29:17 --> 00:29:28 STUDENT: [INAUDIBLE] 471 00:29:28 --> 00:29:29 PROFESSOR: Right, csc has the best cheer. 472 00:29:29 --> 00:29:29 So that's the game. 473 00:29:29 --> 00:29:35 We have to compute, try to compute, that composite. 474 00:29:35 --> 00:29:35 Something wrong with this? 475 00:29:35 --> 00:29:45 STUDENT: [INAUDIBLE] 476 00:29:45 --> 00:29:47 PROFESSOR: What does acceptable mean? 477 00:29:47 --> 00:29:49 Don't you think, that so the question is, isn't this a 478 00:29:49 --> 00:29:51 perfectly acceptable final answer. 479 00:29:51 --> 00:29:53 It's a correct final answer. 480 00:29:53 --> 00:29:56 But this is much more insightful. 481 00:29:56 --> 00:30:00 And after all the original thing was involving square 482 00:30:00 --> 00:30:02 roots and things, this is the kind of thing you might 483 00:30:02 --> 00:30:03 hope for is an answer. 484 00:30:03 --> 00:30:07 This is just a nicer answer for sure. 485 00:30:07 --> 00:30:10 And likely to be more useful to you when you go on and use that 486 00:30:10 --> 00:30:13 answer for something else. 487 00:30:13 --> 00:30:18 OK, so let's try to do this this. 488 00:30:18 --> 00:30:22 Undo a trig substitution that involved a cosecant. 489 00:30:22 --> 00:30:25 And I manipulate around, and I find myself trying to find out 490 00:30:25 --> 00:30:27 what's the tangent of theta. 491 00:30:27 --> 00:30:29 So here's how we go about it. 492 00:30:29 --> 00:30:33 I draw this triangle. 493 00:30:33 --> 00:30:37 Theta is the angle here. 494 00:30:37 --> 00:30:43 This is the adjacent, opposite, hypotenuse. 495 00:30:43 --> 00:30:47 So, the first thing is how can I make the cosecant 496 00:30:47 --> 00:30:49 appear here. csc x. 497 00:30:49 --> 00:30:52 What dimensions should I give to the sides in order for the 498 00:30:52 --> 00:31:01 cosecant of x, sorry, in order for theta to be 499 00:31:01 --> 00:31:02 the cosecant of x. 500 00:31:02 --> 00:31:05 This thing is theta. 501 00:31:05 --> 00:31:18 So, that means that the cosecant of x, that means the 502 00:31:18 --> 00:31:24 cosecant of theta should be x. 503 00:31:24 --> 00:31:28 Theta is the arc cosecant, so x is the cosecant of theta. 504 00:31:28 --> 00:31:31 So, what'll I take the sides to be, to get the cosecant? 505 00:31:31 --> 00:31:37 The cosecant is 1 / sine. 506 00:31:37 --> 00:31:47 And the sine is the opposite / hypotenuse. 507 00:31:47 --> 00:31:49 So I get hypotenuse / opposite. 508 00:31:49 --> 00:31:53 And that's supposed to be what x is. 509 00:31:53 --> 00:31:57 So I could make the opposite anything I want, but the 510 00:31:57 --> 00:31:59 simplest thing is to make it 1. 511 00:31:59 --> 00:32:00 Let's do that. 512 00:32:00 --> 00:32:04 And then what does that mean about the rest of the sides? 513 00:32:04 --> 00:32:06 Hypotenuse had better be x. 514 00:32:06 --> 00:32:07 And then I've recovered this. 515 00:32:07 --> 00:32:12 So here's a triangle that exhibits the correct angle. 516 00:32:12 --> 00:32:14 This remaining side is going to be useful to us. 517 00:32:14 --> 00:32:20 And it is the square root of x^2 - 1. 518 00:32:20 --> 00:32:24 So I've got a triangle of the correct angle theta, and 519 00:32:24 --> 00:32:27 now I want to compute the tangent of that angle. 520 00:32:27 --> 00:32:28 Well, that's easy. 521 00:32:28 --> 00:32:31 That's opposite / adjacent. 522 00:32:31 --> 00:32:38 So I get 1 / square root of x ^2 - 1. 523 00:32:38 --> 00:32:40 Very flexible tool that'll be useful to you in 524 00:32:40 --> 00:32:42 many different times. 525 00:32:42 --> 00:32:45 Whenever you have to undo a trig substitution, this 526 00:32:45 --> 00:32:49 is likely to be useful. 527 00:32:49 --> 00:32:51 OK, that was a good game. 528 00:32:51 --> 00:32:52 No winners in this game. 529 00:32:52 --> 00:32:53 We're all winners. 530 00:32:53 --> 00:33:01 No losers, we're all winners. 531 00:33:01 --> 00:33:02 OK. 532 00:33:02 --> 00:33:06 So, good. 533 00:33:06 --> 00:33:08 So let me make this table of the different trig 534 00:33:08 --> 00:33:11 substitutions, and how they can be useful. 535 00:33:11 --> 00:33:20 Summary of trig substitutions. 536 00:33:20 --> 00:33:30 So over here, we have, if you see, so if your integrand 537 00:33:30 --> 00:33:58 contains, make a substitution to get. 538 00:33:58 --> 00:34:02 So if your integrand contains, I'll write these things out as 539 00:34:02 --> 00:34:07 square roots. if it contains the square root of a ^2 - x ^2, 540 00:34:07 --> 00:34:10 this is what we talked about on Thursday. 541 00:34:10 --> 00:34:14 When I was trying to find the area of that piece of a circle. 542 00:34:14 --> 00:34:18 There, I suggested that we should make the substitution 543 00:34:18 --> 00:34:22 x = a cos theta. 544 00:34:22 --> 00:34:28 Or, x = a sin theta. 545 00:34:28 --> 00:34:31 Either one works just as well. 546 00:34:31 --> 00:34:36 And there's no way to prefer one over the other. 547 00:34:36 --> 00:34:40 And when you make the substitution, x = a cos 548 00:34:40 --> 00:34:45 theta, you get a ^2 - a ^2 cos ^2 theta. 549 00:34:45 --> 00:34:47 1 - cos ^2 is the sin^2. 550 00:34:48 --> 00:34:55 So you get a sine of theta. 551 00:34:55 --> 00:34:59 So this expression becomes equal to this expression 552 00:34:59 --> 00:35:03 under that substitution. 553 00:35:03 --> 00:35:04 And then you go on. 554 00:35:04 --> 00:35:07 Then you've gotten rid of the square root, and you've got a 555 00:35:07 --> 00:35:10 trigonometric integral that you have to try to do. 556 00:35:10 --> 00:35:15 If you made the substitution a sin theta, you'd get a ^2 - a 557 00:35:15 --> 00:35:21 ^2 sin^2, which = a cos theta. 558 00:35:21 --> 00:35:24 And then you can go ahead as well. 559 00:35:24 --> 00:35:26 We just saw another example. 560 00:35:26 --> 00:35:29 Namely, if you have a ^2 + x ^2. 561 00:35:29 --> 00:35:35 That's like the example we had up here. a = 1 in this example. 562 00:35:35 --> 00:35:36 What did we do? 563 00:35:36 --> 00:35:42 We tried the substitution x = a tangent of theta. 564 00:35:42 --> 00:35:44 And the reason is that I can plug into the trig identity 565 00:35:44 --> 00:35:46 up here in the upper left. 566 00:35:46 --> 00:35:55 And replace a ^2 + x ^2 by a sec theta. 567 00:35:55 --> 00:35:57 Square root of the sec^2. 568 00:35:57 --> 00:35:59 569 00:35:59 --> 00:36:03 There's one more thing in this table. 570 00:36:03 --> 00:36:06 Sort of, the only remaining sum or difference 571 00:36:06 --> 00:36:07 of terms like this. 572 00:36:07 --> 00:36:14 And that's what happens if you have x ^2 - a ^2. 573 00:36:14 --> 00:36:21 So there, I think we can make a substitution a sec theta. 574 00:36:21 --> 00:36:32 Because, after all, sec^2 theta, so x ^2 - a ^2. 575 00:36:32 --> 00:36:34 Let's see what happens when I make that substitution. x ^2 - 576 00:36:34 --> 00:36:42 a ^2 = a ^2 sec^2 theta - a ^2. 577 00:36:42 --> 00:36:45 Under this substitution. 578 00:36:45 --> 00:36:48 That's sec ^2 - 1. 579 00:36:48 --> 00:36:51 Well, put the 1 on the other side. 580 00:36:51 --> 00:36:53 And you find tan ^2, coming out. 581 00:36:53 --> 00:36:59 So this is a ^2 ( tan ^2 of theta). 582 00:36:59 --> 00:37:02 And so that's what you get. a tan theta. 583 00:37:02 --> 00:37:07 After I take the square root, I get a tan theta. 584 00:37:07 --> 00:37:13 So these are the three basic trig substitution forms. 585 00:37:13 --> 00:37:15 Where trig substitutions are useful to get rid of 586 00:37:15 --> 00:37:18 expressions like this, and replace them by 587 00:37:18 --> 00:37:22 trigonometric expressions. 588 00:37:22 --> 00:37:25 And then you use this trick, you do the integral if you can 589 00:37:25 --> 00:37:26 and then you use this trick to get rid of the 590 00:37:26 --> 00:37:37 theta at the end. 591 00:37:37 --> 00:37:41 So now, the last thing I want to talk about today is called 592 00:37:41 --> 00:37:59 completing the square. 593 00:37:59 --> 00:38:07 And that comes in because unfortunately, not every square 594 00:38:07 --> 00:38:10 root of a quadratic has such a simple form. 595 00:38:10 --> 00:38:16 You will often encounter things that are not just the square 596 00:38:16 --> 00:38:18 root of something simple. 597 00:38:18 --> 00:38:19 Like one of these forms. 598 00:38:19 --> 00:38:27 Like there might be a middle term in there. 599 00:38:27 --> 00:38:30 I don't actually have time to show you an example of how 600 00:38:30 --> 00:38:33 this comes out in a sort of practical example. 601 00:38:33 --> 00:38:35 But it does happen quite frequently. 602 00:38:35 --> 00:38:38 And so I want to show you how to deal with things like 603 00:38:38 --> 00:38:40 the following example. 604 00:38:40 --> 00:38:48 Let's try to integrate dx / x ^2 + 4x, the square 605 00:38:48 --> 00:38:55 root of x ^2 + 4x. 606 00:38:55 --> 00:38:59 So there's a square root of some square, some quadratic. 607 00:38:59 --> 00:39:02 It's very much like this business. 608 00:39:02 --> 00:39:04 But it isn't of any of these forms. 609 00:39:04 --> 00:39:07 And so what I want to do is show you how to rewrite 610 00:39:07 --> 00:39:11 it in one of those forms using substitution, again. 611 00:39:11 --> 00:39:14 All this is about substitution. 612 00:39:14 --> 00:39:27 So the game is to rewrite quadratic as something like 613 00:39:27 --> 00:39:32 x + something or other. + some other constant. 614 00:39:32 --> 00:39:35 So write it, try to write it, in the form of a square 615 00:39:35 --> 00:39:42 + or - another constant. 616 00:39:42 --> 00:39:45 And then we'll go on from there. 617 00:39:45 --> 00:39:51 So let's do that in this case. x ^2, x ^2 + 4x. 618 00:39:51 --> 00:39:53 Well, if you square this form out, then the middle 619 00:39:53 --> 00:39:57 term is going to be 2ax. 620 00:39:57 --> 00:40:01 So that, since I have a middle term here, I pretty much 621 00:40:01 --> 00:40:03 know what a has to be. 622 00:40:03 --> 00:40:08 The only choice in order to get something like x ^2 + 4x out of 623 00:40:08 --> 00:40:11 this, is to take a to be 2. 624 00:40:11 --> 00:40:16 Because then, this is what you get. 625 00:40:16 --> 00:40:19 This isn't quite right yet, but let's compute what I have here. 626 00:40:19 --> 00:40:26 x ^2 + 4x, so far so good. + 4, and I don't have a + 4 here. 627 00:40:26 --> 00:40:31 So I have to fix that by subtracting 4. 628 00:40:31 --> 00:40:32 So that's what I mean. 629 00:40:32 --> 00:40:33 I've completed the square. 630 00:40:33 --> 00:40:38 The word for this process of eliminating the middle term 631 00:40:38 --> 00:40:41 by using the square of an expression like that. 632 00:40:41 --> 00:40:44 That's called completing the square. 633 00:40:44 --> 00:40:48 And we can use that process to compute this integral. 634 00:40:48 --> 00:40:51 So let's do that. 635 00:40:51 --> 00:40:53 So I can rewrite this integral, rewrite this 636 00:40:53 --> 00:41:00 denominator like this. 637 00:41:00 --> 00:41:01 And then I'm going to try to use one of 638 00:41:01 --> 00:41:03 these forms over here. 639 00:41:03 --> 00:41:08 So in order to get a single variable there, instead of 640 00:41:08 --> 00:41:13 something complicated like x + 2, I'm inclined to come up with 641 00:41:13 --> 00:41:19 another variable name and write it equal, write x + 2 as 642 00:41:19 --> 00:41:20 that other variable name. 643 00:41:20 --> 00:41:29 So here's another little direct substitution. u = x + 2. 644 00:41:29 --> 00:41:31 Figure out what du is. 645 00:41:31 --> 00:41:36 That's pretty easy. 646 00:41:36 --> 00:41:41 And then rewrite the integral in those terms. 647 00:41:41 --> 00:41:44 So dx = du. 648 00:41:44 --> 00:41:47 And then in the denominator I have, well, I have the 649 00:41:47 --> 00:41:49 square root of that. 650 00:41:49 --> 00:41:52 Oh yeah, so I think as part of this I'll write out 651 00:41:52 --> 00:41:56 what x ^2 + 4x is. 652 00:41:56 --> 00:42:03 The point is, it's equal to u ^2 - 4. 653 00:42:03 --> 00:42:12 4. x ^2 + 4x = u ^2 - 4. 654 00:42:12 --> 00:42:20 There's the data box containing the substitution data. 655 00:42:20 --> 00:42:22 And so now I can put that in. 656 00:42:22 --> 00:42:24 I have x ^2 + 4x there. 657 00:42:24 --> 00:42:31 In terms of u, that's u ^2 - 4. 658 00:42:31 --> 00:42:33 Well, now I'm a happier position because I can look for 659 00:42:33 --> 00:42:37 u ^2 - 4 for something like that in my table here. 660 00:42:37 --> 00:42:40 And it actually sits down here. 661 00:42:40 --> 00:42:43 So except for the use of the letter x here 662 00:42:43 --> 00:42:45 instead of u over there. 663 00:42:45 --> 00:42:47 That tells me what I want. 664 00:42:47 --> 00:42:51 So to handle this, what I should use is a 665 00:42:51 --> 00:42:55 trig substitution. 666 00:42:55 --> 00:42:58 And the trig substitution that's suggested is, according 667 00:42:58 --> 00:43:03 to the bottom line with a = 2, so a ^2 = 4. 668 00:43:03 --> 00:43:08 The suggestion is, I should take x but I'd better not 669 00:43:08 --> 00:43:13 use the letter x any more. 670 00:43:13 --> 00:43:15 But I don't have a letter x, I have the letter u. 671 00:43:15 --> 00:43:21 I should take u = 2 secant. 672 00:43:21 --> 00:43:23 And then some letter I haven't used before. 673 00:43:23 --> 00:43:28 And theta is available. 674 00:43:28 --> 00:43:30 This is a look-up table process. 675 00:43:30 --> 00:43:33 I see the square root of u ^2 - 4, I see that 676 00:43:33 --> 00:43:35 that's of this form. 677 00:43:35 --> 00:43:38 I'm instructed to make this substitution. 678 00:43:38 --> 00:43:40 And that's what I just did. 679 00:43:40 --> 00:43:43 Let's see how it works out. 680 00:43:43 --> 00:43:47 So that means the du = 2, OK. 681 00:43:47 --> 00:43:50 What's the derivative of the secant? 682 00:43:50 --> 00:43:52 Secant tangent. 683 00:43:52 --> 00:43:59 So du = 2 secant theta tangent theta. 684 00:43:59 --> 00:44:05 And u ^2 - 4 is, here's the payoff. 685 00:44:05 --> 00:44:07 I'm supposed to be able to rewrite that in 686 00:44:07 --> 00:44:09 terms of the tangent. 687 00:44:09 --> 00:44:19 According to this. u ^2 - 4 = 4 secant ^2 - 4. 688 00:44:19 --> 00:44:22 And secant ^2 - 1 = tangent ^2. 689 00:44:22 --> 00:44:27 So this is 4 tangent ^2. 690 00:44:27 --> 00:44:30 Of theta. 691 00:44:30 --> 00:44:37 Right, yeah? 692 00:44:37 --> 00:44:37 STUDENT: [INAUDIBLE] 693 00:44:37 --> 00:44:42 PROFESSOR: But I squared it. 694 00:44:42 --> 00:44:44 And now I'll square root it. 695 00:44:44 --> 00:44:50 And I'll get a 2 and this tangent will go away. 696 00:44:50 --> 00:44:56 So there's my data box for this substitution. 697 00:44:56 --> 00:45:15 And let's go on to another board. 698 00:45:15 --> 00:45:20 So where I'm at is the integral of du / square 699 00:45:20 --> 00:45:26 root of u ^2 - 4. 700 00:45:26 --> 00:45:30 And I have all the data I need here to rewrite 701 00:45:30 --> 00:45:33 that in terms of theta. 702 00:45:33 --> 00:45:41 So du = 2 sec theta tan theta d theta. 703 00:45:41 --> 00:45:47 And the denominator is 2 tan theta. 704 00:45:47 --> 00:45:48 Ha. 705 00:45:48 --> 00:45:53 Well, so some very nice simplification happens here. 706 00:45:53 --> 00:45:55 The 2's cancel. 707 00:45:55 --> 00:45:58 And the tangents cancel. 708 00:45:58 --> 00:46:01 And I'm left with trying to work with the integral, 709 00:46:01 --> 00:46:03 the secant theta d theta. 710 00:46:03 --> 00:46:07 And luckily enough at the very beginning of the hour, I worked 711 00:46:07 --> 00:46:09 out how to compute the integral of the secant of theta. 712 00:46:09 --> 00:46:11 And there it is. 713 00:46:11 --> 00:46:25 So this is ln ( sec theta + tan theta) + a constant. 714 00:46:25 --> 00:46:27 And we're done with the calculus part. 715 00:46:27 --> 00:46:29 There's no more integral there. 716 00:46:29 --> 00:46:32 But I still am not quite done with the problem, because 717 00:46:32 --> 00:46:37 again I have these two substitutions to try to undo. 718 00:46:37 --> 00:46:40 So let's undo them one by one. 719 00:46:40 --> 00:46:42 Let's see. 720 00:46:42 --> 00:46:44 I have this trig substitution here. 721 00:46:44 --> 00:46:47 And I could use my triangle trick, if I need to. 722 00:46:47 --> 00:46:49 But maybe I don't need to. 723 00:46:49 --> 00:46:51 Let's see, do I know what the secant of theta 724 00:46:51 --> 00:46:53 is in terms of u? 725 00:46:53 --> 00:46:54 Well, I do. 726 00:46:54 --> 00:46:58 So I get ln ( u / 2). 727 00:46:58 --> 00:47:01 Do I know what the tangent is in terms of u? 728 00:47:01 --> 00:47:02 Well, I do. 729 00:47:02 --> 00:47:04 It's here. 730 00:47:04 --> 00:47:06 So I lucked out, in this case. 731 00:47:06 --> 00:47:09 And I don't have to go through and use that triangle trick. 732 00:47:09 --> 00:47:19 So the tangent of theta is the square root of u ^2 - 4 / 2. 733 00:47:19 --> 00:47:20 Good. 734 00:47:20 --> 00:47:24 So I've undone this trig substitution. 735 00:47:24 --> 00:47:28 I'm not quite done yet because my answer is involved with u. 736 00:47:28 --> 00:47:30 And what I wanted originally was x. 737 00:47:30 --> 00:47:33 But this direct substitution that I started with is 738 00:47:33 --> 00:47:35 really easy to deal with. 739 00:47:35 --> 00:47:39 I can just put x + 2 every time I see a u. 740 00:47:39 --> 00:47:46 So this is ln ( x + 2 / 2 + the square root... 741 00:47:46 --> 00:47:51 What's going to happen when I put x + 2 in place for u here? 742 00:47:51 --> 00:47:54 You know what you get. 743 00:47:54 --> 00:47:59 You get exactly what we started with. 744 00:47:59 --> 00:48:00 Right? 745 00:48:00 --> 00:48:04 I put x + 2 in place of the u here. 746 00:48:04 --> 00:48:13 I get x ^2 + 4x. 747 00:48:13 --> 00:48:15 So I've gotten back to a function purely in terms 748 00:48:15 --> 00:48:21 of x OK, that's a good place to quit. 749 00:48:21 --> 00:48:24 Have a great little one-day break. 750 00:48:24 --> 00:48:27 I guess, this class doesn't meet on Monday anyway. 751 00:48:27 --> 00:48:28 Bye. 752 00:48:28 --> 00:48:28