1 00:00:00,000 --> 00:00:02,330 The following content is provided under a Creative 2 00:00:02,330 --> 00:00:03,610 Commons license. 3 00:00:03,610 --> 00:00:05,750 Your support will help MIT OpenCourseWare 4 00:00:05,750 --> 00:00:09,455 continue to offer high quality educational resources for free. 5 00:00:09,455 --> 00:00:12,540 To make a donation or to view additional materials 6 00:00:12,540 --> 00:00:16,150 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:16,150 --> 00:00:21,780 at ocw.mit.edu. 8 00:00:21,780 --> 00:00:24,560 PROFESSOR: So we're going to continue 9 00:00:24,560 --> 00:00:29,420 to talk about trig integrals and trig substitutions. 10 00:00:29,420 --> 00:00:31,520 This is maybe the most technical part 11 00:00:31,520 --> 00:00:34,560 of this course, which maybe is why professor Jerison decided 12 00:00:34,560 --> 00:00:37,520 to just take a leave, go AWOL just now and let me take over 13 00:00:37,520 --> 00:00:38,610 for him. 14 00:00:38,610 --> 00:00:41,960 But I'll do my best to help you learn this technique 15 00:00:41,960 --> 00:00:45,680 and it'll be useful for you. 16 00:00:45,680 --> 00:00:48,880 So we've talked about trig integrals involving 17 00:00:48,880 --> 00:00:51,460 sines and cosines yesterday. 18 00:00:51,460 --> 00:00:53,060 There's another whole world out there 19 00:00:53,060 --> 00:00:58,850 that involves these other trig polynomials-- trig functions, 20 00:00:58,850 --> 00:01:00,520 secant and tangent. 21 00:01:00,520 --> 00:01:03,460 Let me just make a little table to remind you what they are. 22 00:01:03,460 --> 00:01:05,210 Because I have trouble remembering myself, 23 00:01:05,210 --> 00:01:07,630 so I enjoy the opportunity to go back 24 00:01:07,630 --> 00:01:09,321 to remind myself of this stuff. 25 00:01:09,321 --> 00:01:09,820 Let's see. 26 00:01:09,820 --> 00:01:15,860 The secant is one over one of those things, which one is it? 27 00:01:15,860 --> 00:01:18,420 It's weird, it's 1 over the cosine. 28 00:01:18,420 --> 00:01:24,710 And the cosecant is 1 over the sin. 29 00:01:24,710 --> 00:01:26,800 Of course the tangent, we know. 30 00:01:26,800 --> 00:01:31,840 It's the sine over the cosine and the cotangent 31 00:01:31,840 --> 00:01:36,000 is the other way around. 32 00:01:36,000 --> 00:01:37,700 So when you put a "co" in front of it, 33 00:01:37,700 --> 00:01:43,590 it exchanges sine and cosine. 34 00:01:43,590 --> 00:01:46,840 Well, I have a few identities involving tangent and secant 35 00:01:46,840 --> 00:01:50,270 up there, in that little prepared blackboard up above. 36 00:01:50,270 --> 00:01:53,680 Maybe I'll just go through and check that out 37 00:01:53,680 --> 00:01:57,604 to make sure that we're all on the same page with them. 38 00:01:57,604 --> 00:01:59,020 So I'm going to claim that there's 39 00:01:59,020 --> 00:02:00,560 this trig identity at the top. 40 00:02:00,560 --> 00:02:07,310 Secant squared is 1 plus the tangent. 41 00:02:07,310 --> 00:02:10,590 So let's just check that out. 42 00:02:10,590 --> 00:02:12,670 So the secant is 1 over the cosine, 43 00:02:12,670 --> 00:02:14,890 so secant squared is 1 over cosine squared. 44 00:02:14,890 --> 00:02:17,100 And then whenever you see a 1 in trigonometry, 45 00:02:17,100 --> 00:02:23,010 you'll always have the option of writing as cos^2 + sin^2. 46 00:02:27,890 --> 00:02:32,210 And if I do that, then I can divide the cos^2 into that 47 00:02:32,210 --> 00:02:34,300 first term. 48 00:02:34,300 --> 00:02:36,380 And I get 1 + sin^2 / cos^2. 49 00:02:36,380 --> 00:02:37,940 / ^2. 50 00:02:37,940 --> 00:02:42,430 Which is the tangent squared. 51 00:02:42,430 --> 00:02:44,180 So there you go. 52 00:02:44,180 --> 00:02:47,820 That checks the first one. 53 00:02:47,820 --> 00:02:49,320 That's the main trig identity that's 54 00:02:49,320 --> 00:02:51,330 going to be behind what I talk about today. 55 00:02:51,330 --> 00:02:53,460 That's the trigonometry identity part. 56 00:02:53,460 --> 00:02:57,040 How about this piece of calculus. 57 00:02:57,040 --> 00:03:05,490 Can we calculate what the derivative of tan x is? 58 00:03:05,490 --> 00:03:12,350 Actually, I'm going to do that on this board. 59 00:03:12,350 --> 00:03:19,110 So tan x = sin x / cos x. 60 00:03:19,110 --> 00:03:21,250 So I think I was with you when we learned 61 00:03:21,250 --> 00:03:23,500 about the quotient rule. 62 00:03:23,500 --> 00:03:26,030 Computing the derivative of a quotient. 63 00:03:26,030 --> 00:03:30,340 And the rule is, you take the numerator 64 00:03:30,340 --> 00:03:34,330 and you-- sorry, you take the derivative of the numerator, 65 00:03:34,330 --> 00:03:36,510 which is cosine. 66 00:03:36,510 --> 00:03:38,570 And you multiply it by the denominator, 67 00:03:38,570 --> 00:03:41,590 so that gives you cos^2. 68 00:03:41,590 --> 00:03:46,480 And then you take the numerator, take minus the numerator, 69 00:03:46,480 --> 00:03:49,400 and multiply that by the derivative of the denominator, 70 00:03:49,400 --> 00:03:53,240 which is -sin x. 71 00:03:53,240 --> 00:03:57,870 And you put all that over the square of the denominator. 72 00:03:57,870 --> 00:04:02,490 And now I look at that and before my eyes I see the same 73 00:04:02,490 --> 00:04:07,420 trig identity, cos^2 + sin^2 = 1, appearing there. 74 00:04:07,420 --> 00:04:13,850 This is 1 / cos^2(x) which is secant squared. 75 00:04:13,850 --> 00:04:15,022 And, good. 76 00:04:15,022 --> 00:04:16,230 So that's what the claim was. 77 00:04:16,230 --> 00:04:20,420 The derivative of the tangent is the secant squared. 78 00:04:20,420 --> 00:04:22,260 That immediately gives you an integral. 79 00:04:22,260 --> 00:04:25,240 Namely, the integral of secant squared is the tangent. 80 00:04:25,240 --> 00:04:28,510 That's the fundamental theorem of calculus. 81 00:04:28,510 --> 00:04:32,360 So we verified the first integral there. 82 00:04:32,360 --> 00:04:34,490 Well, let's just do the second one as well. 83 00:04:34,490 --> 00:04:40,280 So if I want to differentiate the secant, 84 00:04:40,280 --> 00:04:41,360 derivative of the secant. 85 00:04:41,360 --> 00:04:46,500 So that's d/dx of 1 over the cosine. 86 00:04:46,500 --> 00:04:47,817 And again, I have a quotient. 87 00:04:47,817 --> 00:04:49,900 This one's a little easier because the numerator's 88 00:04:49,900 --> 00:04:51,010 so simple. 89 00:04:51,010 --> 00:04:54,310 So I take the derivative of the numerator, which is 0. 90 00:04:54,310 --> 00:04:56,580 And then I take the numerator, I take 91 00:04:56,580 --> 00:05:00,830 minus the numerator times the derivative of the denominator. 92 00:05:00,830 --> 00:05:05,610 Which is -sin x, and put all that 93 00:05:05,610 --> 00:05:09,900 over the square of the same denominator. 94 00:05:09,900 --> 00:05:12,809 So one minus sign came from the quotient rule, 95 00:05:12,809 --> 00:05:14,350 and the other one came because that's 96 00:05:14,350 --> 00:05:16,580 the derivative of the cosine. 97 00:05:16,580 --> 00:05:21,010 But they cancel, and so I get sin / cos^2, 98 00:05:21,010 --> 00:05:25,640 which is sin / cos times 1 / cos and so that's the secant, 99 00:05:25,640 --> 00:05:32,160 that's 1 / cos, times tan x. 100 00:05:32,160 --> 00:05:34,190 So, not hard. 101 00:05:34,190 --> 00:05:36,300 That verifies that the derivative of secant 102 00:05:36,300 --> 00:05:37,990 is secant tangent. 103 00:05:37,990 --> 00:05:41,800 And it tells you that the integral of that weird thing 104 00:05:41,800 --> 00:05:43,830 in case you ever want to know, the integral 105 00:05:43,830 --> 00:05:47,910 of the secant tangent is the secant. 106 00:05:47,910 --> 00:05:50,470 Well, there are a couple more integrals 107 00:05:50,470 --> 00:05:54,000 that I want to do for you. 108 00:05:54,000 --> 00:05:57,050 Where I can't sort of work backwards like that. 109 00:05:57,050 --> 00:06:07,060 Let's calculate the integral of the tangent. 110 00:06:07,060 --> 00:06:09,380 Just do this straight out. 111 00:06:09,380 --> 00:06:22,610 So the tangent is the sine divided by the cosine. 112 00:06:22,610 --> 00:06:28,520 And now there's a habit of mind, that I hope you get into. 113 00:06:28,520 --> 00:06:33,310 When you see the cosine and you're calculating an integral 114 00:06:33,310 --> 00:06:35,950 like this, it's useful to remember 115 00:06:35,950 --> 00:06:37,700 what the derivative of the cosine is. 116 00:06:37,700 --> 00:06:41,420 Because maybe it shows up somewhere else in the integral. 117 00:06:41,420 --> 00:06:43,540 And that happens here. 118 00:06:43,540 --> 00:06:50,300 So that suggests we make a substitution. u = cos x. 119 00:06:50,300 --> 00:06:55,620 Which means du = -sin x dx. 120 00:06:55,620 --> 00:06:59,020 That's the numerator, except for the minus sign. 121 00:06:59,020 --> 00:07:08,570 And so I can rewrite this as, under the substitution, 122 00:07:08,570 --> 00:07:13,120 I can rewrite this as -du, that's the numerator, 123 00:07:13,120 --> 00:07:19,600 sin x dx is -du, divided by u. 124 00:07:19,600 --> 00:07:22,150 Well, I know how to do that integral too. 125 00:07:22,150 --> 00:07:24,210 That gives me the natural log, doesn't it. 126 00:07:24,210 --> 00:07:31,150 So this is -ln(u) plus a constant. 127 00:07:31,150 --> 00:07:32,360 I'm not quite done. 128 00:07:32,360 --> 00:07:34,750 I have to back-substitute and replace 129 00:07:34,750 --> 00:07:39,080 this new variable that I've made up, called u, with what it is. 130 00:07:39,080 --> 00:07:46,760 And what you get is -ln(cos x). 131 00:07:46,760 --> 00:07:50,410 So the integral of the tangent is minus log cosine. 132 00:07:50,410 --> 00:07:55,590 Now, you find these tables of integrals 133 00:07:55,590 --> 00:07:56,910 in the back of the book. 134 00:07:56,910 --> 00:07:58,320 Things like that. 135 00:07:58,320 --> 00:08:01,060 I'm not sure how much memorization Professor Jerison 136 00:08:01,060 --> 00:08:04,210 is going to ask of you, but there 137 00:08:04,210 --> 00:08:05,770 is a certain amount of memorization 138 00:08:05,770 --> 00:08:06,937 that goes on in calculus. 139 00:08:06,937 --> 00:08:08,520 And this is one of the kinds of things 140 00:08:08,520 --> 00:08:12,380 that you probably want to know. 141 00:08:12,380 --> 00:08:15,450 Let me do one more integral. 142 00:08:15,450 --> 00:08:18,100 I think I'm making my way through a prepared board here, 143 00:08:18,100 --> 00:08:20,350 let's see. 144 00:08:20,350 --> 00:08:23,060 Good. 145 00:08:23,060 --> 00:08:28,940 So the integral of the tangent is minus log cosine. 146 00:08:28,940 --> 00:08:36,090 I'd also like to know what the integral of the secant of x is. 147 00:08:36,090 --> 00:08:41,280 And I don't know a way to kind of go straight at this, 148 00:08:41,280 --> 00:08:45,580 but let me show you a way to think your way through to it. 149 00:08:45,580 --> 00:08:52,720 If I take these two facts, tangent prime is what it is, 150 00:08:52,720 --> 00:08:56,280 and secant prime is what it is, and add them together, 151 00:08:56,280 --> 00:08:57,420 I get this fact. 152 00:08:57,420 --> 00:09:05,400 That the derivative of the sec x + tan x is, 153 00:09:05,400 --> 00:09:08,350 well, it's the sum of these two things. 154 00:09:08,350 --> 00:09:11,060 Secant squared plus secant tangent. 155 00:09:11,060 --> 00:09:15,180 And there's a secant that occurs in both of those terms. 156 00:09:15,180 --> 00:09:17,140 So I'll factor it out. 157 00:09:17,140 --> 00:09:21,440 And that gives me, I'll put it over here. 158 00:09:21,440 --> 00:09:24,516 There's the secant of x that occurs in both terms. 159 00:09:24,516 --> 00:09:26,390 And then in one term, there's another secant. 160 00:09:26,390 --> 00:09:32,740 And in the other term, there's a tangent. 161 00:09:32,740 --> 00:09:36,970 So that's interesting somehow, because this same term appears 162 00:09:36,970 --> 00:09:40,200 on both sides of this equation. 163 00:09:40,200 --> 00:09:49,160 Let's write u, for that sec x + tan x. 164 00:09:49,160 --> 00:09:55,120 And so the equation that I get is u' = u sec x. 165 00:10:00,930 --> 00:10:03,860 I've just made a direct substitution. 166 00:10:03,860 --> 00:10:05,620 Just decide that I'm going to write u 167 00:10:05,620 --> 00:10:07,730 for that single thing that occurs 168 00:10:07,730 --> 00:10:09,600 on both sides of the equation. 169 00:10:09,600 --> 00:10:16,440 So u' is on the left, and u * sec x is on the right. 170 00:10:16,440 --> 00:10:18,560 Well, there's my secant. 171 00:10:18,560 --> 00:10:20,210 That I was trying to integrate. 172 00:10:20,210 --> 00:10:28,560 And what it tells you is that = u' / u. 173 00:10:28,560 --> 00:10:35,550 Just divide both sides by u, and I get this equation. u' / u, 174 00:10:35,550 --> 00:10:36,910 that has a name. 175 00:10:36,910 --> 00:10:38,880 Not sure that professor Jerison's 176 00:10:38,880 --> 00:10:41,604 used this in this class, but u' / u, 177 00:10:41,604 --> 00:10:43,270 we've actually used something like that. 178 00:10:43,270 --> 00:10:45,560 It's on the board right now. 179 00:10:45,560 --> 00:10:47,650 It's a logarithmic derivative. 180 00:10:47,650 --> 00:10:57,750 It is the derivative of the national logarithm of u. 181 00:10:57,750 --> 00:10:59,940 Maybe it's easier to read this from right to left, 182 00:10:59,940 --> 00:11:02,780 if I want to calculate the derivative of the logarithm, 183 00:11:02,780 --> 00:11:06,380 well, the chain rule says I get the derivative of u times 184 00:11:06,380 --> 00:11:11,100 the derivative of the log function, which is 1 / u. 185 00:11:11,100 --> 00:11:25,964 So often u' / u is called the logarithmic derivative. 186 00:11:25,964 --> 00:11:27,130 But it's done what I wanted. 187 00:11:27,130 --> 00:11:31,740 Because it's expressed the secant as a derivative. 188 00:11:31,740 --> 00:11:38,630 And I guess I should put in what u is. 189 00:11:38,630 --> 00:11:46,090 It's the secant plus the tangent. 190 00:11:46,090 --> 00:11:48,500 And so that implies that the integral-- 191 00:11:48,500 --> 00:11:49,510 Integrate both sides. 192 00:11:49,510 --> 00:11:56,840 That says that the integral of sec x dx, is ln(sec x + tan x). 193 00:12:02,280 --> 00:12:09,500 So that's the last line in this little memo that I created. 194 00:12:09,500 --> 00:12:14,500 That we can use now for the rest of the class. 195 00:12:14,500 --> 00:12:19,320 Any questions about that trick? 196 00:12:19,320 --> 00:12:23,740 It's a trick, I have nothing more to say about it. 197 00:12:23,740 --> 00:12:24,540 OK. 198 00:12:24,540 --> 00:12:30,210 So, the next thing I-- oh yes, so now 199 00:12:30,210 --> 00:12:33,060 I want to make the point that using these rules 200 00:12:33,060 --> 00:12:38,350 and some thought, you can now integrate 201 00:12:38,350 --> 00:12:41,650 most trigonometric polynomials. 202 00:12:41,650 --> 00:12:44,850 Most things that involve powers of sines and cosines 203 00:12:44,850 --> 00:12:48,080 and tangents and secants and everything else. 204 00:12:48,080 --> 00:12:56,790 For example, let's try to integrate the integral of sec^4 205 00:12:56,790 --> 00:12:59,180 x. 206 00:12:59,180 --> 00:13:01,914 Big power of the secant function. 207 00:13:01,914 --> 00:13:03,830 Well, there are too many secants there for me. 208 00:13:03,830 --> 00:13:05,680 So let's take some away. 209 00:13:05,680 --> 00:13:09,730 And I can take them away by using that trig identity, 210 00:13:09,730 --> 00:13:12,300 sec^2 = 1 + tan^2. 211 00:13:12,300 --> 00:13:16,680 So I'm going to replace two of those secants by 1 + tan^2. 212 00:13:21,790 --> 00:13:25,190 That leaves me with two left over. 213 00:13:25,190 --> 00:13:27,920 Now there was method to my madness. 214 00:13:27,920 --> 00:13:31,020 Because I've got a secant squared left over there. 215 00:13:31,020 --> 00:13:35,280 And secant squared is the derivative of tangent. 216 00:13:35,280 --> 00:13:40,970 So that suggests a substitution. 217 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. 218 00:13:50,980 --> 00:13:55,700 And I have both terms that occur in my integral 219 00:13:55,700 --> 00:14:01,800 sitting there very nicely. 220 00:14:01,800 --> 00:14:05,020 So this is the possibility of making this substitution 221 00:14:05,020 --> 00:14:08,240 and seeing a secant squared up here as part 222 00:14:08,240 --> 00:14:10,330 of the differential here. 223 00:14:10,330 --> 00:14:13,910 That's why it was a good idea for me to take two 224 00:14:13,910 --> 00:14:16,350 of the secants and write them as 1 + tan^2. 225 00:14:19,170 --> 00:14:22,250 So now I can continue this. 226 00:14:22,250 --> 00:14:25,780 Under that substitution, I get 1. 227 00:14:25,780 --> 00:14:32,810 Oh yeah, and I should add the other fact, that-- 228 00:14:32,810 --> 00:14:38,460 Well I guess it's obvious that tangent squared is u^2. 229 00:14:38,460 --> 00:14:41,070 So I get 1 + u^2. 230 00:14:41,070 --> 00:14:50,110 And then du-- sec^2 sec^2 x dx, that is du. 231 00:14:50,110 --> 00:14:52,430 Well that's pretty easy to integrate. 232 00:14:52,430 --> 00:14:56,690 So I get u + u^3 / 3. 233 00:14:56,690 --> 00:14:57,950 Plus a constant. 234 00:14:57,950 --> 00:15:00,020 And then I just have to back-substitute. 235 00:15:00,020 --> 00:15:03,600 Put things back in terms of the original variables. 236 00:15:03,600 --> 00:15:12,220 And that gives me tan x plus tangent cubed over 3. 237 00:15:12,220 --> 00:15:15,610 And there's the answer. 238 00:15:15,610 --> 00:15:17,070 So we could spend a lot more time 239 00:15:17,070 --> 00:15:21,810 doing more examples of this kind of polynomial trig thing. 240 00:15:21,810 --> 00:15:27,440 It's probably best for you to do some practice on your own. 241 00:15:27,440 --> 00:15:30,020 Because I want to talk about other things, also. 242 00:15:30,020 --> 00:15:32,800 And what I want to talk about is the use 243 00:15:32,800 --> 00:15:39,380 of these trig identities in making really trig substitution 244 00:15:39,380 --> 00:15:51,590 integration. 245 00:15:51,590 --> 00:15:54,320 So we did a little bit of this yesterday, 246 00:15:54,320 --> 00:15:56,990 and I'll show you some more examples today. 247 00:15:56,990 --> 00:15:59,520 Let's start with a pretty hard example right off the bat. 248 00:15:59,520 --> 00:16:07,700 So this is going to be the integral of dx over x^2 times 249 00:16:07,700 --> 00:16:10,200 the square root of 1+x^2. 250 00:16:14,690 --> 00:16:18,120 It's a pretty bad looking integral. 251 00:16:18,120 --> 00:16:20,190 So how can we approach this? 252 00:16:20,190 --> 00:16:25,390 Well, the square root is the ugliest part of the integral, 253 00:16:25,390 --> 00:16:27,240 I think. 254 00:16:27,240 --> 00:16:30,160 What we should try to do is write this square root 255 00:16:30,160 --> 00:16:31,600 in some nicer way. 256 00:16:31,600 --> 00:16:37,180 That is, figure out a way to write 1+x^2 as a square. 257 00:16:37,180 --> 00:16:40,140 That'll get rid of the square root. 258 00:16:40,140 --> 00:16:44,540 So there is an example of a way to write 1 plus something 259 00:16:44,540 --> 00:16:46,140 squared in a different way. 260 00:16:46,140 --> 00:16:47,760 And it's right up there. 261 00:16:47,760 --> 00:16:50,790 sec^2 = 1 + tan^2. 262 00:16:50,790 --> 00:16:53,450 So I want to use that idea. 263 00:16:53,450 --> 00:16:56,650 And when I see this form, that suggests 264 00:16:56,650 --> 00:16:59,520 that we make a trig substitution and write 265 00:16:59,520 --> 00:17:04,682 x as the tangent of some new variable. 266 00:17:04,682 --> 00:17:06,140 Which you might as well call theta, 267 00:17:06,140 --> 00:17:09,060 to because it's like an angle. 268 00:17:09,060 --> 00:17:15,960 Then 1+x^2 is the secant squared. 269 00:17:15,960 --> 00:17:18,620 According to that trig identity. 270 00:17:18,620 --> 00:17:23,970 And so the square root of 1+x^2 is sec(theta). 271 00:17:29,780 --> 00:17:31,070 Right? 272 00:17:31,070 --> 00:17:36,220 So this identity is the reason that the substitution 273 00:17:36,220 --> 00:17:37,120 is going to help us. 274 00:17:37,120 --> 00:17:39,430 Because it gets rid of the square root 275 00:17:39,430 --> 00:17:43,550 and replaces it by some other trig function. 276 00:17:43,550 --> 00:17:46,080 I'd better be able to get rid of the dx, too. 277 00:17:46,080 --> 00:17:48,520 That's part of the substitution process. 278 00:17:48,520 --> 00:17:50,520 But we can do that, because I know 279 00:17:50,520 --> 00:17:52,440 what the derivative of the tangent is. 280 00:17:52,440 --> 00:17:56,070 It's secant squared. 281 00:17:56,070 --> 00:17:58,880 So dx / d theta is sec^2(theta). 282 00:17:58,880 --> 00:18:01,150 So dx is sec^2(theta) d theta. 283 00:18:04,460 --> 00:18:07,350 So let's just substitute all of that stuff 284 00:18:07,350 --> 00:18:10,280 in, and rewrite the entire integral in terms 285 00:18:10,280 --> 00:18:11,750 of our new variable, theta. 286 00:18:11,750 --> 00:18:14,570 So dx is in the numerator. 287 00:18:14,570 --> 00:18:18,210 That's sec^2(theta) d theta. 288 00:18:18,210 --> 00:18:21,197 And then the denominator, well, it has an x^2. 289 00:18:21,197 --> 00:18:22,030 That's tan^2(theta). 290 00:18:24,740 --> 00:18:28,660 And then there's this square root. 291 00:18:28,660 --> 00:18:31,690 And we know what that is in terms of theta. 292 00:18:31,690 --> 00:18:32,730 It's sec(theta). 293 00:18:36,470 --> 00:18:41,920 OK, now. we've done the trig substitution. 294 00:18:41,920 --> 00:18:43,940 I've gotten rid of the square root, 295 00:18:43,940 --> 00:18:46,050 I've got everything in terms of trig functions 296 00:18:46,050 --> 00:18:48,020 of the new variable. 297 00:18:48,020 --> 00:18:49,860 Pretty complicated trig function. 298 00:18:49,860 --> 00:18:52,550 This often happens, you wind up with a complete scattering 299 00:18:52,550 --> 00:18:55,050 of different trig functions in the numerator and denominator 300 00:18:55,050 --> 00:18:56,250 and everything. 301 00:18:56,250 --> 00:18:58,950 A systematic thing to do here is to put everything 302 00:18:58,950 --> 00:19:09,110 in terms of sines and cosines. 303 00:19:09,110 --> 00:19:12,600 Unless you can see right away, how it's going to simplify, 304 00:19:12,600 --> 00:19:15,010 the systematic thing to do is to rewrite 305 00:19:15,010 --> 00:19:17,110 in terms of sines and cosines. 306 00:19:17,110 --> 00:19:19,300 So let's do that. 307 00:19:19,300 --> 00:19:20,260 So let's see. 308 00:19:20,260 --> 00:19:23,520 The secant squared, secant is 1 over cosine. 309 00:19:23,520 --> 00:19:28,920 So I'm going to put a cosine squared in the denominator. 310 00:19:28,920 --> 00:19:32,070 Oh, I guess the first thing I can do is cancel. 311 00:19:32,070 --> 00:19:33,270 Let's do that. 312 00:19:33,270 --> 00:19:34,200 That's clever. 313 00:19:34,200 --> 00:19:35,700 You were all thinking that too. 314 00:19:35,700 --> 00:19:37,220 Cancel those. 315 00:19:37,220 --> 00:19:39,860 So now I just get one cosine denominator 316 00:19:39,860 --> 00:19:44,382 from the secant there in the numerator. 317 00:19:44,382 --> 00:19:46,840 It's still pretty complicated, secant over tangent squared, 318 00:19:46,840 --> 00:19:48,110 who knows. 319 00:19:48,110 --> 00:19:49,410 Well, we'll find out. 320 00:19:49,410 --> 00:19:52,450 Because the tangent is sine over cosine. 321 00:19:52,450 --> 00:19:55,940 So I should put a sine squared where the tangent was, 322 00:19:55,940 --> 00:19:59,330 and a cosine squared up there. 323 00:19:59,330 --> 00:20:02,180 And I still have d theta. 324 00:20:02,180 --> 00:20:04,300 And now you see some more cancellation occurs. 325 00:20:04,300 --> 00:20:07,950 That's the virtue of writing things out in this way. 326 00:20:07,950 --> 00:20:14,940 So now, the square here cancels with this cosine. 327 00:20:14,940 --> 00:20:20,580 And I'm left with cos(theta) d theta / sin^2(theta). 328 00:20:25,340 --> 00:20:26,860 That's a little simpler. 329 00:20:26,860 --> 00:20:33,300 And it puts me in a position to use the same idea I just used. 330 00:20:33,300 --> 00:20:35,100 I see the sine here. 331 00:20:35,100 --> 00:20:37,090 I might look around in this integral 332 00:20:37,090 --> 00:20:39,940 to see if its derivative occurs anywhere. 333 00:20:39,940 --> 00:20:43,890 The differential of the sine is the cosine. 334 00:20:43,890 --> 00:20:49,720 And so I'm very much inclined to make another substitution. 335 00:20:49,720 --> 00:20:54,250 Say, u, direct substitution this time. 336 00:20:54,250 --> 00:20:56,150 And say u is the cosine of theta. 337 00:20:56,150 --> 00:20:59,110 Because then du-- Oh, I'm sorry. 338 00:20:59,110 --> 00:21:01,850 Say, u is the sine of theta. 339 00:21:01,850 --> 00:21:05,980 Because then du is cos(theta) d theta. 340 00:21:13,540 --> 00:21:20,140 And then this integral becomes, well, the numerator just is du. 341 00:21:20,140 --> 00:21:22,660 The denominator is u^2. 342 00:21:22,660 --> 00:21:25,550 And I think we can break out the champagne, 343 00:21:25,550 --> 00:21:28,169 because we can integrate that one. 344 00:21:28,169 --> 00:21:29,710 Finally get rid of the integral sign. 345 00:21:29,710 --> 00:21:30,210 Yes sir. 346 00:21:30,210 --> 00:21:37,842 STUDENT: [INAUDIBLE] 347 00:21:37,842 --> 00:21:40,300 PROFESSOR: OK, how do I know to make u equal to sine rather 348 00:21:40,300 --> 00:21:41,640 than cosine. 349 00:21:41,640 --> 00:21:45,540 Because I want to see du appear up here. 350 00:21:45,540 --> 00:21:49,360 If I'd had a sine up here, that would be a signal to me 351 00:21:49,360 --> 00:21:53,510 that maybe I should say let u be the cosine. 352 00:21:53,510 --> 00:21:54,280 OK? 353 00:21:54,280 --> 00:21:56,980 Also, because this thing in the denominator is something 354 00:21:56,980 --> 00:21:57,870 I want to get rid of. 355 00:21:57,870 --> 00:21:59,050 It's in the denominator. 356 00:21:59,050 --> 00:22:01,030 So I'll get rid of it by wishful thinking 357 00:22:01,030 --> 00:22:04,940 and just call it something else. 358 00:22:04,940 --> 00:22:07,010 It works pretty well in this case. 359 00:22:07,010 --> 00:22:10,400 Wishful thinking doesn't always work so well. 360 00:22:10,400 --> 00:22:18,510 So I integrate u^(-2) du, and I get - -1/u plus a constant, 361 00:22:18,510 --> 00:22:21,340 and I'm done with the calculus part of this problem. 362 00:22:21,340 --> 00:22:22,580 I've done the integral now. 363 00:22:22,580 --> 00:22:25,736 Gotten rid of the integral sign. 364 00:22:25,736 --> 00:22:27,360 But I'm not quite done with the problem 365 00:22:27,360 --> 00:22:29,640 yet, because I have to work my way back 366 00:22:29,640 --> 00:22:31,320 through two substitutions. 367 00:22:31,320 --> 00:22:33,120 First, this one. 368 00:22:33,120 --> 00:22:34,620 And then this one. 369 00:22:34,620 --> 00:22:38,800 So this first substitution isn't so bad to get rid of, to undo, 370 00:22:38,800 --> 00:22:40,200 to back-substitute. 371 00:22:40,200 --> 00:22:43,630 Because u is just sin(theta). 372 00:22:43,630 --> 00:22:46,740 And so 1/u is, I guess a fancy way 373 00:22:46,740 --> 00:22:52,540 to write it is the cosecant of theta. 374 00:22:52,540 --> 00:22:54,890 1 over the sine is the cosecant. 375 00:22:54,890 --> 00:23:00,257 So I get -csc(theta) plus a constant. 376 00:23:00,257 --> 00:23:01,590 Is there a question in the back? 377 00:23:01,590 --> 00:23:02,090 Yes sir? 378 00:23:02,090 --> 00:23:07,010 STUDENT: [INAUDIBLE] 379 00:23:07,010 --> 00:23:09,000 PROFESSOR: I'm sorry, my hearing is so bad. 380 00:23:09,000 --> 00:23:12,500 STUDENT: [INAUDIBLE] 381 00:23:12,500 --> 00:23:16,070 PROFESSOR: How did I know this substitution 382 00:23:16,070 --> 00:23:19,950 in the first place. 383 00:23:19,950 --> 00:23:22,510 It's because of the 1 + x^2. 384 00:23:22,510 --> 00:23:24,510 And I want to make use of the trig identity 385 00:23:24,510 --> 00:23:26,380 in the upper left-hand corner. 386 00:23:26,380 --> 00:23:29,430 I'll make you a table in a few minutes that will put 387 00:23:29,430 --> 00:23:30,870 all this in a bigger context. 388 00:23:30,870 --> 00:23:32,350 And I think it'll help you then. 389 00:23:32,350 --> 00:23:35,880 OK, I'll promise. 390 00:23:35,880 --> 00:23:39,570 So, what I want to try to talk about right 391 00:23:39,570 --> 00:23:43,710 now is how to rewrite a term like this. 392 00:23:43,710 --> 00:23:47,240 A trig term like this, back in terms of x. 393 00:23:47,240 --> 00:23:51,020 So I want to undo this trick substitution. 394 00:23:51,020 --> 00:23:55,190 This is a trig sub. 395 00:23:55,190 --> 00:23:59,350 And what I want to do now is try to undo that trig sub. 396 00:23:59,350 --> 00:24:00,970 And I'll show you a general method 397 00:24:00,970 --> 00:24:03,210 for undoing trig substitutions. 398 00:24:03,210 --> 00:24:05,100 This happens quite often. 399 00:24:05,100 --> 00:24:07,150 I don't know what the cosecant of theta is. 400 00:24:07,150 --> 00:24:11,290 But I do know what the tangent of theta is. 401 00:24:11,290 --> 00:24:14,060 So I want to make a relation between them. 402 00:24:14,060 --> 00:24:20,920 OK, so undoing. 403 00:24:20,920 --> 00:24:24,420 Trig subs. 404 00:24:24,420 --> 00:24:29,420 So let's go back to where trigonometry always comes from, 405 00:24:29,420 --> 00:24:32,130 this right angled triangle. 406 00:24:32,130 --> 00:24:35,540 The theta in the corner, and then these three sides. 407 00:24:35,540 --> 00:24:37,590 This one's called the hypotenuse. 408 00:24:37,590 --> 00:24:41,430 This one is called the adjacent side, 409 00:24:41,430 --> 00:24:45,690 and that one's called the opposite side. 410 00:24:45,690 --> 00:24:52,040 And now, let's find out where x lies in this triangle. 411 00:24:52,040 --> 00:24:55,500 Let's try to write the sides of this triangle in terms of x. 412 00:24:55,500 --> 00:24:58,670 And what I know is, x is the tangent of theta. 413 00:24:58,670 --> 00:25:01,310 So the tangent of theta, tangent of this angle, 414 00:25:01,310 --> 00:25:03,830 is opposite divided by adjacent. 415 00:25:03,830 --> 00:25:05,890 Did you learn SOH CAH TOA? 416 00:25:05,890 --> 00:25:09,520 OK, so it's opposite divided by adjacent. 417 00:25:09,520 --> 00:25:10,580 Is the tangent. 418 00:25:10,580 --> 00:25:13,650 So there are different ways to do that, 419 00:25:13,650 --> 00:25:15,530 but why not just do it in the simplest way 420 00:25:15,530 --> 00:25:21,950 and suppose that the adjacent is 1, and the opposite is x. 421 00:25:21,950 --> 00:25:24,530 This is correct now, isn't it? 422 00:25:24,530 --> 00:25:27,890 I get the correct value for the tangent of theta 423 00:25:27,890 --> 00:25:32,910 by saying that the lengths of those are 1 and x. 424 00:25:32,910 --> 00:25:39,160 And that means that the hypotenuse has length 1 + x^2. 425 00:25:39,160 --> 00:25:40,840 Well, here's a triangle. 426 00:25:40,840 --> 00:25:44,060 I'm interested in computing the cosecant of theta. 427 00:25:44,060 --> 00:25:50,840 Where's that appear in the triangle? 428 00:25:50,840 --> 00:25:51,510 Well, let's see. 429 00:25:51,510 --> 00:25:55,570 The cosecant of theta is 1 over the sine. 430 00:25:55,570 --> 00:26:02,830 And the sine is opposite over hypotenuse. 431 00:26:02,830 --> 00:26:15,240 So the cosecant is hypotenuse over opposite. 432 00:26:15,240 --> 00:26:20,680 And the hypotenuse is the square root of 1 + x^2, 433 00:26:20,680 --> 00:26:24,680 and the opposite is x. 434 00:26:24,680 --> 00:26:26,200 And so I've done it. 435 00:26:26,200 --> 00:26:29,650 I've undone the trig substitution. 436 00:26:29,650 --> 00:26:32,100 I've figured out what this cosecant of theta 437 00:26:32,100 --> 00:26:35,360 is, in terms of x. 438 00:26:35,360 --> 00:26:39,490 And so the final answer is minus the square root of 1+x^2, 439 00:26:39,490 --> 00:26:43,470 over x, plus a constant, and there's an answer 440 00:26:43,470 --> 00:26:53,030 to the original problem. 441 00:26:53,030 --> 00:26:57,160 This took two boards to go through this. 442 00:26:57,160 --> 00:27:00,800 I illustrated several things. 443 00:27:00,800 --> 00:27:02,870 Actually, this three half boards. 444 00:27:02,870 --> 00:27:05,190 I illustrated this use of trig substitution, 445 00:27:05,190 --> 00:27:07,910 and I'll come back to that in a second. 446 00:27:07,910 --> 00:27:10,870 I illustrated patience. 447 00:27:10,870 --> 00:27:15,320 I illustrated rewriting things in terms of sines and cosines, 448 00:27:15,320 --> 00:27:17,080 and then making a direct substitution 449 00:27:17,080 --> 00:27:19,550 to evaluate an integral like this. 450 00:27:19,550 --> 00:27:23,520 And then there's this undoing all of those substitutions. 451 00:27:23,520 --> 00:27:25,670 And it culminated with undoing the trig sub. 452 00:27:25,670 --> 00:27:31,100 So let's play a game here. 453 00:27:31,100 --> 00:27:35,070 Why don't we play the game where you 454 00:27:35,070 --> 00:27:46,970 give me-- So, there's a step in here that I should have done. 455 00:27:46,970 --> 00:27:56,390 I should've said this is -cos(arctan(theta)) plus 456 00:27:56,390 --> 00:27:58,812 a constant. 457 00:27:58,812 --> 00:28:00,520 The most straightforward thing you can do 458 00:28:00,520 --> 00:28:02,760 is to say since x is the tangent of theta, 459 00:28:02,760 --> 00:28:06,950 that means that-- sorry, if x, that means that theta 460 00:28:06,950 --> 00:28:08,970 is the arctangent of x. 461 00:28:08,970 --> 00:28:11,761 And so let's just put in theta as the arctangent of x, 462 00:28:11,761 --> 00:28:12,760 and that's what you get. 463 00:28:12,760 --> 00:28:15,020 So really, what I just did for you 464 00:28:15,020 --> 00:28:19,720 was to show you a way to compute some trig function applied 465 00:28:19,720 --> 00:28:23,900 to the inverse of another trig function. 466 00:28:23,900 --> 00:28:29,860 I computed cosecant of the arctangent by this trick. 467 00:28:29,860 --> 00:28:32,260 So now, let's play the game where you give me 468 00:28:32,260 --> 00:28:35,150 a trig function and an inverse trig function, 469 00:28:35,150 --> 00:28:46,950 and I try to compute what the composite is. 470 00:28:46,950 --> 00:28:47,510 OK. 471 00:28:47,510 --> 00:28:58,530 So who can give me a trig function. 472 00:28:58,530 --> 00:29:03,310 Has to be one of these standard ones. 473 00:29:03,310 --> 00:29:03,900 STUDENT: Tan. 474 00:29:03,900 --> 00:29:04,870 PROFESSOR: Tangent. 475 00:29:04,870 --> 00:29:07,034 Alright. 476 00:29:07,034 --> 00:29:07,950 How about another one? 477 00:29:07,950 --> 00:29:11,120 STUDENT: Sine. 478 00:29:11,120 --> 00:29:12,350 PROFESSOR: Sine. 479 00:29:12,350 --> 00:29:15,180 Do we have agreement on sine. 480 00:29:15,180 --> 00:29:16,910 STUDENT: [INAUDIBLE] 481 00:29:16,910 --> 00:29:17,760 PROFESSOR: Secant? 482 00:29:17,760 --> 00:29:27,802 STUDENT: [INAUDIBLE] 483 00:29:27,802 --> 00:29:29,510 PROFESSOR: Right, csc has the best cheer. 484 00:29:29,510 --> 00:29:30,301 So that's the game. 485 00:29:30,301 --> 00:29:34,867 We have to compute, try to compute, that composite. 486 00:29:34,867 --> 00:29:35,950 Something wrong with this? 487 00:29:35,950 --> 00:29:45,459 STUDENT: [INAUDIBLE] 488 00:29:45,459 --> 00:29:47,000 PROFESSOR: What does acceptable mean? 489 00:29:47,000 --> 00:29:48,810 Don't you think-- so the question is, 490 00:29:48,810 --> 00:29:51,900 isn't this a perfectly acceptable final answer. 491 00:29:51,900 --> 00:29:53,760 It's a correct final answer. 492 00:29:53,760 --> 00:29:56,850 But this is much more insightful. 493 00:29:56,850 --> 00:30:00,919 And after all the original thing was involving square roots 494 00:30:00,919 --> 00:30:02,460 and things, this is the kind of thing 495 00:30:02,460 --> 00:30:03,920 you might hope for is an answer. 496 00:30:03,920 --> 00:30:07,720 This is just a nicer answer for sure. 497 00:30:07,720 --> 00:30:10,210 And likely to be more useful to you when you go on 498 00:30:10,210 --> 00:30:13,830 and use that answer for something else. 499 00:30:13,830 --> 00:30:18,610 OK, so let's try to do this this. 500 00:30:18,610 --> 00:30:22,260 Undo a trig substitution that involved a cosecant. 501 00:30:22,260 --> 00:30:23,910 And I manipulate around, and I find 502 00:30:23,910 --> 00:30:27,540 myself trying to find out what's the tangent of theta. 503 00:30:27,540 --> 00:30:29,240 So here's how we go about it. 504 00:30:29,240 --> 00:30:33,970 I draw this triangle. 505 00:30:33,970 --> 00:30:37,590 Theta is the angle here. 506 00:30:37,590 --> 00:30:43,570 This is the adjacent, opposite, hypotenuse. 507 00:30:43,570 --> 00:30:49,030 So, the first thing is how can I make the cosecant appear here, 508 00:30:49,030 --> 00:30:49,950 csc x. 509 00:30:49,950 --> 00:30:52,640 What dimensions should I give to the sides in order 510 00:30:52,640 --> 00:31:01,400 for the cosecant of x, sorry, in order for theta 511 00:31:01,400 --> 00:31:02,880 to be the cosecant of x. 512 00:31:02,880 --> 00:31:05,280 This thing is theta. 513 00:31:05,280 --> 00:31:17,810 So, that means that the cosecant of x-- 514 00:31:17,810 --> 00:31:24,830 that means the cosecant of theta should be x. 515 00:31:24,830 --> 00:31:28,970 Theta is the arccosecant, so x is the cosecant of theta. 516 00:31:28,970 --> 00:31:31,980 So, what'll I take the sides to be, to get the cosecant? 517 00:31:31,980 --> 00:31:37,050 The cosecant is 1 over the sine. 518 00:31:37,050 --> 00:31:47,960 And the sine is the opposite over the hypotenuse. 519 00:31:47,960 --> 00:31:49,910 So I get hypotenuse over opposite. 520 00:31:49,910 --> 00:31:53,750 And that's supposed to be what x is. 521 00:31:53,750 --> 00:31:56,540 So I could make the opposite anything 522 00:31:56,540 --> 00:31:59,190 I want, but the simplest thing is to make it 1. 523 00:31:59,190 --> 00:32:00,080 Let's do that. 524 00:32:00,080 --> 00:32:04,300 And then what does that mean about the rest of the sides? 525 00:32:04,300 --> 00:32:06,142 Hypotenuse had better be x. 526 00:32:06,142 --> 00:32:07,350 And then I've recovered this. 527 00:32:07,350 --> 00:32:12,350 So here's a triangle that exhibits the correct angle. 528 00:32:12,350 --> 00:32:14,680 This remaining side is going to be useful to us. 529 00:32:14,680 --> 00:32:20,550 And it is the square root of x^2 - 1. 530 00:32:20,550 --> 00:32:24,060 So I've got a triangle of the correct angle theta, 531 00:32:24,060 --> 00:32:27,400 and now I want to compute the tangent of that angle. 532 00:32:27,400 --> 00:32:28,350 Well, that's easy. 533 00:32:28,350 --> 00:32:31,020 That's opposite divided by adjacent. 534 00:32:31,020 --> 00:32:34,690 So I get 1 over the square root of x^2 - 1. 535 00:32:38,270 --> 00:32:40,590 Very flexible tool that'll be useful to you 536 00:32:40,590 --> 00:32:42,200 in many different times. 537 00:32:42,200 --> 00:32:44,170 Whenever you have to undo a trig substitution, 538 00:32:44,170 --> 00:32:49,100 this is likely to be useful. 539 00:32:49,100 --> 00:32:51,740 OK, that was a good game. 540 00:32:51,740 --> 00:32:52,740 No winners in this game. 541 00:32:52,740 --> 00:32:53,700 We're all winners. 542 00:32:53,700 --> 00:33:01,970 No losers, we're all winners. 543 00:33:01,970 --> 00:33:02,740 OK. 544 00:33:02,740 --> 00:33:06,340 So, good. 545 00:33:06,340 --> 00:33:09,810 So let me make this table of the different trig substitutions, 546 00:33:09,810 --> 00:33:11,010 and how they can be useful. 547 00:33:11,010 --> 00:33:20,520 Summary of trig substitutions. 548 00:33:20,520 --> 00:33:28,540 So over here, we have, if you see, 549 00:33:28,540 --> 00:33:58,090 so if your integrand contains, make a substitution to get. 550 00:33:58,090 --> 00:34:01,086 So if your integrand contains, I'll 551 00:34:01,086 --> 00:34:02,710 write these things out as square roots. 552 00:34:02,710 --> 00:34:07,670 If it contains the square root a^2 - x^2, 553 00:34:07,670 --> 00:34:10,560 this is what we talked about on Thursday. 554 00:34:10,560 --> 00:34:14,470 When I was trying to find the area of that piece of a circle. 555 00:34:14,470 --> 00:34:18,130 There, I suggested that we should make the substitution 556 00:34:18,130 --> 00:34:20,260 x = a cos(theta). 557 00:34:22,890 --> 00:34:28,920 Or, x = a sin(theta). 558 00:34:28,920 --> 00:34:31,400 Either one works just as well. 559 00:34:31,400 --> 00:34:36,430 And there's no way to prefer one over the other. 560 00:34:36,430 --> 00:34:41,110 And when you make the substitution, x = a cos(theta), 561 00:34:41,110 --> 00:34:44,030 you get a^2 - a^2 cos^2(theta). 562 00:34:44,030 --> 00:34:45,340 theta. 563 00:34:45,340 --> 00:34:48,280 1 - cos^2 is sin^2. 564 00:34:48,280 --> 00:34:50,620 So you get a sin(theta). 565 00:34:55,270 --> 00:34:59,880 So this expression becomes equal to this expression 566 00:34:59,880 --> 00:35:03,950 under that substitution. 567 00:35:03,950 --> 00:35:04,800 And then you go on. 568 00:35:04,800 --> 00:35:06,626 Then you've gotten rid of the square root, 569 00:35:06,626 --> 00:35:08,250 and you've got a trigonometric integral 570 00:35:08,250 --> 00:35:10,610 that you have to try to do. 571 00:35:10,610 --> 00:35:14,680 If you made the substitution a sin(theta), you'd get a a^2 - 572 00:35:14,680 --> 00:35:21,580 a^2 sin^2, which is a cos(theta). 573 00:35:21,580 --> 00:35:24,530 And then you can go ahead as well. 574 00:35:24,530 --> 00:35:26,430 We just saw another example. 575 00:35:26,430 --> 00:35:29,510 Namely, if you have a^2 + x^2. 576 00:35:29,510 --> 00:35:35,530 That's like the example we had up here. a = 1 in this example. 577 00:35:35,530 --> 00:35:36,510 What did we do? 578 00:35:36,510 --> 00:35:42,150 We tried the substitution x = a tan(theta). 579 00:35:42,150 --> 00:35:44,910 And the reason is that I can plug into the trig identity 580 00:35:44,910 --> 00:35:46,950 up here in the upper left. 581 00:35:46,950 --> 00:35:51,210 And replace a^2 + x^2 by a sec(theta). 582 00:35:55,020 --> 00:35:59,790 Square root of the secant squared. 583 00:35:59,790 --> 00:36:03,410 There's one more thing in this table. 584 00:36:03,410 --> 00:36:06,520 Sort of, the only remaining sum or difference 585 00:36:06,520 --> 00:36:07,544 of terms like this. 586 00:36:07,544 --> 00:36:09,460 And that's what happens if you have x^2 - a^2. 587 00:36:14,050 --> 00:36:17,790 So there, I think we can make a substitution a sec(theta). 588 00:36:21,300 --> 00:36:27,031 Because, after all, sec^2(theta)-- so x^2 - a^2-- 589 00:36:27,031 --> 00:36:27,530 Sorry. 590 00:36:32,310 --> 00:36:34,790 Let's see what happens when I make that substitution. 591 00:36:34,790 --> 00:36:42,850 x^2 - a^2 = a^2 sec^2(theta) - a^2. 592 00:36:42,850 --> 00:36:45,460 Under this substitution. 593 00:36:45,460 --> 00:36:48,690 That's sec^2 - 1. 594 00:36:48,690 --> 00:36:51,390 Well, put the 1 on the other side. 595 00:36:51,390 --> 00:36:53,720 And you find tan^2, coming out. 596 00:36:53,720 --> 00:36:56,710 So this is a a^2 tan^2(theta). 597 00:36:59,440 --> 00:37:02,680 And so that's what you get, a times tan(theta). 598 00:37:02,680 --> 00:37:07,850 After I take the square root, I get a tan(theta). 599 00:37:07,850 --> 00:37:13,080 So these are the three basic trig substitution forms. 600 00:37:13,080 --> 00:37:15,580 Where trig substitutions are useful to get 601 00:37:15,580 --> 00:37:17,920 rid of expressions like this, and replace them 602 00:37:17,920 --> 00:37:22,670 by trigonometric expressions. 603 00:37:22,670 --> 00:37:25,140 And then you use this trick, you do the integral if you can 604 00:37:25,140 --> 00:37:27,265 and then you use this trick to get rid of the theta 605 00:37:27,265 --> 00:37:37,460 at the end. 606 00:37:37,460 --> 00:37:40,630 So now, the last thing I want to talk about today 607 00:37:40,630 --> 00:37:59,180 is called completing the square. 608 00:37:59,180 --> 00:38:03,440 And that comes in because unfortunately, 609 00:38:03,440 --> 00:38:08,350 not every square root of a quadratic 610 00:38:08,350 --> 00:38:10,390 has such a simple form. 611 00:38:10,390 --> 00:38:16,200 You will often encounter things that are not just 612 00:38:16,200 --> 00:38:18,080 the square root of something simple. 613 00:38:18,080 --> 00:38:19,140 Like one of these forms. 614 00:38:19,140 --> 00:38:27,537 Like there might be a middle term in there. 615 00:38:27,537 --> 00:38:29,120 I don't actually have time to show you 616 00:38:29,120 --> 00:38:33,200 an example of how this comes out in a sort of practical example. 617 00:38:33,200 --> 00:38:35,760 But it does happen quite frequently. 618 00:38:35,760 --> 00:38:38,580 And so I want to show you how to deal with things 619 00:38:38,580 --> 00:38:40,550 like the following example. 620 00:38:40,550 --> 00:38:47,380 Let's try to integrate dx over x^2 + 4x, 621 00:38:47,380 --> 00:38:55,090 the square root of x^2 + 4x. 622 00:38:55,090 --> 00:38:59,010 So there's a square root of some square, some quadratic. 623 00:38:59,010 --> 00:39:02,010 It's very much like this business. 624 00:39:02,010 --> 00:39:04,830 But it isn't of any of these forms. 625 00:39:04,830 --> 00:39:06,360 And so what I want to do is show you 626 00:39:06,360 --> 00:39:10,540 how to rewrite it in one of those forms using substitution, 627 00:39:10,540 --> 00:39:11,120 again. 628 00:39:11,120 --> 00:39:14,520 All this is about substitution. 629 00:39:14,520 --> 00:39:26,100 So the game is to rewrite quadratic as something 630 00:39:26,100 --> 00:39:30,950 like x plus something or other. 631 00:39:30,950 --> 00:39:32,470 Plus some other constant. 632 00:39:32,470 --> 00:39:35,690 So write it, try to write it, in the form of a square 633 00:39:35,690 --> 00:39:42,340 plus or minus another constant. 634 00:39:42,340 --> 00:39:45,170 And then we'll go on from there. 635 00:39:45,170 --> 00:39:51,110 So let's do that in this case. x^2, x^2 + 4x. 636 00:39:51,110 --> 00:39:53,390 Well, if you square this form out, 637 00:39:53,390 --> 00:39:57,280 then the middle term is going to be 2ax. 638 00:39:57,280 --> 00:40:00,920 So that, since I have a middle term here, 639 00:40:00,920 --> 00:40:03,640 I pretty much know what a has to be. 640 00:40:03,640 --> 00:40:08,220 The only choice in order to get something like x^2 + 4x out 641 00:40:08,220 --> 00:40:11,440 of this, is to take a to be 2. 642 00:40:11,440 --> 00:40:16,950 Because then, this is what you get. 643 00:40:16,950 --> 00:40:19,790 This isn't quite right yet, but let's compute what I have here. 644 00:40:19,790 --> 00:40:24,490 x^2 + 4x, so far so good, plus 4, 645 00:40:24,490 --> 00:40:26,190 and I don't have a plus 4 here. 646 00:40:26,190 --> 00:40:31,394 So I have to fix that by subtracting 4. 647 00:40:31,394 --> 00:40:32,310 So that's what I mean. 648 00:40:32,310 --> 00:40:33,393 I've completed the square. 649 00:40:33,393 --> 00:40:38,370 The word for this process of eliminating the middle term 650 00:40:38,370 --> 00:40:41,850 by using the square of an expression like that. 651 00:40:41,850 --> 00:40:44,830 That's called completing the square. 652 00:40:44,830 --> 00:40:48,010 And we can use that process to compute this integral. 653 00:40:48,010 --> 00:40:51,040 So let's do that. 654 00:40:51,040 --> 00:40:54,170 So I can rewrite this integral, rewrite this denominator 655 00:40:54,170 --> 00:41:00,100 like this. 656 00:41:00,100 --> 00:41:03,860 And then I'm going to try to use one of these forms over here. 657 00:41:03,860 --> 00:41:07,900 So in order to get a single variable there, 658 00:41:07,900 --> 00:41:12,500 instead of something complicated like x + 2, 659 00:41:12,500 --> 00:41:15,090 I'm inclined to come up with another variable name 660 00:41:15,090 --> 00:41:20,330 and write it equal, write x + 2 as that other variable name. 661 00:41:20,330 --> 00:41:29,950 So here's another little direct substitution. u = x + 2. 662 00:41:29,950 --> 00:41:31,260 Figure out what du is. 663 00:41:31,260 --> 00:41:36,610 That's pretty easy. 664 00:41:36,610 --> 00:41:41,580 And then rewrite the integral in those terms. 665 00:41:41,580 --> 00:41:44,880 So dx = du. 666 00:41:44,880 --> 00:41:47,270 And then in the denominator I have, 667 00:41:47,270 --> 00:41:49,100 well, I have the square root of that. 668 00:41:49,100 --> 00:41:55,100 Oh yeah, so I think as part of this I'll write out what x^2 + 669 00:41:55,100 --> 00:41:56,550 4x is. 670 00:41:56,550 --> 00:41:59,520 The point is, it's equal to u^2 - 4. 671 00:42:04,170 --> 00:42:06,540 x^2 + 4x = u^2 - 4. 672 00:42:12,950 --> 00:42:20,260 There's the data box containing the substitution data. 673 00:42:20,260 --> 00:42:22,640 And so now I can put that in. 674 00:42:22,640 --> 00:42:24,960 I have x^2 + 4x there. 675 00:42:24,960 --> 00:42:31,330 In terms of u, that's u^2 - 4. 676 00:42:31,330 --> 00:42:33,690 Well, now I'm in a happier position because I can look 677 00:42:33,690 --> 00:42:37,060 for u^2 - 4 for something like that in my table here. 678 00:42:37,060 --> 00:42:40,430 And it actually sits down here. 679 00:42:40,430 --> 00:42:43,365 So except for the use of the letter x 680 00:42:43,365 --> 00:42:45,470 here instead of u over there. 681 00:42:45,470 --> 00:42:47,510 That tells me what I want. 682 00:42:47,510 --> 00:42:55,100 So to handle this, what I should use is a trig substitution. 683 00:42:55,100 --> 00:42:58,020 And the trig substitution that's suggested is, 684 00:42:58,020 --> 00:43:03,870 according to the bottom line with a = 2, so a^2 = 4. 685 00:43:03,870 --> 00:43:07,600 The suggestion is, I should take x-- 686 00:43:07,600 --> 00:43:13,909 But I'd better not use the letter x any more. 687 00:43:13,909 --> 00:43:15,950 But I don't have a letter x, I have the letter u. 688 00:43:15,950 --> 00:43:21,310 I should take u equal to 2 secant. 689 00:43:21,310 --> 00:43:23,110 And then some letter I haven't used before. 690 00:43:23,110 --> 00:43:28,820 And theta is available. 691 00:43:28,820 --> 00:43:30,560 This is a look-up table process. 692 00:43:30,560 --> 00:43:33,240 I see the square root of u^2 - 4, 693 00:43:33,240 --> 00:43:35,250 I see that that's of this form. 694 00:43:35,250 --> 00:43:38,230 I'm instructed to make this substitution. 695 00:43:38,230 --> 00:43:40,540 And that's what I just did. 696 00:43:40,540 --> 00:43:43,600 Let's see how it works out. 697 00:43:43,600 --> 00:43:47,690 So that means the du is 2, OK. 698 00:43:47,690 --> 00:43:50,480 What's the derivative of the secant? 699 00:43:50,480 --> 00:43:52,660 Secant tangent. 700 00:43:52,660 --> 00:43:59,050 So du = 2 sec(theta) tan(theta). 701 00:43:59,050 --> 00:44:05,690 And u^2 - 4 is, here's the payoff. 702 00:44:05,690 --> 00:44:07,170 I'm supposed to be able to rewrite 703 00:44:07,170 --> 00:44:09,350 that in terms of the tangent. 704 00:44:09,350 --> 00:44:19,650 According to this. u u^2 - 4 is 4 secant squared minus 4. 705 00:44:19,650 --> 00:44:22,450 And secant squared minus 1 is tangent squared. 706 00:44:22,450 --> 00:44:24,400 So this is 4 tan^2(theta). 707 00:44:30,890 --> 00:44:36,807 Right, yeah? 708 00:44:36,807 --> 00:44:37,640 STUDENT: [INAUDIBLE] 709 00:44:37,640 --> 00:44:42,200 PROFESSOR: But I squared it. 710 00:44:42,200 --> 00:44:44,180 And now I'll square root it. 711 00:44:44,180 --> 00:44:50,650 And I'll get a 2 and this tangent will go away. 712 00:44:50,650 --> 00:44:56,130 So there's my data box for this substitution. 713 00:44:56,130 --> 00:45:15,560 And let's go on to another board. 714 00:45:15,560 --> 00:45:20,980 So where I'm at is the integral of du over the square root 715 00:45:20,980 --> 00:45:22,010 of u^2 - 4. 716 00:45:26,850 --> 00:45:30,060 And I have all the data I need here 717 00:45:30,060 --> 00:45:33,090 to rewrite that in terms of theta. 718 00:45:33,090 --> 00:45:38,170 So du = 2 sec(theta) tan(theta) d theta. 719 00:45:41,720 --> 00:45:44,140 And the denominator is 2 tan(theta). 720 00:45:47,760 --> 00:45:48,590 Ha. 721 00:45:48,590 --> 00:45:53,040 Well, so some very nice simplification happens here. 722 00:45:53,040 --> 00:45:55,940 The 2's cancel. 723 00:45:55,940 --> 00:45:58,260 And the tangents cancel. 724 00:45:58,260 --> 00:46:01,900 And I'm left with trying to work with the integral, sec(theta) d 725 00:46:01,900 --> 00:46:03,920 theta. 726 00:46:03,920 --> 00:46:06,610 And luckily enough at the very beginning of the hour, 727 00:46:06,610 --> 00:46:08,480 I worked out how to compute the integral 728 00:46:08,480 --> 00:46:09,885 of the secant of theta. 729 00:46:09,885 --> 00:46:11,670 And there it is. 730 00:46:11,670 --> 00:46:25,000 So this is ln(sec(theta) + tan(theta)) plus a constant. 731 00:46:25,000 --> 00:46:27,660 And we're done with the calculus part. 732 00:46:27,660 --> 00:46:29,390 There's no more integral there. 733 00:46:29,390 --> 00:46:31,490 But I still am not quite done with the problem, 734 00:46:31,490 --> 00:46:37,180 because again I have these two substitutions to try to undo. 735 00:46:37,180 --> 00:46:40,220 So let's undo them one by one. 736 00:46:40,220 --> 00:46:42,240 Let's see. 737 00:46:42,240 --> 00:46:44,130 I have this trig substitution here. 738 00:46:44,130 --> 00:46:47,210 And I could use my triangle trick, if I need to. 739 00:46:47,210 --> 00:46:49,280 But maybe I don't need to. 740 00:46:49,280 --> 00:46:51,220 Let's see, do I know what the secant of theta 741 00:46:51,220 --> 00:46:53,090 is in terms of u? 742 00:46:53,090 --> 00:46:54,620 Well, I do. 743 00:46:54,620 --> 00:46:55,730 So I get ln(u/2). 744 00:46:58,980 --> 00:47:01,740 Do I know what the tangent is in terms of u? 745 00:47:01,740 --> 00:47:02,730 Well, I do. 746 00:47:02,730 --> 00:47:04,180 It's here. 747 00:47:04,180 --> 00:47:06,500 So I lucked out, in this case. 748 00:47:06,500 --> 00:47:09,930 And I don't have to go through and use that triangle trick. 749 00:47:09,930 --> 00:47:15,700 So the tangent of theta is the square root of u^2 - 4, over 2. 750 00:47:19,000 --> 00:47:20,080 Good. 751 00:47:20,080 --> 00:47:24,220 So I've undone this trig substitution. 752 00:47:24,220 --> 00:47:28,440 I'm not quite done yet because my answer is involved with u. 753 00:47:28,440 --> 00:47:30,780 And what I wanted originally was x. 754 00:47:30,780 --> 00:47:33,780 But this direct substitution that I started with 755 00:47:33,780 --> 00:47:35,250 is really easy to deal with. 756 00:47:35,250 --> 00:47:39,780 I can just put x + 2 every time I see a u. 757 00:47:39,780 --> 00:47:45,590 So this is the natural logarithm of (x+2)/2 plus the square 758 00:47:45,590 --> 00:47:46,090 root... 759 00:47:46,090 --> 00:47:51,330 What's going to happen when I put x + 2 in place for u here? 760 00:47:51,330 --> 00:47:54,490 You know what you get. 761 00:47:54,490 --> 00:47:59,230 You get exactly what we started with. 762 00:47:59,230 --> 00:48:00,260 Right? 763 00:48:00,260 --> 00:48:04,850 I put x + 2 in place of the u here. 764 00:48:04,850 --> 00:48:13,210 I get x^2 + 4x. 765 00:48:13,210 --> 00:48:15,330 So I've gotten back to a function purely 766 00:48:15,330 --> 00:48:21,200 in terms of x OK, that's a good place to quit. 767 00:48:21,200 --> 00:48:24,840 Have a great little one-day break. 768 00:48:24,840 --> 00:48:27,320 I guess this class doesn't meet on Monday anyway. 769 00:48:27,320 --> 00:48:28,477 Bye.