1 00:00:00,000 --> 00:00:01,940 ANNOUNCER: The following content is provided under a 2 00:00:01,940 --> 00:00:03,690 Creative Commons license. 3 00:00:03,690 --> 00:00:06,630 Your support will help MIT OpenCourseWare continue to 4 00:00:06,630 --> 00:00:09,990 offer high quality educational resources for free. 5 00:00:09,990 --> 00:00:12,830 To make a donation or to view additional materials from 6 00:00:12,830 --> 00:00:16,760 hundreds of MIT courses, visit MIT OpenCourseWare at 7 00:00:16,760 --> 00:00:18,010 ocw.mit.edu. 8 00:00:18,010 --> 00:00:33,130 9 00:00:33,130 --> 00:00:35,460 HERBERT GROSS: Hi, today we start a new 10 00:00:35,460 --> 00:00:37,510 segment of our course. 11 00:00:37,510 --> 00:00:41,380 It could be entitled 'Techniques of Integration'. 12 00:00:41,380 --> 00:00:45,000 And a subtitle could be the difference between 'knowing 13 00:00:45,000 --> 00:00:47,810 what to do and knowing how to do it'. 14 00:00:47,810 --> 00:00:51,280 You see, essentially in this chapter, it shall be our aim 15 00:00:51,280 --> 00:00:57,790 to develop various recipes to evaluate the inverse 16 00:00:57,790 --> 00:00:59,010 derivative. 17 00:00:59,010 --> 00:01:00,310 So in fact, let's call it that. 18 00:01:00,310 --> 00:01:02,260 Some basic recipes. 19 00:01:02,260 --> 00:01:03,280 The idea is something like this. 20 00:01:03,280 --> 00:01:04,920 Let's start with an easier problem. 21 00:01:04,920 --> 00:01:08,820 Let's suppose we start with a differentiable function 'G'. 22 00:01:08,820 --> 00:01:11,990 Suppose, for example, that 'G' prime is equal to 'f'. 23 00:01:11,990 --> 00:01:15,170 Then in terms of the language of the indefinite integral, or 24 00:01:15,170 --> 00:01:19,020 the inverse derivative, what we have is that the indefinite 25 00:01:19,020 --> 00:01:22,160 integral of 'f of x' with respect to 'x' is then 'G of 26 00:01:22,160 --> 00:01:23,780 x' plus 'c'. 27 00:01:23,780 --> 00:01:26,430 Now, you see, the harder problem is 28 00:01:26,430 --> 00:01:28,160 the inverse of this. 29 00:01:28,160 --> 00:01:32,800 What we're going to be talking about is not do you find 'f' 30 00:01:32,800 --> 00:01:36,430 once 'G' is given, but techniques for finding 'G' 31 00:01:36,430 --> 00:01:38,200 once 'f' is given. 32 00:01:38,200 --> 00:01:42,700 And as we've seen on several occasions already, that given 33 00:01:42,700 --> 00:01:47,110 'f', it is not that easy, in general, to find a function 34 00:01:47,110 --> 00:01:48,860 whose derivative is 'f'. 35 00:01:48,860 --> 00:01:52,620 In fact, in many cases, we can't do it as explicitly as 36 00:01:52,620 --> 00:01:55,450 we would like to no matter how sharp we are. 37 00:01:55,450 --> 00:01:59,280 For example, oh, I've just reviewed this so that you can 38 00:01:59,280 --> 00:02:00,280 think of this in one block. 39 00:02:00,280 --> 00:02:05,160 Given 'f' to require 'G' may not exist in familiar form. 40 00:02:05,160 --> 00:02:07,740 I don't want to pin down what familiar means any more than 41 00:02:07,740 --> 00:02:08,810 that right now. 42 00:02:08,810 --> 00:02:12,610 For example, the illustration that we've used several times 43 00:02:12,610 --> 00:02:15,700 already in the course, suppose 'f of x' is 'e to 44 00:02:15,700 --> 00:02:17,260 the minus 'x squared''. 45 00:02:17,260 --> 00:02:20,050 And we want a function whose derivative is 'e to the minus 46 00:02:20,050 --> 00:02:20,760 'x squared''. 47 00:02:20,760 --> 00:02:24,100 In terms of areas, the second fundamental theorem of 48 00:02:24,100 --> 00:02:27,640 integral calculus, we can construct such a function in a 49 00:02:27,640 --> 00:02:30,840 rather straightforward way, even though it may be 50 00:02:30,840 --> 00:02:34,490 difficult to name it in terms of familiar functions. 51 00:02:34,490 --> 00:02:38,460 Namely, what we do is we take the curve 'y' equals 'e to the 52 00:02:38,460 --> 00:02:41,520 minus 't squared'' in the yt-plane. 53 00:02:41,520 --> 00:02:45,380 Compute the area bounded above by that curve, below by the 54 00:02:45,380 --> 00:02:49,770 t-axis, on the left by the y-axis, and on the right by 55 00:02:49,770 --> 00:02:51,510 the line 't' equals 'x'. 56 00:02:51,510 --> 00:02:55,080 And the area under the curve is a function of 'x'. 57 00:02:55,080 --> 00:02:58,480 And what we've already seen is that in this particular case, 58 00:02:58,480 --> 00:03:02,090 'A prime of x' is precisely 'e to the minus 'x squared''. 59 00:03:02,090 --> 00:03:05,780 In other words, this area function, which is a function 60 00:03:05,780 --> 00:03:08,950 of 'x', does turn out to be the function whose derivative 61 00:03:08,950 --> 00:03:11,920 with respect to 'x' is 'e to the minus 'x squared''. 62 00:03:11,920 --> 00:03:14,580 Now you see, that will not be primarily what we're 63 00:03:14,580 --> 00:03:17,300 interested in, in this particular 64 00:03:17,300 --> 00:03:18,470 phase of our course. 65 00:03:18,470 --> 00:03:22,620 What we're interested in this phase of the course is given 66 00:03:22,620 --> 00:03:27,480 various classifications for the function 'f', to find 67 00:03:27,480 --> 00:03:31,690 recipes that will give us 'G' in familiar form, where 'G' is 68 00:03:31,690 --> 00:03:34,540 the function whose derivative is 'f'. 69 00:03:34,540 --> 00:03:36,760 Summarize for you to look at here. 70 00:03:36,760 --> 00:03:39,860 The objective of this block of material is to find recipes 71 00:03:39,860 --> 00:03:44,210 for finding 'G of x' for various types of 'f of x'. 72 00:03:44,210 --> 00:03:47,110 Now the best way to illustrate what that means is to actually 73 00:03:47,110 --> 00:03:49,330 pick a few examples. 74 00:03:49,330 --> 00:03:52,970 And by the way, what you'll notice in today's lecture is, 75 00:03:52,970 --> 00:03:56,620 roughly speaking, we do nothing at all that's new. 76 00:03:56,620 --> 00:03:59,770 The new stuff will begin with out next lecture essentially. 77 00:03:59,770 --> 00:04:02,970 But for now what we'll really be doing is giving us an 78 00:04:02,970 --> 00:04:06,620 excuse to pull together all of the recipes that we've 79 00:04:06,620 --> 00:04:07,830 developed so far. 80 00:04:07,830 --> 00:04:10,860 All of the recipes either called indefinite integrals or 81 00:04:10,860 --> 00:04:12,020 inverse derivatives. 82 00:04:12,020 --> 00:04:17,440 For example, the first class of functions which we could 83 00:04:17,440 --> 00:04:21,940 handle were things of the form integral ''u to the n' du'. 84 00:04:21,940 --> 00:04:24,930 Recall that very early in our course, we found out that the 85 00:04:24,930 --> 00:04:29,000 integral of ''u to the n' du' was ''u to the 'n + 1'' over 86 00:04:29,000 --> 00:04:33,960 'n plus 1' plus a constant if 'n' was unequal to minus 1. 87 00:04:33,960 --> 00:04:37,860 And then, in our last block of material we learned to handle 88 00:04:37,860 --> 00:04:40,340 the case where 'n' was equal to minus 1. 89 00:04:40,340 --> 00:04:43,330 Namely, when 'n' is minus 1, integral ''u to 90 00:04:43,330 --> 00:04:44,690 the minus 1' du'. 91 00:04:44,690 --> 00:04:49,010 Namely, integral 'du' over 'u' is natural log absolute value 92 00:04:49,010 --> 00:04:51,490 of 'u' plus a constant. 93 00:04:51,490 --> 00:04:54,200 Now this then, is one of our basic building blocks. 94 00:04:54,200 --> 00:04:57,300 What we're saying is, if an indefinite integral can be 95 00:04:57,300 --> 00:05:00,220 written in this form, we already have a recipe that 96 00:05:00,220 --> 00:05:03,020 allows us to evaluate the integral. 97 00:05:03,020 --> 00:05:06,560 By the way, let us also point out as was mentioned again, 98 00:05:06,560 --> 00:05:10,180 several times in the past, that 'u' is the generic name 99 00:05:10,180 --> 00:05:11,700 for a variable over here. 100 00:05:11,700 --> 00:05:14,930 It is not important in this recipe that we use 'u'. 101 00:05:14,930 --> 00:05:18,050 What is important is that whatever is being raised to 102 00:05:18,050 --> 00:05:21,530 the n-th power inside the integrand must be precisely 103 00:05:21,530 --> 00:05:23,990 the thing that's following the 'd' over here. 104 00:05:23,990 --> 00:05:27,530 For example, instead of 'u', suppose I used 105 00:05:27,530 --> 00:05:28,960 'sine x' over here. 106 00:05:28,960 --> 00:05:30,900 In other words, suppose I had 'sine x' to the n-th 107 00:05:30,900 --> 00:05:33,810 power 'd sine x'. 108 00:05:33,810 --> 00:05:37,210 See, what's being raised to the n-th power is the same 109 00:05:37,210 --> 00:05:39,770 thing as what's following the 'd'. 110 00:05:39,770 --> 00:05:40,770 That's all I really care about. 111 00:05:40,770 --> 00:05:44,850 In other words, if this had been the integrand, the recipe 112 00:05:44,850 --> 00:05:48,750 would have said, this is 1 over 'n + 1' times 'sine x' to 113 00:05:48,750 --> 00:05:50,170 the 'n + 1' power. 114 00:05:50,170 --> 00:05:53,170 Plus a constant if 'n' is not equal to minus 1. 115 00:05:53,170 --> 00:05:56,620 And it's natural log absolute value of 'sine x' plus a 116 00:05:56,620 --> 00:06:00,040 constant if 'n' is equal to minus 1. 117 00:06:00,040 --> 00:06:03,180 Again, the important thing is not the 'u', but the fact that 118 00:06:03,180 --> 00:06:05,610 what's being raised to the n-th power is the same as 119 00:06:05,610 --> 00:06:07,220 what's following the 'd' over here. 120 00:06:07,220 --> 00:06:11,690 By the way, we can write this in more familiar form. 121 00:06:11,690 --> 00:06:16,150 You see, namely, notice that the differential of 'sine x' 122 00:06:16,150 --> 00:06:17,820 is 'cosine x dx'. 123 00:06:17,820 --> 00:06:22,370 And consequently, in this form what our recipe says is that 124 00:06:22,370 --> 00:06:27,010 integral 'sine x' to the n-th power times 'cosine x dx' is 125 00:06:27,010 --> 00:06:28,210 given by this. 126 00:06:28,210 --> 00:06:31,950 In other words, it's sine to the 'n + 1' power 'x over 'n + 127 00:06:31,950 --> 00:06:35,870 1'' plus a constant if 'n' is unequal to minus 1. 128 00:06:35,870 --> 00:06:39,510 And it's the natural log absolute value of 'sine x' 129 00:06:39,510 --> 00:06:43,520 plus 'c' if 'n' is equal to minus 1. 130 00:06:43,520 --> 00:06:53,030 Again, notice this looks more complicated than this. 131 00:06:53,030 --> 00:06:58,390 But to get our problem, to get into the form ''u to the n' 132 00:06:58,390 --> 00:07:01,930 du', notice that the natural substitution here is to let 133 00:07:01,930 --> 00:07:03,780 'u' equal 'sine x'. 134 00:07:03,780 --> 00:07:07,030 In which case, 'du' would be 'cosine x dx'. 135 00:07:07,030 --> 00:07:10,100 In other words, the extra factor of 'cosine x' here 136 00:07:10,100 --> 00:07:13,250 actually is to our advantage over here. 137 00:07:13,250 --> 00:07:15,210 That when we make the substitution, it gives us the 138 00:07:15,210 --> 00:07:16,630 form that we want. 139 00:07:16,630 --> 00:07:19,290 With the factor missing, we're in a little bit of trouble. 140 00:07:19,290 --> 00:07:23,150 By the way, this is what makes this a rather exciting topic. 141 00:07:23,150 --> 00:07:27,820 That there seems to be no really canned ways of finding 142 00:07:27,820 --> 00:07:29,160 indefinite integrals. 143 00:07:29,160 --> 00:07:30,610 That we have recipes. 144 00:07:30,610 --> 00:07:34,180 Frequently, the recipe that we have doesn't cover the 145 00:07:34,180 --> 00:07:35,910 situation that we have. 146 00:07:35,910 --> 00:07:38,630 In fact, I know a lot of times in teaching a course like this 147 00:07:38,630 --> 00:07:41,280 students say, why do we have to learn these recipes and 148 00:07:41,280 --> 00:07:41,820 techniques? 149 00:07:41,820 --> 00:07:44,030 Why can't we just look them up in the tables? 150 00:07:44,030 --> 00:07:48,070 And the answer quite simply, is that in many applications, 151 00:07:48,070 --> 00:07:50,790 the integral that we have doesn't appear in 152 00:07:50,790 --> 00:07:52,110 that form in the table. 153 00:07:52,110 --> 00:07:55,490 That frequently what we must do is a tremendous amount of 154 00:07:55,490 --> 00:07:58,700 algebraic manipulation and the like to even get the given 155 00:07:58,700 --> 00:08:01,970 integral to resemble one that we can find in the table. 156 00:08:01,970 --> 00:08:04,020 Oh, we're starting off on a fairly 157 00:08:04,020 --> 00:08:05,440 straightforward level here. 158 00:08:05,440 --> 00:08:06,710 Nothing really too tricky. 159 00:08:06,710 --> 00:08:09,870 But at least enough so that we get an idea of what kind of 160 00:08:09,870 --> 00:08:11,120 gimmicks are involved. 161 00:08:11,120 --> 00:08:15,080 For example, suppose we had 'cosine x' to the seventh 162 00:08:15,080 --> 00:08:17,030 power 'dx'. 163 00:08:17,030 --> 00:08:19,990 You see the idea here is that when you have cosine raised to 164 00:08:19,990 --> 00:08:23,590 a power, it would have been nice if a 'sine x' factor had 165 00:08:23,590 --> 00:08:24,380 been in here. 166 00:08:24,380 --> 00:08:26,860 Because the differential of 'cosine x' is 167 00:08:26,860 --> 00:08:28,730 minus 'sine x dx'. 168 00:08:28,730 --> 00:08:32,120 And what I'm driving at here is that we often have to have 169 00:08:32,120 --> 00:08:36,440 a certain kind of ingenuity that allows us to reduce a new 170 00:08:36,440 --> 00:08:39,299 integrand to a more familiar one. 171 00:08:39,299 --> 00:08:40,480 And, as I say, there are many 172 00:08:40,480 --> 00:08:42,159 techniques done in the textbook. 173 00:08:42,159 --> 00:08:45,060 There'll be plenty of exercises again, that will 174 00:08:45,060 --> 00:08:46,570 allow you to experiment on this. 175 00:08:46,570 --> 00:08:50,050 All I want to do is hit a few highlights and create the mood 176 00:08:50,050 --> 00:08:51,360 as to what's going on. 177 00:08:51,360 --> 00:08:54,440 For example, you see frequently what happens is, if 178 00:08:54,440 --> 00:08:57,690 we have 'cosine x' raised to an odd power, a very 179 00:08:57,690 --> 00:09:01,520 convenient way of handling this is to split off one of 180 00:09:01,520 --> 00:09:03,870 the factors of 'cosine x' separately. 181 00:09:03,870 --> 00:09:06,780 In other words, write this as 'cosine x' to the sixth power 182 00:09:06,780 --> 00:09:08,700 times 'cosine x dx'. 183 00:09:08,700 --> 00:09:13,180 The idea being that cosine squared can be written as '1 184 00:09:13,180 --> 00:09:14,800 minus 'sine squared''. 185 00:09:14,800 --> 00:09:18,620 In other words, 'cosine x' to the sixth power is really 186 00:09:18,620 --> 00:09:21,150 'cosine squared x' cubed. 187 00:09:21,150 --> 00:09:24,840 But 'cosine squared x' is '1 minus 'sine squared x''. 188 00:09:24,840 --> 00:09:28,140 In other words, notice again the role of the trigonometric 189 00:09:28,140 --> 00:09:29,030 identities. 190 00:09:29,030 --> 00:09:34,160 I take this integrand, write it this way, and now you see 191 00:09:34,160 --> 00:09:38,040 if I use the binomial theorem and expand this, notice that 192 00:09:38,040 --> 00:09:42,160 my typical term will have 'sine x' to a power multiplied 193 00:09:42,160 --> 00:09:44,070 by a 'cosine x'. 194 00:09:44,070 --> 00:09:47,540 And that's precisely the kind of a term that I handled in 195 00:09:47,540 --> 00:09:48,980 the previous case. 196 00:09:48,980 --> 00:09:50,500 And again, I'm not going to bore you 197 00:09:50,500 --> 00:09:51,650 with the details here. 198 00:09:51,650 --> 00:09:53,670 That is the easiest part of the problem. 199 00:09:53,670 --> 00:09:56,850 The hard part is getting to understand why one would want 200 00:09:56,850 --> 00:09:59,750 to use this particular kind of form. 201 00:09:59,750 --> 00:10:02,670 Another variation is, what happens if the cosine is 202 00:10:02,670 --> 00:10:05,790 raised to an even power rather than to an odd power? 203 00:10:05,790 --> 00:10:07,500 How do we handle that? 204 00:10:07,500 --> 00:10:11,010 Again, the same kind of use of trigonometric identities. 205 00:10:11,010 --> 00:10:15,370 For example, it would not be to our advantage in this case, 206 00:10:15,370 --> 00:10:19,040 to replace cosine squared by 1 minus sine squared. 207 00:10:19,040 --> 00:10:21,960 Because then we would wind up with the same kind of 208 00:10:21,960 --> 00:10:25,320 integrands that we started with, only using powers of 209 00:10:25,320 --> 00:10:28,400 sine instead of powers of cosine. 210 00:10:28,400 --> 00:10:31,830 You see, the idea is that if we have a power of sine, we 211 00:10:31,830 --> 00:10:34,080 want a cosine factor to accompany it. 212 00:10:34,080 --> 00:10:37,120 And inversely, if we have a power of cosine, we'd like a 213 00:10:37,120 --> 00:10:39,210 factor of sine to accompany it. 214 00:10:39,210 --> 00:10:39,900 Why? 215 00:10:39,900 --> 00:10:42,355 So that we can get the integral into the form ''u to 216 00:10:42,355 --> 00:10:43,260 the n' du'. 217 00:10:43,260 --> 00:10:46,670 In this particular case, a very common device is to 218 00:10:46,670 --> 00:10:50,910 invoke the identity that cosine squared of an angle is 219 00:10:50,910 --> 00:10:54,350 1 plus cosine twice the angle over 2. 220 00:10:54,350 --> 00:10:57,600 You see again, this is where all of these particular 221 00:10:57,600 --> 00:10:59,430 identities come into play. 222 00:10:59,430 --> 00:11:02,370 It's not a case of saying, let's learn these identities 223 00:11:02,370 --> 00:11:03,910 so that we can recite them. 224 00:11:03,910 --> 00:11:07,750 It's a case of getting into a situation where you want a 225 00:11:07,750 --> 00:11:11,270 certain answer, can't handle it conveniently in the form 226 00:11:11,270 --> 00:11:14,940 that you're in, and hope that you can pick off a synonym 227 00:11:14,940 --> 00:11:18,240 that allows you to tackle the problem more successfully. 228 00:11:18,240 --> 00:11:20,890 That's where this becomes a matter of insight, lucky 229 00:11:20,890 --> 00:11:24,260 guessing, skill, whatever you want to call it. 230 00:11:24,260 --> 00:11:27,370 It's one thing to transform an integral into an equivalent 231 00:11:27,370 --> 00:11:30,920 one, and another thing to have that equivalent integral be 232 00:11:30,920 --> 00:11:32,370 something that you can handle. 233 00:11:32,370 --> 00:11:35,790 Well, at any rate, for example in a problem of this type, a 234 00:11:35,790 --> 00:11:40,730 tendency is to write cosine sixth theta as 235 00:11:40,730 --> 00:11:42,710 cosine squared cubed. 236 00:11:42,710 --> 00:11:46,800 Cosine squared theta is 1 plus cosine 2 theta over 2. 237 00:11:46,800 --> 00:11:50,160 To cube this we can take out the factor of 1/8. 238 00:11:50,160 --> 00:11:53,750 You see, we use the binomial theorem to expand 1 plus 239 00:11:53,750 --> 00:11:55,500 cosine 2 theta cubed. 240 00:11:55,500 --> 00:11:57,750 And notice that of the terms that we get, 241 00:11:57,750 --> 00:11:59,680 this term we can handle. 242 00:11:59,680 --> 00:12:01,640 This term we can handle, it's just cosine 243 00:12:01,640 --> 00:12:03,000 to the first power. 244 00:12:03,000 --> 00:12:05,480 Which involves a sine term when we integrate. 245 00:12:05,480 --> 00:12:07,240 And here's a cosine cubed. 246 00:12:07,240 --> 00:12:10,110 But we've just seen how you can handle something to an odd 247 00:12:10,110 --> 00:12:13,180 power, so that reduces this term to a 248 00:12:13,180 --> 00:12:14,690 type that we can handle. 249 00:12:14,690 --> 00:12:15,630 And then we see what? 250 00:12:15,630 --> 00:12:17,570 That we have another term in here, which is 251 00:12:17,570 --> 00:12:19,350 cosine squared 2 theta. 252 00:12:19,350 --> 00:12:21,280 That's an even power of cosine. 253 00:12:21,280 --> 00:12:22,290 Again, we do what? 254 00:12:22,290 --> 00:12:27,070 We write cosine squared 2 theta as 1 plus cosine of 255 00:12:27,070 --> 00:12:28,260 twice this angle. 256 00:12:28,260 --> 00:12:31,930 That's 1 plus cosine 4 theta over 2. 257 00:12:31,930 --> 00:12:34,510 In other words, we keep reducing these things, always 258 00:12:34,510 --> 00:12:38,850 hoping to break them down into collections of problems that 259 00:12:38,850 --> 00:12:40,230 we solved before. 260 00:12:40,230 --> 00:12:43,900 As I said, these are things that we'll drill on in the 261 00:12:43,900 --> 00:12:47,410 exercises and I think it will become more meaningful there. 262 00:12:47,410 --> 00:12:50,400 Here, as I say, the main aim is to create the mood that 263 00:12:50,400 --> 00:12:54,470 what we're trying to do is to reduce unfamiliar problems to 264 00:12:54,470 --> 00:12:56,750 equivalent familiar ones to which we 265 00:12:56,750 --> 00:12:58,520 already know the answer. 266 00:12:58,520 --> 00:13:02,520 Now again, in the way of review, sometimes an integrand 267 00:13:02,520 --> 00:13:06,190 occurs in the form of the sum or difference of two squares. 268 00:13:06,190 --> 00:13:09,610 We've already handled this in terms of the inverse circular 269 00:13:09,610 --> 00:13:12,830 functions and the inverse hyperbolic functions. 270 00:13:12,830 --> 00:13:16,340 It's covered for the first time in our textbook at this 271 00:13:16,340 --> 00:13:18,040 particular stage of the course. 272 00:13:18,040 --> 00:13:21,100 And I think it's a tough enough concept so that it's 273 00:13:21,100 --> 00:13:22,950 worth emphasizing at this stage. 274 00:13:22,950 --> 00:13:25,280 So I just separate this particular type out. 275 00:13:25,280 --> 00:13:27,790 How do we handle sums and differences of squares? 276 00:13:27,790 --> 00:13:30,690 And what I intend to show is, is that by a suitable 277 00:13:30,690 --> 00:13:33,980 trigonometric function, either circular trigonometric or 278 00:13:33,980 --> 00:13:36,870 hyperbolic trigonometric, we can always reduce that kind of 279 00:13:36,870 --> 00:13:39,550 a problem to the type that we can handle. 280 00:13:39,550 --> 00:13:43,180 Again, to keep computation at a minimum, I have not tried to 281 00:13:43,180 --> 00:13:45,610 pick a rather complicated situation. 282 00:13:45,610 --> 00:13:49,060 In fact, I've picked ones where you can very easily look 283 00:13:49,060 --> 00:13:50,780 them up in the table if that was the 284 00:13:50,780 --> 00:13:52,370 main aim of the exercise. 285 00:13:52,370 --> 00:13:54,510 You see here, it's not so much that I want to get this 286 00:13:54,510 --> 00:13:58,810 answer, as much as it is that I want to emphasize the 287 00:13:58,810 --> 00:14:00,670 technique that one uses. 288 00:14:00,670 --> 00:14:01,950 The idea is something like this. 289 00:14:01,950 --> 00:14:03,630 Here we see the difference of two squares. 290 00:14:03,630 --> 00:14:06,160 That's the square root of a squared minus x squared. 291 00:14:06,160 --> 00:14:10,580 What this should do for us is to suggest a right triangle 292 00:14:10,580 --> 00:14:14,250 whose hypotenuse is a and one of whose sides is 'x'. 293 00:14:14,250 --> 00:14:16,920 That's the diagram I've drawn over here. 294 00:14:16,920 --> 00:14:18,570 This is my reference triangle. 295 00:14:18,570 --> 00:14:20,830 As I say, we've done this before, but I would like to 296 00:14:20,830 --> 00:14:23,030 reinforce this at this particular stage. 297 00:14:23,030 --> 00:14:26,460 Especially now that we're more familiar with both the inverse 298 00:14:26,460 --> 00:14:29,860 circular and inverse hyperbolic functions. 299 00:14:29,860 --> 00:14:32,470 At any rate, we use this reference triangle. 300 00:14:32,470 --> 00:14:35,600 From this triangle, the easiest relationship to pick 301 00:14:35,600 --> 00:14:40,890 off involving 'x' is that sine theta is 'x over a', or 'a 302 00:14:40,890 --> 00:14:42,710 'sine theta'' equals 'x'. 303 00:14:42,710 --> 00:14:45,080 From which it follows that ''a' cosine theta 304 00:14:45,080 --> 00:14:47,630 'd theta'' is 'dx'. 305 00:14:47,630 --> 00:14:49,740 The square root of ''a squared' minus 'x squared'' 306 00:14:49,740 --> 00:14:51,670 over 'a' is cosine theta. 307 00:14:51,670 --> 00:14:53,930 Therefore, the square root of ''a squared' minus 'x 308 00:14:53,930 --> 00:14:56,020 squared'' is 'a cosine theta'. 309 00:14:56,020 --> 00:14:58,330 If we now make this substitution in this 310 00:14:58,330 --> 00:15:02,580 particular integral, we wind up with integral 'dx' over the 311 00:15:02,580 --> 00:15:05,830 square root of ''a squared' minus 'x squared'' is equal to 312 00:15:05,830 --> 00:15:07,880 integral 'd theta'. 313 00:15:07,880 --> 00:15:09,930 Which is theta plus 'c'. 314 00:15:09,930 --> 00:15:15,190 And since theta is the number whose sine is 'x over a', we 315 00:15:15,190 --> 00:15:19,250 have that the answer to this problem is inverse 'sine 'x 316 00:15:19,250 --> 00:15:21,450 over a'' plus a constant. 317 00:15:21,450 --> 00:15:24,370 Again, just a straightforward review, but a 318 00:15:24,370 --> 00:15:25,820 rather important technique. 319 00:15:25,820 --> 00:15:29,180 In our textbook, this is put in a separate section, and the 320 00:15:29,180 --> 00:15:32,470 technique is called trigonometric substitution. 321 00:15:32,470 --> 00:15:34,150 And I just wanted to show you what motivates the 322 00:15:34,150 --> 00:15:35,620 trigonometric substitution. 323 00:15:35,620 --> 00:15:38,930 The harder part as I mentioned in our previous lecture, is 324 00:15:38,930 --> 00:15:42,130 how you motivate trigonometric substitutions when it's the 325 00:15:42,130 --> 00:15:44,200 hyperbolic functions that are involved. 326 00:15:44,200 --> 00:15:46,770 You see, what I'd like to do now is instead of dealing with 327 00:15:46,770 --> 00:15:49,110 the square root of ''a squared' minus 'x squared'', 328 00:15:49,110 --> 00:15:51,300 let me just reverse the order of the terms. 329 00:15:51,300 --> 00:15:55,380 And now let me take integral 'dx' over the 'square root of 330 00:15:55,380 --> 00:15:57,380 ''x squared' minus 'a squared'''. 331 00:15:57,380 --> 00:15:59,690 Notice that the difference between ''x squared' minus 'a 332 00:15:59,690 --> 00:16:02,910 squared'' and ''a squared' minus 'x squared'' is just a 333 00:16:02,910 --> 00:16:04,690 factor of minus 1. 334 00:16:04,690 --> 00:16:08,060 And the fact that that minus 1 is under the square root sign, 335 00:16:08,060 --> 00:16:10,790 really means in a certain manner of speaking that you've 336 00:16:10,790 --> 00:16:15,970 multiplied by i, the square root of minus 1, to transform 337 00:16:15,970 --> 00:16:18,610 the integral that we just had into this 338 00:16:18,610 --> 00:16:19,960 particular form here. 339 00:16:19,960 --> 00:16:23,520 And in terms of our lecture on the hyperbolic functions that 340 00:16:23,520 --> 00:16:27,400 should, in some way, give us a preview of things to come in 341 00:16:27,400 --> 00:16:30,380 the sense that the hyperbolic functions are very strongly 342 00:16:30,380 --> 00:16:32,550 related to the circular functions. 343 00:16:32,550 --> 00:16:34,370 My technique for doing this-- 344 00:16:34,370 --> 00:16:36,630 and again, this is highly subjective. 345 00:16:36,630 --> 00:16:37,630 It's very simple. 346 00:16:37,630 --> 00:16:40,310 I learnt the circular functions long before I learnt 347 00:16:40,310 --> 00:16:41,790 the hyperbolic functions. 348 00:16:41,790 --> 00:16:44,450 Consequently, I feel very much at home with 349 00:16:44,450 --> 00:16:46,350 the circular functions. 350 00:16:46,350 --> 00:16:48,820 As a result, whenever I see the sum or difference of two 351 00:16:48,820 --> 00:16:51,550 squares, I always think in terms of a triangle. 352 00:16:51,550 --> 00:16:54,310 I make a circular trigonometric substitution. 353 00:16:54,310 --> 00:16:56,260 If that works for me, fine. 354 00:16:56,260 --> 00:16:58,240 I haven't lost anything. 355 00:16:58,240 --> 00:17:02,040 If it doesn't work for me, it gives me the hint that I can't 356 00:17:02,040 --> 00:17:05,950 seem to get in my own mind without this hint as to which 357 00:17:05,950 --> 00:17:07,900 hyperbolic function I should use. 358 00:17:07,900 --> 00:17:10,950 Let me show you this in terms of this particular example. 359 00:17:10,950 --> 00:17:14,290 If I saw this problem from scratch and didn't have any 360 00:17:14,290 --> 00:17:16,910 previous knowledge of how to handle this I'd say, this 361 00:17:16,910 --> 00:17:19,319 seems like I should think of a right triangle whose 362 00:17:19,319 --> 00:17:22,760 hypotenuse is 'x' and one of whose sides is 'a'. 363 00:17:22,760 --> 00:17:25,319 So I think of this particular reference triangle. 364 00:17:25,319 --> 00:17:28,010 Now, thinking of this triangle, and in fact, even 365 00:17:28,010 --> 00:17:31,650 look at this triangle, I very easily pick off that 'x' is 366 00:17:31,650 --> 00:17:34,550 equal to 'a cosecant theta'. 367 00:17:34,550 --> 00:17:38,210 See, 'x over a' is 1 over sine theta. 368 00:17:38,210 --> 00:17:40,240 That's cosecant theta. 369 00:17:40,240 --> 00:17:43,610 Therefore and remembering how to differentiate the cosecant, 370 00:17:43,610 --> 00:17:46,310 I take the differential of both sides and I find that 371 00:17:46,310 --> 00:17:49,840 minus 'a cosecant theta cotangent theta 372 00:17:49,840 --> 00:17:52,000 'd theta'' is 'dx'. 373 00:17:52,000 --> 00:17:55,530 Again, from this diagram I see that the square root of ''x 374 00:17:55,530 --> 00:17:58,920 squared' minus 'a squared'' is-- 375 00:17:58,920 --> 00:18:00,080 well, let's see. 376 00:18:00,080 --> 00:18:02,020 The square root of ''x squared' minus 'a squared'' 377 00:18:02,020 --> 00:18:04,450 over 'a' would be cotangent theta. 378 00:18:04,450 --> 00:18:06,710 So the square root of ''x squared' minus 'a squared'' 379 00:18:06,710 --> 00:18:09,720 itself is 'a cotangent theta'. 380 00:18:09,720 --> 00:18:13,090 Making the substitution in the integral 'dx' over the square 381 00:18:13,090 --> 00:18:16,270 root of ''x squared' minus 'a squared'', notice that the 382 00:18:16,270 --> 00:18:20,310 'a's cancel, the cotangent's cancel, and I'm left with 383 00:18:20,310 --> 00:18:22,940 'cosecant theta 'd theta'' and a minus 384 00:18:22,940 --> 00:18:24,690 sign inside my integral. 385 00:18:24,690 --> 00:18:27,830 In other words, somehow or other, this circular 386 00:18:27,830 --> 00:18:31,420 trigonometric function replaces the integral 'dx' 387 00:18:31,420 --> 00:18:34,230 over the 'square root of ''x squared' minus 'a squared''' 388 00:18:34,230 --> 00:18:37,510 by integral minus 'cosecant theta 'd theta''. 389 00:18:37,510 --> 00:18:39,710 And the point I'd like to emphasize at this particular 390 00:18:39,710 --> 00:18:42,470 stage is that we don't say that the 391 00:18:42,470 --> 00:18:44,000 substitution has failed. 392 00:18:44,000 --> 00:18:45,580 The substitution has been made. 393 00:18:45,580 --> 00:18:48,390 We have replaced an integral involving 'x' by 394 00:18:48,390 --> 00:18:50,030 one involving theta. 395 00:18:50,030 --> 00:18:52,450 It's just that in the same way that it's difficult to 396 00:18:52,450 --> 00:18:54,125 integrate secant theta-- 397 00:18:54,125 --> 00:18:57,380 in fact, that will be later in this block of 398 00:18:57,380 --> 00:18:58,320 material we'll do that. 399 00:18:58,320 --> 00:19:01,060 But the point is that secant and cosecant are rather 400 00:19:01,060 --> 00:19:02,780 difficult functions to integrate. 401 00:19:02,780 --> 00:19:05,850 In fact, at this stage of our development, we do not know a 402 00:19:05,850 --> 00:19:09,870 function whose derivative is cosecant theta. 403 00:19:09,870 --> 00:19:12,630 Consequently, we don't know one whose derivative is minus 404 00:19:12,630 --> 00:19:13,830 cosecant theta. 405 00:19:13,830 --> 00:19:16,250 We seemed to have arrived at an impasse here. 406 00:19:16,250 --> 00:19:18,890 We have successfully made the substitution, but the key 407 00:19:18,890 --> 00:19:23,140 point is that the new integral is no easier for us to handle 408 00:19:23,140 --> 00:19:24,610 than the old integral. 409 00:19:24,610 --> 00:19:28,780 Now, here's how I visualize the hyperbolic substitutions. 410 00:19:28,780 --> 00:19:30,330 We did this in the last lecture. 411 00:19:30,330 --> 00:19:32,340 I think it's tough enough, so I'd like to do it a second 412 00:19:32,340 --> 00:19:35,760 time and give you a chance of seeing how easy this is once 413 00:19:35,760 --> 00:19:37,800 you see the structural form. 414 00:19:37,800 --> 00:19:41,600 Notice that the substitution I made here was 'x' equals 'a 415 00:19:41,600 --> 00:19:42,890 cosecant theta'. 416 00:19:42,890 --> 00:19:45,600 The cosecant was the key step here. 417 00:19:45,600 --> 00:19:48,750 The reference triangle, knowing how to draw this 418 00:19:48,750 --> 00:19:51,930 geometrically, allowed me not to consciously have to think 419 00:19:51,930 --> 00:19:53,670 of the trigonometric identities. 420 00:19:53,670 --> 00:19:56,850 But what I really did in evaluating this problem, if I 421 00:19:56,850 --> 00:19:59,870 left the diagram out and wanted to do it analytically, 422 00:19:59,870 --> 00:20:03,370 the identity that I made use of was the one that said 423 00:20:03,370 --> 00:20:07,390 cosecant squared theta minus cotangent squared theta is 424 00:20:07,390 --> 00:20:08,620 identically 1. 425 00:20:08,620 --> 00:20:11,590 In other words, I would like an identity for the hyperbolic 426 00:20:11,590 --> 00:20:14,660 functions that has the structure that the difference 427 00:20:14,660 --> 00:20:17,230 of two squares is identically 1. 428 00:20:17,230 --> 00:20:20,480 The easiest one I can think of is the one that says cosh 429 00:20:20,480 --> 00:20:23,690 squared theta minus sinh squared theta is 1. 430 00:20:23,690 --> 00:20:26,680 You see, structurally, these two are equivalent. 431 00:20:26,680 --> 00:20:30,360 In other words, sinh and cosh are related hyperbolically, 432 00:20:30,360 --> 00:20:34,310 the same way that cotangent and cosecant are in terms of 433 00:20:34,310 --> 00:20:35,650 circular functions. 434 00:20:35,650 --> 00:20:37,210 The identification is this. 435 00:20:37,210 --> 00:20:41,460 Since in the original problem I tried 'x' equals 'a cosecant 436 00:20:41,460 --> 00:20:45,300 theta', and since the identification here is that 437 00:20:45,300 --> 00:20:50,260 cosecant and cosh are matched up, what I try next-- see once 438 00:20:50,260 --> 00:20:53,610 this has failed, I very quickly now say, OK, what I'll 439 00:20:53,610 --> 00:20:57,590 do is instead of 'x' equals 'a cosecant theta', I'll try 'x' 440 00:20:57,590 --> 00:21:01,160 equals 'a cosh theta'. 441 00:21:01,160 --> 00:21:04,060 That's exactly what I did over here. 442 00:21:04,060 --> 00:21:06,110 'a cosh theta' equals 'x'. 443 00:21:06,110 --> 00:21:09,130 From which it follows that 'dx' is 'a 444 00:21:09,130 --> 00:21:11,090 sinh theta 'd theta''. 445 00:21:11,090 --> 00:21:14,570 The square root of ''x squared' minus 'a squared'' is 446 00:21:14,570 --> 00:21:16,720 just ''a squared' cosh squared theta'. 447 00:21:16,720 --> 00:21:17,850 That's 'x squared'. 448 00:21:17,850 --> 00:21:19,220 Minus 'a squared'. 449 00:21:19,220 --> 00:21:22,930 That can be written as the square root of 'a squared' 450 00:21:22,930 --> 00:21:27,040 times the quantity 'cosh squared theta minus 1'. 451 00:21:27,040 --> 00:21:29,910 Now, the interesting point is the 'cosh squared theta minus 452 00:21:29,910 --> 00:21:33,190 1' turns out to be sinh squared theta. 453 00:21:33,190 --> 00:21:35,140 In fact, if we want to just look back here for a moment, 454 00:21:35,140 --> 00:21:36,890 recall that's exactly what we have over here. 455 00:21:36,890 --> 00:21:39,880 That cosh squared theta is 1. 456 00:21:39,880 --> 00:21:44,680 Well, cosh squared theta is 1 plus sinh squared theta. 457 00:21:44,680 --> 00:21:48,680 cosh squared theta minus 1 is sinh squared theta. 458 00:21:48,680 --> 00:21:50,660 And you might, say wasn't that terrific? 459 00:21:50,660 --> 00:21:55,160 What a lucky break it was that this complicated expression 460 00:21:55,160 --> 00:21:57,490 just happened to be sinh squared theta. 461 00:21:57,490 --> 00:22:01,320 And now taking the square root I get a sinh theta. 462 00:22:01,320 --> 00:22:03,850 The thing I'd like to point out is that the element of 463 00:22:03,850 --> 00:22:08,750 luck was removed from this at the instant that I recognized 464 00:22:08,750 --> 00:22:12,580 the fact that cosh and sinh were related by the same 465 00:22:12,580 --> 00:22:15,510 identity that cosecant and cotangent were. 466 00:22:15,510 --> 00:22:19,110 In other words, I rigged this thing so that when I finally 467 00:22:19,110 --> 00:22:22,220 had to take this particular difference, I had to wind up 468 00:22:22,220 --> 00:22:24,750 with an easy square root to extract. 469 00:22:24,750 --> 00:22:26,870 You see, I had the right structure to begin with. 470 00:22:26,870 --> 00:22:30,140 At any rate, to make a long story short here, 'dx' is 'a 471 00:22:30,140 --> 00:22:32,190 sinh theta 'd theta''. 472 00:22:32,190 --> 00:22:35,200 The square root of ''x squared' minus 'a squared'' is 473 00:22:35,200 --> 00:22:36,780 'a sinh theta'. 474 00:22:36,780 --> 00:22:39,640 Therefore, 'dx' over the square root of ''x squared' 475 00:22:39,640 --> 00:22:40,580 minus 'a squared''. 476 00:22:40,580 --> 00:22:42,450 Well, look how nicely this works out. 477 00:22:42,450 --> 00:22:45,950 The a sinh theta cancels from both the numerator and the 478 00:22:45,950 --> 00:22:46,960 denominator. 479 00:22:46,960 --> 00:22:49,690 And it turns out that the integral that I'm looking for, 480 00:22:49,690 --> 00:22:52,770 'dx' over the square root of ''x squared' minus 'a 481 00:22:52,770 --> 00:22:55,730 squared'' is just theta plus a constant. 482 00:22:55,730 --> 00:22:59,570 But now, recalling that a cosh theta equals 'x'. 483 00:22:59,570 --> 00:23:01,970 In other words, cosh theta is 'x over a'. 484 00:23:01,970 --> 00:23:06,100 And therefore, theta is 'inverse cosh 'x over a'', I 485 00:23:06,100 --> 00:23:10,970 arrive at the result that this integral is 'inverse cosh 'x 486 00:23:10,970 --> 00:23:12,850 over a'' plus a constant. 487 00:23:12,850 --> 00:23:14,920 In other words, what I hope that these two little examples 488 00:23:14,920 --> 00:23:19,520 show is that when I have the sum or difference of two 489 00:23:19,520 --> 00:23:24,200 squares, the trigonometric substitutions will usually 490 00:23:24,200 --> 00:23:27,700 give me a crack at getting an equivalent integral that will 491 00:23:27,700 --> 00:23:29,410 be easier to handle. 492 00:23:29,410 --> 00:23:32,120 What may happen is that the circular functions won't help 493 00:23:32,120 --> 00:23:34,120 me, but the hyperbolic ones will. 494 00:23:34,120 --> 00:23:37,330 Or vice versa, and I hope that you see how these are so 495 00:23:37,330 --> 00:23:39,180 delicately connected. 496 00:23:39,180 --> 00:23:43,260 Well, there is one final type of technique that I would like 497 00:23:43,260 --> 00:23:46,190 to show for today's lesson that's related to what we've 498 00:23:46,190 --> 00:23:47,180 already done. 499 00:23:47,180 --> 00:23:50,080 And with that, we'll conclude today's lesson and go into 500 00:23:50,080 --> 00:23:52,910 some more elaborate recipes next time. 501 00:23:52,910 --> 00:23:58,020 But this is a take-off on our old high school construction 502 00:23:58,020 --> 00:23:59,620 of completing a square. 503 00:23:59,620 --> 00:24:03,240 In other words, let's suppose we have an expression of the 504 00:24:03,240 --> 00:24:06,550 form 'ax squared' plus 'bx' plus 'c'. 505 00:24:06,550 --> 00:24:08,540 And this may sound old hat to you. 506 00:24:08,540 --> 00:24:11,430 You may remember this as looking pretty much like the 507 00:24:11,430 --> 00:24:14,650 development of the quadratic equation. 508 00:24:14,650 --> 00:24:16,200 The idea works something like this. 509 00:24:16,200 --> 00:24:17,970 First of all, I can factor an a out of here. 510 00:24:17,970 --> 00:24:19,600 I'm assuming that a is not 0. 511 00:24:19,600 --> 00:24:21,830 In fact, if a were 0, I really wouldn't have a 512 00:24:21,830 --> 00:24:23,150 square term in here. 513 00:24:23,150 --> 00:24:26,470 I factor out an 'a' and I'm left with 'a' times 'x 514 00:24:26,470 --> 00:24:30,730 squared' plus ''b over a'x' plus 'c over a'. 515 00:24:30,730 --> 00:24:34,050 Now it turns out that whenever you have something of the form 516 00:24:34,050 --> 00:24:37,610 'x squared' plus something times 'x', to convert that 517 00:24:37,610 --> 00:24:38,900 into a perfect square. 518 00:24:38,900 --> 00:24:41,970 And again, we'll have ample opportunity to practice this 519 00:24:41,970 --> 00:24:43,350 in the exercises. 520 00:24:43,350 --> 00:24:46,960 You take half the coefficient of 'x' and square it. 521 00:24:46,960 --> 00:24:50,010 Half the coefficient of 'x' here is 'b over 2a'. 522 00:24:50,010 --> 00:24:53,910 If I square that it's 'b squared' over ''4a' squared'. 523 00:24:53,910 --> 00:24:57,060 Now to keep the identity intact here, if I add in 'b 524 00:24:57,060 --> 00:25:00,460 squared' over ''4a' squared', I must subtract that. 525 00:25:00,460 --> 00:25:03,440 In other words, I just add and subtract this term that will 526 00:25:03,440 --> 00:25:05,190 make this a perfect square. 527 00:25:05,190 --> 00:25:09,150 In fact, what is 'x squared' plus ''b over a'x' plus ''b 528 00:25:09,150 --> 00:25:11,720 squared' over '4a' squared''? 529 00:25:11,720 --> 00:25:13,320 You'll notice that this is just ''x' 530 00:25:13,320 --> 00:25:17,460 plus 'b over 2a' squared'. 531 00:25:17,460 --> 00:25:18,800 'x squared'. 532 00:25:18,800 --> 00:25:20,170 Just multiply this thing out. 533 00:25:20,170 --> 00:25:21,540 You see how this particular thing works. 534 00:25:21,540 --> 00:25:24,950 In other words, what I've done is I can write this as 'a' 535 00:25:24,950 --> 00:25:27,020 times a perfect square. 536 00:25:27,020 --> 00:25:29,670 And by the way, what's left over here if I multiply these 537 00:25:29,670 --> 00:25:31,910 two terms through by 'a'? 538 00:25:31,910 --> 00:25:35,980 This is 'c' minus 'b squared over 4a'. 539 00:25:35,980 --> 00:25:39,770 In other words, recognizing that this is ''x' plus 'b over 540 00:25:39,770 --> 00:25:41,310 2a' squared'. 541 00:25:41,310 --> 00:25:44,330 And multiplying these two terms through by 'a', I wind 542 00:25:44,330 --> 00:25:48,460 up with the fact that ''ax' squared' plus 'bx' plus 'c' 543 00:25:48,460 --> 00:25:51,330 can always be written in this particular form. 544 00:25:51,330 --> 00:25:53,340 This is called completing the square. 545 00:25:53,340 --> 00:25:55,450 And notice that this is now a perfect square and this is 546 00:25:55,450 --> 00:25:56,530 some constant. 547 00:25:56,530 --> 00:25:59,480 And the question that comes up is, what does this have to do 548 00:25:59,480 --> 00:26:01,260 with integral calculus? 549 00:26:01,260 --> 00:26:03,630 In other words, with finding antiderivatives. 550 00:26:03,630 --> 00:26:06,720 And again, as is so often the case, and granted 551 00:26:06,720 --> 00:26:08,440 that as this course-- 552 00:26:08,440 --> 00:26:12,250 especially this chapter gets more complicated, the degree 553 00:26:12,250 --> 00:26:14,750 of sophistication will become much greater. 554 00:26:14,750 --> 00:26:17,640 But the basic idea will always remain the same. 555 00:26:17,640 --> 00:26:21,180 Somehow or other, we will always try to reduce a new 556 00:26:21,180 --> 00:26:23,810 situation to a more familiar one. 557 00:26:23,810 --> 00:26:27,100 In other words, let's suppose now we don't have an integrand 558 00:26:27,100 --> 00:26:29,560 that's the sum and difference of two squares, but we have an 559 00:26:29,560 --> 00:26:32,680 integrand that involves a quadratic expression. 560 00:26:32,680 --> 00:26:36,630 Say, for example, we have 'dx' over ''ax' squared' 561 00:26:36,630 --> 00:26:38,210 plus 'bx' plus 'c'. 562 00:26:38,210 --> 00:26:41,780 Well, you see, using hindsight rather than foresight, I 563 00:26:41,780 --> 00:26:45,020 already took care of this problem before we started. 564 00:26:45,020 --> 00:26:47,740 Namely, we just finished completing 565 00:26:47,740 --> 00:26:49,070 the square over here. 566 00:26:49,070 --> 00:26:53,230 Namely, what is ''ax' squared' plus 'bx' plus 'c'? 567 00:26:53,230 --> 00:26:54,780 Notice it can be written in what form? 568 00:26:54,780 --> 00:26:56,720 Let's come back here and take a look. 569 00:26:56,720 --> 00:26:59,820 It's 'a' times ''x' plus 'b over 2a' 570 00:26:59,820 --> 00:27:01,910 squared' plus some constant. 571 00:27:01,910 --> 00:27:03,600 I don't care what it happens to be here. 572 00:27:03,600 --> 00:27:06,240 I'll just call that constant 'c sub 1'. 573 00:27:06,240 --> 00:27:09,500 And by the way, notice depending upon the relative 574 00:27:09,500 --> 00:27:12,740 sizes of 'a', 'b', and 'c', this constant can be either 575 00:27:12,740 --> 00:27:14,330 positive or negative. 576 00:27:14,330 --> 00:27:16,460 Or in fact, even 0. 577 00:27:16,460 --> 00:27:19,860 So to handle that what I'll say is I can always write this 578 00:27:19,860 --> 00:27:24,680 denominator in the form 'a' times ''x' plus 'b over 2a' 579 00:27:24,680 --> 00:27:27,590 squared' plus or minus some constant. 580 00:27:27,590 --> 00:27:30,010 Where I'm assuming now that the constant is positive. 581 00:27:30,010 --> 00:27:31,900 That's why I put the plus or minus sign in. 582 00:27:31,900 --> 00:27:34,660 It's plus some positive constant or else it's minus 583 00:27:34,660 --> 00:27:36,350 some positive constant. 584 00:27:36,350 --> 00:27:39,840 Now what I can do is factor out an a from the denominator. 585 00:27:39,840 --> 00:27:42,920 If I factor out an a from the denominator here, I have 1 586 00:27:42,920 --> 00:27:44,980 over 'a' times a certain integral. 587 00:27:44,980 --> 00:27:46,160 What integral? 588 00:27:46,160 --> 00:27:52,150 'dx' over 'x plus 'b over 2a' squared' plus or minus still 589 00:27:52,150 --> 00:27:53,890 some constant. 590 00:27:53,890 --> 00:27:56,500 In other words, I had 'c1' here. 591 00:27:56,500 --> 00:28:00,740 I factored out on 'a'. 'c2' was actually 'c1 over a'. 592 00:28:00,740 --> 00:28:02,740 I just factored out an 'a' from this thing here. 593 00:28:02,740 --> 00:28:03,900 But it doesn't make any difference. 594 00:28:03,900 --> 00:28:06,750 The important point is I now have this written in this 595 00:28:06,750 --> 00:28:07,960 particular form. 596 00:28:07,960 --> 00:28:11,980 Now if I assume that 'c2' is a positive constant, notice that 597 00:28:11,980 --> 00:28:14,610 any positive number is a perfect square. 598 00:28:14,610 --> 00:28:18,190 Namely, it's the square of its square root. 599 00:28:18,190 --> 00:28:20,960 And again, as I say with the exercises, this will become 600 00:28:20,960 --> 00:28:23,250 clearer when we work with specific numbers. 601 00:28:23,250 --> 00:28:25,550 But the idea here is what? 602 00:28:25,550 --> 00:28:29,200 This is some perfect square, which I'll call 'k'. 603 00:28:29,200 --> 00:28:31,610 And in other words, this integrand can now be written 604 00:28:31,610 --> 00:28:36,540 in the form '1 over a', 'dx' over 'x plus 'b over 2a' 605 00:28:36,540 --> 00:28:39,820 squared' plus or minus 'k squared'. 606 00:28:39,820 --> 00:28:41,910 Well, again, what am I leading up to? 607 00:28:41,910 --> 00:28:46,040 If I now make the substitution that 'u' equals 'x plus 'b 608 00:28:46,040 --> 00:28:50,760 over 2a'', notice that my denominator simply becomes 'u 609 00:28:50,760 --> 00:28:52,880 squared' plus or minus 'k squared'. 610 00:28:52,880 --> 00:28:56,560 In other words, my denominator now has the form of either the 611 00:28:56,560 --> 00:28:58,860 sum or difference of two squares. 612 00:28:58,860 --> 00:29:02,780 On the other hand, if 'u' is 'x plus 'b over 613 00:29:02,780 --> 00:29:05,730 2a'', what is 'du'? 614 00:29:05,730 --> 00:29:07,620 Remember, 'b' and 'a' are constants. 615 00:29:07,620 --> 00:29:09,420 'b over 2a' is a constant. 616 00:29:09,420 --> 00:29:11,680 The derivative of a constant is 0. 617 00:29:11,680 --> 00:29:14,780 So 'du' is equal to 'dx'. 618 00:29:14,780 --> 00:29:18,970 In other words, notice that if I now make this substitution 619 00:29:18,970 --> 00:29:23,850 in here, outside I have a '1 over a', inside I have 'du' 620 00:29:23,850 --> 00:29:27,560 over ''u squared' plus or minus 'k squared''. 621 00:29:27,560 --> 00:29:31,590 In other words, when I have a quadratic I can write that in 622 00:29:31,590 --> 00:29:35,650 the form the sum and/or difference of two squares, 623 00:29:35,650 --> 00:29:39,560 simply by completing the square. 624 00:29:39,560 --> 00:29:41,310 Simply by completing square. 625 00:29:41,310 --> 00:29:46,460 In other words, I can solve the problem of the quadratic 626 00:29:46,460 --> 00:29:50,660 by converting it back into an integral, the type of which 627 00:29:50,660 --> 00:29:52,100 I've solved before. 628 00:29:52,100 --> 00:29:54,920 And this, by and large, becomes the technique that we 629 00:29:54,920 --> 00:29:57,080 will use throughout this chapter. 630 00:29:57,080 --> 00:29:59,820 Well, at any rate, I won't go into that in any more 631 00:29:59,820 --> 00:30:00,930 detail right now. 632 00:30:00,930 --> 00:30:02,070 We will pick up additional 633 00:30:02,070 --> 00:30:03,455 techniques in our next lecture. 634 00:30:03,455 --> 00:30:05,460 And until the next lecture, goodbye. 635 00:30:05,460 --> 00:30:08,200 636 00:30:08,200 --> 00:30:10,740 ANNOUNCER: Funding for the publication of this video was 637 00:30:10,740 --> 00:30:15,460 provided by the Gabriella and Paul Rosenbaum Foundation. 638 00:30:15,460 --> 00:30:19,630 Help OCW continue to provide free and open access to MIT 639 00:30:19,630 --> 00:30:23,830 courses by making a donation at ocw.mit.edu/donate. 640 00:30:23,830 --> 00:30:28,564