1 00:00:00,000 --> 00:00:01,940 FEMALE VOICE: 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 00:00:33,010 --> 00:00:33,980 PROFESSOR: Hi. 9 00:00:33,980 --> 00:00:37,220 Today, we're going to learn a rather powerful technique, 10 00:00:37,220 --> 00:00:40,020 called the technique of partial fractions, that's 11 00:00:40,020 --> 00:00:44,680 particularly applicable for one special type of integrand. 12 00:00:44,680 --> 00:00:48,950 In particular, it's going to be applicable to the situation 13 00:00:48,950 --> 00:00:52,650 where the integral has the form, the quotient, of two 14 00:00:52,650 --> 00:00:53,400 polynomials. 15 00:00:53,400 --> 00:00:57,140 In other words, suppose we have integral ''P of x' dx' 16 00:00:57,140 --> 00:01:00,490 over 'Q of x', where 'P' and 'Q' are both 17 00:01:00,490 --> 00:01:02,740 polynomials in 'x'. 18 00:01:02,740 --> 00:01:05,080 And, by the way, for reasons that will become apparent 19 00:01:05,080 --> 00:01:10,200 later, we'll assume that the degree of 'P' is less than the 20 00:01:10,200 --> 00:01:11,300 degree of 'Q'. 21 00:01:11,300 --> 00:01:13,940 In other words, that the highest exponent in the 22 00:01:13,940 --> 00:01:17,110 numerator is less than the highest exponent in the 23 00:01:17,110 --> 00:01:17,920 denominator. 24 00:01:17,920 --> 00:01:21,470 By the way, if that's not the case, we can always carry out 25 00:01:21,470 --> 00:01:26,850 long division first and carry out enough terms until we get 26 00:01:26,850 --> 00:01:29,800 a term in which this is the particular case. 27 00:01:29,800 --> 00:01:32,420 You see, I'll make that clearer as we go along. 28 00:01:32,420 --> 00:01:35,230 But, for the time being, all we really care about in terms 29 00:01:35,230 --> 00:01:36,660 of outlining this form-- 30 00:01:36,660 --> 00:01:38,900 and, again, as we will often do in these types of 31 00:01:38,900 --> 00:01:39,620 techniques-- 32 00:01:39,620 --> 00:01:42,290 the place that we'll pick up the real fine computational 33 00:01:42,290 --> 00:01:44,710 points are in the exercises. 34 00:01:44,710 --> 00:01:48,730 We'll use the lectures to just outline what the technique is 35 00:01:48,730 --> 00:01:49,760 and how its used. 36 00:01:49,760 --> 00:01:52,160 But suppose, now, that we have the quotient of two 37 00:01:52,160 --> 00:01:54,560 polynomials and we want to integrate it. 38 00:01:54,560 --> 00:02:00,920 Now, the idea is this, there are two key steps that dictate 39 00:02:00,920 --> 00:02:02,510 this particular method. 40 00:02:02,510 --> 00:02:06,470 The first thing is, we saw it in the last lecture, that we 41 00:02:06,470 --> 00:02:10,419 can handle denominators if they involve nothing worse 42 00:02:10,419 --> 00:02:13,700 than linear and quadratic polynomials. 43 00:02:13,700 --> 00:02:16,250 In other words, we know how to integrate something like 44 00:02:16,250 --> 00:02:18,740 'dx/x' plus a constant. 45 00:02:18,740 --> 00:02:23,480 We know now how to integrate 'dx / ''ax squared' plus 'bx' 46 00:02:23,480 --> 00:02:27,020 plus 'c'', so we know how to handle that type, OK? 47 00:02:27,020 --> 00:02:28,680 And the second important thing-- 48 00:02:28,680 --> 00:02:30,970 and I'll amplify this, too, as we go along-- 49 00:02:30,970 --> 00:02:34,560 the second important thing is that, theoretically, every 50 00:02:34,560 --> 00:02:37,470 real polynomial can be factored into linear and 51 00:02:37,470 --> 00:02:38,960 quadratic terms. 52 00:02:38,960 --> 00:02:41,460 Now, this is a little bit misleading, if you try to read 53 00:02:41,460 --> 00:02:43,560 more into this than what it really says. 54 00:02:43,560 --> 00:02:48,640 It doesn't mean that we can always find a factorization 55 00:02:48,640 --> 00:02:49,260 quite simply. 56 00:02:49,260 --> 00:02:51,240 In other words, we may, at best, be able to 57 00:02:51,240 --> 00:02:52,720 approximate a root. 58 00:02:52,720 --> 00:02:54,290 We'll have to use things like we've talked about in the 59 00:02:54,290 --> 00:02:57,630 notes so far: Newton's method, and things like this, linear 60 00:02:57,630 --> 00:03:00,290 approximations, tangential approximations. 61 00:03:00,290 --> 00:03:02,260 We may have to approximate the roots. 62 00:03:02,260 --> 00:03:05,810 But that, theoretically, if a polynomial with real 63 00:03:05,810 --> 00:03:09,280 coefficients has degree greater than 2, it has, at 64 00:03:09,280 --> 00:03:11,040 least, one real root. 65 00:03:11,040 --> 00:03:13,940 In other words, we can keep factoring things out this way. 66 00:03:13,940 --> 00:03:16,950 The only thing we can't do is to guarantee that, once we get 67 00:03:16,950 --> 00:03:20,530 down to a quadratic, that we can get real roots out of this 68 00:03:20,530 --> 00:03:21,220 particular thing. 69 00:03:21,220 --> 00:03:24,190 In fact, the classic example is to visualize 'x 70 00:03:24,190 --> 00:03:25,670 squared plus 1'. 71 00:03:25,670 --> 00:03:29,680 In other words, we can't factor 'x squared plus 1' 72 00:03:29,680 --> 00:03:32,580 unless we introduce non-real numbers. 73 00:03:32,580 --> 00:03:36,330 Remember the technique you can write, that 'x squared plus 1' 74 00:03:36,330 --> 00:03:40,330 is 'x squared' minus 'i squared' where 'i' is the 75 00:03:40,330 --> 00:03:41,890 square root of minus 1. 76 00:03:41,890 --> 00:03:46,265 And this factors into 'x plus i' times 'x minus i', et 77 00:03:46,265 --> 00:03:48,880 cetera, but these are non-real numbers. 78 00:03:48,880 --> 00:03:52,130 In other words, you can't always factor a quadratic to 79 00:03:52,130 --> 00:03:52,830 get real numbers. 80 00:03:52,830 --> 00:03:57,750 In fact, if you recall the quadratic formula of this: 81 00:03:57,750 --> 00:04:01,280 square root of 'b squared' minus '4 ac', if 'b squared' 82 00:04:01,280 --> 00:04:04,740 is less than '4 ac', what's inside the square root sign is 83 00:04:04,740 --> 00:04:08,320 negative, and that leads to non-real roots. 84 00:04:08,320 --> 00:04:09,810 So, you see, we can't always factor 85 00:04:09,810 --> 00:04:12,160 quadratics into real factors. 86 00:04:12,160 --> 00:04:15,670 By the way, don't identify things that were difficult to 87 00:04:15,670 --> 00:04:18,290 factor with things that can't be factored. 88 00:04:18,290 --> 00:04:20,519 You know, a lot of times, we think of, say-- here's an 89 00:04:20,519 --> 00:04:23,380 example I thought you might enjoy seeing here-- 90 00:04:23,380 --> 00:04:26,070 take, for example, ''x' to the fourth plus 1'. 91 00:04:26,070 --> 00:04:28,620 That looks something like it belongs to the family 'x 92 00:04:28,620 --> 00:04:30,270 squared plus 1'. 93 00:04:30,270 --> 00:04:32,570 And 'x squared plus 1' can't be factored. 94 00:04:32,570 --> 00:04:34,020 You might think that ''x' to the fourth 95 00:04:34,020 --> 00:04:35,660 plus 1' can't be factored. 96 00:04:35,660 --> 00:04:38,940 Now, again, it's not important that you understand what made 97 00:04:38,940 --> 00:04:40,890 me think of these tricks or what have you. 98 00:04:40,890 --> 00:04:43,920 What I do want to show is that even situations where the 99 00:04:43,920 --> 00:04:46,660 polynomials might look like a known factor, they really do. 100 00:04:46,660 --> 00:04:49,870 For example, with ''x' to the fourth plus 1', we can write 101 00:04:49,870 --> 00:04:53,980 this in the disguised form ''x squared plus 1' squared', see? 102 00:04:53,980 --> 00:04:56,150 That's, what? ''x' to the fourth plus 1'. 103 00:04:56,150 --> 00:04:58,820 But there's a middle term here of 2 'x squared', so I 104 00:04:58,820 --> 00:05:00,970 subtract off the 2 'x squared'. 105 00:05:00,970 --> 00:05:03,500 Now, this has the form, the sum and difference of 2 106 00:05:03,500 --> 00:05:07,290 squares, namely, ''x squared plus 1' squared' and, also, 107 00:05:07,290 --> 00:05:10,400 the square root of 2 times 'x squared'. 108 00:05:10,400 --> 00:05:14,530 In other words, I can write this as this plus the square 109 00:05:14,530 --> 00:05:18,240 root of 2 times 'x' times this minus the square 110 00:05:18,240 --> 00:05:19,990 root of '2 times x'. 111 00:05:19,990 --> 00:05:23,710 Observing, that even though the square root of 2 is an 112 00:05:23,710 --> 00:05:27,480 irrational number, it is, nonetheless, a real number. 113 00:05:27,480 --> 00:05:29,420 The important point that I want to point out, though, as 114 00:05:29,420 --> 00:05:31,720 far as setting up this technique called partial 115 00:05:31,720 --> 00:05:35,020 fractions, is that, whether it's easy or not easy, the 116 00:05:35,020 --> 00:05:38,070 fact remains that, when we have a polynomial in our 117 00:05:38,070 --> 00:05:42,760 denominator, it can always be factored into a combination of 118 00:05:42,760 --> 00:05:47,350 linear and quadratic factors using real numbers. 119 00:05:47,350 --> 00:05:50,610 Well, it's difficult to factor some of these things, so, by 120 00:05:50,610 --> 00:05:52,610 way of illustration, let me pick out 121 00:05:52,610 --> 00:05:54,470 one that comes factored. 122 00:05:54,470 --> 00:05:57,000 Let me start with a particular problem here. 123 00:05:59,500 --> 00:06:04,670 Let's take the integral 'dx/ ''x minus 1' times 'x 124 00:06:04,670 --> 00:06:06,010 squared plus 1''. 125 00:06:06,010 --> 00:06:08,120 So, what is the integrand in this case? 126 00:06:08,120 --> 00:06:12,620 It's 1/ ''x minus 1' times 'x squared plus 1''. 127 00:06:12,620 --> 00:06:13,790 Now, the idea is this. 128 00:06:13,790 --> 00:06:15,490 See, here's a quadratic factor, 129 00:06:15,490 --> 00:06:17,900 here's a linear factor. 130 00:06:17,900 --> 00:06:21,560 What fractions do I have to add to wind up with 131 00:06:21,560 --> 00:06:22,380 something like this? 132 00:06:22,380 --> 00:06:25,220 Well, again-- and this is going to be another example of 133 00:06:25,220 --> 00:06:28,190 our old adage that it's easier to scramble an egg than to 134 00:06:28,190 --> 00:06:29,540 unscramble one. 135 00:06:29,540 --> 00:06:32,430 You see, given two fractions, it's one thing 136 00:06:32,430 --> 00:06:34,050 to find their sum. 137 00:06:34,050 --> 00:06:36,130 Given the sum, it's quite another thing, 138 00:06:36,130 --> 00:06:39,550 computationally, to find what fractions you had to add to 139 00:06:39,550 --> 00:06:40,520 get that sum. 140 00:06:40,520 --> 00:06:41,560 The idea is this. 141 00:06:41,560 --> 00:06:44,570 If you wind up with a denominator that has an 'x 142 00:06:44,570 --> 00:06:48,210 minus 1' term and an 'x squared plus 1' term, it 143 00:06:48,210 --> 00:06:52,860 appears that you must have started with terms, say, what? 144 00:06:52,860 --> 00:06:55,380 In other words, you must have had one term which had a 145 00:06:55,380 --> 00:06:58,640 denominator of 'x minus 1' and a term which had a denominator 146 00:06:58,640 --> 00:06:59,980 of 'x squared plus 1'. 147 00:06:59,980 --> 00:07:01,800 Because, you see, if I start with these kind of 148 00:07:01,800 --> 00:07:04,860 denominators when I cross-multiply and put things 149 00:07:04,860 --> 00:07:08,580 over a common denominator, I will wind up with 'x minus 1' 150 00:07:08,580 --> 00:07:10,500 times 'x squared plus 1'. 151 00:07:10,500 --> 00:07:13,930 The question that comes up is what shall our numerators be? 152 00:07:13,930 --> 00:07:16,970 And here's the main reason why we picked the degree of the 153 00:07:16,970 --> 00:07:19,800 numerator to be less than the degree of the denominator. 154 00:07:19,800 --> 00:07:23,660 Notice, for example, in this particular case, the numerator 155 00:07:23,660 --> 00:07:26,650 has degree 0, namely, the highest power of 'x' to the 156 00:07:26,650 --> 00:07:29,260 peers is the 'x' to the 0 term. 157 00:07:29,260 --> 00:07:32,050 On the other hand, the denominator has degree 3. 158 00:07:32,050 --> 00:07:35,130 See, there's an 'x cubed' term in the denominator. 159 00:07:35,130 --> 00:07:36,820 You see, if they're doing 'x' to the fourth in the 160 00:07:36,820 --> 00:07:38,740 numerator, I could have multiplied out the 161 00:07:38,740 --> 00:07:42,990 denominator, divided it into the numerator, and just kept 162 00:07:42,990 --> 00:07:45,930 carrying out the division long enough until I wound up with a 163 00:07:45,930 --> 00:07:50,820 remainder which was less than a degree, less than a cubic. 164 00:07:50,820 --> 00:07:51,780 In other words, less than a third 165 00:07:51,780 --> 00:07:53,570 degree polynomial remainder. 166 00:07:53,570 --> 00:07:54,540 That's not the point. 167 00:07:54,540 --> 00:07:57,210 The point is that as long as the degree of the numerator is 168 00:07:57,210 --> 00:07:59,850 less than the degree of the denominator, it means that the 169 00:07:59,850 --> 00:08:02,290 terms that we're adding must have the degree of the 170 00:08:02,290 --> 00:08:04,940 numerator less than the degree of the denominator. 171 00:08:04,940 --> 00:08:06,710 See, it's like adding fractions. 172 00:08:06,710 --> 00:08:09,850 If you start with one fraction whose numerator is greater 173 00:08:09,850 --> 00:08:12,900 than the denominator then, certainly, any sum that you 174 00:08:12,900 --> 00:08:15,410 get is going to be bigger than 1. 175 00:08:15,410 --> 00:08:18,960 In other words, if you want to wind-up, dealing with positive 176 00:08:18,960 --> 00:08:21,640 numbers, and you want to wind-up with a fraction which 177 00:08:21,640 --> 00:08:24,470 is less than 1, it stands to reason that all of the 178 00:08:24,470 --> 00:08:27,740 fractions that you're adding must be less than 1. 179 00:08:27,740 --> 00:08:30,740 So, if I'm going to wind-up adding quotients of 180 00:08:30,740 --> 00:08:34,570 polynomials to get a sum in which the degree of the 181 00:08:34,570 --> 00:08:37,850 numerator is less than the degree of the denominator, it 182 00:08:37,850 --> 00:08:40,679 means that every one of the terms in my sum must have this 183 00:08:40,679 --> 00:08:42,100 particular property. 184 00:08:42,100 --> 00:08:45,180 In other words, with this as a hint, I say look at, my 185 00:08:45,180 --> 00:08:48,790 denominator here is 'x minus 1', that's degree one. 186 00:08:48,790 --> 00:08:52,720 That means my numerator can't be greater than degree 0. 187 00:08:52,720 --> 00:08:55,150 But degree 0 means a constant. 188 00:08:55,150 --> 00:08:59,400 So I say OK, that means that this has to form some constant 189 00:08:59,400 --> 00:09:01,950 over 'x minus 1'. 190 00:09:01,950 --> 00:09:03,910 Now, I look at this denominator. 191 00:09:03,910 --> 00:09:05,160 It's quadratic. 192 00:09:05,160 --> 00:09:09,500 And I say to myself I'm starting out with a quadratic, 193 00:09:09,500 --> 00:09:12,020 the degree of the numerator can't be greater than the 194 00:09:12,020 --> 00:09:13,470 degree of the denominator. 195 00:09:13,470 --> 00:09:15,980 Since the degree of the denominator is 2, that means 196 00:09:15,980 --> 00:09:20,040 the degree of the numerator can't be more than 1. 197 00:09:20,040 --> 00:09:23,300 And the most general first degree polynomial has the 198 00:09:23,300 --> 00:09:24,040 form, what? 199 00:09:24,040 --> 00:09:27,240 Some constant times 'x' plus a constant. 200 00:09:27,240 --> 00:09:30,290 So what I'm saying is, OK, to wind-up with 201 00:09:30,290 --> 00:09:31,690 something of the form-- 202 00:09:31,690 --> 00:09:35,670 well, to wind-up with 1/ ''x minus 1' times 'x squared plus 203 00:09:35,670 --> 00:09:39,030 1'', I had better start with something of the form 'A/ 'x 204 00:09:39,030 --> 00:09:44,330 minus 1'' plus ''Bx plus C' over 'x squared plus 1'' where 205 00:09:44,330 --> 00:09:46,650 'A', 'B', and 'C' are constants. 206 00:09:46,650 --> 00:09:49,870 The key point is that if we weren't sure that the degree 207 00:09:49,870 --> 00:09:51,880 of the numerator were less than the degree of the 208 00:09:51,880 --> 00:09:54,380 denominator, we would not know where 209 00:09:54,380 --> 00:09:55,930 to stop in our numerator. 210 00:09:55,930 --> 00:09:59,130 In other words, by this convention, we're sure that 211 00:09:59,130 --> 00:10:02,090 the degree of the numerator in any one of these terms can't 212 00:10:02,090 --> 00:10:04,580 be greater than the degree of the corresponding 213 00:10:04,580 --> 00:10:06,700 denominator, you see? 214 00:10:06,700 --> 00:10:09,280 And by the way, notice, for example, if it turns out that 215 00:10:09,280 --> 00:10:10,630 we put too much in-- 216 00:10:10,630 --> 00:10:13,230 for example, suppose it turns out that this numerator here 217 00:10:13,230 --> 00:10:14,620 should only be a constant-- 218 00:10:14,620 --> 00:10:18,170 there's no law against having 'B' turn out to be 0. 219 00:10:18,170 --> 00:10:20,660 By the way, again, what these things are called-- 220 00:10:20,660 --> 00:10:23,540 just to increase our vocabulary-- 221 00:10:23,540 --> 00:10:25,740 what I'm going to do now is called the method of 222 00:10:25,740 --> 00:10:27,370 undetermined coefficients. 223 00:10:27,370 --> 00:10:31,060 You see, what I know is that I must have the form 'A over 'x 224 00:10:31,060 --> 00:10:35,050 minus 1' plus ''Bx plus C' over 'x squared plus 1''. 225 00:10:35,050 --> 00:10:37,560 What I don't know is, specifically, how to choose 226 00:10:37,560 --> 00:10:39,650 the values of 'A', 'B', and 'C'. 227 00:10:39,650 --> 00:10:41,620 And the technique works something like this. 228 00:10:41,620 --> 00:10:45,730 What we do is we put this over a common denominator. 229 00:10:45,730 --> 00:10:47,540 What will the common denominator be? 230 00:10:47,540 --> 00:10:50,960 It'll be 'x minus 1' times 'x squared plus 1'. 231 00:10:50,960 --> 00:10:53,200 How will I put this over a common denominator? 232 00:10:53,200 --> 00:10:58,780 It'll be 'A' times 'x squared plus 1', plus 'Bx plus C' 233 00:10:58,780 --> 00:11:00,520 times 'x minus 1'. 234 00:11:00,520 --> 00:11:03,130 Now, this is supposed to be an identity. 235 00:11:03,130 --> 00:11:07,700 Now, if two fractions are identical and the denominators 236 00:11:07,700 --> 00:11:10,350 are the same, which is what they will be after I put this 237 00:11:10,350 --> 00:11:14,180 over a common denominator, the only way they can be identical 238 00:11:14,180 --> 00:11:16,140 is for the numerators to be identical. 239 00:11:16,140 --> 00:11:19,330 So, you see, what I'm going to do is to cross-multiply here 240 00:11:19,330 --> 00:11:21,740 to obtain the numerator of the right-hand side. 241 00:11:21,740 --> 00:11:24,690 And I will equate that to the numerator on the left-hand 242 00:11:24,690 --> 00:11:26,380 side, which is 1. 243 00:11:26,380 --> 00:11:26,820 All right. 244 00:11:26,820 --> 00:11:29,230 Now, what is the numerator on the right-hand side? 245 00:11:29,230 --> 00:11:32,520 It's 'A' times 'x squared plus 1', that's 'A x 246 00:11:32,520 --> 00:11:34,240 squared plus A'. 247 00:11:34,240 --> 00:11:38,770 Then it's going to be 'x minus 1' times 'Bx plus C'. 248 00:11:38,770 --> 00:11:44,610 That's going to be 'B x squared' minus 'Bx plus C' 249 00:11:44,610 --> 00:11:48,110 times 'x minus C'. 250 00:11:48,110 --> 00:11:51,170 And now the idea something like this, and I'll come back 251 00:11:51,170 --> 00:11:53,730 to this in a few moments to hammer this home from a 252 00:11:53,730 --> 00:11:54,780 different point of view. 253 00:11:54,780 --> 00:11:57,720 What is the coefficient of 'x squared' on the right-hand 254 00:11:57,720 --> 00:11:59,390 side of the equation? 255 00:11:59,390 --> 00:12:01,940 The coefficient of 'x squared' on the right-hand side of the 256 00:12:01,940 --> 00:12:04,600 equation is 'A plus B'. 257 00:12:04,600 --> 00:12:08,060 What is the coefficient of 'x squared' on the left-hand side 258 00:12:08,060 --> 00:12:09,160 of the equation? 259 00:12:09,160 --> 00:12:12,080 And at first glance you say there is no 'x squared' on the 260 00:12:12,080 --> 00:12:13,460 left-hand side of the equation. 261 00:12:13,460 --> 00:12:16,070 What that means, of course, is that the coefficient of 'x 262 00:12:16,070 --> 00:12:19,220 squared' on the left-hand side of the equation is 0. 263 00:12:19,220 --> 00:12:22,080 So what we say is OK, the coefficients of 'x squared' 264 00:12:22,080 --> 00:12:25,110 must match up, therefore, 'A plus B', which is a 265 00:12:25,110 --> 00:12:27,930 coefficient of 'x squared' on the right-hand side, must 266 00:12:27,930 --> 00:12:30,780 equal 0, which is the coefficient of 'x squared' on 267 00:12:30,780 --> 00:12:32,170 the left-hand side. 268 00:12:32,170 --> 00:12:36,590 In a similar way, the coefficient of 'x' is 'C minus 269 00:12:36,590 --> 00:12:38,870 B' on the right-hand side. 270 00:12:38,870 --> 00:12:42,410 The coefficient of 'x' on the left-hand side is 0. 271 00:12:42,410 --> 00:12:46,080 Consequently, 'C minus B' must be 0. 272 00:12:46,080 --> 00:12:49,620 And, finally, the constant term on the right-hand side is 273 00:12:49,620 --> 00:12:52,140 given by 'A minus C'. 274 00:12:52,140 --> 00:12:55,650 On the left-hand side, the right-hand term is 1. 275 00:12:55,650 --> 00:12:59,050 Consequently, 'A minus C' must equal 1. 276 00:12:59,050 --> 00:13:01,660 What do I wind-up with? 277 00:13:01,660 --> 00:13:04,470 Three equations with three unknowns. 278 00:13:04,470 --> 00:13:06,730 Well, there are a number of ways of handling these things. 279 00:13:06,730 --> 00:13:09,670 The easiest one I see, off-hand, is I notice if I add 280 00:13:09,670 --> 00:13:13,750 the first two equations, I wind-up with 'A plus C' is 0. 281 00:13:13,750 --> 00:13:17,720 Knowing that 'A plus C' is 0 and, also, that 'A minus C' is 282 00:13:17,720 --> 00:13:22,310 1, it's easy for me to conclude that 'A' must be 1/2 283 00:13:22,310 --> 00:13:24,260 and 'C' must be minus 1/2. 284 00:13:24,260 --> 00:13:27,180 And, by the way, now knowing what 'A' and 'C' are, I can 285 00:13:27,180 --> 00:13:30,510 use either of these two equations to determine 'B', 286 00:13:30,510 --> 00:13:33,430 and 'B' turns out to be minus 1/2. 287 00:13:33,430 --> 00:13:35,190 In other words, what does this tell me? 288 00:13:35,190 --> 00:13:39,490 It tells me that if I replace 'A' here by 1/2, and 'B' and 289 00:13:39,490 --> 00:13:44,220 'C' each by minus 1/2, the right-hand side here will be 290 00:13:44,220 --> 00:13:47,310 an identity for the left-hand side, that there'll be two 291 00:13:47,310 --> 00:13:49,850 different ways of naming the same number for 292 00:13:49,850 --> 00:13:51,470 each value of 'x'. 293 00:13:51,470 --> 00:13:54,980 In fact, doing this now out in more detail, what we've really 294 00:13:54,980 --> 00:13:59,110 shown here, in other words, if I replace 'A' by 1/2 and 'B' 295 00:13:59,110 --> 00:14:03,320 and 'C' each by minus 1/2, then I factor the 1/2 out. 296 00:14:03,320 --> 00:14:08,830 What I've shown is that one over the quantity 'x minus 1' 297 00:14:08,830 --> 00:14:13,040 times 'x squared plus 1' is equal to 1/2 times the 298 00:14:13,040 --> 00:14:18,006 quantity '1 over 'x minus 1'' minus the quantity ''x plus 1' 299 00:14:18,006 --> 00:14:20,870 over 'x squared plus 1''. 300 00:14:20,870 --> 00:14:23,670 By the way, I can separate this into two terms, each of 301 00:14:23,670 --> 00:14:25,840 which has a denominator of 'x squared plus 302 00:14:25,840 --> 00:14:27,420 1', and I get, what? 303 00:14:27,420 --> 00:14:34,220 1/2 times 1/ 'x minus 1' minus 1/ ''2 times x'/ 'x squared 304 00:14:34,220 --> 00:14:41,230 plus 1'' minus 1/2 times '1/ 'x squared plus 1''. 305 00:14:41,230 --> 00:14:43,110 Now, the key is this. 306 00:14:43,110 --> 00:14:45,460 I didn't really want this, what I wanted was, what? 307 00:14:45,460 --> 00:14:47,340 This was to be my integrand. 308 00:14:47,340 --> 00:14:51,080 What I wanted was to integrate this with respect to 'x'. 309 00:14:51,080 --> 00:14:55,880 Well, if given this identity over here, the integral of 310 00:14:55,880 --> 00:14:59,850 'dx'/ ''x minus 1' times 'x squared plus 1'', recalling 311 00:14:59,850 --> 00:15:02,410 that the integral of a sum is a sum of the integrals, et 312 00:15:02,410 --> 00:15:04,940 cetera, can now be written as, what? 313 00:15:04,940 --> 00:15:11,600 It's 1/2 integral 'dx'/ 'x minus 1' minus 1/2 integral 314 00:15:11,600 --> 00:15:18,680 'x'/ 'x squared plus 1' times 'dx' minus 1/2 integral 1/ 'x 315 00:15:18,680 --> 00:15:21,050 squared plus 1' times 'dx'. 316 00:15:21,050 --> 00:15:22,200 Now, here's the point. 317 00:15:22,200 --> 00:15:25,010 Notice that every one of my denominators, now, is either 318 00:15:25,010 --> 00:15:27,030 linear or quadratic. 319 00:15:27,030 --> 00:15:30,470 In fact, without going through the details here again, if I 320 00:15:30,470 --> 00:15:34,830 let 'u' equal 'x minus 1' in this example, this reduces 321 00:15:34,830 --> 00:15:38,510 this to the form 'du/u', in other words, a ''u 322 00:15:38,510 --> 00:15:40,710 to the n' du' form. 323 00:15:40,710 --> 00:15:46,330 If I let 'u' equal 'x squared plus 1' over here, if 'u' is 324 00:15:46,330 --> 00:15:50,990 'x squared plus 1', notice that 'du' is '2 xdx'. 325 00:15:50,990 --> 00:15:55,480 So my numerator becomes a constant multiple of 'du' and, 326 00:15:55,480 --> 00:15:58,280 again, I have a ''u of the n' du' form. 327 00:15:58,280 --> 00:16:01,910 And finally, if I look at my last integral here, notice 328 00:16:01,910 --> 00:16:05,230 that this is the sum of two squares, which suggests the 329 00:16:05,230 --> 00:16:07,280 circular trigonometric functions that we were talking 330 00:16:07,280 --> 00:16:08,840 about last time. 331 00:16:08,840 --> 00:16:11,320 Or, if you wish, you can go to tables and look 332 00:16:11,320 --> 00:16:11,940 these things up. 333 00:16:11,940 --> 00:16:12,780 They're all in there. 334 00:16:12,780 --> 00:16:15,870 But, again, without going through the details because 335 00:16:15,870 --> 00:16:18,650 this is the easy part, again, it turns out that once you 336 00:16:18,650 --> 00:16:22,480 have this relationship here we can integrate this to obtain 337 00:16:22,480 --> 00:16:25,400 log absolute value 'x minus 1'. 338 00:16:25,400 --> 00:16:30,080 The integral here is log, natural log, absolute value of 339 00:16:30,080 --> 00:16:31,490 'x squared plus 1'. 340 00:16:31,490 --> 00:16:34,170 I can leave the absolute value signs out because 'x squared 341 00:16:34,170 --> 00:16:36,380 plus 1' has to be, at least, as big as 1. 342 00:16:36,380 --> 00:16:37,430 It can't be negative. 343 00:16:37,430 --> 00:16:40,330 And finally, either by trigonometric substitution, or 344 00:16:40,330 --> 00:16:43,610 by memorization, or what have you, integral of 'dx'/ 'x 345 00:16:43,610 --> 00:16:48,200 squared plus 1' is just the inverse tangent of 'x'. 346 00:16:48,200 --> 00:16:51,810 In other words, by using partial fractions and reducing 347 00:16:51,810 --> 00:16:56,510 a complicated polynomial denominator into a sum of 348 00:16:56,510 --> 00:17:01,840 linear and quadratic terms, I was able, by knowing my 349 00:17:01,840 --> 00:17:04,670 techniques of last time, to integrate the given 350 00:17:04,670 --> 00:17:05,839 expression. 351 00:17:05,839 --> 00:17:08,460 And, by the way, I would be a little remiss at the stage of 352 00:17:08,460 --> 00:17:12,780 the game if I did not take the time to, once again, reinforce 353 00:17:12,780 --> 00:17:16,960 a very important concept, and that is it may be difficult, 354 00:17:16,960 --> 00:17:19,819 starting with this, to get this. 355 00:17:19,819 --> 00:17:23,980 What should not be difficult is, starting with this, to be 356 00:17:23,980 --> 00:17:27,087 able to differentiate it and show that you wind-up with 1/ 357 00:17:27,087 --> 00:17:31,200 ''x minus 1' times 'x squared plus 1''. 358 00:17:31,200 --> 00:17:34,200 In other words, as usual, with the inverse derivative, once 359 00:17:34,200 --> 00:17:36,460 you find an answer and you want to see whether your 360 00:17:36,460 --> 00:17:39,060 answer is correct or not, all you have to do is 361 00:17:39,060 --> 00:17:43,070 differentiate the answer and see if you get the integrand. 362 00:17:43,070 --> 00:17:46,130 But, at any rate, this is how the technique called partial 363 00:17:46,130 --> 00:17:47,330 fractions works. 364 00:17:47,330 --> 00:17:50,220 It works for the quotient of two polynomials. 365 00:17:50,220 --> 00:17:52,820 And to make the undetermined coefficients technique work 366 00:17:52,820 --> 00:17:56,170 right, you must assume that the degree of the numerator is 367 00:17:56,170 --> 00:17:57,900 less than the degree of the denominator. 368 00:17:57,900 --> 00:18:00,420 And, obviously, in the exercises, I'll give you some 369 00:18:00,420 --> 00:18:03,010 where the degree of the numerator is greater than the 370 00:18:03,010 --> 00:18:04,570 degree of the denominator. 371 00:18:04,570 --> 00:18:07,110 And, if you don't perform long division first, you're going 372 00:18:07,110 --> 00:18:08,690 to get into trouble trying to find the 373 00:18:08,690 --> 00:18:09,690 answer to the problem. 374 00:18:09,690 --> 00:18:12,800 But all I want to emphasize here is the technique. 375 00:18:12,800 --> 00:18:15,450 And, by the way, what I want to do before I go any further, 376 00:18:15,450 --> 00:18:20,440 also, is to emphasize a rather special property of polynomial 377 00:18:20,440 --> 00:18:21,650 identities. 378 00:18:21,650 --> 00:18:24,070 You recall that undetermined coefficients 379 00:18:24,070 --> 00:18:25,260 hinged on the following. 380 00:18:25,260 --> 00:18:27,600 And I'll pick a quadratic to illustrate it with. 381 00:18:27,600 --> 00:18:31,720 Suppose you have two quadratic expressions in 'x' identically 382 00:18:31,720 --> 00:18:34,920 equal, in other words, 'a sub-2 'x squared'' plus 'a 383 00:18:34,920 --> 00:18:39,380 sub-1 x' plus 'a sub-0' was identically equal to 'b sub-2 384 00:18:39,380 --> 00:18:43,190 'x squared'' plus 'b sub-1 x' plus 'b0'. 385 00:18:43,190 --> 00:18:45,900 Notice the technique that we used was is we said look at, 386 00:18:45,900 --> 00:18:48,470 let's compare the coefficients of 'x squared'. 387 00:18:48,470 --> 00:18:51,480 Let's equate the coefficients of 'x', and let's equate the 388 00:18:51,480 --> 00:18:53,750 constant terms. 389 00:18:53,750 --> 00:18:57,350 How do we know that you're allowed to do this? 390 00:18:57,350 --> 00:19:01,780 Well, let's see if we can show that this must be the case. 391 00:19:01,780 --> 00:19:04,460 For example, let's do this without any calculus at all. 392 00:19:04,460 --> 00:19:06,430 Suppose this is an identity. 393 00:19:06,430 --> 00:19:08,830 If this is an identity, it must be true for 394 00:19:08,830 --> 00:19:10,240 all values of 'x'. 395 00:19:10,240 --> 00:19:13,980 In particular, it must be true when 'x' is 0. 396 00:19:13,980 --> 00:19:16,750 Notice that, when 'x' is 0, the left-hand side is 'a 397 00:19:16,750 --> 00:19:21,170 sub-0', the right-hand side is 'B sub-0', and we wind-up with 398 00:19:21,170 --> 00:19:23,860 'a sub-0' equals 'b sub-0'. 399 00:19:23,860 --> 00:19:25,870 See, the constant terms are equal. 400 00:19:25,870 --> 00:19:29,800 Well, if 'a sub-0' equals 'b sub-0', we can cancel 'a 401 00:19:29,800 --> 00:19:32,330 sub-0' and 'b sub-0' from this equation. 402 00:19:32,330 --> 00:19:35,460 That leaves us with this equaling this. 403 00:19:35,460 --> 00:19:38,930 From this, we can factor out an 'x', and get to 'x' times 404 00:19:38,930 --> 00:19:43,800 'a sub-2 x' plus 'a1' is identical with 'x' times 'b 405 00:19:43,800 --> 00:19:46,630 sub-2 x' plus 'b1'. 406 00:19:46,630 --> 00:19:50,070 If 'x' is not 0, we can cancel 'x' from both sides of the 407 00:19:50,070 --> 00:19:51,900 equation, and that shows, what? 408 00:19:51,900 --> 00:19:56,390 That, for any non-zero value of 'x', 'a 2x' plus 'a1' must 409 00:19:56,390 --> 00:20:00,840 be identically equal to 'b 2x' plus 'b1'. 410 00:20:00,840 --> 00:20:04,460 Once we have this formula established, let's let 'x' 411 00:20:04,460 --> 00:20:06,850 equal 0 in here, and we see, what? 412 00:20:06,850 --> 00:20:10,670 With 'x' equal to 0, then 'a1' equals 'b1', in other words, 413 00:20:10,670 --> 00:20:13,240 that the coefficient of 'x' on the left-hand side equals the 414 00:20:13,240 --> 00:20:15,500 coefficient of 'x' on the right-hand side. 415 00:20:15,500 --> 00:20:18,410 You can keep on this way, but a very nice technique to use 416 00:20:18,410 --> 00:20:20,760 here is a reinforcement of something that we 417 00:20:20,760 --> 00:20:22,120 talked about before. 418 00:20:22,120 --> 00:20:23,710 I think it was when we were doing implicit 419 00:20:23,710 --> 00:20:25,660 differentiation, and we talked about 420 00:20:25,660 --> 00:20:27,690 identities verses equations. 421 00:20:27,690 --> 00:20:31,790 You see, if this is an identity, and what I mean by 422 00:20:31,790 --> 00:20:33,780 an identity, that this-- 423 00:20:33,780 --> 00:20:35,800 we're not saying find what values of 'x' 424 00:20:35,800 --> 00:20:36,860 this is true for. 425 00:20:36,860 --> 00:20:39,840 By an identity, we're saying look at, these two expressions 426 00:20:39,840 --> 00:20:41,880 are the same for all values of 'x'. 427 00:20:41,880 --> 00:20:43,150 They're synonyms. 428 00:20:43,150 --> 00:20:45,340 And what we're saying is, if these two things are synonyms, 429 00:20:45,340 --> 00:20:47,620 their derivatives must be synonyms. 430 00:20:47,620 --> 00:20:49,400 And all we're saying is look at, if you want to use 431 00:20:49,400 --> 00:20:53,320 calculus here, differentiate both sides of this expression. 432 00:20:53,320 --> 00:20:57,740 And you wind-up with '2a sub-2 x' plus 'a1' is identically 433 00:20:57,740 --> 00:21:01,330 equal to '2b sub-2 x' plus 'b1'. 434 00:21:01,330 --> 00:21:04,680 Since these two things are identical, let's equate their 435 00:21:04,680 --> 00:21:05,950 derivatives again. 436 00:21:05,950 --> 00:21:09,240 See, the derivative of identities are identical, and 437 00:21:09,240 --> 00:21:13,580 we wind-up with '2 a2' is identically equal to '2 b2' 438 00:21:13,580 --> 00:21:16,590 and, therefore, 'a2' must equal 'b2'. 439 00:21:16,590 --> 00:21:20,420 Knowing that 'a2' equals 'b2', we can come back to this step 440 00:21:20,420 --> 00:21:22,700 and show that 'a1' equals 'b1'. 441 00:21:22,700 --> 00:21:26,270 And now, knowing that 'a2' equals 'b2', and 'a1' equals 442 00:21:26,270 --> 00:21:29,415 'b1', we can come back to the original equation and show 443 00:21:29,415 --> 00:21:32,010 that 'a0' must equal 'b0'. 444 00:21:32,010 --> 00:21:37,410 Now, you may wonder why are we making all of this ado over 445 00:21:37,410 --> 00:21:39,860 what appears to be a very obvious thing? 446 00:21:39,860 --> 00:21:42,310 And I would like to give you a caution here. 447 00:21:42,310 --> 00:21:44,200 I'd like you to beware of something. 448 00:21:44,200 --> 00:21:47,180 It's something which works very nicely for polynomials, 449 00:21:47,180 --> 00:21:49,530 but doesn't always have to work all the time. 450 00:21:49,530 --> 00:21:52,840 In fact, later, when one studies differential 451 00:21:52,840 --> 00:21:56,330 equations, this becomes a very important concept, which later 452 00:21:56,330 --> 00:21:58,910 gets the name linear dependence and linear 453 00:21:58,910 --> 00:22:00,020 independence. 454 00:22:00,020 --> 00:22:01,820 We're not going to go into that now, but 455 00:22:01,820 --> 00:22:03,180 the key idea is this. 456 00:22:03,180 --> 00:22:08,740 In general, knowing that 'a1 u1' plus 'a2 u2' is equal to 457 00:22:08,740 --> 00:22:15,750 'b1 u1' plus 'b2 u2', you cannot say ah, therefore, the 458 00:22:15,750 --> 00:22:18,950 coefficients of 'u1' must be equal, and the coefficients of 459 00:22:18,950 --> 00:22:20,890 'u2' must be equal. 460 00:22:20,890 --> 00:22:25,140 Don't get me wrong, if 'a1' equals 'b1', and 'a2' equals 461 00:22:25,140 --> 00:22:30,700 'b2', then, certainly, 'a1 u1' plus 'a2 u2' is equal to 'b1 462 00:22:30,700 --> 00:22:32,670 u1' plus 'b2 u2'. 463 00:22:32,670 --> 00:22:33,750 I'm not saying that. 464 00:22:33,750 --> 00:22:36,190 What I'm saying is, conversely, if you start, 465 00:22:36,190 --> 00:22:40,010 knowing that this is true, it does not follow that they must 466 00:22:40,010 --> 00:22:43,470 match up coefficient by coefficient, OK? 467 00:22:43,470 --> 00:22:45,630 And you say well, why doesn't it have to follow? 468 00:22:45,630 --> 00:22:47,180 And I think the best way to do that is 469 00:22:47,180 --> 00:22:48,870 by means of an example. 470 00:22:48,870 --> 00:22:51,910 For example, in this general expression, let 'u1' equal 471 00:22:51,910 --> 00:22:55,160 'x', and let 'u2' to be 'x/2'. 472 00:22:55,160 --> 00:22:58,360 Look at the expression '5x' plus '6 times 473 00:22:58,360 --> 00:23:00,160 x/2', that's, what? 474 00:23:00,160 --> 00:23:03,790 It's '5x' plus '3x' is '8x'. 475 00:23:03,790 --> 00:23:07,280 Look at '3x' plus '10 times x/2'. 476 00:23:07,280 --> 00:23:11,090 That's '3x' plus '5x', which is also '8x'. 477 00:23:11,090 --> 00:23:15,890 In other words, 5 times 'x' plus 6 times 'x/2' is 478 00:23:15,890 --> 00:23:20,860 identically equal to 3 times 'x' plus 10 times 'x/2'. 479 00:23:20,860 --> 00:23:24,270 Yet, you can't say, therefore, the coefficients of 'x' must 480 00:23:24,270 --> 00:23:28,980 be equal, and the coefficients of 'x/2' must be equal. 481 00:23:28,980 --> 00:23:31,570 In fact, if you said that, you'd be saying that 5 is 482 00:23:31,570 --> 00:23:35,670 equal to 3 and 6 is equal to 10, which, of course, 483 00:23:35,670 --> 00:23:38,620 is not true, OK? 484 00:23:38,620 --> 00:23:42,030 So, at any rate, I just wanted to show you here the kind of 485 00:23:42,030 --> 00:23:45,800 mathematical rigor and cautions that have to be taken 486 00:23:45,800 --> 00:23:48,090 if one is going to use a technique called partial 487 00:23:48,090 --> 00:23:51,420 fractions, what the key ingredients are. 488 00:23:51,420 --> 00:23:54,590 Now, it turns out that not only are partial fractions 489 00:23:54,590 --> 00:23:58,190 important in their own right, it also turns out that partial 490 00:23:58,190 --> 00:24:03,750 fractions handles a rather difficult type of integral: 491 00:24:03,750 --> 00:24:10,260 one that uses polynomials in 'sine x' and 'cosine x'. 492 00:24:10,260 --> 00:24:12,890 And the reason I wanted to mention this was not so much 493 00:24:12,890 --> 00:24:14,770 because the technique is nice. 494 00:24:14,770 --> 00:24:17,130 The technique, by the way, is in the text. 495 00:24:17,130 --> 00:24:19,620 But there's something very interesting in the text, the 496 00:24:19,620 --> 00:24:21,520 way the author introduces a topic. 497 00:24:21,520 --> 00:24:23,250 And I thought that that was worth an 498 00:24:23,250 --> 00:24:24,680 aside in its own right. 499 00:24:29,610 --> 00:24:33,550 In the section where this thing appears, it says, "It 500 00:24:33,550 --> 00:24:38,210 has been discovered that." The author makes no attempt to 501 00:24:38,210 --> 00:24:41,330 show logically why one would expect that a certain thing is 502 00:24:41,330 --> 00:24:42,590 going to work. 503 00:24:42,590 --> 00:24:46,240 All the author says is, "It has been discovered that." And 504 00:24:46,240 --> 00:24:50,850 this tells a long story that, in many cases, we wind up with 505 00:24:50,850 --> 00:24:53,090 an integrand that we don't know how to handle. 506 00:24:53,090 --> 00:24:56,440 We make all sorts of substitutions in the hope that 507 00:24:56,440 --> 00:24:59,720 we can reduce the given integrand to a form that we 508 00:24:59,720 --> 00:25:01,200 know how to handle. 509 00:25:01,200 --> 00:25:04,300 Sometimes we're successful, sometimes we're not. 510 00:25:04,300 --> 00:25:08,310 In the cases where we're not successful, somebody, either 511 00:25:08,310 --> 00:25:11,490 by clever intuition or what have you, maybe it's just 512 00:25:11,490 --> 00:25:14,200 blind luck, stumbles across a technique 513 00:25:14,200 --> 00:25:16,040 that happens to work. 514 00:25:16,040 --> 00:25:18,980 And I, sort of, liked this particular example in the text 515 00:25:18,980 --> 00:25:21,360 where the author says, "It has been discovered that," 516 00:25:21,360 --> 00:25:24,320 because, to me, it's not at all self-evident, and yet, 517 00:25:24,320 --> 00:25:26,550 it's a rather pretty result. 518 00:25:26,550 --> 00:25:30,520 The result says this: suppose you make the substitution 'z' 519 00:25:30,520 --> 00:25:33,690 equals tangent 'x/2'. 520 00:25:33,690 --> 00:25:35,920 Now, where do you pull this out of the hat from? 521 00:25:35,920 --> 00:25:39,650 'z' equals tangent 'x/2', that's the ingenuity, the 522 00:25:39,650 --> 00:25:41,330 experience, the luck. 523 00:25:41,330 --> 00:25:44,850 But the idea is, let's suppose that we stumbled across this 524 00:25:44,850 --> 00:25:46,120 one way or the other. 525 00:25:46,120 --> 00:25:52,420 If we translate this equation into a reference triangle, we 526 00:25:52,420 --> 00:25:53,560 have, what? 527 00:25:53,560 --> 00:25:56,070 We'll call the angle 'x/2'. 528 00:25:56,070 --> 00:25:58,870 And the tangent is 'z', so we'll make the side opposite 529 00:25:58,870 --> 00:26:01,070 'z', the side-adjacent 1. 530 00:26:01,070 --> 00:26:03,280 That makes the hypotenuse the square root 531 00:26:03,280 --> 00:26:05,410 of '1 plus z squared'. 532 00:26:05,410 --> 00:26:07,470 Now, watch what happens when you do this. 533 00:26:07,470 --> 00:26:12,500 See, 'z' is equal to '10x/2', therefore, 'dz' is the 534 00:26:12,500 --> 00:26:15,180 differential of '10x/2'. 535 00:26:15,180 --> 00:26:19,950 Remember, the differential of '10x', with respect to 'x', is 536 00:26:19,950 --> 00:26:21,500 secant squared. 537 00:26:21,500 --> 00:26:24,710 But we also, by the chain rule, have to multiply by a 538 00:26:24,710 --> 00:26:26,980 derivative of 'x/2' with respect to 'x'. 539 00:26:26,980 --> 00:26:31,960 In other words, if 'z' is '10x/2', 'dz' is not secant 540 00:26:31,960 --> 00:26:37,750 squared 'x/2', it's 1/2 'secant squared 'x/2' dx'. 541 00:26:37,750 --> 00:26:40,460 But what's 'secant squared 'x/2''? 542 00:26:40,460 --> 00:26:43,540 Let's go back and look at our diagram. 543 00:26:43,540 --> 00:26:47,790 The secant of 'x/2' is the hypotenuse over side-adjacent. 544 00:26:47,790 --> 00:26:52,010 That's the square root of '1 plus 'z squared'' over 1 and, 545 00:26:52,010 --> 00:26:56,990 therefore, the square of the secant is just '1 plus 'z 546 00:26:56,990 --> 00:27:00,280 squared'' over 1, see, '1 plus 'z squared''. 547 00:27:00,280 --> 00:27:03,510 So, with the 1/2 in here, this becomes '1 plus 'z squared'' 548 00:27:03,510 --> 00:27:05,710 over 2 times 'dx'. 549 00:27:05,710 --> 00:27:08,680 And, consequently, if I compare these two now, notice 550 00:27:08,680 --> 00:27:13,390 that 'dx' is just twice 'dz'/ '1 plus 'z squared''. 551 00:27:13,390 --> 00:27:15,300 In other words, what's happened to 'dx'? 552 00:27:15,300 --> 00:27:18,350 It's been replaced by a differential in 'z', which 553 00:27:18,350 --> 00:27:21,100 involves the quotient of two polynomials. 554 00:27:21,100 --> 00:27:24,140 So, 'dx' comes out very nicely this way, in terms of, what? 555 00:27:24,140 --> 00:27:26,680 The quotient of two polynomials. 556 00:27:26,680 --> 00:27:28,840 How about 'sine x' and 'cosine x'? 557 00:27:28,840 --> 00:27:32,690 And, again, notice the dependency on identities. 558 00:27:32,690 --> 00:27:37,490 'Sine x' is twice 'sine 'x/2' cosine 'x/2''. 559 00:27:37,490 --> 00:27:41,020 But, from my reference triangle, I can pick off the 560 00:27:41,020 --> 00:27:44,150 trigonometric functions of 'x/2' very easily. 561 00:27:44,150 --> 00:27:48,400 Namely, the sine of 'x/2' is just 'z' over the square root 562 00:27:48,400 --> 00:27:53,300 of '1 plus 'z squared'', and the cosine of 'x/2' is just 1 563 00:27:53,300 --> 00:27:56,280 over the square root of '1 plus 'z squared''. 564 00:27:56,280 --> 00:28:00,390 Plugging that in here, I find that 'sine x' is twice 'z' 565 00:28:00,390 --> 00:28:03,960 over the square root of '1 plus 'z squared'' times 1 over 566 00:28:03,960 --> 00:28:06,190 the square root of '1 plus 'z squared''. 567 00:28:06,190 --> 00:28:08,500 Multiplying this out, I find, what? 568 00:28:08,500 --> 00:28:12,610 That 'sine x' is twice 'z', and the square root of '1 plus 569 00:28:12,610 --> 00:28:16,020 'z squared'' times itself is just '1 plus 'z squared''. 570 00:28:16,020 --> 00:28:20,420 In other words, 'sine x' is '2z'/ '1 plus 'z squared''. 571 00:28:20,420 --> 00:28:22,700 In other words, with the substitution, what 572 00:28:22,700 --> 00:28:25,320 happens to 'sine x'? 573 00:28:25,320 --> 00:28:29,870 It becomes '2z'/ '1 plus 'z squared'', which is also the 574 00:28:29,870 --> 00:28:33,530 quotient of two polynomials in 'z', OK? 575 00:28:33,530 --> 00:28:36,610 Finally, how about 'cosine x'? 576 00:28:36,610 --> 00:28:40,650 What identity can we use to reduce 'cosine 577 00:28:40,650 --> 00:28:43,360 x' in terms of 'x/2'? 578 00:28:43,360 --> 00:28:45,270 Why do we want the 'x/2'? 579 00:28:45,270 --> 00:28:48,870 Again, notice that even though we may not have invented this 580 00:28:48,870 --> 00:28:52,740 substitution by ourself, once it's invented, the 581 00:28:52,740 --> 00:28:56,260 relationship to the angle 'x/2' becomes very apparent. 582 00:28:56,260 --> 00:28:59,200 At any rate, notice the identity that says that 583 00:28:59,200 --> 00:29:03,500 'cosine 2x' is 'cosine squared x' minus 'sine squared x' 584 00:29:03,500 --> 00:29:08,010 translates into 'cosine x' is cosine squared of half the 585 00:29:08,010 --> 00:29:11,330 angle minus sine squared of half the angle. 586 00:29:11,330 --> 00:29:17,400 Well, 'cosine squared 'x/2'', well, 'cosine 'x/2'' is just 1 587 00:29:17,400 --> 00:29:20,570 over the square root of '1 plus 'c squared'', so 'cosine 588 00:29:20,570 --> 00:29:24,600 squared 'x/2'' is just 1/ '1 plus 'z squared''. 589 00:29:24,600 --> 00:29:28,620 Similarly, 'sine squared 'x/2'' is just 'z squared' 590 00:29:28,620 --> 00:29:30,630 over '1 plus 'z squared''. 591 00:29:30,630 --> 00:29:34,450 And what we find is that 'cosine x' is '1 minus 'z 592 00:29:34,450 --> 00:29:38,500 squared'' over '1 plus 'z squared''. 593 00:29:38,500 --> 00:29:40,020 And, again, what's happened? 594 00:29:40,020 --> 00:29:43,620 'Cosine x' is now expressable as the quotient of two 595 00:29:43,620 --> 00:29:45,520 polynomials in 'z'. 596 00:29:45,520 --> 00:29:48,840 In fact, as an application of this, let's come back to an 597 00:29:48,840 --> 00:29:51,340 integrand that's been giving us some trouble for 598 00:29:51,340 --> 00:29:52,800 quite some time now. 599 00:29:52,800 --> 00:29:57,350 Let's look at the integral 'secant x dx' and see if we 600 00:29:57,350 --> 00:30:00,750 can't use this technique that we've just learned to solve 601 00:30:00,750 --> 00:30:02,780 this particular problem. 602 00:30:02,780 --> 00:30:06,750 Notice that secant is 1 over cosine, therefore, integral 603 00:30:06,750 --> 00:30:11,020 'secant x dx' is integral 'dx' over 'cosine x'. 604 00:30:11,020 --> 00:30:13,010 Now, let's come back here for a moment just 605 00:30:13,010 --> 00:30:14,280 to refresh our memories. 606 00:30:14,280 --> 00:30:20,290 We saw that 'dx' was two 'dz'/ '1 plus 'z squared''. 607 00:30:20,290 --> 00:30:26,320 We saw that 'cosine x' was '1 minus 'z squared'' over '1 608 00:30:26,320 --> 00:30:30,515 plus 'z squared'', therefore, '1 over cosine x' will just be 609 00:30:30,515 --> 00:30:31,840 the reciprocal of this. 610 00:30:31,840 --> 00:30:35,740 In other words, coming back here now, we can replace 'dx' 611 00:30:35,740 --> 00:30:40,920 by '2 dz' over '1 plus 'z squared'' and '1 over cosine 612 00:30:40,920 --> 00:30:43,950 x' by '1 plus 'z squared'' over '1 minus 613 00:30:43,950 --> 00:30:46,840 'z squared'', OK? 614 00:30:46,840 --> 00:30:50,710 Therefore, in terms of this substitution, integral 'secant 615 00:30:50,710 --> 00:30:57,290 x dx' is just twice integral 'dz'/ '1 minus 'z squared''. 616 00:30:57,290 --> 00:30:59,010 But, what is this integrand? 617 00:30:59,010 --> 00:31:04,030 This integrand is the quotient of two polynomials in 'z', 618 00:31:04,030 --> 00:31:07,590 therefore, I could use partial fractions here, write this as 619 00:31:07,590 --> 00:31:13,600 a term: something over '1 plus z' plus something over '1 620 00:31:13,600 --> 00:31:17,270 minus z', et cetera, solve the problem in terms of 'z', and 621 00:31:17,270 --> 00:31:22,460 then, remembering that 'z' is 'tangent 'x/2'', I can then 622 00:31:22,460 --> 00:31:25,596 replace 'z' by what its equal to in terms of 'x', and in 623 00:31:25,596 --> 00:31:27,460 that way solve the problem. 624 00:31:27,460 --> 00:31:30,020 Now, you see, what I want you to see again, that comes true 625 00:31:30,020 --> 00:31:33,590 here all the time, is how we are continually looking for 626 00:31:33,590 --> 00:31:39,650 ways of reducing integrals to equivalent integrals, but, 627 00:31:39,650 --> 00:31:43,290 hopefully, ones that are more familiar to us, meaning, what? 628 00:31:43,290 --> 00:31:45,670 Ones that we are able to handle. 629 00:31:45,670 --> 00:31:49,000 In the next lesson, we're going to find a very powerful 630 00:31:49,000 --> 00:31:51,950 technique, which is far more general than partial 631 00:31:51,950 --> 00:31:55,560 fractions, a technique which is used over, and over, again, 632 00:31:55,560 --> 00:31:58,920 which is probably the single most important technique. 633 00:31:58,920 --> 00:32:00,970 But, I won't say any more about that right now. 634 00:32:00,970 --> 00:32:03,560 We'll continue with this discussion next time. 635 00:32:03,560 --> 00:32:07,400 And, until next time, good bye. 636 00:32:07,400 --> 00:32:10,390 MALE VOICE: Funding for the publication of this video was 637 00:32:10,390 --> 00:32:15,110 provided by the Gabriella and Paul Rosenbaum Foundation. 638 00:32:15,110 --> 00:32:19,280 Help OCW continue to provide free and open access to MIT 639 00:32:19,280 --> 00:32:23,740 courses by making a donation at ocw.mit.edu/donate.