WEBVTT 00:00:00.040 --> 00:00:01.940 NARRATOR: The following content is provided under a 00:00:01.940 --> 00:00:03.690 Creative Commons license. 00:00:03.690 --> 00:00:06.630 Your support will help MIT OpenCourseware continue to 00:00:06.630 --> 00:00:09.980 offer high-quality educational resources for free. 00:00:09.980 --> 00:00:12.830 To make a donation or to view additional materials from 00:00:12.830 --> 00:00:16.760 hundreds of MIT courses, visit MIT OpenCourseWare at 00:00:16.760 --> 00:00:18.010 ocw.mit.edu. 00:00:34.750 --> 00:00:35.500 PROFESSOR: Hi. 00:00:35.500 --> 00:00:39.790 The aim of our lecture today is to present what is perhaps 00:00:39.790 --> 00:00:43.840 the singly most powerful recipe for evaluating 00:00:43.840 --> 00:00:47.160 indefinite or, for that matter, definite integrals. 00:00:47.160 --> 00:00:50.920 It's called integration by parts, and therefore, today's 00:00:50.920 --> 00:00:53.420 lecture is called 'Integration by Parts'. 00:00:53.420 --> 00:00:56.940 And what I want to emphasize is essentially that this is a 00:00:56.940 --> 00:01:00.960 technique which shows us how to find indefinite integrals, 00:01:00.960 --> 00:01:03.510 but then in theory, it changes nothing over 00:01:03.510 --> 00:01:04.319 what we've had before. 00:01:04.319 --> 00:01:07.530 This is essentially, as we've mentioned, a recipe for being 00:01:07.530 --> 00:01:12.000 able to simplify or reduce indefinite integrals to things 00:01:12.000 --> 00:01:12.960 that we can handle. 00:01:12.960 --> 00:01:18.390 Well, it centers around the rule for 00:01:18.390 --> 00:01:19.660 differentiating a product. 00:01:19.660 --> 00:01:22.150 And to show you what I mean by that, let's just barge right 00:01:22.150 --> 00:01:23.050 into this thing. 00:01:23.050 --> 00:01:26.460 In differential notation, we call the 'u' and 'v' our 00:01:26.460 --> 00:01:29.260 differentiable functions of 'x'. 00:01:29.260 --> 00:01:32.690 The differential of 'uv' is 'u' times the differential of 00:01:32.690 --> 00:01:35.830 'v', plus 'v' times the differential of 'u'. 00:01:35.830 --> 00:01:39.190 And solving for 'udv', just transposing, we have that 00:01:39.190 --> 00:01:44.820 'udv' is the differential of 'u' times 'v' minus 'vdu'. 00:01:44.820 --> 00:01:46.960 Now, what we do is we integrate both 00:01:46.960 --> 00:01:49.580 sides of this equality. 00:01:49.580 --> 00:01:52.920 On the left-hand side, we have interval udv. 00:01:52.920 --> 00:01:56.270 On the right-hand side, recall that the integral of the 00:01:56.270 --> 00:02:00.145 derivative just gives us back the original function. 00:02:00.145 --> 00:02:02.580 In other words, if we integrate the differential of 00:02:02.580 --> 00:02:05.210 'uv' up to an arbitrary constant, we get 00:02:05.210 --> 00:02:07.150 back 'u' times 'v'. 00:02:07.150 --> 00:02:11.480 The integral of 'vdu' we'll just denote by integral 'vdu'. 00:02:11.480 --> 00:02:14.740 And observe that there's really an arbitrary constant 00:02:14.740 --> 00:02:17.210 coming from this term, an arbitrary constant coming from 00:02:17.210 --> 00:02:19.130 this term, an arbitrary constant 00:02:19.130 --> 00:02:20.280 coming from this term. 00:02:20.280 --> 00:02:22.920 But since the sum and difference of arbitrary 00:02:22.920 --> 00:02:25.460 constants is still an arbitrary constant, we can 00:02:25.460 --> 00:02:28.950 neglected all the constants here, and just lump them 00:02:28.950 --> 00:02:31.730 together, and at the end write plus 'c'. 00:02:31.730 --> 00:02:35.330 Or we can leave the 'c' off, making it implied that the 00:02:35.330 --> 00:02:38.330 arbitrary constant is contained in the symbol 00:02:38.330 --> 00:02:39.820 integral 'vdu'. 00:02:39.820 --> 00:02:42.390 All right, now we've done this, but a much more 00:02:42.390 --> 00:02:45.060 important question would be-- or exclamation, for that 00:02:45.060 --> 00:02:47.550 matter-- would be so what? 00:02:47.550 --> 00:02:51.210 See, who cares whether this happens to be true or not? 00:02:51.210 --> 00:02:52.280 The answer is we care. 00:02:52.280 --> 00:02:54.270 That's why we're making up a lecture on it. 00:02:54.270 --> 00:02:56.510 And the reason that we care is this. 00:02:56.510 --> 00:03:01.490 That frequently, for a given 'u' and 'v', 'udv' may have a 00:03:01.490 --> 00:03:06.040 completely different nature about it than does 'vdu' 00:03:06.040 --> 00:03:09.230 insofar as being able to find the integral. 00:03:09.230 --> 00:03:12.490 Now again, the best way to show what we mean by this is 00:03:12.490 --> 00:03:15.040 to choose a particular example. 00:03:15.040 --> 00:03:18.680 Let me start, just arbitrarily here, picking a couple of 00:03:18.680 --> 00:03:21.930 simple functions for us to differentiate or to integrate. 00:03:21.930 --> 00:03:26.300 Let's pick 'u' to equal 'x' and 'v' to equal 'sine x'. 00:03:26.300 --> 00:03:29.330 Now, it's simple to see from this example that in this 00:03:29.330 --> 00:03:33.970 case, if 'u' equals 'x' and 'v' is 'sine x', then 'dv' is 00:03:33.970 --> 00:03:35.356 ''cosine x' dx'. 00:03:35.356 --> 00:03:40.380 In fact, let's write that down so we can see it here. 00:03:40.380 --> 00:03:43.790 Notice that in this particular case, integral 'udv' is 00:03:43.790 --> 00:03:46.670 integral of 'x 'cosine x' dx'-- 00:03:46.670 --> 00:03:48.260 'udv'. 00:03:48.260 --> 00:03:51.500 On the other hand, what is 'vdu'? 00:03:51.500 --> 00:03:53.350 'v' is 'sine x'. 00:03:53.350 --> 00:04:01.790 Then since 'u' equals 'x', 'du' is 'dx'. 00:04:01.790 --> 00:04:06.550 Therefore, 'vdu' is just integral ''sine x' dx'. 00:04:06.550 --> 00:04:10.200 Now, look at these two integrals. 00:04:10.200 --> 00:04:13.640 This one we can integrate just by looking at it. 00:04:13.640 --> 00:04:15.760 It's minus 'cosine x'. 00:04:15.760 --> 00:04:19.649 This one, well, offhand, we don't know a function whose 00:04:19.649 --> 00:04:23.720 derivative with respect to 'x' is 'x 'cosine x''. 00:04:23.720 --> 00:04:29.360 However, by this recipe, we now have a way of showing how 00:04:29.360 --> 00:04:36.070 to 'udv' once we know integral 'vdu'. 00:04:36.070 --> 00:04:39.140 In other words, in this particular problem, letting 00:04:39.140 --> 00:04:46.890 'u' equal 'x' and 'dv' being ''cosine x' dx', we can 00:04:46.890 --> 00:04:48.240 proceed as follows. 00:04:48.240 --> 00:04:51.480 Namely, by this recipe, we get what? 00:04:51.480 --> 00:04:57.130 Integral 'x 'cosine x' dx' is equal to-- this is 'udv' 00:04:57.130 --> 00:04:58.790 equals 'u' times 'v'. 00:04:58.790 --> 00:05:02.370 Recall that 'u' was 'x', 'v' is 'sine x', so 'u' times 'v' 00:05:02.370 --> 00:05:08.010 is 'x sine x', minus integral 'vdu'. 00:05:08.010 --> 00:05:12.460 But 'v' is 'sine x', 'du' is 'dx', so we wind up with this. 00:05:12.460 --> 00:05:15.160 But now we know that the integral of 'sine x' with 00:05:15.160 --> 00:05:17.740 respect to 'x' is minus the 'cosine x'. 00:05:17.740 --> 00:05:20.640 The minus and the minus cancel to give me a plus. 00:05:20.640 --> 00:05:24.330 And I wind up with that the integral of 'x 'cosine x' dx' 00:05:24.330 --> 00:05:27.770 is 'x sine x' plus 'cosine x' plus an 00:05:27.770 --> 00:05:30.790 arbitrary constant, OK. 00:05:30.790 --> 00:05:33.760 That's all integration by parts really is. 00:05:33.760 --> 00:05:37.970 All we do here is we evaluate one integral by knowing 00:05:37.970 --> 00:05:38.970 another integral. 00:05:38.970 --> 00:05:42.600 And by the way, what I meant here when I said that one did 00:05:42.600 --> 00:05:46.710 not have to change any of the theory, what does the integral 00:05:46.710 --> 00:05:47.720 mean here anyway? 00:05:47.720 --> 00:05:49.600 What we mean by this is what? 00:05:49.600 --> 00:05:53.240 Find all functions whose derivative with respect to 'x' 00:05:53.240 --> 00:05:55.100 is 'x cosine x'. 00:05:55.100 --> 00:05:58.780 Now, what integration by parts did for us was it showed us 00:05:58.780 --> 00:06:00.330 how to get this result. 00:06:00.330 --> 00:06:03.620 But if we didn't know how to get this result, observe that 00:06:03.620 --> 00:06:06.030 we check the answer the same way as always. 00:06:06.030 --> 00:06:09.100 Namely, one thing that hopefully we can do is 00:06:09.100 --> 00:06:10.600 differentiate this thing. 00:06:10.600 --> 00:06:12.040 And if we differentiate it, let's see. 00:06:12.040 --> 00:06:13.140 This is a product. 00:06:13.140 --> 00:06:15.490 It's the first factor times the derivative of the second. 00:06:15.490 --> 00:06:17.380 That's 'x cosine x'. 00:06:17.380 --> 00:06:20.430 Plus the second factor times the derivative of the first. 00:06:20.430 --> 00:06:22.100 That's just 'sine x'. 00:06:22.100 --> 00:06:24.720 The derivative of 'cosine x' is minus 'sine x'. 00:06:24.720 --> 00:06:26.350 The derivative of a constant is 0. 00:06:26.350 --> 00:06:29.440 And if we now sum up the pieces that we have here, we 00:06:29.440 --> 00:06:31.900 wind up with 'x cosine x'. 00:06:31.900 --> 00:06:35.650 Exactly what we were supposed to get, OK? 00:06:35.650 --> 00:06:38.190 Now, you see again, nothing has happened in theory. 00:06:38.190 --> 00:06:42.680 All we have now is a technique that allows us to explicitly 00:06:42.680 --> 00:06:45.210 solve certain problems that we couldn't have solved 00:06:45.210 --> 00:06:48.420 otherwise, even though we could've stated them before. 00:06:48.420 --> 00:06:51.570 For example, let's go back to a problem that technically 00:06:51.570 --> 00:06:54.150 speaking we could have presented before we got to 00:06:54.150 --> 00:06:56.340 this unit of our material. 00:06:56.340 --> 00:06:59.930 Let's support that the region 'R' is the region bounded 00:06:59.930 --> 00:07:04.240 above by 'y' equals 'cosine x', below by the x-axis 00:07:04.240 --> 00:07:07.490 between the y-axis at 'x' equals pi/2. 00:07:07.490 --> 00:07:11.040 What we want to do is take this region 'R' and revolve it 00:07:11.040 --> 00:07:12.530 about the y-axis. 00:07:12.530 --> 00:07:16.100 And what we'd like to find is what volume is generated when 00:07:16.100 --> 00:07:19.160 the region 'R' is revolved about the y-axis. 00:07:19.160 --> 00:07:22.020 Now, when we talked about volumes and cylindrical 00:07:22.020 --> 00:07:24.420 shells, we already had the recipe. 00:07:24.420 --> 00:07:27.630 Recall that to rotate this about the y-axis, just to 00:07:27.630 --> 00:07:30.940 refresh memories here, if we call this 'x' and we call this 00:07:30.940 --> 00:07:33.620 'y', the element of volume was what? 00:07:33.620 --> 00:07:38.010 Well, you swung out a circumference '2 pi x' times 00:07:38.010 --> 00:07:40.460 height 'y' times thickness 'dx'. 00:07:40.460 --> 00:07:43.590 In other words, that the volume when this was rotated 00:07:43.590 --> 00:07:50.620 about the y-axis was just 2 pi integral 'x 'cosine x' dx' as 00:07:50.620 --> 00:07:52.850 'x' goes from 0 to pi/2. 00:07:52.850 --> 00:07:55.300 You see, I rigged this particular problem so we 00:07:55.300 --> 00:07:57.800 wouldn't have to waste the fact that we've already found 00:07:57.800 --> 00:08:00.040 out what the indefinite integral of 'x 00:08:00.040 --> 00:08:01.650 'cosine x' dx' is. 00:08:01.650 --> 00:08:03.870 In other words, we get down to this step here. 00:08:03.870 --> 00:08:07.180 Now, by the first fundamental theorem of integral calculus, 00:08:07.180 --> 00:08:10.490 we know that whatever this integral is, it's also equal 00:08:10.490 --> 00:08:15.830 to 'G of pi/2' minus 'G of 0', where 'G' is any function of 00:08:15.830 --> 00:08:18.670 'x' whose derivative with respect to 'x' is the 00:08:18.670 --> 00:08:21.690 integrand here, which is 'x cosine x'. 00:08:21.690 --> 00:08:25.040 Now, you see, this part, or up to here, we could have got 00:08:25.040 --> 00:08:27.610 this result without today's lecture. 00:08:27.610 --> 00:08:30.370 The only thing that we learned to do today that we couldn't 00:08:30.370 --> 00:08:34.059 do before was the idea of how we find a function whose 00:08:34.059 --> 00:08:35.900 derivative is 'x cosine x'. 00:08:35.900 --> 00:08:38.030 In fact, what we found was what? 00:08:38.030 --> 00:08:42.289 A function whose derivative is 'x cosine x' is just 'x sine 00:08:42.289 --> 00:08:44.390 x' plus 'cosine x'. 00:08:44.390 --> 00:08:47.220 So by the first fundamental theorem, the answer to our 00:08:47.220 --> 00:08:51.810 volume problem is that it's 2 pi times the quantity 'x sine 00:08:51.810 --> 00:08:56.330 x' plus 'cosine x' evaluated between 0 and pi/2. 00:08:56.330 --> 00:09:00.660 Putting in 'x' equals pi/2, this term is pi/2. 00:09:00.660 --> 00:09:02.240 This term is 0. 00:09:02.240 --> 00:09:05.890 Putting in the lower limit when 'x' is 0, this term is 0. 00:09:05.890 --> 00:09:07.030 This term is 1. 00:09:07.030 --> 00:09:08.740 We subtract the lower limit. 00:09:08.740 --> 00:09:10.310 That gives us a minus 1. 00:09:10.310 --> 00:09:14.860 And therefore, the volume that we get is simply 2 pi times 00:09:14.860 --> 00:09:16.860 pi/2 minus 1. 00:09:16.860 --> 00:09:19.830 Again, notice then, that the only new thing that we had to 00:09:19.830 --> 00:09:23.890 do here was devise a way for finding a function whose 00:09:23.890 --> 00:09:27.510 derivative with respect to 'x' was 'x cosine x'. 00:09:27.510 --> 00:09:30.330 By the way, I think you'll observe over here that I 00:09:30.330 --> 00:09:34.210 started in kind of a rigged fashion, namely, I started 00:09:34.210 --> 00:09:36.040 with a specific 'u' and 'v'. 00:09:36.040 --> 00:09:39.290 It might be interesting to see how would one tackle this 00:09:39.290 --> 00:09:43.500 problem if one had just been given the 'x 'cosine x' dx' 00:09:43.500 --> 00:09:45.040 and had to work from there. 00:09:45.040 --> 00:09:47.960 In other words, instead of working backwards now, let's 00:09:47.960 --> 00:09:50.310 present the problem as it would come up in a real 00:09:50.310 --> 00:09:53.430 problem, namely, all of a sudden, we find that we have 00:09:53.430 --> 00:09:55.990 to integrate 'x cosine x'. 00:09:55.990 --> 00:09:58.100 And the idea go something like this. 00:09:58.100 --> 00:09:59.120 You see, we say lookit. 00:09:59.120 --> 00:10:01.790 If this 'x' weren't in here, this would be 00:10:01.790 --> 00:10:02.940 pretty easy to handle. 00:10:02.940 --> 00:10:04.130 So why don't we do this? 00:10:04.130 --> 00:10:07.260 Why don't we call this piece here 'u' and call 00:10:07.260 --> 00:10:10.060 this piece here 'dv'. 00:10:10.060 --> 00:10:14.730 Then, you see, 'du' would just be 'dx'. 00:10:14.730 --> 00:10:17.200 'v' would just be 'sine x'. 00:10:17.200 --> 00:10:20.230 In other words, the differential of 'sine x' is 00:10:20.230 --> 00:10:21.970 'cosine x dx'. 00:10:21.970 --> 00:10:24.640 See, if 'u' is 'x', 'du' is 'dx'. 00:10:24.640 --> 00:10:28.480 If 'dv' is 'cosine x dx', 'v' is equal to 'sine x'. 00:10:28.480 --> 00:10:31.070 I suppose, technically speaking here, I should put in 00:10:31.070 --> 00:10:35.060 plus a constant, but I'll show later why we can disregard 00:10:35.060 --> 00:10:36.010 this constant. 00:10:36.010 --> 00:10:38.290 Roughly speaking, it just hinges on the fact that we 00:10:38.290 --> 00:10:39.770 want only one solution. 00:10:39.770 --> 00:10:41.330 But we'll talk about that later. 00:10:41.330 --> 00:10:44.380 For the time being, I'll just leave out the fact that 00:10:44.380 --> 00:10:45.590 there's a whole family of functions. 00:10:45.590 --> 00:10:48.170 All I want is one function whose differential 00:10:48.170 --> 00:10:49.140 is 'cosine x dx'. 00:10:49.140 --> 00:10:54.520 At any rate, I now apply my recipe, namely, integral 'udv' 00:10:54.520 --> 00:10:57.260 is equal to 'u' times 'v'-- 00:10:57.260 --> 00:10:59.100 that's 'x sine x'-- 00:10:59.100 --> 00:11:03.580 minus integral 'v', which is 'sine x', times 'du', 00:11:03.580 --> 00:11:05.050 which is just 'dx'. 00:11:05.050 --> 00:11:08.740 In other words, integral 'x 'cosine x' dx' is just 'x sine 00:11:08.740 --> 00:11:11.210 x' minus 'sine x dx'. 00:11:11.210 --> 00:11:13.620 But I know how to integrate 'sine x'. 00:11:13.620 --> 00:11:15.500 It's just minus 'cosine x'. 00:11:15.500 --> 00:11:19.720 Therefore, replacing integral 'sine x dx' by minus 'cosine 00:11:19.720 --> 00:11:21.500 x' plus a constant. 00:11:21.500 --> 00:11:22.860 The minus cancels here. 00:11:22.860 --> 00:11:26.830 I wind up with the same result as before, only now hopefully 00:11:26.830 --> 00:11:29.410 in terms of a smooth flow not that didn't 00:11:29.410 --> 00:11:30.950 presuppose the answer. 00:11:30.950 --> 00:11:33.860 Namely, this integral is 'x sine x' plus 00:11:33.860 --> 00:11:36.000 'cosine x' plus a constant. 00:11:36.000 --> 00:11:38.400 And by the way, I should mention here that this 00:11:38.400 --> 00:11:42.060 technique does require a certain amount of ingenuity. 00:11:42.060 --> 00:11:45.300 In other words, in a given problem, where you're not told 00:11:45.300 --> 00:11:48.580 what to call 'u' and what to call 'v', where you have the 00:11:48.580 --> 00:11:51.820 choice of picking what's going to be 'u' and what's going to 00:11:51.820 --> 00:11:55.610 be 'dv', there is a certain element of insight that one 00:11:55.610 --> 00:11:57.880 needs to guess wisely. 00:11:57.880 --> 00:12:00.470 In this particular problem, it's kind of obvious 00:12:00.470 --> 00:12:01.360 which way to guess. 00:12:01.360 --> 00:12:04.370 But let me, just to give you the experience, show you what 00:12:04.370 --> 00:12:06.270 happens when you guess incorrectly. 00:12:06.270 --> 00:12:09.600 For example, given the same problem, why couldn't I have 00:12:09.600 --> 00:12:14.100 said gee, I think I'll let 'cosine x' be 'u', and what's 00:12:14.100 --> 00:12:16.290 left over, namely, 'xdx'. 00:12:16.290 --> 00:12:18.790 That will be 'dv'. 00:12:18.790 --> 00:12:20.850 Now, notice that mathematically, no harm is 00:12:20.850 --> 00:12:21.760 done this way. 00:12:21.760 --> 00:12:26.080 Namely, if 'u' is 'cosine x', 'du' is minus 'sine x dx'. 00:12:26.080 --> 00:12:30.040 If 'dv' is 'xdx', 'v' is 1/2 'x squared'. 00:12:30.040 --> 00:12:34.260 And now, applying the recipe, integral 'udv' is what? 00:12:34.260 --> 00:12:35.460 It's 'uv'. 00:12:35.460 --> 00:12:40.050 That's 1/2 'x squared' times 'cosine x' minus 00:12:40.050 --> 00:12:42.140 the integral 'vdu'. 00:12:42.140 --> 00:12:47.780 But 'vdu' is already minus 1/2 'x squared' 'sine x dx', so 00:12:47.780 --> 00:12:51.020 minus that is plus 1/2 integral ''x 00:12:51.020 --> 00:12:52.670 squared' 'sine x' dx'. 00:12:52.670 --> 00:12:55.920 And I wind up with the result that the integral 'x 'cosine 00:12:55.920 --> 00:13:00.820 x' dx' is 1/2 'x squared' 'cosine x' plus 1/2 integral 00:13:00.820 --> 00:13:02.690 ''x squared' 'sine x' dx'. 00:13:02.690 --> 00:13:05.460 And there is absolutely nothing at all wrong with 00:13:05.460 --> 00:13:12.420 this, except that the chances are if I didn't know how to 00:13:12.420 --> 00:13:15.640 find the function whose derivative was 'x cosine x', 00:13:15.640 --> 00:13:18.330 what is the likelihood that I know a function whose 00:13:18.330 --> 00:13:20.760 derivative is ''x squared' 'sine x' dx'? 00:13:20.760 --> 00:13:26.140 In other words, here's a case where replacing 'udv' by 'vdu' 00:13:26.140 --> 00:13:29.930 gave me an equivalent integral to evaluate which was more 00:13:29.930 --> 00:13:31.830 cumbersome than the first. 00:13:31.830 --> 00:13:35.140 In fact, somehow or other, it would seem that by multiplying 00:13:35.140 --> 00:13:39.560 through here by 2 and solving for ''x squared' 'sine x' dx', 00:13:39.560 --> 00:13:42.940 that it would seem that somehow or other this would be 00:13:42.940 --> 00:13:45.560 a way of simplifying the integral ''x 00:13:45.560 --> 00:13:47.210 squared' 'sine x' dx'. 00:13:47.210 --> 00:13:50.170 In other words, if you had started with this, I think 00:13:50.170 --> 00:13:53.110 you'd accept the fact, well, this still isn't as nice as 00:13:53.110 --> 00:13:57.430 I'd like it, but it does seem less complicated than this. 00:13:57.430 --> 00:13:59.730 You see, at least the power of 'x' is only to the first power 00:13:59.730 --> 00:14:02.050 here, whereas here it's the second power. 00:14:02.050 --> 00:14:04.790 In fact, this leads to a technique 00:14:04.790 --> 00:14:07.170 called a reduction formula. 00:14:07.170 --> 00:14:10.270 Again, using hindsight rather than foresight, having 00:14:10.270 --> 00:14:13.850 motivated the integral ''x squared' 'sine x' dx', let's 00:14:13.850 --> 00:14:16.640 see how I could have solved this particular problem if I 00:14:16.640 --> 00:14:20.380 didn't know what the answer was going to be in advance. 00:14:20.380 --> 00:14:23.000 You see, I say to myself, you know, this x squared here is 00:14:23.000 --> 00:14:23.980 bothering me. 00:14:23.980 --> 00:14:26.270 If that 'x squared' weren't here, this would be pretty 00:14:26.270 --> 00:14:27.280 easy to handle. 00:14:27.280 --> 00:14:30.560 Well, the best way to knock the 'x squared' down is to let 00:14:30.560 --> 00:14:32.840 'u' equal 'x squared', in which case, 00:14:32.840 --> 00:14:34.910 'du' will be '2xdx'. 00:14:34.910 --> 00:14:38.710 In other words, your second power will now be replaced by 00:14:38.710 --> 00:14:39.660 a first power. 00:14:39.660 --> 00:14:42.810 It's not perfect, but at least there's hope here. 00:14:42.810 --> 00:14:46.660 At any rate, what I'm saying is motivated into calling 'u' 00:14:46.660 --> 00:14:50.670 'x squared' and therefore letting 'dv' be the rest of 00:14:50.670 --> 00:14:53.870 this integrand, which is 'sine x dx', we wind up with the 00:14:53.870 --> 00:14:59.190 fact that 'du' is '2xdx' and 'v' is minus 'cosine x'. 00:14:59.190 --> 00:15:03.105 Well, you see now, remembering that 'u' is 'x squared', 'dv' 00:15:03.105 --> 00:15:07.420 is 'sine x dx', the recipe tells us that integral 'udv' 00:15:07.420 --> 00:15:09.360 is 'u' times 'v'. 00:15:09.360 --> 00:15:16.290 That's minus 'x cosine x' minus integral 'vdu'. 00:15:16.290 --> 00:15:21.480 But integral 'dvu' is integral twice 'x 'cosine x' dx' with a 00:15:21.480 --> 00:15:22.550 minus sign. 00:15:22.550 --> 00:15:25.040 The minus sign in the recipe makes it plus. 00:15:25.040 --> 00:15:28.350 And we therefore wind up with that integral ''x squared' 00:15:28.350 --> 00:15:33.470 'sine x' dx' is equal to minus ''x squared' 'cosine x'' plus 00:15:33.470 --> 00:15:37.650 twice integral 'x 'cosine x' dx'. 00:15:37.650 --> 00:15:40.640 By the way, just to refresh your memories on this, if we 00:15:40.640 --> 00:15:44.880 come back and compare this with what we had over here, 00:15:44.880 --> 00:15:47.820 it's exactly the same thing. 00:15:47.820 --> 00:15:50.080 The only difference now is that we arrived 00:15:50.080 --> 00:15:51.980 at this thing directly. 00:15:51.980 --> 00:15:56.220 Now you see the point is, having reduced this to this, I 00:15:56.220 --> 00:16:00.190 can now use integration by parts again to simplify this. 00:16:00.190 --> 00:16:03.150 In other words, what I would do next is let 'u' equal 'x' 00:16:03.150 --> 00:16:05.490 and 'dv' be 'cosine x dx'. 00:16:05.490 --> 00:16:07.730 But let's face it, I've already done this problem. 00:16:07.730 --> 00:16:08.950 Time is getting short. 00:16:08.950 --> 00:16:10.390 Why do it again? 00:16:10.390 --> 00:16:14.670 The point is that integral 'x cosine x' using integration by 00:16:14.670 --> 00:16:18.540 parts again is going to come out to be 'x sine x' plus 00:16:18.540 --> 00:16:22.060 'cosine x' plus 'c' because that's what we got before when 00:16:22.060 --> 00:16:24.310 we used integration by parts to do this. 00:16:24.310 --> 00:16:28.190 At any rate, what I would now do is replace integral 'x 00:16:28.190 --> 00:16:32.200 cosine x' in here by this, OK? 00:16:32.200 --> 00:16:34.060 And if I do that, I have what? 00:16:34.060 --> 00:16:40.750 Integral ''x squared' 'sine x' dx' is equal to what? 00:16:40.750 --> 00:16:45.470 Minus ''x squared' 'cosine x'' plus twice times this, which 00:16:45.470 --> 00:16:46.600 is twice this. 00:16:46.600 --> 00:16:47.340 That's what? 00:16:47.340 --> 00:16:53.050 2 'x sine x' plus 2 'cosine x' plus a constant. 00:16:53.050 --> 00:16:56.310 And by the way, again notice, if I wanted to check whether 00:16:56.310 --> 00:17:00.840 this was right, the one thing I can do if somebody gives me 00:17:00.840 --> 00:17:02.320 this is what? 00:17:02.320 --> 00:17:05.690 I can differentiate it with respect to 'x'. 00:17:05.690 --> 00:17:08.560 When I differentiate it with respect to 'x', if I get ''x 00:17:08.560 --> 00:17:12.040 squared' 'sine x'' as my result, this is correct. 00:17:12.040 --> 00:17:13.630 Otherwise, I've made a mistake. 00:17:13.630 --> 00:17:16.740 But the idea is keep in mind that these can always be 00:17:16.740 --> 00:17:19.250 checked because they're inverse derivatives. 00:17:19.250 --> 00:17:22.670 They can always be checked by actual differentiation. 00:17:22.670 --> 00:17:25.849 And again, notice the beauty of this technique. 00:17:25.849 --> 00:17:29.310 If somebody said construct the function whose derivative with 00:17:29.310 --> 00:17:32.730 respect to 'x' is ''x squared' 'sine x'', notice how 00:17:32.730 --> 00:17:37.630 difficult it would be to try to piece this thing together. 00:17:37.630 --> 00:17:39.800 See, notice the difference between being able to check 00:17:39.800 --> 00:17:43.060 that this is a right answer and how one would find this. 00:17:43.060 --> 00:17:46.980 This is the technique of a power of integration by parts. 00:17:46.980 --> 00:17:50.190 In fact, in making up exam questions, this is exactly how 00:17:50.190 --> 00:17:52.000 professor sometimes do this. 00:17:52.000 --> 00:17:55.610 That what you start with is you start with this, then give 00:17:55.610 --> 00:17:58.300 a person this problem and tell them to differentiate this. 00:17:58.300 --> 00:18:00.140 And that whole mess simply comes out to be ''x 00:18:00.140 --> 00:18:01.270 squared' 'sine x''. 00:18:01.270 --> 00:18:03.830 And that's why a lot of times when you have a messy problem 00:18:03.830 --> 00:18:06.190 to differentiate and you look at the answer in the back of 00:18:06.190 --> 00:18:09.060 the book, you become amazed to see that the thing really 00:18:09.060 --> 00:18:09.910 simplified. 00:18:09.910 --> 00:18:11.330 Again, notice the idea. 00:18:11.330 --> 00:18:14.630 The integrand here is relatively simple. 00:18:14.630 --> 00:18:18.350 What we have to differentiate to get that relatively simple 00:18:18.350 --> 00:18:20.960 integrand is not quite so simple. 00:18:20.960 --> 00:18:23.150 But enough about that. 00:18:23.150 --> 00:18:25.970 Let me just go on with a few more examples. 00:18:25.970 --> 00:18:29.450 See, I have plenty of examples in the exercises to keep you 00:18:29.450 --> 00:18:30.980 busy and to give you drill on this. 00:18:30.980 --> 00:18:34.450 All I'm trying to do now is to motivate places where one 00:18:34.450 --> 00:18:36.520 would want to use integration by parts. 00:18:36.520 --> 00:18:39.700 Well, let me show you one place it might be used. 00:18:39.700 --> 00:18:42.820 One place that it's always worth trying is when you look 00:18:42.820 --> 00:18:45.680 at the integrand, haven't the vaguest notion as to how to 00:18:45.680 --> 00:18:49.210 integrate it, but you are able to differentiate it. 00:18:49.210 --> 00:18:50.600 You see, you've got nothing to lose. 00:18:50.600 --> 00:18:53.370 The worst that will happen is that the resulting integral 00:18:53.370 --> 00:18:57.660 vdu will be no easier to handle than the original udv, 00:18:57.660 --> 00:18:59.800 but at least there's a chance that something will work. 00:18:59.800 --> 00:19:01.720 But let me give you a for instance. 00:19:01.720 --> 00:19:03.290 I see I've written this rather badly. 00:19:03.290 --> 00:19:04.570 This is not 'Stan'. 00:19:04.570 --> 00:19:10.080 This is the integral of inverse 'tangent x dx', OK? 00:19:10.080 --> 00:19:12.210 Integral inverse 'tangent x dx'. 00:19:12.210 --> 00:19:15.780 What function has its derivative equal to inverse 00:19:15.780 --> 00:19:16.720 'tangent x'? 00:19:16.720 --> 00:19:19.870 Well, you see, the thing I do know hopefully is what the 00:19:19.870 --> 00:19:22.320 derivative of inverse 'tangent x' is. 00:19:22.320 --> 00:19:25.100 The derivative of inverse 'tangent x' with respect to 00:19:25.100 --> 00:19:27.430 'x' is '1 over '1 plus x squared''. 00:19:27.430 --> 00:19:29.760 And by the way, if you don't know this, there's no hope of 00:19:29.760 --> 00:19:32.360 even beginning the integration by parts. in other words, we 00:19:32.360 --> 00:19:34.440 have to assume that we can do something in a 00:19:34.440 --> 00:19:35.280 problem like this. 00:19:35.280 --> 00:19:38.070 At any rate, knowing this, what do I do next? 00:19:38.070 --> 00:19:41.180 I say, well, since I know how to differentiate the inverse 00:19:41.180 --> 00:19:44.500 tangent, that's a very natural thing to call 'u' because I 00:19:44.500 --> 00:19:46.170 can certainly find 'du'. 00:19:46.170 --> 00:19:50.030 Moreover, the nice thing is that calling 'dv' 'dx', if I 00:19:50.030 --> 00:19:52.870 can integrate anything, it's certainly 'dx'. 00:19:52.870 --> 00:19:56.280 Namely, what I'm saying is if 'u' equals inverse 'tan x' and 00:19:56.280 --> 00:20:00.790 'dv' is 'dx', then 'du' is 'dx' over '1 plus x squared', 00:20:00.790 --> 00:20:02.980 and 'v' is just equal to 'x'. 00:20:02.980 --> 00:20:06.320 Now, therefore, integral inverse 'tan x dx', in other 00:20:06.320 --> 00:20:08.870 words, integral 'udv', is just what? 00:20:08.870 --> 00:20:11.150 It's still just 'u' times 'v'. 00:20:11.150 --> 00:20:18.200 That's 'x' times inverse 'tan x' minus 'vdu'. 00:20:18.200 --> 00:20:21.730 That's minus integral 'x' times 'dx' 00:20:21.730 --> 00:20:23.790 over '1 plus x squared'. 00:20:23.790 --> 00:20:25.530 Now, look what happens in this case. 00:20:25.530 --> 00:20:28.580 Obviously, with time being short and having other places 00:20:28.580 --> 00:20:31.520 to show you my failures, namely in the exercises, I 00:20:31.520 --> 00:20:33.120 picked one that worked nicely. 00:20:33.120 --> 00:20:37.260 Here's the case, you see, that the integral inverse 'tan x 00:20:37.260 --> 00:20:43.590 dx', by using parts, becomes replaced by integral 'xdx' 00:20:43.590 --> 00:20:45.300 over '1 plus x squared'. 00:20:45.300 --> 00:20:47.880 This is one I can handle from before. 00:20:47.880 --> 00:20:50.500 This is one where the substitution 'u' equals '1 00:20:50.500 --> 00:20:53.150 plus x squared' works very nicely for me. 00:20:53.150 --> 00:20:56.160 Or if you're getting glib with these things, just observe 00:20:56.160 --> 00:20:59.970 that this integrand is 1/2 natural 00:20:59.970 --> 00:21:01.630 log '1 plus x squared'. 00:21:01.630 --> 00:21:03.810 Because how would you differentiate natural log '1 00:21:03.810 --> 00:21:05.010 plus x squared'? 00:21:05.010 --> 00:21:09.760 It would just be what? '1 over '1 plus x squared'' times the 00:21:09.760 --> 00:21:11.370 derivative of this, which is '2x'. 00:21:11.370 --> 00:21:14.750 The 2 and the 1/2 would cancel, and we just have left 00:21:14.750 --> 00:21:17.280 'x over '1 plus x squared''. 00:21:17.280 --> 00:21:21.540 In other words, again, notice that the answer here is what? 00:21:21.540 --> 00:21:24.710 This is the function that one would have to differentiate 00:21:24.710 --> 00:21:29.730 with respect to 'x' to wind up with inverse 'tan x'. 00:21:29.730 --> 00:21:31.810 I'll show you another example in a few minutes. 00:21:31.810 --> 00:21:34.800 What I did just remember is that I mentioned why we can 00:21:34.800 --> 00:21:36.370 drop the arbitrary constant. 00:21:36.370 --> 00:21:39.250 Let me just take a minute here and show how this problem 00:21:39.250 --> 00:21:40.600 should have really been worked. 00:21:40.600 --> 00:21:44.180 If we wanted to be precise here, we'd say OK, 'u' equals 00:21:44.180 --> 00:21:45.620 inverse 'tan x'. 00:21:45.620 --> 00:21:48.850 Therefore, 'du' is 'dx over '1 plus x squared''. 00:21:48.850 --> 00:21:50.220 That part's fine. 00:21:50.220 --> 00:21:52.810 Then we say, OK, 'dv' is 'dx'. 00:21:52.810 --> 00:21:55.850 Now, if 'dv' is 'dx', what must 'v' be? 00:21:55.850 --> 00:21:59.280 Well, 'v' could be 'x', but it must be something of the form 00:21:59.280 --> 00:22:01.950 'x' plus some constant, which I'll call 'c1'. 00:22:01.950 --> 00:22:05.240 Could I have chosen 'v' to be 'x plus c1'? 00:22:05.240 --> 00:22:06.240 Why not? 00:22:06.240 --> 00:22:08.490 In fact, what would have happened if I did that? 00:22:08.490 --> 00:22:10.620 Notice that what would have happened in this case is that 00:22:10.620 --> 00:22:15.490 'u' times 'v' now would simply have been 'x plus c1' times 00:22:15.490 --> 00:22:20.060 inverse 'tan x', whereas integral 'vdu', instead of 00:22:20.060 --> 00:22:23.610 being 'xdx' over '1 plus x squared' would have been ''x 00:22:23.610 --> 00:22:26.840 plus c1' dx' over '1 plus x squared'. 00:22:26.840 --> 00:22:30.090 In other words, leaving the constant in, I would get this. 00:22:30.090 --> 00:22:32.810 And now all I want you to see here, and I've done this just 00:22:32.810 --> 00:22:35.860 for this particular example, and in the exercises we'll 00:22:35.860 --> 00:22:37.100 work it out in general. 00:22:37.100 --> 00:22:38.430 The whole idea is this. 00:22:38.430 --> 00:22:40.880 This, of course, can be written as two terms, one of 00:22:40.880 --> 00:22:44.880 which is 'x inverse tan x', and the other of which is 'c1 00:22:44.880 --> 00:22:46.990 inverse tan x'. 00:22:46.990 --> 00:22:49.770 This also can be written as two integrals. 00:22:49.770 --> 00:22:53.150 Namely, with the minus sign kept aside here, this is 00:22:53.150 --> 00:22:58.390 integral of 'xdx' over '1 plus x squared' and plus integral 00:22:58.390 --> 00:23:02.500 'c1 dx' over '1 plus x squared', the minus making 00:23:02.500 --> 00:23:03.440 this minus. 00:23:03.440 --> 00:23:07.270 But we know that the integral 'dx' over '1 plus x squared' 00:23:07.270 --> 00:23:10.050 is inverse 'tan x'. 00:23:10.050 --> 00:23:12.350 In other words, if we now separate these into 00:23:12.350 --> 00:23:14.030 pieces, we have what? 00:23:14.030 --> 00:23:17.600 'x inverse tan x' minus integral 'xdx' 00:23:17.600 --> 00:23:19.660 over '1 plus x squared'. 00:23:19.660 --> 00:23:24.040 And notice that the 'c1' term drops out because we have 'c1 00:23:24.040 --> 00:23:26.670 inverse tan x' occurring twice. 00:23:26.670 --> 00:23:32.160 It's this term here, and it's also 00:23:32.160 --> 00:23:33.340 this part of our integral. 00:23:33.340 --> 00:23:36.050 And see, with the minus sign here, they just cancel out. 00:23:36.050 --> 00:23:39.020 And that's why we really don't have to keep the arbitrary 00:23:39.020 --> 00:23:42.400 constant in there, but if we wish to, we may. 00:23:42.400 --> 00:23:45.080 At any rate, let me summarize with just one more problem to 00:23:45.080 --> 00:23:49.570 make sure that we have the technique down fairly pat. 00:23:49.570 --> 00:23:53.660 Let's suppose we were given integral 'log x dx'. 00:23:53.660 --> 00:23:56.890 What we certainly do know I hope by this time is that the 00:23:56.890 --> 00:24:00.580 derivative of 'log x' with respect to 'x' is '1/x'. 00:24:00.580 --> 00:24:04.090 Since I can differentiate 'log x', it becomes very meaningful 00:24:04.090 --> 00:24:07.270 to try letting 'u' equal 'log x'. 00:24:07.270 --> 00:24:09.620 If I do this, 'u' is 'log x'. 00:24:09.620 --> 00:24:11.520 'dv' is 'dx'. 00:24:11.520 --> 00:24:15.340 'du' then becomes ''1 over x' dx', and 'v' is 'x'. 00:24:15.340 --> 00:24:18.240 And now I put in plus 'c' in parentheses here to indicate 00:24:18.240 --> 00:24:20.080 that if you wanted to use it, you could. 00:24:20.080 --> 00:24:21.900 At any rate, what happens now? 00:24:21.900 --> 00:24:26.150 Integral udv, integral 'log x dx' in this case is 'u' times 00:24:26.150 --> 00:24:30.530 'v', which is 'x log x', minus integral 'vdu'. 00:24:30.530 --> 00:24:33.830 Well, you see the 'x' times the '1/x' is just 1. 00:24:33.830 --> 00:24:38.920 Therefore, 'vdu' is just 'dx', so minus that 00:24:38.920 --> 00:24:39.870 is just minus 'dx'. 00:24:39.870 --> 00:24:43.170 In other words, minus the integral 'vdu' is just minus 00:24:43.170 --> 00:24:44.450 integral 'dx'. 00:24:44.450 --> 00:24:47.100 And now, you see I have what? 'x log x'. 00:24:47.100 --> 00:24:49.020 And this is very easy to integrate. 00:24:49.020 --> 00:24:51.890 It just happens to be minus 'x', then I tack on the 00:24:51.890 --> 00:24:53.600 arbitrary constant. 00:24:53.600 --> 00:24:56.830 Again, to check whether this was the right answer, all I 00:24:56.830 --> 00:25:00.720 have to do is differentiate 'x log x' minus 'x' with respect 00:25:00.720 --> 00:25:05.070 to 'x' and see if I get 'log x', and I will, OK? 00:25:05.070 --> 00:25:07.220 Now, how would I use a problem like this, say, 00:25:07.220 --> 00:25:08.740 in a practical situation? 00:25:08.740 --> 00:25:12.280 Well, for example, suppose I wanted to find the area of the 00:25:12.280 --> 00:25:16.060 region 'R' where 'R' was the region bounded above by the 00:25:16.060 --> 00:25:20.540 curve 'y' equals 'natural log x', below by the x-axis, on 00:25:20.540 --> 00:25:23.710 the left by 'x' equals 1, and on the right by 'x' equals 00:25:23.710 --> 00:25:26.580 'b', where 'b' can be any number you want right now as 00:25:26.580 --> 00:25:28.630 long as it's bigger than 1 in this diagram. 00:25:28.630 --> 00:25:31.820 Notice that the definite integral says that the area of 00:25:31.820 --> 00:25:34.110 the region 'R' is the definite integral from 1 00:25:34.110 --> 00:25:36.250 to 'b', 'log x dx'. 00:25:36.250 --> 00:25:37.930 And the first fundamental theorem of 00:25:37.930 --> 00:25:39.320 calculus says lookit. 00:25:39.320 --> 00:25:41.720 If you happen to know a function whose derivative with 00:25:41.720 --> 00:25:44.100 respect to 'x' is 'log x', you can evaluate 00:25:44.100 --> 00:25:45.460 this thing very quickly. 00:25:45.460 --> 00:25:49.140 Namely, it's just 'G of b' minus 'G of 1' where 'G prime 00:25:49.140 --> 00:25:50.470 of x' is 'log x'. 00:25:50.470 --> 00:25:54.050 Well, you see, we do know a function now whose derivative 00:25:54.050 --> 00:25:57.160 with respect to 'x' is 'log x', namely, 'x 00:25:57.160 --> 00:25:59.050 log x' minus 'x'. 00:25:59.050 --> 00:26:02.550 Consequently, to solve this problem, all I have to do is 00:26:02.550 --> 00:26:06.010 evaluate this between the limits of 1 and 'b', which I 00:26:06.010 --> 00:26:09.780 won't bother doing right now, OK? 00:26:09.780 --> 00:26:11.180 Now, here's the idea. 00:26:11.180 --> 00:26:14.940 What we've really learned here is one more powerful technique 00:26:14.940 --> 00:26:16.870 for evaluating integrals. 00:26:16.870 --> 00:26:20.620 How we arrive at integrals in no way changes in this 00:26:20.620 --> 00:26:21.710 particular lecture. 00:26:21.710 --> 00:26:23.260 All we have now is what? 00:26:23.260 --> 00:26:28.500 A new technique whereby we evaluate udv by being able to 00:26:28.500 --> 00:26:32.080 evaluate equivalently vdu. 00:26:32.080 --> 00:26:36.040 At any rate, that finishes up what we had in mind for today, 00:26:36.040 --> 00:26:37.760 and so until next time, goodbye. 00:26:40.530 --> 00:26:43.070 NARRATOR: Funding for the publication of this video was 00:26:43.070 --> 00:26:47.780 provided by the Gabriella and Paul Rosenbaum Foundation. 00:26:47.780 --> 00:26:51.960 Help OCW continue to provide free and open access to MIT 00:26:51.960 --> 00:26:56.160 courses by making a donation at ocw.mit.edu/donate.