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These are the user uploaded subtitles that are being translated: 1 00:00:00,570 --> 00:00:05,550 Today we're going to be talking about how to use substitution to evaluate an indefinite integral. 2 00:00:05,550 --> 00:00:11,310 And in this particular problem we've been given the integral of 4 X divided by the square root of x 3 00:00:11,310 --> 00:00:13,080 squared plus 1. 4 00:00:13,080 --> 00:00:17,440 Now we've been told already that we need to use use substitution to evaluate this integral. 5 00:00:17,550 --> 00:00:21,510 But even if we hadn't been told that specifically that would be the first thing that we want to try 6 00:00:21,510 --> 00:00:28,150 to use to evaluate this integral and the reason is because we have kind of an obvious giveaway for you 7 00:00:28,150 --> 00:00:34,830 substitution or at least for trying it which is we have two values of x and our integral and they differ 8 00:00:34,860 --> 00:00:41,220 by one degree and what I mean by that is we have this x value in the numerator and this is essentially 9 00:00:41,310 --> 00:00:47,130 X to the first power right because it's just this a single value of x Here x x to the first power. 10 00:00:47,130 --> 00:00:55,170 We also have in our denominator here X squared or X to the second power and therefore in our numerator 11 00:00:55,380 --> 00:01:01,500 we have a first degree x value in our denominator we have a second degree x value and so we have two 12 00:01:01,500 --> 00:01:04,990 different x values and they differ by 1 degree. 13 00:01:05,040 --> 00:01:13,640 When that's the case it's often very convenient to set you equal to the higher degree x value or the 14 00:01:13,650 --> 00:01:16,110 higher degree x value plus whatever goes with it. 15 00:01:16,110 --> 00:01:21,930 So in this particular case because we have x squared plus one under the square root sign it's probably 16 00:01:21,930 --> 00:01:28,980 going to be helpful for us to set you equal to x squared plus 1 because then we'll be able to put you 17 00:01:28,980 --> 00:01:30,270 underneath the square root sign. 18 00:01:30,270 --> 00:01:35,460 We'll just have the square root of you as opposed to the square root of you plus 1 which would be a 19 00:01:35,460 --> 00:01:37,980 little bit messier and harder to deal with. 20 00:01:37,980 --> 00:01:42,250 So that's one way that we can determine what we're going to set you equal to. 21 00:01:42,420 --> 00:01:48,330 Now remember that a huge substitution problem basically all we're doing here is using a substitution 22 00:01:48,570 --> 00:01:54,240 to simplify our integral so that it's easier to evaluate because for X divided by the square of X plus 23 00:01:54,240 --> 00:02:00,710 one is a little bit of a complicated integral it's not just some simple polynomial that we use. 24 00:02:00,710 --> 00:02:03,200 Power rule with to evaluate the integral. 25 00:02:03,330 --> 00:02:08,640 It's a little bit more complicated and we can't see very clearly how we're going to evaluate this integral 26 00:02:09,000 --> 00:02:11,040 as it is in its original state. 27 00:02:11,040 --> 00:02:16,830 So we want to try you substitution first because that's our quickest method to evaluate the integral 28 00:02:16,830 --> 00:02:17,600 if it works. 29 00:02:17,760 --> 00:02:22,350 If you substitution doesn't where we might want to try integration by parts and if that doesn't work 30 00:02:22,350 --> 00:02:27,330 we might want to consider partial fractions and then maybe some other methods after that. 31 00:02:27,330 --> 00:02:31,750 So we've identified for you this x squared plus one value. 32 00:02:31,890 --> 00:02:36,840 Once we identify this value for you which is the value we're going to substitute we need to take the 33 00:02:36,840 --> 00:02:39,110 derivative of you and the derivative of you. 34 00:02:39,240 --> 00:02:44,750 We're going to denote as d you and all we're going to do is take the derivative of the right hand side. 35 00:02:44,970 --> 00:02:51,210 So the derivative of x squared is just 2 x the derivative of 1 is zero. 36 00:02:51,240 --> 00:02:54,190 So we don't need to include that in our derivative here. 37 00:02:54,360 --> 00:02:59,700 We just have d equals to X and once we're done taking the derivative here we just multiply this right 38 00:02:59,700 --> 00:03:06,230 hand side here by d X. That's really important because it allows us to make a proper substitution. 39 00:03:06,360 --> 00:03:08,310 Later on in this integral. 40 00:03:08,310 --> 00:03:13,440 So once we have you NDU then we can start substituting back into our original integral we'll go head 41 00:03:13,440 --> 00:03:16,050 here and rewrite our integral. 42 00:03:16,110 --> 00:03:21,690 Keep for x in the numerator because we haven't identified a substitution value at all for 4 x so that's 43 00:03:21,690 --> 00:03:22,620 going to stay. 44 00:03:22,990 --> 00:03:26,670 But in our denominator we said that was equal to x squared plus 1. 45 00:03:26,940 --> 00:03:32,160 So we're going to take the square root of you instead of x squared plus 1 because we set x squared plus 46 00:03:32,160 --> 00:03:33,210 1 equal to you. 47 00:03:33,220 --> 00:03:39,480 So that's our denominator then for d x what we want to do is solve this equation we wrote down for you. 48 00:03:39,660 --> 00:03:44,700 We want to solve it for Dayaks So we're going to come back here like this and we're going to divide 49 00:03:44,700 --> 00:03:52,230 both sides by 2 X which will solve the right hand side for Dayaks and we're going to get you over to 50 00:03:52,230 --> 00:03:55,680 x is equal to x. 51 00:03:55,680 --> 00:04:01,230 Now we have a value for X that we can plug into our original integral So instead of putting d x in here 52 00:04:01,590 --> 00:04:08,010 to our substitution integral we're going to put in d u over to x. 53 00:04:08,010 --> 00:04:12,020 Now once we've gotten to this point we want to simplify the integral as much as we can. 54 00:04:12,030 --> 00:04:17,310 So as you can see we've got x to the first power and x to the first power in our numerator and denominator 55 00:04:17,310 --> 00:04:17,670 here. 56 00:04:17,700 --> 00:04:19,860 Those are going to cancel with one another. 57 00:04:19,890 --> 00:04:24,260 We also have four and two in our numerator and denominator respectively. 58 00:04:24,480 --> 00:04:25,580 That too is going to cancel. 59 00:04:25,590 --> 00:04:32,700 And this four here is going to become just 2 for over two is two we can pull that two out in front of 60 00:04:32,700 --> 00:04:38,780 our integral because it's a constant coefficient on this one over the square of you term here. 61 00:04:39,060 --> 00:04:46,710 And so what will be left with is two times the integral of one over the square root of you do you. 62 00:04:46,710 --> 00:04:52,170 Now from here the simplest way to evaluate the integral is to transform this fraction that we have right 63 00:04:52,170 --> 00:04:57,190 here 1 divided by the square of you into just a polynomial a power term. 64 00:04:57,190 --> 00:05:02,800 What we're going to do is we're going to recognize that this of you is equal to you to the one half 65 00:05:02,800 --> 00:05:05,680 power of the same thing as a squirt of you. 66 00:05:05,680 --> 00:05:09,860 So essentially we have one divided by you to the one half power. 67 00:05:09,910 --> 00:05:14,260 What we're going to do is we're going to move that you to the one half power to the numerator we're 68 00:05:14,260 --> 00:05:16,800 going to take it out of the denominator and move it to the numerator. 69 00:05:16,960 --> 00:05:20,290 When we do that the sign on the exponent is going to flip. 70 00:05:20,290 --> 00:05:28,080 And so what we're going to be left with is two times the integral of you to the negative one half power. 71 00:05:28,130 --> 00:05:29,980 Do you like this. 72 00:05:29,980 --> 00:05:34,990 Now it's really easy for us to take the integral because we can just use power rule like we have before 73 00:05:35,260 --> 00:05:37,150 with simple indefinite integrals. 74 00:05:37,330 --> 00:05:42,700 So I'll be left with here is two and then we're going to multiply by whatever we get when we evaluate 75 00:05:42,700 --> 00:05:44,140 the integral here. 76 00:05:44,140 --> 00:05:49,540 Remember that the first thing we'll do is add one to the exponent so we're going to get you to the negative 77 00:05:49,630 --> 00:05:53,140 1 1/2 plus 1 we're going to add 1 to the exponent. 78 00:05:53,500 --> 00:05:58,570 Then we're going to divide our coefficient by our new exponent so our coefficient currently is just 79 00:05:58,600 --> 00:05:59,370 1. 80 00:05:59,380 --> 00:06:04,780 We're going to divide by our new exponent which is negative 1 1/2 plus 1 like this. 81 00:06:04,780 --> 00:06:09,150 And then we're going to add to this our cost of integration plus C. 82 00:06:09,280 --> 00:06:14,410 And the reason that we don't have to add the C inside of the parentheses and therefore multiply it by 83 00:06:14,410 --> 00:06:20,650 the two is because we just need C to act as a placeholder constant for whatever constant may have been 84 00:06:20,650 --> 00:06:25,700 there before we took the derivative of our original function and got this for X divided by the square 85 00:06:25,720 --> 00:06:26,780 of X plus 1. 86 00:06:26,860 --> 00:06:29,480 So C can say outside here by itself. 87 00:06:29,480 --> 00:06:32,480 All we need to do now is simplify what we've got in the parentheses. 88 00:06:32,620 --> 00:06:39,640 So we're going to have two times one over negative one half plus one is positive one half sort and get 89 00:06:39,640 --> 00:06:41,530 one over positive. 90 00:06:41,540 --> 00:06:46,120 One half you the positive one half like this plus. 91 00:06:46,140 --> 00:06:54,530 See now you divided by one half is the same thing as 1 times 2 over 1. 92 00:06:54,670 --> 00:07:00,070 We just take the reciprocal of this fraction in the denominator and we multiply by the reciprocal. 93 00:07:00,070 --> 00:07:08,290 Instead of dividing by the original fraction so we get 1 times 2 over 1 you to the 1 1/2 plus C and 94 00:07:08,290 --> 00:07:15,070 now you can see we're going to get 2 times 2 you to the 1 1/2 plus C. 95 00:07:15,250 --> 00:07:20,390 And at this point we can multiply the twos together what we'll get is four and we'd have four you to 96 00:07:20,390 --> 00:07:25,650 the one half but remember that you to the 1 1/2 is equal to the square root of you it's the same thing. 97 00:07:25,810 --> 00:07:32,740 So we're going to get 4 times the square root of you plus see at this point we'd be done except that 98 00:07:33,040 --> 00:07:38,010 we want our answer back in terms of X since we started our original problem in terms of x. 99 00:07:38,010 --> 00:07:41,290 You always want to finish with the original variable that you started with. 100 00:07:41,410 --> 00:07:43,440 So we're going to back substitute for you. 101 00:07:43,480 --> 00:07:45,960 And we're going to plug in x squared plus 1 again. 102 00:07:46,000 --> 00:07:51,790 We're going to bring that back and we're going to say that our final answer is four times the square 103 00:07:51,790 --> 00:07:58,780 root of x squared plus 1 plus C to account for that constant of integration. 104 00:07:58,960 --> 00:07:59,950 And that's our final answer. 105 00:07:59,950 --> 00:08:03,700 That's how you use your substitution to evaluate an indefinite integral. 12309