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Integration by Substitution

by Batool Akmal
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    00:01 So we’ve just learned how to integrate trig, e of x's and ln of x's I’m now going to show you some techniques when functions become fairly complicated, the things that we can do to simplify it.

    00:15 In mathematics, there’s this thing called substitution, which we quite like when things get a bit too complicated.

    00:22 We can make it easier by making substitutions dealing with shorter functions, dealing with our own substitution and then changing it back at the end.

    00:31 It’s a very useful technique especially when things get more complex.

    00:35 So, what I mean by that is if you have a function f of x dx, and imagine it’s not very straightforward, it has two or three things multiplying together, or some functions, you can change this entire function to a u function.

    00:50 So the idea is that we change everything in that function to u functions.

    00:54 We then differentiate it with respect to u rather than with respect to x, but we also change the limits.

    01:02 Now, this may look a bit bizarre as to why should we do this.

    01:05 You’ll see this is useful especially when we do the more complicated functions.

    01:10 It probably isn’t a good idea doing this when your functions are straightforward and you can just integrate.

    01:15 Add one to the power and divide by new power.

    01:17 But you’ll see that the functions were about to integrate.

    01:20 It makes life a lot easier if we use substitution methods.

    01:25 Keep this definition in mind, but I’ll teach you this by using numerical examples because it’s a lot easier and it makes a lot of sense when we do it with functions.

    01:37 So let’s look at this example.

    01:38 We’re going to work through this slowly because with it, I’m going to teach you integration by substitution.

    01:45 So, our integral says that we’ve got x multiplied by cos, x squared plus 1 dx.

    01:58 We also have the limits 0 and 1.

    02:02 Now, if I asked you just to integrate it, you probably won’t be able to because it’s a function that’s being multiplied with another function, and we haven’t really spoken about any techniques here.

    02:12 So, we do need to find a way to make this easier for us in order for us to integrate it.

    02:18 To do this, we use this method called substitution.

    02:21 So, anything that look too complicated, we can substitute easier letters for them, and we can change the whole equation in terms of that letter.

    02:30 We usually use the letter u.

    02:33 So, here’s what I’m going to do.

    02:34 I’m going to change this angle into something a little bit easier.

    02:38 So, I’m going to say here on the side, let u equals to x squared plus 1.

    02:44 So, this equation or this integral now becomes x cos of u dx.

    02:53 Now, it’s a very mixed statement here.

    02:56 There’s lots going on.

    02:57 I’ve got an x, I’ve got a u, I’ve got a dx, and I’ve got these limits which are my x limits.

    03:02 So be clear about those.

    03:04 These are your x limits and they can only go into x functions.

    03:09 So, in some sense, I’ve made it a little bit messier, but let’s continue.

    03:13 I am now keeping a clear objective that I want to make this entire equation in terms of u’s.

    03:19 I want to change everything into u.

    03:22 So the next step I’m going to start with is changing my dx to du.

    03:27 I can do that by using my substitution here.

    03:31 So, I said u was x squared plus 1.

    03:35 I can now do du by dx.

    03:36 I can differentiate this to give me 2x.

    03:40 I can get this dx by itself just by rearranging this equation.

    03:43 So I can write du equals to 2x dx.

    03:47 I’ve multiplied this on the other side.

    03:49 And remember, I want to take this by itself, so make this the subject of the equation so I can rewrite this as du over 2x equals to dx.

    04:01 So hopefully, this is straightforward enough.

    04:02 I took my u substitution, I differentiated du with dx, and then I rearranged to get dx by itself.

    04:10 Why did I do it? Because I can replace this dx with this dx here.

    04:16 So, watch what happens when I do this.

    04:18 I do 1 0 x cos of u.

    04:22 And instead of my dx, I can now write du over 2x.

    04:29 Still looks like a mixture, but if you notice here, this x is cancelling with this x.

    04:36 Let’s see what our integral looks like now.

    04:38 0 1 cos of u du.

    04:43 And this I’m a lot happier with because I know that cos of anything integrated with respect to that thing will give me a straightforward answer.

    04:51 I know my identities.

    04:54 There’s one little thing I need to do here though.

    04:57 Remember I said that these were your x limits.

    05:01 So, I need to change those limits.

    05:03 Those are my x limits and those can’t go into this function because this is all in terms of u.

    05:09 But again, I come back to my substitution, u substitution.

    05:14 I said u was x squared plus 1.

    05:18 So, we want to know what u is when x equals to 1 and when x equals to 2 -- when x equals to 0, I mean.

    05:29 So, you can find out u when we put 1 in; 1 squared plus 1 is just 2.

    05:34 And when you put 0 in, that gives you 0 squared plus 1 is just 1.

    05:39 So, what you’ve done here is you’ve changed your x limit; x equals to 1, x equals to 0.

    05:44 You’ve changed your x limit to u equals to 2, and your x limit of 0 to u equals to 1.

    05:51 And now, if I change my limits, remember, the 1 has become 2, and the 0 has become 1.

    05:57 I have cos u du, and life is a lot easier integrating this as compared to the first function.

    06:06 How did you integrate cos? Remember the little tables that we did.

    06:10 Sine goes to cos, cos goes to -sine when you differentiate; when you integrate, you go backwards.

    06:19 So cos, and if you look at this here, we’re going in that direction here.

    06:22 Cos will integrate to sine.

    06:25 So, we now end up with sine u, nice and easy integrated.

    06:30 We have to put our limits in.

    06:31 So we’ll put 1 and 2 in.

    06:33 So, 2 goes in first and then 1 later.

    06:36 So, we can just leave it as is because we’re not using calculators.

    06:39 So the answer to this as a statement is sine2 minus sine1.

    06:45 And that here is your integral.

    06:48 If you’re using calculators, it’s just a simple job of you putting it into your calculator to find your answer.


    About the Lecture

    The lecture Integration by Substitution by Batool Akmal is from the course Advanced Integration.


    Included Quiz Questions

    1. f(u) = 2u + cos(u)
    2. f(u) = u + cos(u)
    3. f(u) = x + cos(x²)
    4. f(u) = x + cos(x)
    5. f(u) = x + cos(u²)
    1. u = x³
    2. u = x²
    3. u = x
    4. u = x + 1
    5. u = x³ + 1
    1. -cos(x²) + c
    2. cos(x²) + c
    3. -cos (x) + c
    4. cos(x) + c
    5. -cos(u²) + c
    1. 3/2
    2. 1/2
    3. 1
    4. 4/3
    5. 3/4

    Author of lecture Integration by Substitution

     Batool Akmal

    Batool Akmal


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