# 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, 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

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