# Integration by Substitution: Examples

by Batool Akmal

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00:01 So now, it’s your chance to try some integration with substitution.

00:05 You may be thinking that this is getting fairly complicated calculus.

00:10 But remember that you are developing extremely strong analytical skills.

00:14 You are developing extremely strong problem solving skills, which as medical students, you will need all through your course.

00:21 You are looking at finding areas under curves, and although these curves look very complicated and very mathematical, this is a skill that you will be able to use at any point within your medical field.

00:33 So, let’s now try and have a go at this exercise.

00:37 I’ll give you some time to do this and then we’ll go through this together.

00:41 Let’s look at our first example that we set for you to do.

00:45 We have then an example fairly similar to this, but we’re asking you now to use substitution.

00:51 So, the first example states that you have x that’s been multiplying with 2x squared minus 5 to the power of 3 dx.

01:03 If we use substitution, go for the more complicated part of this integral and make a substitution for that.

01:10 So you can say here, let u equals to 2x squared minus 5.

01:16 So just the inside part here.

01:19 We can now replace this with x, and then that becomes u to the power of 3, because I’m saying that all of this is u, and then I still have dx at the end.

01:30 Remember the second step from the example.

01:32 You now want to change your dx.

01:35 So, when we change our dx, we need to do du by dx equals to 4x.

01:41 Rearrange to get dx by itself.

01:43 So, du over 4x equals to dx.

01:48 Let’s replace our dx.

01:49 We’ve got x multiplied by u cube, and then you have du over 4x.

01:56 So from here, and good things are happening, the x is cancelled out.

02:02 Leaving you with a nice, easy integral, we have u cubed over 4 and the integral, du.

02:10 Now, the 4 at the bottom of that confuses you, it’s just the number, you can take it out as a fraction outside of the integral.

02:17 So, we have u cubed du.

02:20 nd now when we integrate, we have no limits to change, so that’s good news.

02:24 We have 1 over 4.

02:26 Add 1 to the power, and add 1 to the power, I mean, u to the 4 divided by 4.

02:34 You can rewrite this as u to the 4 over 16.

02:38 Very last bit, and I remember, that the question asks you the question in terms of x’s.

02:44 We can now change this in terms of x’s again.

02:48 So we said that u was 2x squared minus 5.

02:51 So we can rewrite this u as 2x squared minus 5 to the power of 4 over 16 plus c.

03:01 Let’s now look at our next example.

03:03 You can see that it looks fairly complicated.

03:05 We have cos and sine and we have square roots.

03:08 But let’s see if we can try and make it easier now by using some sensible substitutions.

03:14 So the question is asking us to find square root of 5 cos(x) minus 1 multiplied by sine(x)dx.

03:26 Now, I sense that we’ll be doing a lot of trig differentiation and integration.

03:31 So let’s just write that out here.

03:33 Sine goes to cos, differentiates to cos, and cos differentiates to -sine.

03:39 And similarly, if you want to integrate, cos integrates to sine when I go in that direction, and sine integrates to -cos.

03:49 Right.

03:49 I’m going to try and do this in a fewer steps than before.

03:53 So, let’s make this our substitution.

03:56 So let’s say let u equals to 5cos(x) minus 1.

04:01 I know that I’m going to have to differentiate it so I may as well do it now.

04:04 So, du by dx.

04:06 Now, look at cos.

04:08 Cos is differentiating.

04:09 So, differentiation in that direction.

04:11 Cos is differentiating to -sine.

04:14 So this becomes -5sine(x).

04:18 Rearrange to get dx by itself.

04:20 And it’s only because we know we’re going to do that, anyway.

04:25 So, we’re kind of working a little bit ahead of ourselves.

04:29 Now, come back to our integral.

04:31 So our integral is saying that we have square root of 5cos(x) minus 1.

04:35 I said that to be u just to make an easier substitution.

04:40 We’ve still got the sine(x) there, and the dx becomes du over -5sine(x).

04:52 So, some good things are happening here.

04:54 You can see that you have a sine(x) and a sine(x).

04:58 So that sine(x) cancels with that.

05:01 You have this -5 here.

05:03 So, you can rewrite this as square root of u over -5 du.

05:09 Or alternatively, you can take the -5 outside like this to make it easier to comprehend.

05:17 So we’ve got -1 over 5 out of the integral and we’re integrating square root of u du.

05:23 Before we integrate, we can change the u to u to the half du.

05:30 Remember that to integrate u, add 1 to the power, divided by new power.

05:35 So I have 1 over 5.

05:36 When you add 1 to the power, that goes to u to the 3 over 2, and then you divide it by 3 over 2.

05:44 We can tidy this up a little bit so it becomes 1 over 5.

05:48 Take the two up from here.

05:50 We’ve discussed why we do this earlier.

05:52 So that becomes 2u to the 3 over 2 over 3.

05:58 And lastly, the question was asked in terms of x’s.

06:04 So you can change your u back to x.

06:07 So I can rewrite this as -1 over -- multiply these together -- 15.

06:12 You still have the 2 at the top, so the 2 can come out here.

06:15 And then the u value, we can change back to our x.

06:19 So that was 5 cos(x) minus 1 to the power of 3 over 2 plus c.

06:26 So you can see that this looked fairly complicated but we’ve made it a little bit easier by substitution, and the integral at the end was fairly straightforward just by adding 1 to the power and dividing by the new power.

06:43 And the last question in your exercise, which wants you to do substitution.

06:47 Let’s have a go.

06:51 We have x squared over 2 minus x cubed squared dx.

06:57 You’ll see that this works a little bit different to the others that we’ve done because you have a denominator.

07:03 So you have this function here at the bottom.

07:06 Now, remember what we could do and what used to do in differentiation.

07:09 If there’s anything like that, you can always take it up.

07:12 So we can rewrite this as 0 to 1 x squared, and you can bring the entire bracket up and rewrite it as 2 minus x cubed to the minus 2 dx.

07:25 Now, we can make a substitution to make things a little bit easier.

07:29 So let’s go through this part here and say that let u equals to 2 minus x cubed.

07:36 If I do everything else here, so find dy by dx as well so I can do du by dx equals to -3x squared.

07:45 And if I rearranged to get my dx, so that gives me du over -3x squared equals to dx.

07:53 Now, when I put this all in into my integral, you’ll see that it becomes a little bit easier than it was before.

07:59 So I have x squared.

08:00 Instead of 2 minus x cube, I can rewrite this as u to the minus 2.

08:05 And instead of the dx, I can rewrite du over minus 3 x squared.

08:12 It looks a little bit scary, but when you cancel the x squares, you see that it becomes a nice and straightforward integral.

08:19 If I rewrite this now, I’ve got u to the minus 2.

08:23 Let’s not forget the -3 at the bottom.

08:25 I have du and then I have the integral 01.

08:29 So remember that this -3 can go anywhere? You can put it under here or you can take it out of the integral.

08:34 It really is up to you.

08:37 The next thing to do is to change these limits.

08:39 So, don’t forget that these are your x limits.

08:42 So in order for this to go into a u function, you need to change them to your u limits.

08:47 And for that, we can use this function here.

08:51 So we can say that when x equals to 1, u equals to 2 minus 1 cubed, which just gives you 1.

08:59 And when x equals to 0, u equals to 2 minus 0 to give you an answer of two.

09:06 So we can now change our limits.

09:08 Remember that these are your u limits here.

09:11 So we’ve changed them to u's, which makes this entire integral in terms of u’s.

09:15 Now, I’ll take this -3 out, so we can write it as -1 over 3 on the outside.

09:20 My new u limits, when I put 1 in, I get 1.

09:23 When I put x equals to 0 in, I got a u limit of 2.

09:27 I still have u to the -2du.

09:31 And now, we integrate.

09:32 So it’s fairly straightforward from here minus 1 over 3.

09:36 When you integrate, add 1 to the power, divided by new power, where limits -- we’ve got our limits 2 and 1.

09:44 I’m just going to neaten this up a little bit.

09:46 So I have 1 over 3, and then I’ve got -1 over u when I bring the u down, and the minus is just making everything negative.

09:55 And then I’ve got 2 and 1 as my limits Last step, we just have to put our limits in and get our final answer.

10:01 So the top limit goes in first.

10:05 So, my top limit is -1 over 1.

10:08 So the u changes to 1.

10:10 And then you minus the second part, which is -1 over 2.

10:15 So notice what I’ve done.

10:16 I’ve just changed this u first to 1 and then to 2.

10:20 So here, I put 1 in, here I put 1 in, and then I put 2 in.

10:24 And I’ve just put brackets around this so that I don’t forget that there’s two minuses.

10:30 Almost there.

10:31 I’ve got -1 over 3.

10:33 I’ve got -1 as over 1 over 1 as -1 plus a half.

10:37 So this minus and a minus becomes positive.

10:41 And then I have -1 over 3.

10:43 That gives me -1 a half on the inside.

10:46 Multiply them together to give you 1 over 6.

10:49 So the area underneath x squared over 2 minus x cubed, all squared, fairly complicated graph.

10:57 Between the limits 0 and 1 is 1 over 6 unit squared.

### About the Lecture

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

### Included Quiz Questions

1. x³ ln(x) / 3
2. x² ln(x) / 3
3. x ln(x) / 3
4. ln(x)
1. [(x³ + 4)³ / 9] + c
2. [(x³ + 4)³ / 6] + c
3. [2(x³ + 4)³ / 9] + c
4. (x³ + 4)³ + c
5. [(x³ + 4)² / 6] + c
1. Chain rule
2. Power rule
3. Product rule
4. Integration by parts
5. Fundamental theorem of calculus
1. u = ln(x)
2. u = (1 / x)
3. u = x
4. u = x²
5. u = ln(x) / x
1. [( 1 / 2) (lnx)²] + c
2. [(1 / 2) (lnx)] + c
3. (lnx)² + c
4. [(1 / 4) (lnx)²] + c
5. [(2) (lnx)²] + c
1. -cos(x² + x) + c
2. cos(x² + x) + c
3. -cos(x) + c
4. -cos(x²) + c
5. -cos(2x + 1) + c

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