Integration by Parts: Example 2

by Batool Akmal

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    00:01 So after all that substitution, we are now going to move to our last type of integration method.

    00:07 This is probably the most complicated method that we will study in this course and it’s called integration by parts.

    00:14 Now, this is similar or the kind of functions that you look at are similar to the functions that you can use substitution on.

    00:20 Basically, it’s two functions multiplying together.

    00:24 But it really is up to you whatever method you choose now, whether you use substitution or integration by parts.

    00:30 You will see that the questions that require integration by parts are usually a bit more complicated than the once that you can solve by substitution.

    00:38 It might just be a process of elimination where you look at a question, you started off with substitution and realize maybe this will be easier if I do this with integration by parts.

    00:48 But when I do the calculations, I usually leave this method as my last resort.

    00:52 I try out all the other methods, see if I can simplify the equation first to do it easily, just basic integration.

    00:59 If not, I then move to integration by substitution and if that doesn’t work or give me a nice answer, I move to integration by parts.

    01:06 But a lot of the times, you can do integration by parts questions by substitution and you can do the substitution questions by integration by parts because they have the same format.

    01:17 They are two functions multiplying together.

    01:20 Anyway, if I give you the definition for this, when you’re looking at integration by parts, you have two functions.

    01:26 So, you have a function of x and a function of x.

    01:28 But this time we split it into two parts.

    01:31 So that’s why we call it integration by parts.

    01:33 We split it into a u-term and we split it into a v-term, so into dv by dx.

    01:40 Now this maybe a little bit complicated, but we will do this as a numerical example and then you’ll be able to understand.

    01:46 Once we have done that, we differentiate one of the terms and we integrate to the other terms and we use the other half of this formula which says UV minus the integral of vdu dx or vu dx.

    02:01 Now have a little look at this formula.

    02:03 Firstly, we do need to know this in order to apply integration by parts.

    02:07 But as we do more numerical examples, you will see that it’s fairly straightforward as long as you know the formula.

    02:13 We just have to find the different parts and put them in.

    02:18 So let’s look at our first example.

    02:20 Now, you may look at this and think, “I’m going to try and do this by substitution.” Think about it.

    02:26 If you take x as your u, this function will become ue to the u something.

    02:32 Now, that doesn’t make this function any easier.

    02:35 When we use substitution, we’re usually trying to cancel things out to make it easier and make it in terms of one function effects.

    02:42 So in this case, we'd have to use integration by parts.

    02:46 Now, we’ll break it down into several steps and then hopefully by the end of this example, you will feel more confident applying integration by parts to these questions, but just notice that there is no other way that we can do this.

    02:58 It’s not straightforward integration, so you’re not just adding 1 to the power dividing by a new power.

    03:04 It is two functions that are timesing together, so we need a more complicated method.

    03:09 So, let’s try this once again by integration by parts.

    03:13 I have my integral x, e to the power of x, dx.

    03:19 I’m just going to remind you of integration by parts here on the side, so you have the integral of UV dashed equals to uv minus the integral of vdu dx.

    03:30 That means that if you have any integral in this form, you can break it into u and v ', so that’s what we are going to do.

    03:36 We are going to break this function down into two parts.

    03:40 One of them, we will call u and the other part, we will call v dashed.

    03:44 Now v ' dashed is a differentiated version of v ' dv/dx.

    03:49 If I call the first one u equals to x and the second one e of x, I now need to do two different things to it.

    03:57 Remember what I need for the formula, I need u, which is u differentiated, and I also need v by itself.

    04:04 So, I will have to change this to u dashed, which means I have to differentiate it and in order to go from v dashed to v, think about it, what is the opposite of differentiation? So if something has been differentiated to go to the original function, we are going to have to integrate it.

    04:23 So let me just write that down here.

    04:24 We are going to differentiate this function and we are going to integrate this function.

    04:29 Now, that’s probably the only complicated step here, and it often gets quite confusing because in this question, you’re differentiating and integrating at the same time and then putting it together.

    04:39 But as long as you’re clear, you write it out in a methodical way, it should be fairly straightforward.

    04:45 The first step, I’m going to differentiate my x.

    04:48 X differentiates -- not integrate -- differentiates to 1.

    04:52 So you bring the power down and you decrease the power by 1.

    04:56 The e of x, I now need to integrate.

    04:58 So I’m going to integrate e of x.

    05:00 Remember, e of x integrates to just e of x, so it doesn’t change, so it’s nice and easy.

    05:07 Okay, I have everything.

    05:08 I have u.

    05:09 I have u dashed.

    05:11 I have v dashed and I have v.

    05:12 So I have every component that I need for this formula.

    05:16 So let’s put this in, I don’t need to re-write this part because that’s what I’m doing.

    05:21 I’m integrating this function.

    05:23 I’m just going to go straight into writing the equals part, so I’m trying to find the integral.

    05:30 So in the first part, I do u times v.

    05:33 So here’s my u and here is my v, so I’m going to multiply those two terms.

    05:38 So, u is x and my V is E to the x, minus the integral of vdu/dx.

    05:45 So this is my v function and this is my du/dx.

    05:50 So I’m going to multiply those together, so I have 1 multiplied by e of x dx.

    05:56 I should have a dx here at the end and dx there.

    05:59 Okay, observe why we've done this.

    06:02 This, we are not having to do anything with, so that's done, that's solved.

    06:06 But look at my integral this time and see how it’s different to this question here.

    06:11 So look at my integral here, I am just integrating e of x dx.

    06:15 As compared to the question, which asks me to integrate x multiplied by e of x.

    06:22 I have made it a lot easier for myself.

    06:24 I made the integral a lot easier by using integration by parts.

    06:28 I now have a very straightforward integral that I can easily do and I’ve simplified it, and that’s the idea behind integration by parts.

    06:35 You have broken into parts, so that it’s easier rather than more complicated to solve these equations.

    06:42 Remember this one thing in mathematics, when we’re solving equations, we are always trying to make solutions easier for ourselves.

    06:48 If at any point things are getting more and more complicated and the equation is growing, you’ll probably need to go back and rethink your solution to make it easier.

    06:58 So I’m just going to rewrite to show you.

    07:00 I have xe of x minus -- the one, obviously, doesn’t make any difference.

    07:05 So I have e of x dx and all I need to do is work this out now.

    07:09 xe of x were not doing anything, not differentiating or integrating, so that can stay and all I have to do now is integrate e of x.

    07:17 e of x integrates to just e of x and then you can put a plus C at the end to say that this is the end of your integral.

    07:24 So once again, I have made this function a lot easier by integrating by parts.

    07:28 We started off with two functions multiplying.

    07:32 We split them into u and v dashed, and then we put them into our formula, hoping that our integral would become easier.

    07:40 If you’re getting to a point where this integral is not becoming easier, it’s becoming bigger, you probably need to go back to this stage and need to swap your u and v dashed to see if it gets any easier.

    07:52 Remember that when you differentiate something, you’re making the power smaller, which is good.

    07:57 We like that.

    07:59 When you integrate something, you’re making the power bigger because you are adding 1 to the power.

    08:03 So keep that in mind when you’re doing these questions, whether you want to simplify it and make it smaller or whether you want your function to become bigger with the powers of x.

    08:12 There is one last little thing that I can do here perhaps.

    08:15 I can take a common factor of e of x, leaving me with x minus 1 and then, don’t forget your plus C is just that final constant that we put at the end of indefinite integrals.

    About the Lecture

    The lecture Integration by Parts: Example 2 by Batool Akmal is from the course Advanced Integration.

    Included Quiz Questions

    1. the technique only performs a part of the original integration
    2. it is the inverse of the product rule for differentiation
    3. the integrand is split into parts
    4. it is the inverse of the chain rule for differentiation
    5. it is partial fraction method
    1. ln(x)
    2. x
    3. xln(x)
    4. 1
    5. 1/x
    1. u= tan⁻¹(x) and v'=x
    2. u=x and v'=tan⁻¹(x)
    3. u=tan(x) and v'=x
    4. u=tan(x) and v'=x²
    5. u=x² and v'=tan⁻¹(x)
    1. (-x/6)cos(6x)+(1/36)sin(6x)+c
    2. xcos(6x)+(1/36)sin(6x)+c
    3. (x/6)cos(6x)+(1/36)sin(6x)+c
    4. 6xcos(6x)+(1/36)sin(6x)+c
    5. -6xcos(6x)+36sin(6x)+c

    Author of lecture Integration by Parts: Example 2

     Batool Akmal

    Batool Akmal

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