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Integration: Exercise 2

by Batool Akmal
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    00:01 Have a look at our next example where you have a function 2x minus 1 that’s being multiplied with a sin of something.

    00:09 So, it’s a sin function.

    00:11 We have decided that we are going to use integration by parts, but usually when you have 2 functions multiplying together, you have only 2 options.

    00:18 You either use the substitution method or integration by parts.

    00:22 So we have sine of 3x dx.

    00:27 Since I have decided that we are going to be doing integration by parts here, I’m going to write my formula.

    00:33 So I have uv' dx equals to uv minus the integral of vdu dx.

    00:42 I’m also going to remind you a little bit about trig and differentiating and integrating.

    00:47 So, sine goes to cos and cos goes to minus sine.

    00:51 Differentiate in that direction, integrate in that direction.

    00:54 So I’ve got all the information or the things my mind could perhaps slip up on, on the side here before I start this question.

    01:01 So let’s split this into 2.

    01:03 U equals to 2X minus 1.

    01:06 V' equals the sine of 3X.

    01:10 This, we’re going to differentiate.

    01:12 This, we’re going to integrate to get your u' and v.

    01:16 U' gives me 2 and v gives me -- remember we are integrating that, so sine integrates to minus cos.

    01:25 So that’s going to go to minus cos 3X and when you integrate, you divide by the differential.

    01:33 We put this all into the formula.

    01:34 So you have u times v from here firstly.

    01:38 You multiply u with v to give you minus and we just put this minus on the outside and then I have 2x minus 1 multiplied by cos of 3x over 3, minus the integral of vdu dx So, the integral of those two terms here.

    01:54 So if I do u' first, I’ve got 2 multiplied by -cos of 3x over 3, dx.

    02:04 Again, look at your integral, that’s looking good.

    02:08 We have only one function of X.

    02:10 There are few things we need to tidy up before, but the good news is, that there is just 1 function of X which we can deal with.

    02:18 So, you can move that 3 here.

    02:20 So we can re-write this as minus 2x minus 1 over 3 multiplied by cos of 3x.

    02:26 There is no reason you need to do that, but 3 can stay anywhere, it’s dividing that entire term.

    02:32 Let’s take this 2 over 3 out of the integral or let’s do this minus and a minus first.

    02:38 So, let’s do the plus and out of the integral, we are also taking 2 over 3 out, leaving us with just the integral of cos 3X.

    02:46 So it’s easy as you can make it before you start to integrate.

    02:49 So I’ve done minus times a minus to give me a plus, this sign here.

    02:53 And then, I’ve also taken this 2 and 3 out.

    02:56 So, 2 is multiplying here and dividing there.

    02:59 So I’ve written as a fraction 2 over 3.

    03:02 Now, I just need to integrate.

    03:04 If I re-write this as minus 2x minus 1 over 3, cos of 3x not much more I can do on the outside.

    03:12 We got a plus 2 over 3 here.

    03:15 I’m integrating cos of 3x, so cos integrates to sine.

    03:20 But remember, do this 1 step at a time firstly.

    03:23 You have sine of 3x and then divide it by the differential which is 3.

    03:28 And then once I’ve multiplied it through, I’ll add my plus C.

    03:31 So this gives me minus 2x minus 1 over 3, cos of 3X.

    03:38 These 2 numbers can multiply at the bottom, so that gives me plus 2 over 9, sine of 3X plus C.


    About the Lecture

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


    Author of lecture Integration: Exercise 2

     Batool Akmal

    Batool Akmal


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