# Integration: Exercise 2

by Batool Akmal
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DLM Advanced Integration Exercise Calculus Akmal.pdf
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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.

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

### Included Quiz Questions

1. u=x² and v'=e^(x)
2. u=x² and v=e^(x)
3. u=e^(x) and v=x²
4. u=e^(x) and v'=x²
5. u=1 and v'=x²
1. [ (x² - 2x +2)e^(x) ]+c
2. [ (x² + 2x +2)e^(x) ]+c
3. [ (x²- 2x - 2)e^(x) ]+c
4. [ (x²- 2x +1)e^(x) ]+c
5. [ (x²- x +2)e^(x) ]+c
1. [ (-1/3)(x+1)cos(3x) ] + (1/9)sin(3x) + c
2. [ (-1/3)(x+1)cos(3x) ] + (1/3)sin(3x) + c
3. [ (-1/3)(x+1)cos(3x) ] - (1/9)sin(3x) + c
4. [ (-1/3)(x+1)cos(3x) ] - (1/3)sin(3x) + c
5. [ (-1/9)(x+1)cos(3x) ] - (1/9)sin(3x) + c

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