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.