Have a look at our next example
where you have a function 2x minus 1
that’s being multiplied
with a sin of something.
So, it’s a sin function.
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.
You either use the substitution
method or integration by parts.
So we have sine of 3x dx.
Since I have decided that we are going
to be doing integration by parts here,
I’m going to write my formula.
So I have uv' dx equals to
uv minus the integral of vdu dx.
I’m also going to remind you
a little bit about trig
So, sine goes to cos and
cos goes to minus sine.
Differentiate in that direction,
integrate in that direction.
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.
So let’s split this into 2.
U equals to 2X minus 1.
V' equals the sine of 3X.
This, we’re going to differentiate.
This, we’re going to integrate
to get your u' and v.
U' gives me 2 and v gives me --
remember we are integrating that,
so sine integrates to minus cos.
So that’s going to go to minus
cos 3X and when you integrate,
you divide by the differential.
We put this all into the formula.
So you have u times v from here firstly.
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.
So if I do u' first, I’ve got 2
multiplied by -cos of 3x over 3, dx.
Again, look at your integral,
that’s looking good.
We have only one function of X.
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.
So, you can move that 3 here.
So we can re-write this as minus 2x
minus 1 over 3 multiplied by cos of 3x.
There is no reason you need to
do that, but 3 can stay anywhere,
it’s dividing that entire term.
Let’s take this 2 over
3 out of the integral
or let’s do this minus and a minus first.
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.
So it’s easy as you can make it
before you start to integrate.
So I’ve done minus times a minus
to give me a plus, this sign here.
And then, I’ve also taken this 2 and 3 out.
So, 2 is multiplying
here and dividing there.
So I’ve written as a fraction 2 over 3.
Now, I just need to integrate.
If I re-write this as
minus 2x minus 1 over 3,
cos of 3x not much more I
can do on the outside.
We got a plus 2 over 3 here.
I’m integrating cos of 3x,
so cos integrates to sine.
But remember, do this 1
step at a time firstly.
You have sine of 3x and then divide
it by the differential which is 3.
And then once I’ve multiplied
it through, I’ll add my plus C.
So this gives me minus 2x
minus 1 over 3, cos of 3X.
These 2 numbers can
multiply at the bottom,
so that gives me plus 2 over 9,
sine of 3X plus C.