www.mathcentre.ac.uk/.../9.4%20Integration%20by%20Parts.mp4

When we have a
product of two functions.

And we need to integrate them.

Then the question is how do we

do it? It's very tricky X
times by Cos X. There's no

obvious substitution. There's no
standard form that looks like

that. But there may be some
help to be had from

Differentiation, because for
all we don't know how to

integrate a product of two
functions. One of the things

we do know is how to
differentiate it.

So let's take Y equals U times

by V. Where U&V A2
functions of X and let's.

Differentiated Sode why by DX is
equal to the derivative of UV.

Which we do as you.

Stevie by the X Plus V.

Do you buy the X?

Now I'm going to take this bit.

And Rearrange it, going to focus
on this as the subject of my

formula. The major interest, so
to speak. So I'm going to take

this bit away from both sides.
So that's U.

TV by the X is
equal to the derivative of

you times by V with
respect to X minus VDU

by DX. Now, this
is still a product

of two functions.

So let's integrate it.

And see what happens.

So we have the integral

of U. TV by
the X with respect to X is

equal to the integral.

Of the derivative of UV with
respect to X minus the integral

of VDU by the X with respect to
X. Now this is easy. This bit

we've differentiated it and
we've integrated it. So we must

get back to what we started
with. So we have the integral.

Of you TV by DX with
respect to X is equal to

UV minus the integral of
the du by DX with respect

to X this.

Is formula which we
call integration.

By parts

And we can see why it gets that
name, because in essence we're

integrating a part of the
product at a time.

So we take DVD by DX, we
integrate it and we put it there

and we put it there.

We take you, we put it there and
we differentiate it here, and

our hope is that starting from
one product you times DV by the

X, we end up with an integral
here of another product which is

simpler because we always have a
choice. So let's go back to the

original example that we started

with. We asked ourselves the
question, could we integrate X

cause X? With respect to
X, remember we've got a choice.

We can take this to be you or we
can take this function to be

you. Now let's just write down
the formula again so we can see

what happens. You DV by The X.

Integrated with respect to X is
equal to UV minus the integral

of VDU by the X integrated
with respect to X. Let's examine

our choice we could take.

Cause X to be you. Let's just
track through what would happen.

We have kozaks there.

We have cause X there.

When we differentiate it, the
derivative of causes minus sign.

This will take to be divvied by
DX, so when we integrate it to

get V, we'd get X squared over
two, we get X squared over 2. So

we think about it here.

We have X squared over 2 times
by minus sign X. That is no

simpler. In fact it's more
difficult than this is.

Let's change our choice round
and let's take you to be X.

Do you buy the X?

Is just one.

Let's take cause X to be
divvied by DXODV by DX is

equal to cause X, and if
we integrate that, V is equal

to sign X.

So let's just check again what's
going to happen.

This will be 1 here.

This will be sign X so we have
sine X times by one and we can

integrate that no problem. We
can do that so this is what our

choice is going to look like.
We're going to choose X to be

you and cause X to be DV by DX.

OK, let's do this again on a
clean piece of paper. So we've

got the integral of X.

Cause XDX

We're going to choose this to be

you. And this one to be
DV by The X.

Let's write down the formula so
we can see what we're doing.

This will be you.

V minus the integral of
VDU by the X with

respect to X.

You is X.

Now V is the integral of Cos X,
which is just sign X.

Minus the integral of V, which
we know is sign X times by due

by the ex which we know is one.

So this is X Sign X
minus the integral of sine XDX.

And this bit stays the same, and
the integral of sign is minus

cause, which with that minus
sign there gives us plus cause X

Plus C. And so we've achieved
what we wanted to achieve. We've

integrated this product, but
notice how important this choice

is. OK, let's take a second
example. Let's say we want to

integrate X squared times by E
to the 3X with respect to X.

It's a product to functions, so
we can know that we're going to

want to use our integration by

parts. Formula.

So now we gotta make our choice
between U and DV by DX. Now

we've already learned that it
probably makes sense to take

this to be you, because when we
differentiate it to put it here,

it's going to be simpler.

So this is going to go
from X squared to 2X

when it goes in there.

That is simply therefore than
what we had here. So let's make

that choice, and we're probably
going to have to recognize we

going to have to do this second
integral again by parts, so will

choose this to be you.

And will choose this one to be
DV by The X.

So equals U which is X squared.
Now we want V. So we need to

integrate E to the 3X, which
will be 1/3 E to the 3X.

Minus. The integral of.

V is the same again 1/3 of
E to the three X this time

times by due by DX times by
two X with respect to X. Tidy

this bit from top 1/3 X squared
E to the three X minus the

integral of 2/3.

XE to the three X the X and we
can see that this integral is

going in the right direction.
This is simpler than the one we

started off with, so we're going
to do this again, and we're

going to make similar choices.
This is going to be you.

And this is going to be DV by
the X, so 1/3 X squared E to

the three X minus. Now it's
helpful here to put in a bracket

so that we keep all of this
together and we take it all away

and not just little bits of it
and forget exactly what it is

we're doing. So we've made our
choice that this here will be

you. We want UV, so that's 2/3
of X now. This is now DV

by DX. We need to integrate it,
so that's 1/3 E to the three

X minus the integral of.

1/3 E to the three X times by
and we want the derivative of

2/3 of X, which is just 2/3 the
X and close that bracket.

Now let's tidy this up. 1/3 of
X squared E to the three X

minus, and I'm going to keep the
bracket there for the time

being. 1/3 * 1/3 is 2. Ninths
XE to the three X minus the

integral of. Two ninths.

1/3 times by 2/3 E to the
three XDX and close the bracket.

So I'm now in a position to
finish this off, so I'll turn

the page, write it down again,
1/3 of X squared E to the three

X minus. And then the
bracket two ninths of XE to

the three X minus the integral
of two ninths E to the

three XDX and close the bracket.

So.
What will do is will remove this

bracket first, one third X
squared E to the three X minus

two ninths of XE to the 3X and
then minus minus gives us a

plus, and that's why I advised
putting in the bracket. So you

keep all this together so that
when it comes to doing the

simplification, we get the

signed right. The integral of
two ninths of E to the three X

the X and we just have this last
integration to do to finish it

off 1/3 of X squared E to the
three X minus two ninths of XE

to the 3X.

And then finally we integrate E
to the three X. That's 1/3 of E

to the three X. We have two
ninths here to multiply it by,

so that's 220 sevenths of E to
the 3X. And that's the last

integral. So we want plus C for
the constant of integration.

So we've seen 2 examples now.
One where we have to use

integration by parts once and
the second one where we used

integration by parts twice.
Sometimes you may have to do it

3 times, but that would be a
little unkind. But let's have a

look at a somewhat different

example. The integral of
X times by the

log of X.

It's a product of two functions,
X and log of X.

But The choice here
is important. Let's write down

the formula. But we've
got for integration by parts

and think it through.

Our natural instinct would be to
choose X to be you.

So let's do that. So X is
U and the log function the log

of X to be DV by The

X. OK, we know we're going to
have to differentiate Xu equals

X, so do you buy the X is one.

And we've got DVD by the X
that's equal to the log of X,

and we've got to get V out of
it, so we've got to be able to

integrate the log of X with
respect to X.

Equals what? Do we know
what the integral of log of

access? Well, no, we don't.

But we do know how to
differentiate X the log of X.

Rather we do know how to do that
differentiation. If we were to

change our minds and do you
equals log of X, then du by the

X would be one over X.

So here. The choices made by
which of the functions we can

actually integrate and
differentiate. We can't

integrate log of X, but we can
differentiate it. That means

that our choice has to be you is
log of X.

Select start this one
off again X log

of XDX. Equals our
formula says this is UV minus

the integral of the du by
DX with respect to X.

And we'll make the choice this
time that this is DV by The X.

And this is you.

So. You, that's
just log of X.

Times by the we have to
integrate X well we can do that.

That's straightforward X squared

over 2. Minus the integral
of the again X squared

over 2 times by.

The derivative of this, which we
can do. It's one over XX.

So let's put this in front of
the log XXX squared log of X

minus. We can do a little bit of
Simplification here. We can

divide by X and by X and that
would leave us with X over 2.

To be integrated with respect to
X. So now we can complete that.

The integral of X is X squared
over two. It's already over 2,

so that would be X squared
over 4 plus. Again, our

constant of integration.

So just to hammer home that
point here, our choice was

governed by what we could
actually integrate and what we

could actually differentiate.
Let's take that a stage further.

You remember that when we looked
at this, one of the things that

we couldn't do was we couldn't
integrate log X when we first

looked at it. That's what
prompted is to change our.

Choice, so to speak, to move
from one to the other.

So what if we said,
well, come on, what is

the integral of log X?

How can we do it?

Or first, let's make it into a
product and let's make it into a

product of two functions by
saying it's one times log X.

Then make the same choices as

before. The Log X becomes you,
but this one is what becomes DV

by the X because we can

integrate one. So writing down
our formula again, this is

UV minus the integral of
VDUIDX with respect to X.

Equals you.
The log of X times by V now

DV by DX is one, so I have
to integrate one that's just X.

Minus the integral of V, so

that's X. Times by du by
DX the derivative of log X

that's just one over X.

Equals X times by the log of X
minus the integral of X Times

one over X is just one.

And we can do that integration.

The integral of one is
just X plus a constant

of integration, see.

And so we've integrated log X
now. That's not one to remember,

but it is one to remember how to
work it out because it works

with other functions that one
comes across that you don't know

how to integrate. You turn them
into a product like this.

And away we go I want
to take one final example.

The integral of E
to the X sign

of X. Now we look at this
and say, well, we've got a

choice to make one of them's got
to be you one of them's got to

be divvied by the X, but which?

But one of the things that we
know from the examples that

we've done is that if it was X
Times E to the X, we choose the

X to be you. So we could
differentiate it because the E

to the X doesn't get any

simpler. But we do the same with
the trick function because that

doesn't get any simpler.

So it doesn't look good. We
don't seem able to make a choice

between these two, because
neither of these two functions

when we differentiate them get
any simpler. But we still gotta

make a choice. So the choice
that I'm going to make really

doesn't matter, but I'm going to
do it anyway, so I'm going to

call this you.

E to the X and I'm going to call
the sign XDV by The X.

Equals let's write down the

formula. UV minus the
integral of VDU by the

X with respect to X.

You is E to the X,
it stays the same.

The is got by integrating sign
and the integral of sign is

minus cause X.

Minus the integral of minus

cause. X times by the
derivative of you, which is just

E to The xx. Let me just
tidy that up a bit. Lots of

minus signs knocking around, so
we've minus cause X times by E

to the X plus the integral of.

E to the X cause.

XDX

And back to something that looks
like this. Again, let me make

the same choices. Then let me
call this U let me call this DV

by X. So this is
minus costs X times by

E to the X Plus.

The bracket.

UVU is E to the XVV
is now the integral of Cos

X and the integral of Cos
X is sign X.

Minus the integral of the which
is just sign X times. Do you

buy the X? Do you buy the
access E to the XDX?

And close the bracket.

So we've got this term at
the front, which is minus

E to the X cause X
Plus E to the X Sign

X minus.

E to the X sign

XDX. But this is exactly what I
started out with.

Exactly what I started out with.

It's the integral I began with,
so if it's the integral I began

with, let's call it I. Let's say
I equals this integral. So what

I've got here is a sort of
equation for I.

Minus

I. Define ad
I to both sides. I'll end up

with two I equals, minus E to
the X cause X Plus E to

the X sign X now.

I can take out half of each
side, divide both sides by two

1/2 of minus E to the X cause
X Plus E to the X Sign X.

And so I found my integral
rather remarkably, are found. It

without actually apparently
integrating it at all. It came

out of solving an equation.

So. That concludes integration
by parts. You need to remember

the formula. Let's just write
that down to finish with that.

If we are integrating you DV
by The X, the result we

have is UV minus the integral

of. The ubaidi X with respect to
X and I need to put that in

there as well.

We always have a choice which
of the two functions where

multiplying together to be you
and which to be divvied by the

X. Our first choice is usually
governed by. If we

differentiate EU, will it make
it simpler? Will it make this

second integral simpler? If
that doesn't work?

It's governed by whether we can
actually integrate or

differentiate certain of the

functions. And sometimes we need
to repeat it so that we can

solve an equation to get the

integral. This.

Needs to be
remembered. The

formula for
integration by parts.