www.mathcentre.ac.uk/.../9.5%20Integration%20by%20substitution.mp4

Sometimes and apparently
complicated integral can

be evaluated by first of
all making a substitution.

The effect of this is to change
the variable, say from X to

another variable U.

It changes the integrand.
That's the quantity that

we're trying to integrate.

And if we're dealing with
definite integrals, those

with limits than the limits
may change too.

Before we look at our first
example, I want to give you a

preliminary result which will be
very useful in all of the

examples that we look at.

Suppose we have a variable U.

Which is a function of X, so you
is a function of X.

Suppose we can
differentiate this

function of X and
workout du DX.

Then a quantity
called the

differential du is
given by du DX.

Multiply by DX and this will be
a particularly important result

in all of the examples that we
do. So, for example, if I had

you equals 1 - 2 X, so there's
my function of X.

We can differentiate this du DX.

Well, in this case du
DX will be minus 2.

Then the differential du is
given by du DX.

Minus 2 multiplied by
DX.

So this formula will be very
useful in all the examples that

we do. Let's have a look at our
first example of integration by

substitution. Suppose we want to
perform this integration.

Suppose we want to find the
integral of X +4.

All raised to the power 5 and we
want to integrate this with

respect to X.

Now the problem here in this
example is the X +4.

This is the complicated bit
and the reason why it's

complicated is because if this
was just a single variable,

say X would be integrating X
to the power 5, and we all

know how to do that already.
So this is the problem. We'd

like this to be a single
variable, and so we make a

substitution and let you be
this quantity X +4.

Let's see what happens when we
make this substitution.

This integral is going to become
the integral X +4 is going to

become just simply you.

We've got you to the power 5.

And then we have to be a little
bit careful because we've got to

take care of this quantity an
appropriate way as well. Now we

do that using the results I've
just given you. We workout the

differential du. Now remember
that du is du DX.

Multiplied by DX.

In this case, du DX is
just one. If you equals X

+4 du DX, the derivative
of you will be just one.

So in this first example with
a nice simple case in which du

is just equal to 1, DX or du
is equal to DX. So when we

make the substitution of, the
DX can become simply.

See you.

Now this is a much simpler
integral than the one we started

with and will be able to finish
it off in the usual way just by

increasing the power by one, so
will get you to the Palace 6.

We divide by the new power.

And we don't forget to add the
constant of integration.

Now, for all intents and
purposes, that's the problem

solved. However, it's normal to
go back to our original

variables. Remember, we
introduced this new variable

you, so we go back to the
variable of the original

problem, which was X. Remember

that you. Is equal to X +4.

So replacing you by X +4.

We have X +4 to the power
six or divided by 6 plus a

constant of integration.

And that's our first example of
integration by substitution.

Let's have a look at another
example. Suppose this time

that we want to integrate a
trigonometric function and

the one I'm going to look at
is the cosine.

Of three X +4.

And we want to integrate this.

With respect to X.

I'm going to assume that we
already know how to integrate

cosine X Sign X, the standard
trig functions. This is a bit

more complicated, and the reason
it's more complicated is the 3X

plus four in here.

So as before, we make a
substitution to change this

into just a single variable,
so I'm going to write you

equals 3X plus 4.

The effect of doing that is to
change the integral to this one.

The integral of the cosine.

And instead of 3X plus
four, we're now going to

have just simply. You are
much simpler problem than

the one we started with.

As before, we need to take
particular care with this term.

Here the DX. And we do that with
our standard results that du.

Is equal to du DX.

DX

now what's du DX? In this
example? Well, you equals 3X

plus four, and if we
differentiate this function with

respect to X, will get that du
DX is just three.

So our differential du is simply
3D XDU is 3D X if we just

rearrange this will be able to
replace the DX by something in

terms of du over here. So if du
is 3 DX then dividing both sides

by three, we can write down that
DX is 1/3.

Of du

so the DX in here is replaced by
1/3 of du. Now I'll put the du

there and the factor of 1/3 can
be put outside the integral sign

like that. A constant factor can
immediately be moved outside the

integral. Now this is very
straightforward to finish. We

all know that the integral of
the cosine of you with respect

to you is sign you.

Plus a constant of integration.

And as before, we revert to our
original variables using the

given substitution. You was 3X
plus four, so this becomes 1/3

of the sign of three X +4.

Plus a constant of integration.

And that is our second example
of integration by substitution.

Now before I go on, I want to
generalize this a little bit.

This particular example we had
the cosine of a constant times

X plus another constant and
let's just generalize that so

we can deal with any
situations where we have a

cosine of a constant times X
plus a constant. So let's

suppose we look at this
example.

Suppose we want to integrate the
cosine of AX plus B.

With respect to X.

When A&B are constants.

Again, the problem is the
quantity in the brackets

here, and we avoid the
problem by making a

substitution. We let you be
all this quantity X plus B.

That changes the integral to the
integral of the cosine and X

Plus B becomes simply you.

We've got to deal with the DX,
and we do that by calculating

the Differentials Du.

Du will be du DX, which in this
case is simply A.

DX

and this result will allow us to
replace the DX in here or DX

will be one over a du.

So the DX here will become one
over a DUI, right? The du there

under one over a being a
constant factor. I'll bring

straight outside here.

This is straightforward to
finish off because the

integral of the cosine you is
just the sign of you plus a

constant of integration.

In terms of our original
variables, will have sign and

you is a X plus B.

Plus a constant of integration.

And that's a general result.
We can use any time that we

have to integrate the cosine
of a quantity like this. This

is an example of a linear
function X plus be, so we've

got the cosine of a linear
linear function, and we can

use this results anytime we
want to integrate it. So for

example, if I write down,
what's the integral of the

cosine of Seven X +3?

Then immediately we recognize
that the A is 7.

The B is 3 and we use this
general result to state that

this integral is one over A.

Which will be one over 7.

The sign.

Of the original content in the
brackets, which was Seven X +3.

Plus a constant of
integration. So anytime we get

a function to integrate, which
is the cosine of a linear

function, we can use this
result.

There's a very similar
results for integrating sign,

and I won't prove it. I'll
leave it for you to to prove

for yourself, but the result
is this one that the integral

of the sign.

AX plus B.

With respect to X will be minus
one over A.

Cosine of X plus B.

Plus a constant of
integration, and that's

another standard result you
should be aware of, and you

should now be able to prove
for yourself.

OK.

Let's have a look at another

example. Suppose we want to find
the integral of 1 / 1 - 2 X.

With respect to X.

Now I'm going to assume that
if you had a single letter

down here, if they've just
been an X down here, so we

we're integrating one over X,
you'd know how to do that just

by the. By that the integral
of one over X with respect to

X is the natural logarithm of
the modulus of X. So I'm going

to assume that you know how to
do that already.

So if we can convert this
integral here into one where we

just got one over a single

variable. Perhaps we will have
to proceed, so will make a

substitution, and the
substitution will make is U

equals 1 - 2 X.

What will that do to the
integral? Well, the integral

will become the integral of
1 divided by.

1 - 2 X will become you much
simpler. We have to take care of

the DX in an appropriate way.

As before DU is
du DX.

Multiply by DX.

What's du DX in this case?

Well, you is 1 - 2 X.

So do you DX the derivative of
this quantity is just minus 2.

DX

so when we want to replace the
DX in the substitution process,

we can replace DX by du.

Divided by minus two, which is
minus 1/2 of DS.

So the DX here becomes du and
the factor of a minus 1/2. I can

right outside being a constant
factor. Now this is

straightforward to finish off
because the integral of one over

you do. You will be just the

logarithm. Of the
modulus of you.

So we've minus half natural
logarithm of the modulus of you.

Plus a constant of integration.

Nearly finished or we do is go
back to our original variables.

Now. Remember that the original
variable was U equals 1 - 2 X,

so this will become minus 1/2
natural logarithm of the modulus

of 1 - 2 X.

Plus a constant of integration.

And that's another example.

It's very easy to generalize
this to any cases of the form

where you have one divided by a
linear function of X. Let's just

see how we can do that, and then
we'll get a very useful general

result. Suppose we want to
integrate 1 divided by a X

plus B.

We want to integrate
with respect to X.

The substitution we make is U
equals a X plus B.

Do you?

Will be du DX, which in this
case differentiating this

function will just leave us
the constant a do you will be

equal to a times DX. So when
we want to replace the DX we

will do so by replacing it
with one over a du.

What will that do to our
integral? Well, the integral

will become the integral of
1 divided by.

AX plus B becomes you.

And the DX becomes one over a
du. I'll write the du here

and the one over a being a
constant factor. Uh, bring

straight outside.

Now this is very familiar,
straightforward for us to finish

this the integral of one over U.

Will be the natural logarithm of
the modulus of you plus a

constant of integration.

Nearly finished, we just return
to our original variables, one

over a natural logarithm of the
modulus U was X plus B.

Plus a constant of integration.

Now, this is a particularly
important general results, which

I'd like to get very familiar
with because it's going to crop

up over and over again, and you
want this sort of thing at your

fingertips. For example, if I
ask you to integrate one over X

plus one with respect to XI,
want you to be able to almost

just write these things down if
we identify the one over X plus

one with the one over X Plus B,
then we see that a is one.

And B is one.

Got a is one and be as one. This
general result will give us one

over one which is just one
natural logarithm of the modulus

of X Plus B, which in this case
is X plus one.

So our integral of one over X
Plus One is just the logarithm

of the denominator. Another
example, the integral of one

over 3X minus two with
respect to X.

Well, using our general result
we see that a is 3.

Be is minus two. Put all these
into the formula and we'll get

one over A.

Play being 3 means
we get one over 3.

Natural logarithm. Of the
modulus of the quantity we

started with, which was 3X
minus 2.

Let's see, as I say, this is
a particularly important

result and you should have it
at your fingertips. You'll

need it over and over again
in lots of situations as you

perform integration.

All the examples that we've
looked at so far have been

examples of indefinite
integrals. They've not been any

limits on the integral
whatsoever, so let's now have a

look at how we do an integration
by substitution when there are

limits on the integral as well,
and the example that I want to

look at is this one. Suppose we
want to integrate from X equals

1 to X equals 3, the function 9

plus X. All raised to the power
two and we want to integrate

that with respect to X.

Now have this just being an X to
the power two. There would be no

problem. Increase the power by
one divided by the new power and

you be finished. The problem is
the 9 plus X. So as before we

make a substitution, we make a
substitution to try to simplify

what we're doing and the
substitution we make is U is

equal to 9 plus X.

Let's see what that
does to the integral.

Our integral will become, well,
the 9 plus X becomes U so will

have you squared.

Let's deal next with the DX, as
we've seen already, do you is

equal to du DX?

Multiplied by DX and in this
particular case, this is a

nice simple case. Du DX will
be just one.

So whenever we see a DX, we
can replace it immediately

by du to USD X, so this DX
becomes a deal.

So far it's the same as before,
but now we've got to deal

specifically with these limits.
We have these limits.

A lower limit of one, an upper
limit of three, and these are

limits on the variable X.

We're integrating from X equals
1 to X equals 3. When we change

the variable. It's very
important that we changed the

limits as well, and the way we
do that is we use the given

substitution. We use this to get
new limits limits on you.

Now when X is one.

The lower limit you is going to
be 9 + 1.

10 so when X is one US 10.

At the upper limit, when X is 3.

Our substitution tells
us that you is going to

be 9 + 3, which is 12.

We're going to use the
substitution to change from

limits on X to limits on you.

So these limits on you are now

you Ekwe. Tools 10 to you

equals 12. And I've
explicitly written the

variable in here so that we
know we've changed from X to

you.

This is straightforward to
finish off because the integral

of you squared is you cubed over
3 plus, plus no constant of

integration because it's a
definite integral and we write

square brackets around here.

And we write the limits
on the right hand side.

We finish this off by putting
the upper limit in first, so you

being 12 we want 12 cubed.

Divided by three.

That's the upper limit gone
in. We want to put the lower

limit in you being 10, so we
want 10 cubed.

Divided by three and as
usual, we find the difference

of these two quantities.

Now 12 cubed.

Get your Calculator out
12 cubed.

Is 1728 so we have 1728
/ 3 subtract 10 cubed which

is 1000. Divided by three,
so our final answer is

going to be 728.

Divided by three.

And that's our first example of
a definite integral using

integration by substitution.
Always remember that you use the

given substitution to change the
limits on the integral as well.

Don't forget to do that.

OK, all the examples that we've
looked at so far have been

fairly straightforward, and
they've required a substitution

of the form you equals X plus be
a linear substitution. We're

going to now have a look at some

more complicated ones. All
the examples I'm going to

look at or of this form.

So all going to take the form of
the integral of F of G of X.

G dash devex.

DX now this looks horrendous. So
what will do is will try and

take it apart a little bit to
see what's going on here.

Suppose we have a function
G of X which is equal to 1

plus X squared.

That's going to be this
function in here, and you'll

notice from this expression
that the G of X is used as

input to another function F.

Suppose the function F is the

square root function. The
function which takes the

square root of the input.
So I'm going to write that

like this. Suppose F of you
is the square root of you.

Then what does this mean? F of G
of X? Well, this is called a

composite function or the
composition of the functions F&G

the Function G is used as the
input to F, so F of G of X.

Well, the F function square
roots the input, so we want

to square root the input
which is G of X, which is one

plus X squared.

So in this particular case,
we're looking at an example

that's going to be something
like this, but the integral of F

of G of X.

Is the square root
of 1 plus X squared.

What about the G dash of XG dash
trivex is the derivative of G

with respect to X.

Well, his GG of X is one plus X
squared. What's its derivative?

The derivative G Dash Devex.

Is just 2X.

So if I substitute for G, Dash,
Devex as 2X, I'll be dealing

with an integral like that.

So as a first example, we're
going to look at an integral of

this particular form, and I hope
that you can now see it's of

this family of integrals. We've
got a function G of X in here,

one plus X squared, that's that,
bit its derivative G dash,

devex. The two X appears out
here and the G itself is input

to another function, which is
the square routing function F.

So it looks a bit
complicated, but I hope we

can see what all the all the
different ingredients are.

Now, the way we tackle a problem
like this is to always make the

substitution you equals G of X
whatever the G of X was.

Well, in this particular case, G
of X was one plus X squared, so

I'm going to make this
substitution. You equals 1 plus

X squared. Let's see what that
will do to this integral.

Letting you be one plus X
squared will just have the

square root of you in here.

Now we've got it handled all
this. The two X DX. Well in

terms of Differentials, we know
already that du is du DX DX.

Well do you DX in this case is

just 2X. So do you. DX DX
is 2X DX.

Now this is very fortunate
because we see that the

whole of two X DX.

Can be replaced by.

Do you so the whole of two X
DX becomes the du and

integrals of this form? This
will always happen when we

make a substitution like
this. Now this is very

straightforward to finish,
provided you know a little

bit of algebra. The square
root of U is due to the

power half, so we're
integrating you to the half

with respect to you.

How do we do this?

We increase the power by one.

That will give us you to
the half plus one that's

1 1/2 or three over 2.

And we divide by the gnu power
and the new power. In this case

is 3 over 2, so we're dividing
by three over 2, and we add a

constant of integration.

Nearly finished. Dividing by
three over 2 is like multiplying

by 2/3, so I'll write that as

2/3. And in terms of our

original variable. You was one
plus X squared.

And all that needs to be raised
to the power three over 2.

And we need a constant of
integration.

So that's the first example of
a much more complicated

integral, and I want to go
back and just point a few

things out. Remember, a
crucial step was to recognize

that the 2X in here was the
derivative of this quantity,

one plus X squared in there.

The G dash of X in the general
case is the derivative of the

input to the F function here.

What will do is will have a
look at another example and

then we'll try and start to do
some of this purely by

inspection because at the end
of the day we want you to be

able to get to the stage where
you're happy, almost just

recognizing some of these
integrals.

OK, let's have another
look at another example

of one of these
complicated things. F. Of

G. Of X.

G dash devex.

And in this case I'm going to
take as my G of X function. The

following G of X is going to be
2 X squared plus one.

This is going to be used as
input to the F function, and in

this case I'm going to choose F
as this function F of you is

going to be the function which
takes an input, finds it square

roots, and then finds its
reciprocal. So F of you is one

over the square root of you.

What would that look like in
terms of this integral?

Well, this integral will be
F of G of X.

All this means that the G of X
function is used as input to

F. So if we use two X squared
as one as input to this

function, F will get one over
the square root of 2 X squared

plus one. So I'll be
integrating something like

this.

We also need to look
at the G Dash Devex.

Well, because G of X is
2 X squared plus one, if

we differentiate GG
dash, devex will be just

simply to choose of 4X.

So for G Dash Devex, I'm going
to write 4X.

And with the DX.

Look in this example. It's an
integral like this now this

looks very complicated. It might
have actually been given to you

in a form that looked more like
this because we might have

written the 1 * 4 X all as the
single term. We might have

actually written it down like 4X
divided by the square root of 2

X squared at one, all integrated
with respect to X.

Just look a bit complicated, but
we'll see by making a

substitution that we can make
some progress and the

substitution that will make as
before is we let you be the

function G of X.

We let you be this
function in here.

So if you.

Is G of X or G of X? In this
case was two X squared plus one.

What's going to happen?

Well, let's put all this
into this integral.

Don't worry about the 4X for
a minute, but the two X

squared plus one here in the
denominator becomes a U, so

we'll have a square root of
you in the denominator.

We've got to. Look after the
four X DX as well, but in terms

of differentials do you remember
is du DX, which is 4X times DX

and you'll see because of the
nature of this sort of problem

that the four X DX quantity is
exactly what we've got over here

for X DX. So the whole of four X
DX becomes du the whole of that.

Becomes a du.

That's the nice thing about
problems like this. That

substitution will always make
them drop out in this way.

Again, provided that you know a
bit of algebra, you can finish

this off. Let's just do that.
The square root of U is due to

the power half. One over the
square root will give us you to

the minus 1/2. So the problem
that we're integrating here is

you to the minus 1/2.

With respect to you.

You to the minus 1/2 integrated
we increase the power by one.

So we increase minus 1/2 by one
will give us plus 1/2.

And we divide by the new power.

And we need a constant of
integration and for all intents

and purposes that's finished,
but we can revert to our

original variable X through the
substitution. What did we have?

We had you.

What's 2 X squared plus one? So
this is going to become two X

squared plus one all raised to
the power half plus C and

division by 1/2 is like
multiplication by two, so I have

a two there at the beginning and
that's the solution of this

fairly complicated integral done
through a substitution.

So let me try and extract from
all that are fairly general

results and the general result
is this that if we have an

integral of F of G of X.

He dashed of X DX.

Then the
substitution U

equals G of X.

Do you is G dashed
of XDX?

That substitution will
always transform this

integral into simply F of G
of X, which is F of U.

And the whole of G dash DX.

Will become simply do youth.

And this hopefully will
be an integral which is

much simpler to evaluate.

Now really what we want to be
able to do is get you into the

habit of spotting some of these
and being able to write down the

answer straight away. So let me
just give you one or two more

examples where hopefully we can
actually spot what's going on.

Supposing we looking at
the integral of E to the X

squared multiplied by two
X and we want to integrate

that with respect to X.

Let's try and compare what we've
got here with what's up here.

If you think of the X squared as
being the G of X.

So this bit is the G of X.

If we differentiate, X squared
would get 2X and you see that

appears out here, so this, but
in here is the G Dash, Devex.

And there's another function
involved in here as well.

It's the F function, and in
this particular example, the

F function is the exponential
function, and we see that G

is input to F because X
squared is input to the

exponential function.

So if we want to tackle this
particular integral, the

substitution U equals whatever
the G of X was, which in this

case is X squared will simplify

this integral. You being X
squared do you will be 2X DX.

And immediately this integral
will become the integral of.

E to the power X squared, which
was E to the power you.

And automatically the two X DX
is taken care of in the du.

Now this is just the integral of
the exponential function E to EU

with respect to you and we can
write the answer straight down

as the same thing eater, the you
and a constant of integration.

If we return to our original
variables, you was X squared, so

will hav E to the X squared plus
a constant. So this quantity E

to the X squared plus a constant
is the integral of two XE to the

X squared DX.

And as I say, we want to get
into the habit of being able

to almost spotless, and you
should get into the habit of

spotting that the quantity
here the two X is the

derivative of this function
in this other composite

function. Here, let's see if
we can do one straightaway.

Supposing, supposing that I
ask you to integrate the

cosine of three X to the
power 4 * 12 X cubed. Suppose

we want to integrate this
horrible looking thing.

Now the thing I want you to spot
is that if you differentiate

this function in here 3X to the
power four, you'd actually get 4

threes or 12X cubed. You get the
concert if it's out here.

So this is like the G of X.

And this is like the G Dash,
Devex, and making the

substitution you equals 3X plus
four would immediately reduce

you to a much simpler integral,
just in terms of you. I'm hoping

that you by now you can spot
that if we integrate, this will

actually just get the sign.

Of three X to the four plus a
constant of integration. Suggest

you go back and look at that
again if you not too happy about

it. And that's integration
by substitution.