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.