So we’ve just learned how to
e of x's and ln of x's
I’m now going to show you some techniques
when functions become fairly complicated,
the things that we can do to simplify it.
In mathematics, there’s this
thing called substitution,
which we quite like when things
get a bit too complicated.
We can make it easier by making substitutions
dealing with shorter functions,
dealing with our own substitution and
then changing it back at the end.
It’s a very useful technique especially
when things get more complex.
So, what I mean by that is if
you have a function f of x dx,
and imagine it’s not
it has two or three things multiplying
together, or some functions,
you can change this entire
function to a u function.
So the idea is that we change everything
in that function to u functions.
We then differentiate it with respect
to u rather than with respect to x,
but we also change the limits.
Now, this may look a bit bizarre
as to why should we do this.
You’ll see this is useful especially when
we do the more complicated functions.
It probably isn’t a good idea doing this
when your functions are straightforward
and you can just integrate.
Add one to the power and
divide by new power.
But you’ll see that the functions
were about to integrate.
It makes life a lot easier if
we use substitution methods.
Keep this definition in mind,
but I’ll teach you this by
using numerical examples
because it’s a lot easier and it makes a
lot of sense when we do it with functions.
So let’s look at this example.
We’re going to work through
this slowly because with it,
I’m going to teach you
integration by substitution.
So, our integral says that
we’ve got x multiplied by cos,
x squared plus 1 dx.
We also have the limits 0 and 1.
Now, if I asked you just to integrate
it, you probably won’t be able to
because it’s a function that’s being
multiplied with another function,
and we haven’t really spoken
about any techniques here.
So, we do need to find a way
to make this easier for us
in order for us to integrate it.
To do this, we use this
method called substitution.
So, anything that look too complicated, we
can substitute easier letters for them,
and we can change the whole
equation in terms of that letter.
We usually use the letter u.
So, here’s what I’m going to do.
I’m going to change this angle into
something a little bit easier.
So, I’m going to say here on the side,
let u equals to x squared plus 1.
So, this equation or this integral
now becomes x cos of u dx.
Now, it’s a very mixed
There’s lots going on.
I’ve got an x, I’ve got a u, I’ve got a dx,
and I’ve got these limits
which are my x limits.
So be clear about those.
These are your x limits and they
can only go into x functions.
So, in some sense, I’ve made it a
little bit messier, but let’s continue.
I am now keeping a clear
objective that I want to
make this entire equation
in terms of u’s.
I want to change everything into u.
So the next step I’m going to start
with is changing my dx to du.
I can do that by using
my substitution here.
So, I said u was x squared plus 1.
I can now do du by dx.
I can differentiate
this to give me 2x.
I can get this dx by itself just
by rearranging this equation.
So I can write du
equals to 2x dx.
I’ve multiplied this
on the other side.
And remember, I want to
take this by itself,
so make this the subject of the equation
so I can rewrite this as
du over 2x equals to dx.
So hopefully, this is
I took my u substitution, I
differentiated du with dx,
and then I rearranged
to get dx by itself.
Why did I do it?
Because I can replace this
dx with this dx here.
So, watch what happens when I do this.
I do 1 0 x cos of u.
And instead of my dx, I
can now write du over 2x.
Still looks like a mixture,
but if you notice here, this
x is cancelling with this x.
Let’s see what our integral looks like now.
0 1 cos of u du.
And this I’m a lot happier with
because I know that cos of anything
integrated with respect to that thing
will give me a straightforward answer.
I know my identities.
There’s one little thing I
need to do here though.
Remember I said that these
were your x limits.
So, I need to change
Those are my x limits and those
can’t go into this function
because this is all
in terms of u.
But again, I come back to my
substitution, u substitution.
I said u was x squared plus 1.
So, we want to know what u is
when x equals to 1 and when x equals
to 2 --
when x equals to 0, I mean.
So, you can find out u when we put
1 in; 1 squared plus 1 is just 2.
And when you put 0 in, that gives
you 0 squared plus 1 is just 1.
So, what you’ve done here is
you’ve changed your x limit;
x equals to 1, x equals to 0.
You’ve changed your x
limit to u equals to 2,
and your x limit of
0 to u equals to 1.
And now, if I change my limits,
remember, the 1 has become
2, and the 0 has become 1.
I have cos u du,
and life is a lot easier integrating
this as compared to the first function.
How did you integrate cos?
Remember the little tables that we did.
Sine goes to cos, cos goes to
-sine when you differentiate;
when you integrate,
you go backwards.
So cos, and if you look at this here,
we’re going in that direction here.
Cos will integrate to sine.
So, we now end up with sine u,
nice and easy integrated.
We have to put our limits in.
So we’ll put 1 and 2 in.
So, 2 goes in first and then 1 later.
So, we can just leave it as is
because we’re not using calculators.
So the answer to this as a
statement is sine2 minus sine1.
And that here is your integral.
If you’re using calculators,
it’s just a simple job of you putting it
into your calculator to find your answer.