Indefinite Integration (part III)

Integration by doing the chain rule in reverse.

Welcome back.

Well I'm now going to do a
presentation on how to

essentially invert the chain
rule or reverse the chain

rule, because we're doing
integration, which is the

opposite of taking
the derivative.

So let's just take a
review of what the chain

rule told us before.

If I were to take the
derivative of f of g of x--

hopefully this doesn't
confuse you too much.

I'll give another example
with a concrete f of x

and a concrete g of x.

If I want to take the
derivative of that, the chain

rule just says the derivative
of this composite function is

just the derivative of
the inside function.

g prime of x times the
derivative of the outer, or

kind of the parent function,
but still having g

of x in at times.

f prime of g of x.

I know this might seem
complicated if you aren't too

comfortable with this type of
notation, but done in kind

of an example form it
makes a lot of sense.

If I said what is the
derivative of let's

say sin of x squared.

Well in this situation, f
of x is sin of x, right?

And then g of x is
x squared, right?

And sin of x squared is
essentially f of g of x.

And this review of chain rule.

You could go watch the video on
the chain rule as well, but I

don't mind doing a couple
of problems here.

All this is saying that the
derivative of this is you take

the derivative of the inside
function-- g of x in this

example, which is 2x-- and you
multiply it times the

derivative of the
outer function or the

parent function.

And we memorize I guess that
the derivative of sin of x

is cosine of x, so it's
times cosine of g of x.

So we keep the x squared there.

If it confuses you, just
think about the inside

and the outside function.

If you take the derivative of
kind of this composite

function, it's the same thing
that equals the derivative of

the inside function, which is
2x times the derivative

of the outside function.

But we keep this inside
function in it, and we

keep this x right there.

So that's a review
of the chain rule.

So what happens if we want
to reverse the chain rule?

Well if we wanted to reverse
it, we're essentially saying

that we want to take the
integral of something where we

have the derivative of kind of
the inner function, and then we

have the derivative of a
larger composite function.

I'm just rewriting the chain
rule, but I'm writing in

an integral form that this
is equal to f of g of x.

This statement up here is
the exact same thing as

the statement down here.

All I did is I took the
integral of both sides.

I'm saying the integral
of this is equal to the

integral of this right here.

I probably shouldn't switched
equal signs like that

with you, but let's use
this formula I guess.

But all you have to know is
this the reverse of the chain

rule to solve some problems.

Image invert colors.

Let me rewrite that.

The integral-- if I have g
prime of x times f prime of

g of x dx, then that is
equal to f of g of x.

This is just the chain
rule in reverse.

And I know it's very
complicated sometimes when you

have it in this notation,
but I'll give you a

couple of examples.

What if I had the integral of
let's say-- this is actually

one that's often kind of viewed
as a trick, but you'll see

it's actually not that
tricky of a trick.

OK.

So let's say I have the natural
log squared over x dx.

And if you saw an integral
like this, you'd probably be

daunted, and you'd be
surprised, many people well

into college calculus courses
are still daunted

by this problem.

But all you have to
recognize is this is

the reverse chain rule.

Why is this the
reverse chain rule?

Well, this is the same thing as
the integral of 1/x times the

natural log-- whoops, this
should be nlx, right-- the

natural log of x squared dx.

These are the same thing,
I just took the 1/x out.

Now this might look
a little familiar.

Well, what's the derivative
of the natural log of x?

If you remember from
the derivative module,

it's 1/x, right?

Let me write that down
in the corner here.

The derivative of the natural
log of x is equal to 1/x.

So right here we have
the derivative of the

natural log of x.

So now we can just say that we
could essentially treat this

natural log of x as kind
of a variable by itself.

And essentially what I'm
going to be doing if I could

actually substitute for.

Actually let's do that.

Well no, no, no I don't do that
now, that'll confuse you.

Although my flip-flopping
is probably confusing

you even more.

I have the derivative of the
natural log of x, so I can then

say well I have the derivative
there, so this is a

composite function.

This is essentially
f prime of g of x.

So then I can say well
that integral must be

equal to this thing.

This is something
squared, right?

So what's the integral
of something squared?

Well the integral of
something squared is 1/3.

That's something to
the third power.

We learned in the previous
indefinite integral

module, right?

And then it's 1/3 something to
the third power, and then we

know from the chain rule that
something is the ln of x.

And I don't know if I've
already forgotten to do it

once, but don't forget
to do the plus c.

Now you say, Sal, this
completely confused me,

because it probably did.

And if it completely confused
you, let's just take the

derivative of this and I think
you'll see it happening the

other way around and it
might make a little sense.

When you take the derivative,
we just use the chain rule.

You take the derivative
of the inside first.

The derivative of the inside is
1/x and you multiply that times

the derivative of the outside
function, and then you

keep the inside the same.

So the derivative of the
outside function is 3 times it

coefficient, so it's 3 times
1/3 times the whole thing

to one less exponent.

So the whole thing is ln of x.

And then of course
plus 0, right.

The derivative of c is 0.

Well this is just equal
3, 3 cancel out.

This is equal to 1/x times
the ln of x squared, which

is our original problem.

Let me do another problem
because I probably started

off with something a
little bit too hard.

What is the integral of
let's say sin of x to

the third power dx.

That's often written like this.

That's often written
like sin of x.

Same thing, but I like to
think of it this way because

it's not a new notation.

Actually this is a mistake.

Clearly I'm making up these
problems on the fly.

Actually I don't
want to do that.

That is the wrong problem.

I want to take the integral--
and actually you can see kind

of how I'm thinking about these
problems-- I'm going to take

the integral of cosine of x
times the sin of x to

the third power dx.

Well, we have this kind of more
complicated part, the sin of x,

and we have the derivative sin
of x because we learned the

derivative sin of
x is cosine of x.

So if we have a function inside
of a larger composite function,

and we have it's derivative, we
can just treat this function as

kind of like a single entity.

Like if this was just one
variable and then we

take integral of it.

So this just equal to sin of x
and we raise this one more

power to the fourth and
we multiply times 1/4.

And how did we do that?

Because we know that the
integral of say x to the fourth

dx is equal to-- I mean x to
the third dx-- is equal

to 1/4 x to the fourth.

Instead of an x we
had a sin here.

And remember the reason why we
did that is because the

derivative of the sin function
is sitting right here.

In the next presentation, I'll
show you why this can also be

done using substitution, or
why they're the same thing.

I'll see you in the
next presentation.