Definite integral with substitution

Solving a definite integral with substitution (or the reverse chain rule)

So I've been sent this definite
integral problem and it seemed

as good as any, and I think the
key with this is just to

see a lot of examples.

So let's do it.

This definite integral is from
pi over 2 to pi of minus cosine

squared of x times sin of x dx.

So before we just chug
through the math and do the

antiderivatives and use the
fundamental theorem of calculus

to evaluate the definite
integral, let's think about

what we're even doing.

So I've graphed this function
right here, minus cosine

squared of x times sin of x.

And what we care about,
we're defining the definite

integral between pi over 2,
which is roughly here.

Let me see if I can make
this a little bit bigger.

So between pi over 2 which is
right there, and between pi.

So the definite integral of
this function between here and

here is essentially the area of
the curve between the

curve and the x-axis.

And since the a curve is below
the x-axis here, this area is

going to be a negative number.

So that gives us
immediately an intuition.

We should be getting a negative
number when we evaluate this

and just to prove this I
actually typed it in up here.

So let's now evaluate
this definite integral.

Now I'll rearrange some of the
terms here just to make it a

little bit easier to read.

But the way I always think
about it is, well I have

a cosine and a sin.

The cosine is squared, so all
these crazy things are

happening to it, so it seems
like I could use substitution

or the reverse chain
rules some out here.

And what was the chain rule?

The chain rule said if I took
the derivative of f of g of x

that this is equal to f prime
of g of x times g prime of x.

That might completely confuse
you, but I just wrote that here

because we could say, well what
if g of x is cosine of x. f

prime of g of x is the cosine
of x squared, and then the

derivative of g of x or the
derivative of cosine

of x is sin of x.

Well it's actually minus sin of
x, and we have a minus sin

here, so that works
out well too.

If this confuse you, ignore it.

Well essentially we're just
going to do the same thing,

but we're going to do
it with substitution.

So let me do it
with substitution.

Let me a erase this if
this confuses people.

I want to do it however it
is least confusing to you.

OK let me erase that.

Actually, let me do it with
substitution, just because the

way I was just doing is kind
of my shortcut back in the

day when I was a mathlete.

But it's good to be able to do
it with substitution, helps you

from making careless mistakes.

So let me rewrite
this first of all.

This is the same thing as the
integral from pi over 2 to

pi of cosine squared of x.

Actually let me write that
as cosine of x squared.

Same thing, right.

Times minus sin of x dx.

And now it should
be clearer to you.

What's the derivative
of cosine of x?

It's minus sin of x.

So I have a function and it's
being squared, and I have its

derivative, so I can figure out
its antiderivative by using

substitution or the
reverse chain rule.

So let's make a substitution.

u is equal to cosine of x.

How did I know to substitute
u is equal to cosine of x?

Well because I say, well
the derivative of this

function is here.

So when I find du, this whole
thing is going to end up

being du, and let me
show that to you.

So what is du/dx?

du/dx is equal to
minus sin of x.

That hopefully we've
learned already.

So what is du?

If we multiply both sides by
the differential d of x get du

is equal to minus sin of x dx.

So if we look at the original
equation, this right here we

just showed is equal to du,
and this right here is what?

Cosine of x is u.

That was our original
substitution.

So we have u squared.

So now let's take the integral.

And I will arbitrarily
switch colors.

And now this is a very
important thing.

If you're going to do
substitution, if we're going to

say u is equal to cosine of x,
we're going to have to actually

make this substitution
on the boundaries.

Or we could do the substitution
and reverse the substitution

and then evaluate the
boundaries, but let's do that.

So if this is going from x is
equal to pi over 2 to pi,

what is u going from?

Well when x is equal to
pi over 2, u is equal

to cosine of pi over 2.

Because u is just cosine of x.

And then when x is pi, i is
going to be cosine of pi.

And now the fun part.

Cosine of x squared.

Well that's just the same
thing is u squared.

And minus sin of x dx,
that's the same thing

is du-- did that here.

This is pretty straightforward
and I'm just going

to rewrite it.

What's cosine of pi?

Cosine of pi is minus 1.

Cosine of pi over 2,
well that's a 0.

So we have the integral from u
is equal to 0 to u is equal to

negative 1 of u squared du.

And now this seems like
a simple problem.

So this is equal to the
antiderivative of u

squared, which is
fairly straightforward.

u cubed over 3.

You could just take the
derivative of this,

you get this.

All I did is I increased this
exponent to get the third, and

I divided by that exponent.

Now we're going to have to
evaluate it at minus 1 and

subtract from that,
it evaluated at 0.

So this is equal to minus 1
to the third over 3 minus

0 over 3, and so this
is equal to minus 1/3.

And we are done.

And if we look at this area
from our original graph, what

we just solved, as we said the
area of the curve between

here and here is minus 1/3.

Or if we wanted the absolute
area because you can't really

have a negative area, it's 1/3
but we know it's negative

because this curve is
below the x-axis here.

And that looks about right,
that looks about 1/3.

I mean if this cube right
here is 1, then that

looks about 1/3.

The intuition all
works out at least.

So hopefully you found
that vaguely useful.

Actually let me-- since we
have a little bit of time.

Hopefully you understood this,
and if you did, don't worry

about what I'm going to do now,
but I want to show you how I

tend to do it where I
just think of it is the

reverse chain rule.

It's a little bit
quicker sometimes.

But it's really the same
thing as what we just

did with substitution.

So if we erase all of this.

And so if we have this integral
right here, what I do is I say,

well I have cosine
of x squared.

I have cosine of x squared, and
then I have minus sin of x.

This is the derivative of this.

Since the derivative is here,
I can just treat this whole

thing like an x term.

So this is the same thing.

So the antiderivative is cosine
of x to the third over 3, and I

evaluate it at pi
and pi over 2.

And remember, how
did I do this?

What allowed me to treat this
cosine of x just like an x or

like a u when I did it
with the substitution?

Well I had its derivative
sitting right here,

minus sin of x.

So that's what gave me the
license to just take the

antiderivative, pretend like
this cosine of x is just an x,

is just a u, you could even
say, and just take it's

exponent, raise it by 1 and
divide it by 3 and then

evaluate it from
pi to pi over 2.

So this is equal to cosine of
pi cubed over 3 minus cosine

of pi over 2 cubed over 3.

This is minus 1 to the third,
so this is equal to minus

1/3 minus this is 0.

So we get the same answer.

I just wanted to show you that.

It's exactly the same
with substitution.

It's just I didn't formally do
the substitution, but it's

the exact same thing.

Anyway, hope you
found that helpful.