Indefinite Integration (part 7)

Another example using integration by parts.

I'm now going to do it
integration by parts problems.

I think it's just a fun problem
to see because one, it's the

example a lot of people use,
sometimes even a trick problem

that's given on a really hard
math exam, or if you go to

calculus competitions like
I used to in high school.

Not to make myself too-- I was
actually not that geeky as a

high school student, but I have
to admit, I was a mathlete.

But anyway, this is just a fun
integration by parts problem

because you actually never have
to evaluate the final integral.

So let's say we want to
take the integral-- it's

a bit of a classic.

I wouldn't be surprised if your
math teacher does the same

problem for you, just to show
you integration by parts.

Let's take the integral of e to
the x-- you probably never

heard of someone call a math
problem a classic, but

hopefully I will instill in you
this love for mathematics and

you will also consider this
to be a classic problem.

e to the x times cosine of x.

I think you might already see
where I'm going with this,

because these are both fun
functions, because e to the x

you can take the derivative,
you could take the

anti-derivative and it
still stays e to the x.

Cosine of x you take the
derivative, you go to minus

sign of x, you take the
derivative again then you to

minus cosine of x, then you
take the derivative again then

you get a plus sign of x.

It's like this cycle.

The same thing happens when
you take the anti-derivative.

It's not as cool as e to the x,
it doesn't stay exactly the

same, but it kind of cycles.

If you take two
anti-derivatives you get back

to the negative of itself.

And if you take two
derivatives, you get back

to the negative of itself.

It's also a pretty cool
function and I think you can

start to see how integration
by parts might be cool here.

Whenever I do integration by
parts I always like to assume

that this is the g prime of x.

That e to the x is g prime
of x, because e to the x

literally doesn't change.

Although we could do this
problem the other way.

Maybe I'll experiment
doing it the other way.

but let's assume this is
g prime of x, and let's

assume this f of x.

So this is derivative.

So integration by parts, as we
take the original functions,

g of x and f of x.

If this is g prime of
x, what's go of x.

What's the anti-derivative
of e to the x.

It's just use e to the x.

I'm going to switch colors,
I don't like this blue.

So this is g of x.

I actually took the
anti-derivative of it, but

it's the same exact thing.

And then times f of x.

Then I want to subtract
the indefinite integral

of f prime of x.

One, g of x.

This is the same as this, which
are both the anti-derivative of

this, although they
are all the same.

So this is g of x and then I
would take the derivative

of f of x. f prime of x.

What's the derivative
of cosine of x?

It's minus sine of x.

So sine of x d x, it's
minus sine of x.

I could put the minus here,
that'll make it look messy, I

could put the minus here
that'll make it messy or I

could just put minus here and
make these minuses cancel

out and I get a plus here.

So I get the integral of e to
the x cosine of x d x is equal

to e to the x cosine of x plus
the integral of e to

the x sine of x d x.

Hopefully I haven't
confused you too much.

I should actually do some
integration by parts problems

without e to the x.

It's very hard to keep track
of what I've done here.

This is the anti-derivative.

This is the anti-derivative
and this is also the

anti-derivative.

This is g prime of
x, this is g of x.

So once again we are not
clear whether we've

made any progress.

We've

gone from e to the x cosine of
x to e to the x sine of x.

Let's take integration by parts
again, see what happens.

I'm just going to write on the
right side of the equal sign,

because this might
get a little long.

I'm just going to write this
first part x to the x cosine

of x plus-- and now let's do
integration by parts again.

For this round of integration
by parts this was g of x, but

now, for this around, I'm going
to assume it's g prime of x.

Which doesn't really make a
difference because whenever I

take the anti-derivative of it
to g of x, it stays the same.

And then I'm going to assume
that this is f of x.

So integration by parts tells
us we take f of x times g of x,

so I take this function and the
anti-derivative of

this function.

The anti-derivative of this
function is once again just e

to the x and then f times that
function unchanged

time sine of x.

From that I subtract the
integral of the anti-derivative

of this or I take g of x which
is e to the x, and then the

derivative of f of
x, f prime of x.

What's the derivative
of sine of x?

It's cosine of x.

Cosine of x d of x.

Let's see if we're
getting anywhere.

It seems like I just keep
adding terms, making it

more and more complicated.

In order to see if we're
getting anywhere, let me just

rewrite the whole thing and
maybe get rid of these

parenthesis, because it's
just a plus, so we can get

rid of the parenthesis.

Let me use a new color.

OK.

So this is the original
problem, e to the x cosine of x

d x equals, and now let me
switch back to this color, it

equals e to the x cosine of x,
and then I can just-- this

parentheses doesn't matter
because I'm just adding

everything in the parentheses--
e to x cosine of x plus e to

the x sine of x minus e to
the x cosine x access d x.

Now you might think that I
arbitrarily switched colors

here when I rewrote this,
but if you look you might

see why I actually did
switch colors here.

See anything interesting?

Exactly.

This is the same thing as
this, just a minus right?

So we're going to do something
what I think to be fairly cool.

Let's add this term to both
sides of the equation.

Let's take this and
let's put it on to this

side of the equation.

If I take this and put
it on this side of the

equation, what happens?

I then have two of these on the
left side equation, so that

becomes-- I mean I could write
it out it's e to the x cosine

of x d x plus, right?

Because I'm taking this and I'm
putting it on that side of the

equation, e to the
x cosine of x d x.

That's just the same thing as
2 times the integral of e

to the x cosine of x d x.

And then that equals this term.

Which equals e to the x cosine
of x plus e to the x sine of x.

I know it's really messy.

All I have to do now to solve
this integral is divide both

sides by 2 and I'm done.

So let me write it out, this
is very exciting, it's

the home stretch.

If I divide both sides by 2, I
get-- and I'm going to try to

write it so you can see
everything-- e to the x cosine

of x d x equals and on that
side I have e to the x cosine

of x plus e to the x
sine of x over 2.

I think that's pretty neat.

It's neat how integration by
parts allowed us to do this.

We actually never even have
to evaluate this integral.

We said, this integral is just
the original problem again.

And you can think about
why that happened, right?

Because these trick
functions cycle.

So we had to do integration
by parts twice to get back

to where we were before.

And then we use that to solve
it without actually having

to evaluate the integral.

And what I also think is cool
is even if you just look

at this solution, it's
kind of neat, right?

The anti-derivative of e to the
x and-- actually never forget

the plus c, that would've given
me minus 1 point on the exam.

What's kind of cool, the
integral of e to the x cosine

of x is this expression that's
e to the x cosine of x plus e

to the x sine of
x divided by 2.

It's the average of e to
the x cosine of x and

e to the x sine of x.

I think that's a pretty neat
property, and you might want

to graph them and play with
them, but it's kind of neat.

Hopefully I have convinced you
that is a classic of a problem,

and you also find it neat, and
I'll see you in the

next presentation.