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PROFESSOR: So
again, welcome back.
And today's topic
is a continuation
of what we did last time.
We still have a little
bit of work and thinking
to do concerning
polar coordinates.
So we're going to talk
about polar coordinates.
And my first job today is to
talk a little bit about area.
That's something we
didn't mention last time.
And since we're all
back from Thanksgiving,
we can certainly talk
about it in terms of a pie.
Which is the basic idea for
area in polar coordinates.
Here's our pie, and
here's a slice of the pie.
The slice has a piece of
arc length on it, which
I'm going to call delta theta.
And the area of that
shaded-in slice,
I'm going to call delta A.
And let's suppose
that the radius is a.
Little a.
So this is a pie of radius a.
That's our picture.
Now, it's pretty easy
to figure out what
the area that slice of pie is.
The total area is,
of course, pi a^2.
We know that.
And to get this fraction,
delta A, all we have to do
is take the percentage of the
arc of the total circumference.
That's (delta theta) / This is
the fraction of area-- sorry,
fraction of the
total circumference,
the total length around the rim.
And then we multiply
that by pi a^2.
And that's giving
us the total area.
And if you work that out,
that's delta A is equal
to, the pi's cancel and we
have 1/2 a^2 delta theta.
So here's the basic formula.
And now what we need to do is to
talk about a variable pie here.
That would be a pie with
a kind of a wavy crust.
Which is coming
around like this.
So r = r(theta).
The distance from
the center is varying
with the place where we are, the
angle where we're shooting out.
And now I want to subdivide
that into little chunks here.
Now, the idea for adding
up the area, the total area
of this piece
that's swept out, is
to break it up into
little slices whose areas
are almost easy to calculate.
Namely, what we're
going to do is to take,
and I'm going to
label it this way.
I'm going to take these little
circular arcs, which go--
So I'm going to extend
past where this goes.
And then I'm going to take
each circular arc here.
So here's a circular arc.
And then here's
another circular arc.
And here's another circular arc.
It's just right on
the nose in that case.
Now, in these two
cases, so basically
the picture that I'm trying
to draw for you is this.
I have some sector.
And then I have
some circular arc.
And maybe it takes
a little extra.
There's a little extra area,
I'm making an error in the area.
This is a little extra area.
And maybe to draw
it the other way.
I'm a little short on this one.
And let's say on this one
I'm right on the nose.
I have the same arc as
the curve of the surface.
Now this is a little bit
like the step functions
that we used in Riemann sums.
It's practically the same.
Eventually, this little band
of stuff that we're missing by,
if we take very, very
narrow little slices here,
is going to be negligible.
It'll get closer and
closer to the curve itself.
So that area will tend
to 0 in the limit.
So we don't have
to worry about it.
And the approximate
relationship is sitting here.
Where this distance now is r.
So this radius is r.
And this is this delta theta.
And so in the approximate case,
what we have is that delta A is
approximately 1/2
r^2 delta theta.
Which is practically the
same thing we had here.
Except that that r is
replacing the constant there.
And it's approximately
true, because r is varying.
And then in the limit,
we have the exact formula
for the differential.
Which is this one.
So this is the main
formula for area.
And if you like, the total
area then is going to be
the integral from some starting
place to some end place of 1/2
r^2 d theta.
Now, this is only useful in
the situation that we're in.
Namely-- So this is the
other important formula.
And this is only useful when
r is a function of theta.
When this is the way in which
the region is presented to us.
So that's the setup.
And that's our main formula.
Let's do what example.
The example that
I'm going to take
is the one that we did
at the end of last time,
or near the end of last time.
Which was this formula
here. r = 2a cos(theta).
Remember, that was the same
as (x-a)^2 + y^2 = a^2.
So this is what
we did last time.
We connected this
rectangular representation
to that polar representation.
And the picture is of a circle.
Where this is the point (2a, 0).
So let's figure out
what the area is.
Well, first of all,
we have to figure out
when we sweep out
the area, we have
to realize that we only
go from -pi/2 to pi/2.
So that's something we
can get from the picture.
You can also get it
directly from this formula
if you realize that cosine is
positive in this range here.
And at the ends, it's 0.
So the thing encloses
a region at these ends.
So at the ends, cosine of plus
or minus pi/2 is equal to 0.
That's what cinches this up
like a little sack, if you like.
So the area is now going to be
the integral from -pi/2 to pi/2
of 1/2 times the square of
r, that's (2a cos(theta))^2,
d theta.
Question.
STUDENT: [INAUDIBLE]
PROFESSOR: How do I know
from looking at the picture
that I'm going from -pi/2
to pi/2, is the question.
I do it with my whole body.
I say, here I am pointing down.
That's -pi/2.
I sweep up, that's 0.
And I get all the way
up to here. that's pi/2.
So that's the way I do it.
That's really the way I
do it, I'm being honest.
Now if you're a machine,
you can't actually look.
And you don't have a body,
so you can't point your arms.
Then you would have
to go by the formulas.
And you'd have to actually use
something like this formula
here.
The fact that this is
where the loop cinches up.
This is where the
radius comes into 0.
At pi/2.
So you need to know that in
order to understand the range.
Another question.
STUDENT: [INAUDIBLE]
PROFESSOR: So when
we're doing these,
should we just guess that
it's going to be a loop?
I'm probably going to give
you some clues as to what's
going on.
Because it's very hard to
figure these things out.
Sometimes it'll be bounded by
one curve and another curve,
and I'll say it's the thing
in between those two curves.
That's the kind of
thing that I could do.
Here, you really should
know this one in advance.
This is by far the
most-- or this is one
of the typical cases, anyway.
I'm going to give you
a couple more examples.
Don't get too
worked up over this.
You will somehow be
able to visualize it.
I'll give you some examples
to help you out with it later.
So here's the situation.
Here's my integral.
And now we're faced
with a trig integral.
Which we have to
remember how to do.
Now, the trig integral
here-- so first
let me factor out the constants.
This is 4a 4a^2 / 2, so it's
2a^2 integral from -pi/2
to pi/2 of cos^2(theta) d theta.
And now you have to
remember what you're
supposed to do at this point.
So think, if you
haven't done it yet,
this is practice you need to do.
This trig integral is handled
by a double angle formula.
As it happens, I'm
going to be giving you
these formulas on
the review sheet.
You'll see they're written
on the review sheet.
At least in some form.
So for example,
there's a formula,
and this will be
on the exam, too.
So this is the correct
formula to use here.
Is that this is + (1 +
cos(2theta)) / 2 d theta.
So that's the substitution that
you use for the cosine squared
in order to integrate it.
That serves as a little
review of trig integrals.
And now, this is quite easy.
This integral now is easy.
Why is it easy?
Well, because it's
the antiderivative
of a constant, cos(2theta),
its antiderivative
you're supposed to be
able to write down.
So the antiderivative
of 1 is theta.
And the antiderivative
of the cosine
is 1/2 the sine
when it's 2 theta.
And that is a ^2 a^2
(pi/2 - (-pi/2)).
And the signs go away
because they're both 0.
So all told we get pi a^2, which
is certainly what we would like
it to be.
It's the area of the circle.
Another question?
STUDENT: [INAUDIBLE]
PROFESSOR: The
question, so I'm not
sure which question
you're asking.
I pivoted my arm around (0, 0).
This point, this is the point
we're talking about, (0, 0),
is a key point.
It's where I guess you could
say I stuck my elbow there.
Now, the reason is that
it's the place where r = 0.
So it's more or less the center
of the universe from the point
of view of this problem.
So it's the reference
point and if you like,
when you're doing this, it's a
little bit like a radar screen.
Everything is
centered at the origin
and you're taking rays
coming out from it.
And seeing where
they're going to go.
So for example, this
is the theta = 0 ray,
this is the theta = pi/4 ray.
This the theta = pi/2 ray.
And indeed, if my elbow is
right at this center here,
I'm pointing in those
various directions.
So that's what I had in
mind when I did that.
You can always get these
formulas, by the way,
from the original business, x =
r cos(theta), y = r sin(theta).
But it's useful to have the
geometric picture as well.
In other words, if
you were a machine
you'd have to rely
on these formulas.
And plot things using these.
Always.
Now, in terms of plotting
I want to expand your brain
a little bit.
So we need just a little bit
more practice with plotting.
In polar coordinates.
And so, the first question
that I want to ask you
is, what happens outside
of this range of theta?
In other words, what happens
if theta's beyond pi/2?
Can somebody see
what's happening
to the formulas in that case?
So what I'm looking at
now, let's go back to it.
What I'm looking at
is this formula here.
But to use the
elbow analogy here,
I'm swept around like this.
But now I'm going
to point this way.
I'm going to point
out over there.
My hand is up here in
the northwest direction.
So what's going to happen?
Somebody want to tell me?
STUDENT: [INAUDIBLE]
PROFESSOR: It goes
around itself.
That's right.
What happens is that when r
crosses this vertical, r = 0,
when it crosses over
here it goes negative.
So although my theta is
pointing me this way,
the thing is going
to go backwards.
And there's another clue.
Which is very important.
How far backwards is it going?
Well, you don't actually need to
know anything but this equation
here, to understand that it
has to be on the same circle.
So when I'm pointing
this way, the things
points backwards to
this point over there.
So what happens is,
it goes around once.
And then when I
point out this way,
it sweeps around a second time.
It just keeps on going
around the same circle.
So over here it's empty.
Because it's pointing
the other way and it's
sweeping around the same curve.
A second time.
Now, if you were foolish
enough to integrate, say,
from 0 to 2pi or some wider
range, what would happen
is you would just
double the area.
Because you would have
swept it out twice.
So that's the mistake
that you'll make.
Sometimes you'll count things
as negative and positive.
But because there's
a square here,
it's always a positive quantity.
And you'll always over-count
if you go too far.
So that's what happens.
Again, it sweeps
out the same region.
That's because these
two equations really
are equivalent to each other.
It's just that this one
sweeps it out twice.
And this one doesn't say
how it's sweeping it out.
Yeah, another question.
STUDENT: Doesn't this
equation also work
if you just go from 0 to pi?
PROFESSOR: Does the
integration work
if you just go from 0 to pi?
The answer is yes.
That's a very weird
object, though.
Let me just show
you what that is.
If you started from 0 to 2pi.
So I'll illustrate it on here.
The first thing that you
swept out between 0 and pi/2
is this part here.
That was swept out.
And then, when you're going
around this next quadrant here,
you're actually sweeping
out this underside here.
So actually, you're
getting it because you're
getting half of it on one half,
and getting the other half
on the other quadrant.
So it's actually giving
you the right answer.
That turns out to be OK.
It's a little weird way
to chop up a circle.
But it's legal.
But of course,
that's an accident
of this particular figure.
You can't count
on that happening.
It's much better to
line it up exactly
with what the figure does.
So don't do that too often.
You might run into troubles.
So I'm going to give you
a couple more examples
of practice with these pictures.
And maybe I'm going to get
rid of this one up here.
So here's another favorite.
Here's another favorite.
So this, if you
like, is Example 2.
I guess we had an
Example 1 up there.
And now we're really
not going to try
to do any more area examples.
The area examples are
actually straightforward.
It's really just figuring out
what the picture looks like.
So this is examples of drawings.
So this one is one
that's kind of fun to do.
This is r = sin(2theta).
Something like this
is on your homework.
And so what happens
here is the following.
What happens here is
that at theta = 0,
that's the first place.
So let's just plot
a few places here.
I'm not going to plot very many.
Theta = 0, I get r is 1.
Whoops, I get r is 0.
Sorry.
And then pi/4, that's where I
get sin(pi/2), I get 1 here.
For this.
And then again, at
pi/2 I get sin(pi),
which is back at 0 again.
So it's-- And the other thing
to say is in between here
it's positive.
In between.
So what it does is,
it starts out at 0
and it goes out to the
radius 1 over here.
And then it comes back.
So it does something like this.
It goes out, and it comes back.
Now because of the symmetries
of the sine function,
this is pretty much
all you need to know.
It does something similar
in all of the quadrants.
But in order to see what it's
doing, it's useful for you
to watch me draw it.
Because the order is very
important for understanding
what it's doing.
It's similar to this weird
business with the circle here.
So watch me draw this guy.
I'll draw it in red because
it usually has a name.
So here it is.
It does this thing.
And then it does this.
And then it does this.
And then it does that.
So it's called a four-leaf rose.
I drew it in pink because
it's kind of a rose here.
So it started out over here.
This is Step 1.
And this is the range
0 < theta < pi/4.
It did this part here.
And then it went 2 here.
So I should draw these
in white, because they're
harder to read in red.
But now look at what it did.
It did not make a
right angle turn.
It was nice and smooth.
It went around here and
then it went down here.
This is 3.
Back here, that's 4.
And then over here, that's 5.
Back up here, that's 6.
And then around here, that's 7.
And down here, that's 8.
And then back where it
started and goes around again.
And this is because
actually it's switching sign
when it crosses the origin.
When it was over in this
quadrant the first time,
it actually was tracing
what's directly behind it.
So this is kind of amusing.
From this little tiny formula
you get this pretty diagram
here.
Anyway that's, as I
say, an old favorite.
And here if you want to
do the area of one leaf,
you've got to make sure
you understand that it's
a small piece of the whole.
OK, now I have one
last drawing example
that I want to discuss with you.
And it involves another
skill that we haven't quite
gotten enough practice with.
So I'm going to do that one.
And it's also preparation
for an exercise.
But one that we're going
to do after the test.
So here's my last example.
We're going to discuss what
happens with this function
here.
Sorry, that's not
legible, is it.
That's a cosine. r = r
= 1/(1 + 2cos(theta)).
Now, the first
thing I want to do
is just take our time a little
bit and plot a few points.
So here's the values of theta
and here are the values of r,
and we'll see what happens.
And we'll try to figure
out what it's doing.
When theta = 0, cosine is 1.
So r = 1/3.
The denominator is
1 + 2, so it's 1/3.
If theta-- I'm going
to make it easy,
we're not going to do so many.
I'm going to do pi/2, that's
an easy value of the cosine.
That's cos(pi/2) = 0.
So that value of r = 1.
And now I'm going to
back up and do -pi/2.
-pi/2, again, cosine is 0.
And r = 1.
So now I'd like to just plot
those points anyway, and see
what's going on with
this expression here.
The first one is a
rectangular-- I'm
going to write the
rectangular coordinates here,
not the polar coordinates.
The rectangular coordinates here
are 1/3 out at the horizontal,
so it's (1/3, 0).
The polar coordinates
is (1/3, 0),
but the rectangular
coordinate is also that.
And over here, at pi/2,
the distance is 1.
So this is the point (0,
1) in x-y coordinates.
And then down here at,
-pi/2, it's (0, -1).
Let me just emphasize.
You should be able to
think of this visually
if you can crank your
arm around and think it.
Or if you're right-handed
you'll bend that way, no.
Anyway.
Or you'll have to
use-- But this also
works using this formulas x =
r cos(theta), y = r sin(theta).
Notice that in this case, r
was 1 but the cosine was 0.
So you plug in theta = -pi/2.
And r = 1.
And lo and behold,
you get 0 here.
And here you get
-1 here you get 1.
So this is -1.
So this is an example.
I did it purely visually
or sort of organically.
But you can also do it by
plugging in the numbers.
Now in between, the
denominator is positive.
And it's something in between.
It's going to sweep around
something like this.
That's what happens in between.
As theta increases
from -pi/2 to pi/2.
And now something
interesting happens
with this particular
function, which
is that we notice that
the denominator is
0 at a certain place.
Namely, if I solve
2 cos(theta) = -1,
then the denominator
is going to be 0 there.
That's cos(theta) =
-1/2, so theta is equal
to, it turns out,
plus or minus 2pi/3.
Those are the values here.
So when we're out here
somewhere, in these directions,
there's nothing.
It's going infinitely far out.
Those ways.
OK that's about as
much as we'll be
able to figure out
of this diagram
without doing some
analytic work.
And that's the other little
piece that I want to explain.
Namely, going backwards
from polar coordinates
to rectangular coordinates.
Which is one thing
that we haven't done.
So let's do that.
So what is the
rectangular equation?
That means the (x, y)
equation for this r = 1 /
(1 + 2cos(theta)).
And let's see what it is.
Well, first I'm going to
clear the denominator here.
This is r + 2r cos(theta) = 1.
And now I'm going to rewrite
it as r = 1 - 2r cos(theta).
And the reason for that
is that in a minute
I'll explain to you why.
This is 1 - 2x.
And this guy, I'm
going to square now.
I'm going to make
this r^2 = (1 - 2x)^2.
And now, with an r^2, I
can plug in x^2 + y^2.
So this is a
standard thing to do.
And it's basically
what you're going
to do any time you're faced
with an equation like this.
Is try to work it out.
And, in these situations
where you have 1 /
(a + b cos(theta)),
or sin(theta),
you'll always come out with some
quadratic expression like this.
Now, I'm going to combine terms.
So here I have -3x^2 + y^2, and
put everything on the the left
side.
So that's this.
And we recognize, well
you're supposed to recognize,
that this is what's
known as a hyperbola.
If the signs are the
same, it's an ellipse.
If the the signs are
opposite it's a hyperbola.
And in between, if one of the
coefficients on the quadratic
is 0, it's a parabola.
So now we see that the picture
that we drew there is actually,
turns out it's going
to have asymptotes,
it's going to be a hyperbola.
So now, let me ask you the last
little mind-bending question
that I want to ask.
Which is, what happens-- So
now I'm using my right arm,
I guess.
But my elbow's at
the origin here.
What happens if I pass
outside, to the range where
this denominator is negative.
It crossed 0 and it
went to negative.
It's sweeping out
something over here.
Is it sweeping out
the same curve?
Anybody have any
idea what it's doing?
Yeah.
STUDENT: [INAUDIBLE]
PROFESSOR: Yeah, exactly.
Good answer.
It's the other branch
of the hyperbola.
So what's actually
happening is in disguise,
there's another branch
of the hyperbola
which is being swept up by
the other piece of this thing.
Now, that is consistent with
these algebraic equations.
The algebraic equation
that I got here
doesn't say which branch
of the hyperbola I've got.
It's actually got two branches.
And the curve really
was, in disguise,
capturing both of them.
I want to make
the connection now
with the basic
formula for area here.
Because this is a really
beautiful connection.
And I want to make that
connection in connection
also with this example.
The hyperbolas, as
you probably know,
are the trajectories of comets.
And ellipses, which is what you
would get if maybe you put 1/2
here instead of a 2,
would be the trajectories
of planets or asteroids.
But there's actually
something much more important,
physically that goes on.
That's special about this
particular representation
of the hyperbola.
And what happens when you
get the ellipses as well.
Which is that in
this case, r = 0
is the focus of the hyperbola.
And what that means
is that it's actually
the place where the sun is.
So this is the right
representation,
if you want the
center of gravity
in the center of your picture.
And pretty much any other.
I mean, you can't
tell that at all
from the algebraic
equations here.
So this hyperbola is
going to be the trajectory
of some comet going by here.
And this formula
here is actually
a rather central
formula in astronomy.
Namely, there's something
called Kepler's Law.
Which says that the rate
of change of area which
is swept out is constant.
The rate of change of area
relative to the center of mass,
relative to the sun.
So in equal areas,
this is amount of area.
So this tells you now that when
a comet goes around the sun
like this, its speed varies.
And it's speed varies according
to a very specific rule.
Namely, this one here.
And this rule was
observed by Kepler.
But if you have this
connection here,
we also have something else.
We also know that dA/dt
= 1/2 r^2 d theta / dt.
So that's this formula here,
formally dividing by t.
That's the rate of
change with respect
to this time parameter, which
is the honest to goodness time.
Real physical time.
And that means, this
quantity here is constant.
And this is one of
the key insights
that physicists had,
long after Kepler
made his physical
observations, they realized
that he had managed to get the
best physics experiment of all,
because it's a
frictionless setup.
Outer space, there's no air.
Nothing is going on.
This is what's known
nowadays as conservation
of angular momentum.
This is the expression
for angular momentum.
And what Kepler was
observing, it turns out,
is what we see all
the time in real life.
Which is when you start
something spinning around it
continues to spin at
roughly the same rate.
Or, if you're an
ice skater and you
get yourself scrunched
together a little bit more,
you can spin faster.
And there's an exact
quantitative rule
that does that.
And it's exactly this
polar formula here.
So that's a neat thing.
And we will do a little
exercise on this rate of change
after the exam.
So that's it for generalities
and a little pep talk
on what's coming
up to you when you
learn a little more physics.
Right now we need to
talk about the exam.
So first of all, let me tell
you what the topics are.
They're the same as
last year's test.
Which you can take a look at.
And let's see.
So what did we do?
One of the main
topics of this unit
were techniques of integration.
And there are three,
which we will test.
One is trig substitution.
One is integration by parts.
And one is partial fractions.
So that's more than half
of the exam, right there.
The other half of the
exam is parametric curves.
Arc length.
These are all interrelated.
And area of surfaces
of revolution.
Those are the only kind
that we can handle.
Just as we did with volume
of surfaces of revolution.
And then there's a final topic,
which is polar coordinates.
And area in polar
coordinates, including area.
That's it.
That's what's on the test,
there are six problems.
They're very similar.
Well, they're not
actually that similar.
But they're somewhat
similar to last year's.
I'd say the test is similar.
Maybe a tiny bit more
difficult. We'll see.
We'll see.
Yeah.
STUDENT: [INAUDIBLE]
PROFESSOR: The question
was, we didn't do arc length
in polar coordinates, did we?
And the answer is
no, we did not.
We did not do arc length
in polar coordinates.
When I give you an exercise,
I'm going to ask you about,
if you know the speed
of a comet here, what's
the speed of the comet there.
And we'll have to know
about arc length for that.
But we're not doing
it on this exam.
Other questions.
STUDENT: [INAUDIBLE]
PROFESSOR: The question is,
will I expect you to know r
equals--- so let's see if I
can formulate this question.
It's related to this
four-leaf rose here.
So the question is, suppose
I gave you something
that looked like this.
Would I expect you to be
able to know what it is.
I think the answer, the
fair answer to give you,
is if it's this complicated,
I only have two possibilities.
I can give you a long
time to sketch this out.
And think about what it does.
Or I can tell you
that it happens
to be a three-leaf rose.
And then you have some
clue as to what it's doing.
It doesn't have six.
Because of some weird thing,
having to do with repetitions.
But the odds and evens
work differently.
So, in fact I would
have to tell you
what the picture
looks like, if it's
going to be this complicated.
Similarly, so this is an
important point to make,
when we come to techniques
of integration, any integral
that you have, I'm not going to
tell you which of these three
techniques to use
on the ones which
are straightforward integrals.
But if it's an
integral that I think
you're going to get
stuck on, either I'm
going to give you a hint,
tell you how to do it.
Or I'm going to tell
you, don't do it.
If I tell you don't do
it, don't try to do it.
It may be impossible.
And even if it's possible,
it's going to be very long.
Like an hour.
So don't do it
unless I tell you to.
On the other hand,
all of these setups
in this second
half of this unit,
they involve somehow
setting something up.
And they're basically
three issues.
One is what the integrand is.
One is what the lower limit
is, what is the upper limit.
They're just three
things, three inputs,
to setting up an integral.
All integrals, this is going to
be the setup for all of them.
And then the second
step is evaluating.
Which really is what we
did in the first half here.
And, unfortunately, we don't
have infinitely many techniques
and indeed there's
some integrals
that can't be evaluated
and some that are too long.
So we'll just try
to avoid those.
I'm not trying to give you
ones which are hopelessly long.
Alright, other questions.
Yes.
STUDENT: [INAUDIBLE]
PROFESSOR: The question is, will
the percentages be the same.
And the answer is, no.
I'll tell you exactly.
This is 55 points.
Unless I change
the point values.
This is 55, and this is 45.
That's what it came out to be.
You are going to want to
know about all of the things
that I've written down here.
You're definitely
going to want to know,
for example, surfaces
of revolution.
How to set those up.
Yes. there was another
question I saw.
Yes.
STUDENT: [INAUDIBLE]
PROFESSOR: So if you have a
partial fraction with something
like (x+2)^2 and maybe
an x and maybe an x+1,
and you're interested in what
happens with this denominator
here?
So what's going to
happen is, you're
going to need a coefficient
for each degree of this.
So altogether, the setup
is going to be this.
Plus one for x.
And one for x+1.
This is the setup.
So you need--
STUDENT: [INAUDIBLE]
PROFESSOR: So if I change this
to being a 3 here, then I need,
I guess I'll have to
call it E, (x+2)^3.
I need that.
Now, it gets harder and harder.
The more repeated
roots there are,
the more repeated factors
there are, the harder it is.
Because the ones you can pick
off by the cover-up method are,
is just the top one here.
And these two.
So C, D, and E you can get.
But B and A you're
going to have to do
by either plugging in or some
other, more elaborate, algebra.
So the more of these lower
terms there are, the worse off
you are.
STUDENT: [INAUDIBLE]
PROFESSOR: The question is,
does this x^3 + 21 affect this
setup.
And the answer is almost no.
That is, not at all.
It's the same setup exactly.
But, there's one thing.
If the degree gets
too big, then you've
got to use long division
first to knock it down.
I'll give you an example
of this type of practice.
Unless there are more question.
Yes.
STUDENT: [INAUDIBLE]
PROFESSOR: Are you
going to have to know
how to do reduction formulas?
Anything that's a little out of
the ordinary like a reduction
formula, I will have
to coach you to do.
So, in other words,
what you'll have
to be able to do in that
situation is follow directions.
If I tell you OK,
you're faced with this,
then do an integration by parts.
And do that, then get
the reduction formula.
STUDENT: [INAUDIBLE]
PROFESSOR: Yeah.
OK, so the question had do with
the partial fractions method.
And what happens if
you have a quadratic.
So, for instance,
if it were this,
this one's too disgusting.
I'm going to just do
it with two of them.
So the parts with x
and x+1 are the same.
But now you have
linear factors here.
(Ax + B) / (x^2 + 2).
And A-- maybe I'll call them
1, and A_2 x + (A_2 x + B_2) /
(x^2 + 2)^2 + C / x + D / (x+1).
This is the way it works.
OK, I'm going to give you
one more quick example
of an integration technique
just to liven things up.
Let's see.
So here's a somewhat
tricky example.
This is just a little
trickier than I
would give you on a test.
But it's the same principle, and
I may do this on a final exam.
So suppose you're faced
with this integral.
What are you going to do?
Integration by parts, great.
That's right, that's
because this guy is
begging to be differentiated,
to be made simpler.
So that means that I
want this one to be u,
and I want this one to be v'.
And I want to use
integration by parts.
And then u ' = 1 /
(1+x^2), and v = x^2 / 2.
So the answer is now, x^2
/ x^2/2 tan^(-1) x minus
the integral of this guy.
Which is going to be x^2 / 2.
And then I have
1 / (1 + x^2) dx.
Now, you are not
done at this point.
You're still in
slightly hot water.
You're in tepid water, anyway.
So what is it that
you have to do here?
You're faced with
this integral, which
I'll put on the next board.
It's a lot simpler
than the other one,
but as I say you're not
quite out of the woods.
You're faced with
the integral of 1/2--
-1/2 x^2 / (1 + x^2) dx. dx.
STUDENT: [INAUDIBLE] PROFESSOR:
Trig substitution actually,
interestingly, will work.
But that wasn't what
I wanted you to do.
I wanted you to, yeah, go ahead.
STUDENT: [INAUDIBLE]
PROFESSOR: Add and subtract
1 to the numerator.
So now, that's the
correct answer.
This is the case where the
numerator and the denominator
are tied.
And so you have to
use long division.
But a shortcut is
just to observe
that the result of
long division is
the same thing as doing this.
And then noticing that
this is 1 - 1 / (1 + x^2).
So this is the same as long
division, in this case.
Because when you divide in, it
goes in with a quotient of 1.
And so this guy turns out to
be -1/2 the integral of 1 -
1/(1+x^2) dx.
Which is 1/2 x -
1/2 tan^(-1) x + c.
So this is one extra step that
you may be faced with someday
in your life.
And just keep that in mind.
