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Welcome back to recitation.

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In this video, I'd like us to
work on an example of something

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you actually saw in one
of the lecture videos,

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and that's about
reduction formulas.

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So what I'd like us to
do is to find a reduction

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formula for the integral of
sine to the n x dx, which

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I'm denoting by F sub n of x.

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That's actually, F sub n
of x denotes the integral

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of sine x raised to the n dx.

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So the object here,
just to remind you

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of what we'd like
to do, is to be

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able to write F sub
n of x in a form

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where maybe it involves
some functions,

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but any integral
that it involves

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is an integral of sine
to a lower power than n.

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So an integral of sine x
to a lower power than n.

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So that's the object.

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The object is to write
this down in such a way

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that this integral is equal
to some functions maybe,

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product of some functions,
added to another function that's

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like F, but with
a lower subscript.

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So I'm going to give you a
little time to work on it.

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Why don't you pause the video,
take some time to work on it,

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and when I come back,
I'll show you how I do it.

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OK.

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Welcome back.

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Well, hopefully you were
able to make some headway

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on this problem.

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So let's take a look again
at what we want to do,

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what the goal is, and then
we'll figure out a good strategy

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to do that.

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So the goal is to be able to
write F sub n of x as something

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that involves a lower
numbered subscript than n.

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Right?

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So we'd like to be able to
take this integral, that's

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a power of sine x, and
reduce the number of powers.

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You know, it would be great
if we could just write down

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a final formula, but
if we're able to reduce

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the power of sine x, then we're
able to head in a direction

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where we could
start to write down,

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start to write this
down as functions

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that don't involve integrals.

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So the first step in
these kinds of things

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is finding a reduction formula.

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So a couple of things
I want to remind us.

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We're going to make
the common substitution

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that we make for powers of
sine or powers of cosine

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that we see a lot.

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Well, I guess,
actually there's two.

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When you just see
a sine squared,

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sometimes you do this half
angle formula, or double angle

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formula, that Joel has
mentioned in some videos.

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But in this one, we're more
interested in manipulating

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the identity sine squared x plus
cosine squared x equals one.

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In particular, we're going to
replace one of the sine squared

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x's, or we're going to
replace two of the sines

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by 1 minus cosine squared x.

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So that'll be our first step.

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So we're going to do
that substitution.

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We'll break it into some
pieces, we'll see where we get.

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OK?

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So let me come over here,
and I will start the problem.

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So I'm starting
with f sub n of x,

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and I'm going to rewrite
it as the integral

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of sine to the n minus
2 x quantity 1 minus

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cosine squared x dx.

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So let me take a step back.

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So what did we do here?

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We took two powers of
sine, and we replaced them

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by what I said earlier.

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We replaced them by 1
minus cosine squared x.

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Right?

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So this is a sine squared x.

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Here I have sine x
to the n minus 2.

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When multiply those together,
I get sine x to the n.

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So everything's still
working out so far.

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Why is this good?

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Well, notice the terms we have.

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Sine to the n minus 2
x dx minus the integral

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of sine to the n minus
2 x cosine squared x dx.

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OK.

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This we like.

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Why do we like this?

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Because it's actually is
just like the F sub n of x,

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except it has a lower power.

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So this piece is actually
F sub n minus 2 of x.

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So that's something we like.

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Because it's already reduced
the power of sine x by 2.

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So that's good.

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That's part of the
reduction idea.

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This one, obviously we can't
write it in terms of capital F

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with some subscript.

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And it maybe doesn't look so
easy to integrate right away.

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But let's think about
it for a second, how

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we could split this thing up.

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And the goal was to use
an integration by parts.

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So we're going to examine
this integral in particular.

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So let me just write
that one down again here.

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So we need to analyze--
I'm going to drop the minus

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sign for the moment, n minus 2.

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So our goal, again, is to
try and write this somehow.

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Either manipulate it so it
looks like a capital F function,

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or manipulate it so that
the integral sign is gone.

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OK?

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That's our main goal here.

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So let's look at what
we can do with this.

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Well, I can split this up.

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What I'm going to do,
it's kind of tricky,

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but what I'm going to do, is I'm
going to take all of the sine

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x's, and I'm going to take
one of the cosine x's.

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And I'm going to
make that one thing,

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and I'm going to make the
other cosine x another thing.

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And I'm going to
write them down first,

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and we'll figure out
which is a good u

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and which is a good v prime.

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So that's going to be one thing.

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And then the other
cosine x is going

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to be-- does that look
like an equals sign?

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Equals is going to
be the other thing.

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So again, sine to
the n minus 2 x times

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cosine x times cosine
x gives me what's

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up here, inside the integral.

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And now what I
really want to see,

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I'm going to use an
integration by parts.

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Well, this is easy to integrate
or take a derivative of.

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But the point is that
now I can integrate this.

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And why can I integrate this?

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Well, I can do a
substitution on this.

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So this is kind of complicated.

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Because there's first,
you have to notice,

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an integration by parts is
a good way to go with this.

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And then you have to
see, oh my goodness,

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one of the steps of integration
by parts requires substitution.

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So it's a little tricky.

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But I'm going to make
this, then, my v prime,

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and I'm going to make this my u.

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Now, I'm going to come over to
the other board to figure out,

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just to make sure I
do the v prime right.

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OK?

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And then I'm going to
finish off this problem.

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So we'll come back
here in a second.

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So right now, this
is a question mark.

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We're going to make sure we
do the v prime part right,

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and then we're
going to write down,

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this should be u*v minus
the integral of v du.

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OK?

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So let me give
myself some space.

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So I'm trying to integrate now
sine the n minus 2 x cosine x

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dx.

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All right?

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That's my integral
of v prime dx.

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OK, so this will give me v.

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Well, let's see what we do.

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We're going to
now make-- I don't

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want to use a u for
my substitution,

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because I already
have in my mind a u.

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So I'm going to let w
equal sine x, and then dw

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is equal to cosine x dx.

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Right?

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What's the point here?

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Why do we see this
as easy to integrate?

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Because this is powers of sine,
and the derivative of sine

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is cosine.

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So that's why this lends itself
to a substitution so easily.

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So that should be
fairly familiar by now,

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but just to make
sure we're clear.

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So then this is now the integral
of w to the n minus 2 dw.

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All right?

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I have sine x to the n minus 2.

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So that's replaced by
w to the n minus 2.

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And then dw replaces
cosine x dx.

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And that is pretty easy
to integrate, I think.

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That's w to the n
minus 1 over n minus 1.

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I'm not going to worry
about my c's right now.

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That'll come up
right at the end.

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We'll put in a plus
c if we needed to.

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But, so don't worry
about your plus c.

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I don't want to
carry that around.

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OK.

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So then when I when I look at
this, what does this tell me?

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This tells me that if this
is the integral of v prime,

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this tells me that v is equal
to sine x to the n minus 1,

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sine to the n minus
1 x, over n minus 1.

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OK?

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This whole bit was to help us
in the middle of the problem.

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So this was the v we needed.

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All right?

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So let's come back, fill in
a little piece we needed,

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that little gap we had,
and then we'll finish

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the problem on the far right.

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So let's come back.

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Where was I?

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OK.

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So I was integrating
this quantity.

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Sine to the n minus 2
x cosine squared x dx.

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And I was trying to use
integration by parts.

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So I had u and v prime,
and I've now calculated v,

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and I know what du is.

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So I'm going to write
down what this equals.

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This is going to be
u, which is cosine x.

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And I'll just write down v here,
so we'll see what it is again.

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Times sine to the n
minus 1 x over n minus 1.

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And then I have
to subtract v du.

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So this is v. And what's du?

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du is, or I should say, u prime
is how you saw it in class.

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u prime is negative sine x.

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So what am I going to get?

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I'm going to get negative
sine x times the v that I had.

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So when I subtract, I'm going
to end up with a plus sign.

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And I'm going to
have plus, I'm going

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to pull out the constant,
1 over n minus 1,

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integral, now let's look at
the magic that happens here.

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OK.

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Let's make sure you
agree with this power.

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It's the nth power of sine x.

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So let's double check.

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I said u was cosine x, so
u prime is negative sine x.

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00:10:06,550 --> 00:10:11,380
So I have to take u prime v.
u prime is negative sine x.

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If we come over here, we see
v is sine x to the n minus 1,

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so u prime v is indeed
negative-- well, I should say,

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v is the sine x to the n
minus 1 over n minus 1.

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So u prime v, in fact,
has the power n on sine.

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So that's maybe a little bit
complicated and nerve-racking,

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but that's OK.

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We're actually headed
in the right direction.

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So I'm going to
rewrite the stuff

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we know, and we'll see how we're
headed in the right direction.

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OK.

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So again, we were trying
to find F sub n of x.

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And I'm going to write out
the pieces we already knew.

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So we had an F sub
n minus 2 of x.

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That was the first term we had.

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And then we analyzed
this second piece.

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And the second piece
is the one that we just

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spent a little while on.

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And that gave us a cosine
x sine to the n minus 1

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of x over n minus 1, and
then a plus 1 over n minus 1.

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And what was our thing here?

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Our thing here was
F sub n of x, again.

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OK.

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So this might be a
little complicated.

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So I'm going to show us where
all the pieces come from,

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just remind us.

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So we started with F
sub n of x, and I'm

245
00:11:31,290 --> 00:11:33,790
claiming there are two
pieces that are of interest.

246
00:11:33,790 --> 00:11:34,420
OK?

247
00:11:34,420 --> 00:11:38,030
The first piece--
let's come back over.

248
00:11:38,030 --> 00:11:40,705
The first piece, this
was our F sub n of x.

249
00:11:40,705 --> 00:11:41,580
We wrote it this way.

250
00:11:41,580 --> 00:11:44,890
The first piece was the
F sub n minus 2 of x.

251
00:11:44,890 --> 00:11:46,280
Sine to the n minus 2 of x.

252
00:11:46,280 --> 00:11:47,490
No problem.

253
00:11:47,490 --> 00:11:51,017
Then we subtracted
this integral.

254
00:11:51,017 --> 00:11:53,100
And so I spent a little
while trying to figure out

255
00:11:53,100 --> 00:11:54,490
how to write that integral.

256
00:11:54,490 --> 00:11:57,650
And that integral was
written right down here.

257
00:11:57,650 --> 00:12:00,430
So I have to subtract
off all of that.

258
00:12:00,430 --> 00:12:02,850
And the point is that this
integral here, integral

259
00:12:02,850 --> 00:12:05,570
of sine x to the n, the
quantity of sine x to the n,

260
00:12:05,570 --> 00:12:07,580
is F sub n of x, again.

261
00:12:07,580 --> 00:12:09,580
Now you might be nervous,
because again, we just

262
00:12:09,580 --> 00:12:11,760
said, we see the
same thing over-- we

263
00:12:11,760 --> 00:12:14,144
see what we want to reduce
on the right-hand side.

264
00:12:14,144 --> 00:12:16,060
But now all we have to
do is a little algebra.

265
00:12:16,060 --> 00:12:20,200
So I'm going to come over and
show you the magic of algebra.

266
00:12:20,200 --> 00:12:22,540
OK?

267
00:12:22,540 --> 00:12:25,630
Let's, to make this
as easy as possible,

268
00:12:25,630 --> 00:12:29,900
I'm going to erase the
parentheses here, and put

269
00:12:29,900 --> 00:12:33,680
that minus sign here, so I don't
have to rewrite the whole line.

270
00:12:33,680 --> 00:12:35,350
OK?

271
00:12:35,350 --> 00:12:39,410
So the parentheses are gone, and
I've distributed the negative 1

272
00:12:39,410 --> 00:12:40,680
to both terms.

273
00:12:40,680 --> 00:12:43,120
Now notice, I want to
figure out what this is.

274
00:12:46,630 --> 00:12:49,120
I'm going to figure out
what that blue thing equals.

275
00:12:49,120 --> 00:12:51,470
Well, we're back to sort
of an algebra problem.

276
00:12:51,470 --> 00:12:55,500
We can isolate the
blue things, and solve

277
00:12:55,500 --> 00:12:57,200
for what's in the blue box.

278
00:12:57,200 --> 00:12:59,139
So that's what
we're going to do.

279
00:12:59,139 --> 00:13:00,180
So what do we have to do?

280
00:13:00,180 --> 00:13:04,059
Well, we have to add this
to the left-hand side.

281
00:13:04,059 --> 00:13:04,850
And what do we get?

282
00:13:04,850 --> 00:13:05,766
Let's see what we get.

283
00:13:05,766 --> 00:13:12,120
We get 1 plus 1 over n minus 1.

284
00:13:12,120 --> 00:13:22,690
F sub n of x is equal to F sub
n minus 2 of x minus cosine

285
00:13:22,690 --> 00:13:29,170
x sine x quantity to the
n minus 1 over n minus 1.

286
00:13:29,170 --> 00:13:33,900
So all I did was I added
1 over n minus 1 times

287
00:13:33,900 --> 00:13:37,580
F sub n of x to the left-hand
side-- well, to both sides,

288
00:13:37,580 --> 00:13:39,310
so it moved to the
left-hand side--

289
00:13:39,310 --> 00:13:42,360
and there was one of them here,
and there's 1 over n minus 1

290
00:13:42,360 --> 00:13:44,230
of them here.

291
00:13:44,230 --> 00:13:46,000
And let's make sure
we can simplify that.

292
00:13:46,000 --> 00:13:46,500
Let's see.

293
00:13:46,500 --> 00:13:48,970
We get n minus 1 over
n minus 1 plus 1.

294
00:13:48,970 --> 00:13:51,780
I think it gives me
n over n minus 1.

295
00:13:51,780 --> 00:13:55,660
So that gives me
n over n minus 1.

296
00:13:55,660 --> 00:13:59,060
I guess I'll write it again.

297
00:13:59,060 --> 00:13:59,995
We're really close.

298
00:14:05,140 --> 00:14:06,579
And then this whole mess again.

299
00:14:11,270 --> 00:14:13,220
And now, we only
have one more step

300
00:14:13,220 --> 00:14:15,850
to figure out what
F sub n of x is.

301
00:14:15,850 --> 00:14:22,290
And that is to multiply both
sides by n minus 1 over n.

302
00:14:22,290 --> 00:14:25,630
n minus 1 over n
multiplied here gives me 1,

303
00:14:25,630 --> 00:14:33,460
and so then I get n minus 1
over n, F sub n minus 2 of x,

304
00:14:33,460 --> 00:14:35,680
minus-- the n
minus 1's kill off,

305
00:14:35,680 --> 00:14:46,790
and I just get cosine x sine
to the n minus 1 x over n.

306
00:14:46,790 --> 00:14:49,497
So this is our answer.

307
00:14:49,497 --> 00:14:50,580
Because what is the point?

308
00:14:50,580 --> 00:14:52,620
Why can I say, well,
I've reduced it?

309
00:14:52,620 --> 00:14:55,710
I've reduced it because
I have-- the only

310
00:14:55,710 --> 00:14:59,570
integral I have is an integral
that was a power of sine x.

311
00:14:59,570 --> 00:15:02,100
And that's the goal, is
that I have it in the form.

312
00:15:02,100 --> 00:15:06,010
This is the power of sine
x to the n integrated.

313
00:15:06,010 --> 00:15:09,360
This is sine x to the
n minus 2 integrated.

314
00:15:09,360 --> 00:15:11,581
and this is just a function.

315
00:15:11,581 --> 00:15:12,080
OK?

316
00:15:12,080 --> 00:15:14,440
And, you know, if
we want to say,

317
00:15:14,440 --> 00:15:19,970
what's the true antiderivative,
we have to allow for a plus c

318
00:15:19,970 --> 00:15:21,640
here, for a true antiderivative.

319
00:15:21,640 --> 00:15:28,351
But we're really interested
in this part here.

320
00:15:28,351 --> 00:15:28,850
OK?

321
00:15:28,850 --> 00:15:31,410
Because if we evaluate
this at bounds,

322
00:15:31,410 --> 00:15:33,450
that's going to go away.

323
00:15:33,450 --> 00:15:35,540
But I want to just make
sure we understand--

324
00:15:35,540 --> 00:15:37,520
I think we understand the goal.

325
00:15:37,520 --> 00:15:40,040
Maybe the hardest
part of this problem

326
00:15:40,040 --> 00:15:45,310
is once you come over here,
was once you have these two

327
00:15:45,310 --> 00:15:46,250
components.

328
00:15:46,250 --> 00:15:49,180
So we did the substitution
that seems natural.

329
00:15:49,180 --> 00:15:51,145
You have these two components.

330
00:15:51,145 --> 00:15:52,895
This one, you should
be able to recognize,

331
00:15:52,895 --> 00:15:54,480
is in the form you want.

332
00:15:54,480 --> 00:15:57,390
This one is the hard
one to deal with.

333
00:15:57,390 --> 00:16:00,340
But the trick is to see that
under the right splitting up

334
00:16:00,340 --> 00:16:02,966
of these functions, one of
them is easy to integrate,

335
00:16:02,966 --> 00:16:05,340
and one of them, you know,
it's easy to integrate or take

336
00:16:05,340 --> 00:16:06,250
a derivative.

337
00:16:06,250 --> 00:16:08,196
So I have a little
bit of freedom.

338
00:16:08,196 --> 00:16:09,570
You don't want to
substitute back

339
00:16:09,570 --> 00:16:13,047
in for cosine squared
with 1 minus sine squared,

340
00:16:13,047 --> 00:16:15,380
because it will get you right
back to where you started.

341
00:16:15,380 --> 00:16:16,980
You would be
substituting one way,

342
00:16:16,980 --> 00:16:18,710
and then substituting
the other way.

343
00:16:18,710 --> 00:16:20,269
So the goal at this
point is to sort

344
00:16:20,269 --> 00:16:21,810
of-- the trick with
this one, I mean,

345
00:16:21,810 --> 00:16:24,090
is this integration
by parts, which

346
00:16:24,090 --> 00:16:26,520
requires some substitution.

347
00:16:26,520 --> 00:16:30,110
So hopefully this was a
good exercise for you,

348
00:16:30,110 --> 00:16:33,280
and you see some
of the strategies

349
00:16:33,280 --> 00:16:36,810
that one needs to reduce
the powers of sine x.

350
00:16:36,810 --> 00:16:40,750
You can do a very similar
type of thing with cosine x

351
00:16:40,750 --> 00:16:41,250
to the n.

352
00:16:41,250 --> 00:16:44,524
So if you felt like, I'm not
quite sure about this method,

353
00:16:44,524 --> 00:16:46,440
you might want to try
and do something similar

354
00:16:46,440 --> 00:16:50,530
with cosine x to the
n as your function

355
00:16:50,530 --> 00:16:53,290
that you're integrating, and
see if you can reduce that.

356
00:16:53,290 --> 00:16:54,950
And I guarantee you
can find a formula

357
00:16:54,950 --> 00:16:57,340
for that in a book
or online, if you

358
00:16:57,340 --> 00:16:59,040
wanted to check that answer.

359
00:16:59,040 --> 00:17:01,188
So I guess this is
where I'll stop.