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PROFESSOR: Well, because our
subject today is trig integrals

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substitutions, Professor
Jerison called in his

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substitute teacher for today.
 

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That's me.
 

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Professor Miller.
 

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And I'm going to try to tell
you about trig substitutions

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and trig integrals.
 

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And I'll be here tomorrow to
do more of the same, as well.

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So, this is about trigonometry,
and maybe first thing I'll do

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is remind you of some basic
things about trigonometry.

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So, if I have a circle,
trigonometry is all based on

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the circle of radius 1, and
centered at the origin.

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And so if this is an angle of
theta, up from the x-axis, then

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the coordinates of this point
are cosine theta

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and sine theta.
 

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And so that leads right away
to some trig identities,

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which you know very well.
 

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But I'm going to put them up
here because we'll use them

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over and over again today.
 

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Remember the convention
sin^2 theta secretly

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means (sin theta)^2.
 

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It would be more sensible to
write a parenthesis around

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the sign of theta and
then say you square that.

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But everybody in the world puts
the 2 up there over the sin,

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and so I'll do that too.
 

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So that follows just because
the circle has radius 1.

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But then there are some
other identities too, which

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I think you remember.
 

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I'll write them down here.
 

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Cos 2 theta, there's this
double angle formula that

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says cos 2 theta = cos
^2 theta - sin ^2 theta.

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And there's also the
double angle formula

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for the sin 2 theta.
 

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Remember what that says?
 

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2 sin theta cos theta.
 

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I'm going to use these trig
identities and I'm going to

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use them in a slightly
different way.

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And so I'd like to pay a little
more attention to this one and

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get a different way of
writing this one out.

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So this is actually the
half angle formula.

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And that says, I'm going to try
to express the cos theta in

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terms of the cos 2 theta.
 

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So if I know the cos 2 theta,
I want to try to express the

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cos theta in terms of it.
 

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Well, I'll start with a cos
2 theta and play with that.

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OK.
 

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Well, we know what this
is, it's cos ^2 theta

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- sin ^2 theta.
 

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But we also know what the
sin square root of theta

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is in terms of the cosine.
 

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So I can eliminate the
sin^2 from this picture.

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So this is equal to the cosine
^2 theta - (the quantity

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1 - cos ^2 theta).
 

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I put in what sin^2 is
in terms of cos^2.

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And so that's 2 cos
^2 of theta - 1.

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There's this cos^2,
which gets a plus sign.

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Because of these
two minus signs.

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And there's the one that was
there before, so altogether

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there are two of them.
 

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I want to isolate
what cosine is.

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Or rather, what cos^2 is.
 

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So let's solve for that.
 

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So I'll put the 1
on the other side.

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And I get 1 + cos 2 theta.
 

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And then, I want to divide by
this 2, and so that puts a

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2 in this denominator here.
 

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So some people call that
the half angle formula.

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What it really is for us is
it's a way of eliminating

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powers from sines and cosines.
 

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I've gotten rid of this square
at the expense of putting

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in a 2 theta here.
 

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We'll use that.
 

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And, similarly, same
calculation shows that the sin

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^2 theta = 1 cos 2 theta / 2.
 

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Same cosine, in that formula
also, but it has a minus sign.

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For the sin^2.
 

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OK. so that's my little review
of trig identities that we'll

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make use of as this
lecture goes on.

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I want to talk about
trig identity.

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Trig integrals, and you
know some trig integrals,

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I'm sure, already.
 

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Like, well, let me write the
differential form first.

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You know that d sin theta, or
maybe I'll say d sin x, is,

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let's see, that's the
derivative of sin x dx, right.

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The derivative of
sin x = cos x dx.

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And so if I integrate both
sides here, the integral form

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of this is the integral
of the cos x dx.

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Is the sin x + a constant.
 

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And in the same way, d
cos x = - sin x dx.

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Right, the derivative of
the cosine is - sine.

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And when I integrate that,
I find the integral of the

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sin x dx = - cos x + c.
 

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So that's our starting point.
 

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And the game today, for the
first half of the lecture, is

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to use that basic, just those
basic integration formulas,

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together with clever use of
trig identities in order to

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compute more complicated
formulas involving

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trig functions.
 

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So the first thing, the first
topic, is to think about

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integrals of the form
sin^n (x) cos^n(x).

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Where here I have
in mind m and n.

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Are non-negative integers.
 

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So let's try to
integrate these.

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I'll show you some applications
of these pretty soon.

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Looking down the road a little
bit, integrals like this show

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up in Fourier series and many
other subjects in mathematics.

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It turns out they're quite
important to be able to do.

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So that's why we're
doing them now.

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Well, so there are two
cases to think about here.

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When you're integrating
things like this.

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There's the easy case, and
let's do that one first.

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The easy case is when at
least one exponent is odd.

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That's the easy case.
 

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So, for example, suppose that
I wanted to integrate, well,

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let's take the case m = 1.
 

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So I'm integrating
sin^n (x) cos x dx.

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I'm taking -- oh.
 

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I could do that one.
 

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Let's see if that's
what I want to take.

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Yeah.
 

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My confusion is that I meant to
have this a different power.

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You were thinking that.
 

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So let's do this
case when m = 1.

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So the integral I'm trying to
do is any power of sin cos.

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Well, here's the trick.
 

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Recognize, use this formula up
at the top there to see cos x

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dx as something that we already
have on the blackboard.

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So, the way to exploit that
is to make a substitution.

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And substitution is
going to be u = sin x.

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And here's why.
 

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Then this integral that
I'm trying to do is

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the integral of u ^ n.
 

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That's already a
simplification.

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And then there's that cos x dx.
 

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When you make a substitution,
you've got to go all the way

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and replace everything in
the expression by things

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involving this new variable
that I've introduced.

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So I'd better get rid of the
cosine of x dx and rewrite it

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in terms of du or
in terms of u.

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And I can do that because
du, according to that

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formula, is the cos x dx.
 

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Let me put a box around that.
 

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That's our substitution.
 

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00:09:55 --> 00:09:55
When
 

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you make a substitution, you
also want to compute the

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differential of the variable
that you substitute in.

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So the cos x dx that appears
here is just, exactly, du.

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And I've replaced this trig
integral with something that

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doesn't involve trig
functions at all.

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This is a lot easier.
 

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We can just plug into
what we know here.

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This is (u ^ (n + 1) / n
+ 1) + a constant, and

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I've done the integral.
 

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But I'm not quite done
with the problem yet.

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Because to be nice to your
reader and to yourself, you

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should go back at this point,
probably, go back and get rid

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of this new variable
that you introduced.

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You're the one who introduced
this variable, you.

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Nobody except you, really,
knows what it is.

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But the rest of the world
knows what they asked for the

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first place that involved x.
 

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So I have to go back
and get rid of this.

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And that's not hard to do in
this case, because u = sin x.

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And so I make this
back substitution.

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And that's what you get.
 

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So there's the answer.
 

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OK, so the game was, I use this
odd power of the cosine here,

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and I could see it appearing as
the differential of the sine.

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So that's what made this
substitution work.

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Let's do another example to
see how that works out in

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00:11:25 --> 00:11:36
a slightly different case.
 

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So here's another example.
 

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Now I do have an odd power.
 

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One of the exponents is odd,
so I'm in the easy case.

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But it's not 1.
 

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The game now is to use this
trig identity to get rid of the

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largest even power that you
can, from this odd power here.

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So use sin^2 x = 1 - cos^2
x, to eliminate a lot of

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powers from that odd power.
 

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Watch what happens.
 

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So this is not really a
substitution or anything, this

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is just a trig identity.
 

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This sin^3 = sin^2 sin.
 

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And sin^2 = (1 - cos^2
x) And then I have

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the remaining sin x.
 

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And then I have cos^2 x dx.
 

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So let me rewrite that a little
bit to see how this works out.

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This is the integral of (cos^2
(x) -, and then there's

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the product of these two.
 

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That's cos^ 4 x) sin of x dx.
 

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So now I'm really exactly
in the situation that

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I was in over here.
 

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I've got a single power
of a sine or cosine.

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It happens that
it's a sine here.

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But that's not going to cause
any trouble, we can go ahead

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and play the same game
that I did there.

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So, so far I've just been
using trig identities.

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00:13:28 --> 00:13:41
But now I'll use a
trig substitution.

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00:13:41 --> 00:13:45
And I think I want to write
these as powers of a variable.

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And then this is going
to be the differential

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00:13:47 --> 00:13:47
of that variable.
 

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So I'll take u to be cosine
of x, and that means

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00:13:52 --> 00:13:58
that du = - sin x, dx.
 

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00:13:58 --> 00:14:04
There's the substitution.
 

214
00:14:04 --> 00:14:09
So when I make that
substitution, what do we get.

215
00:14:09 --> 00:14:11
Cos^2 becomes u^2.
 

216
00:14:11 --> 00:14:15
 


217
00:14:15 --> 00:14:24
Cos^ 4 becomes u ^ 4, and sin x
dx becomes not quite du, watch

218
00:14:24 --> 00:14:27
for the signum, watch for
this minus sign here.

219
00:14:27 --> 00:14:32
It becomes - du.
 

220
00:14:32 --> 00:14:32
But that's OK.
 

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00:14:32 --> 00:14:34
The minus sign comes outside.
 

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00:14:34 --> 00:14:37
And I can integrate both
of these powers, so

223
00:14:37 --> 00:14:43
I get - u ^3 / 3.
 

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00:14:43 --> 00:14:46
And then this 4th power
gives me a 5th power,

225
00:14:46 --> 00:14:48
when I integrate.
 

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00:14:48 --> 00:14:53
And don't forget the constant.
 

227
00:14:53 --> 00:14:55
Am I done?
 

228
00:14:55 --> 00:14:55
Not quite done.
 

229
00:14:55 --> 00:14:58
I have to back substitute
and get rid of my choice

230
00:14:58 --> 00:15:00
of variable, u, and
replace it with yours.

231
00:15:00 --> 00:15:01
Questions?
 

232
00:15:01 --> 00:15:06
STUDENT: [INAUDIBLE]
 

233
00:15:06 --> 00:15:07
PROFESSOR: There should indeed.
 

234
00:15:07 --> 00:15:10
I forgot this minus sign
when I came down here.

235
00:15:10 --> 00:15:12
So these two gang up
to give me a plus.

236
00:15:12 --> 00:15:14
Was that what the other
question was about, too?

237
00:15:14 --> 00:15:16
Thanks.
 

238
00:15:16 --> 00:15:18
So let's back substitute.
 

239
00:15:18 --> 00:15:23
And I'm going to put
that over here.

240
00:15:23 --> 00:15:27
And the result is, well, I just
replace the u by cosine of x.

241
00:15:27 --> 00:15:38
So this is - cos^3 x / 3 +,
thank you, cos^5 x / 5 + c.

242
00:15:38 --> 00:15:44
And there's the answer.
 

243
00:15:44 --> 00:15:47
By the way, you can remember
one of the nice things about

244
00:15:47 --> 00:15:49
doing an integral is
it's fairly easy to

245
00:15:49 --> 00:15:51
check your answer.
 

246
00:15:51 --> 00:15:54
You can always differentiate
the thing you get, and see

247
00:15:54 --> 00:15:56
whether you get the right
thing when you go back.

248
00:15:56 --> 00:16:01
It's not too hard to use
the power rules and the

249
00:16:01 --> 00:16:06
differentiation rule for the
cosine to get back to this if

250
00:16:06 --> 00:16:09
you want to check the work.
 

251
00:16:09 --> 00:16:14
Let's do one more example, just
to handle an example of this

252
00:16:14 --> 00:16:18
easy case, which you might
have thought of at first.

253
00:16:18 --> 00:16:22
Suppose I just want
to integrate a cube.

254
00:16:22 --> 00:16:29
Sin^3 x.
 

255
00:16:29 --> 00:16:32
No cosine in sight.
 

256
00:16:32 --> 00:16:36
But I do have an odd power
of a trig function,

257
00:16:36 --> 00:16:37
of a sine or cosine.
 

258
00:16:37 --> 00:16:39
So I'm in the easy case.
 

259
00:16:39 --> 00:16:44
And the procedure that I was
suggesting says I want to take

260
00:16:44 --> 00:16:47
out the largest even power
that I can, from the sin^3.

261
00:16:48 --> 00:16:51
So I'll take that out,
that's a sin^2 and

262
00:16:51 --> 00:16:53
write it as 1 - cos^2.
 

263
00:16:53 --> 00:16:57
 


264
00:16:57 --> 00:16:58
Well, now I'm very happy.
 

265
00:16:58 --> 00:17:05
Because it's just like the
situation we had somewhere

266
00:17:05 --> 00:17:06
on the board here.
 

267
00:17:06 --> 00:17:07
It's just like the
situation we had up here.

268
00:17:07 --> 00:17:11
I've got a power of
a cos sin x dx.

269
00:17:11 --> 00:17:16
So exactly the same
substitution steps in.

270
00:17:16 --> 00:17:19
You get, and maybe you
can see what happens

271
00:17:19 --> 00:17:20
without doing the work.
 

272
00:17:20 --> 00:17:22
Shall I do the work here?
 

273
00:17:22 --> 00:17:24
I make the same substitution.
 

274
00:17:24 --> 00:17:33
And so this is (1
- u ^2 )( - du).

275
00:17:33 --> 00:17:40
Which is u - u ^3 / 3.
 

276
00:17:40 --> 00:17:42
But then I want to put this
minus sign in place, and so

277
00:17:42 --> 00:17:47
that gives me - u + u
^3 / 3 + a constant.

278
00:17:47 --> 00:17:58
And then I back substitute
and get cos x + cos^3 x / 3.

279
00:17:58 --> 00:17:59
So this is the easy case.
 

280
00:17:59 --> 00:18:02
If you have some odd power
to play with, then you can

281
00:18:02 --> 00:18:07
make use of it and it's
pretty straightforward.

282
00:18:07 --> 00:18:10
OK the harder case is when
you don't have an odd power.

283
00:18:10 --> 00:18:11
So what's the program?
 

284
00:18:11 --> 00:18:13
I'm going to do the harder
case, and then I'm going to

285
00:18:13 --> 00:18:19
show you an example of how
to integrate square roots.

286
00:18:19 --> 00:18:26
And do an application, using
these ideas from trigonometry.

287
00:18:26 --> 00:18:30
So I want to keep
this blackboard.

288
00:18:30 --> 00:18:34
Maybe I'll come back
and start here again.

289
00:18:34 --> 00:18:55
So the harder case is when
they're only even exponents.

290
00:18:55 --> 00:18:58
I'm still trying to
integrate the same form.

291
00:18:58 --> 00:19:00
But now all the
exponents are even.

292
00:19:00 --> 00:19:03
So we have to do some game.
 

293
00:19:03 --> 00:19:10
And here the game is use
the half angle formula.

294
00:19:10 --> 00:19:16
Which I just erased, very
sadly, on the board here.

295
00:19:16 --> 00:19:18
Maybe I'll rewrite them
over here so we have

296
00:19:18 --> 00:19:23
them on the board.
 

297
00:19:23 --> 00:19:44
I think I remember
what they were.

298
00:19:44 --> 00:19:46
So the game is I'm going to use
that half angle formula to

299
00:19:46 --> 00:19:50
start getting rid of
those even powers.

300
00:19:50 --> 00:19:54
Half angle formula written like
this, exactly, talks about, it

301
00:19:54 --> 00:19:57
rewrites, even powers
of sines and cosines.

302
00:19:57 --> 00:20:00
So let's see how that
works out in an example.

303
00:20:00 --> 00:20:08
How about just the
cos^2 for a start.

304
00:20:08 --> 00:20:09
What to do?
 

305
00:20:09 --> 00:20:11
I can't pull anything out.
 

306
00:20:11 --> 00:20:15
I could rewrite this as 1 -
sin^2, but then I'd be faced

307
00:20:15 --> 00:20:19
with integrating the sin^2,
which is exactly as hard.

308
00:20:19 --> 00:20:23
So instead, let's use
this formula here.

309
00:20:23 --> 00:20:29
This is really the same
as 1 + cos 2 theta / 2.

310
00:20:29 --> 00:20:32
And now, this is easy.
 

311
00:20:32 --> 00:20:34
It's got two parts to it.
 

312
00:20:34 --> 00:20:38
Integrating one half gives
me theta over -- oh.

313
00:20:38 --> 00:20:42
Miraculously, the x
turned into a theta.

314
00:20:42 --> 00:20:44
Let's put it back as x.
 

315
00:20:44 --> 00:20:47
I get x / 2 by integrating 1/2.
 

316
00:20:47 --> 00:20:49
So, notice that something
non-trigonometric occurs

317
00:20:49 --> 00:20:54
here when I do these even
integrals. x / 2 appears.

318
00:20:54 --> 00:20:57
And then the other one, OK, so
this takes a little thought.

319
00:20:57 --> 00:21:03
The integral of the cosine
is the sine, or is it

320
00:21:03 --> 00:21:11
minus the sine. - sine.
 

321
00:21:11 --> 00:21:13
Shall we take a vote?
 

322
00:21:13 --> 00:21:13
I think it's positive.
 

323
00:21:13 --> 00:21:18
And so you get the sin
2x, but is that right?

324
00:21:18 --> 00:21:19
Over 2.
 

325
00:21:19 --> 00:21:24
If I differentiate the sin
2x, this 2 comes out.

326
00:21:24 --> 00:21:25
And would give me
an extra 2 here.

327
00:21:25 --> 00:21:29
So there's an extra 2 that
I have to put in here

328
00:21:29 --> 00:21:34
when I integrate it.
 

329
00:21:34 --> 00:21:37
And there's the answer.
 

330
00:21:37 --> 00:21:39
This is not a substitution.
 

331
00:21:39 --> 00:21:41
I just played with
trig identities here.

332
00:21:41 --> 00:21:45
And then did a simple trig
integral, getting your help

333
00:21:45 --> 00:21:46
to get the sign right.
 

334
00:21:46 --> 00:21:49
And thinking about what
this 2 is going to do.

335
00:21:49 --> 00:21:52
It produces a 2 in
the denominator.

336
00:21:52 --> 00:21:59
But it's not applying
any complicated thing.

337
00:21:59 --> 00:22:03
It's just using this identity.
 

338
00:22:03 --> 00:22:05
Let's do another example
that's a little bit harder.

339
00:22:05 --> 00:22:07
This time, sin^2 cos^2.
 

340
00:22:07 --> 00:22:35
 


341
00:22:35 --> 00:22:37
Again, no odd powers.
 

342
00:22:37 --> 00:22:40
I've got to work a
little bit harder.

343
00:22:40 --> 00:22:42
And what I'm going to
do is apply those

344
00:22:42 --> 00:22:44
identities up there.
 

345
00:22:44 --> 00:22:48
Now, what I recommend doing
in this situation is going

346
00:22:48 --> 00:22:51
over to the side somewhere.
 

347
00:22:51 --> 00:22:55
And do some side work.
 

348
00:22:55 --> 00:22:58
Because it's all just playing
with trig functions.

349
00:22:58 --> 00:23:06
It's not actually doing any
integrals for a while.

350
00:23:06 --> 00:23:11
So, I guess one way to get rid
of the sin^2 and the cos^2 is

351
00:23:11 --> 00:23:14
to use those identities
and so let's do that.

352
00:23:14 --> 00:23:20
So the sin = 1 - cos 2x / 2.
 

353
00:23:20 --> 00:23:27
And the cos = 1 + cos 2x / 2.
 

354
00:23:27 --> 00:23:29
So I just substitute them in.
 

355
00:23:29 --> 00:23:31
And now I can
multiply that out.

356
00:23:31 --> 00:23:38
And what I have is a
difference times a sum.

357
00:23:38 --> 00:23:40
So you know a formula for that.
 

358
00:23:40 --> 00:23:42
Taking the product of these
two things, well there'll

359
00:23:42 --> 00:23:44
be a 4 in the denominator.
 

360
00:23:44 --> 00:23:47
And then in the numerator, I
get the square of this minus

361
00:23:47 --> 00:23:57
the square of this. (a -
b)( a + b) = a ^2 - b^2.

362
00:23:57 --> 00:23:59
So I get that.
 

363
00:23:59 --> 00:24:02
Well, I'm a little bit
happier, because at

364
00:24:02 --> 00:24:03
least I don't have 4.
 

365
00:24:03 --> 00:24:07
I don't have 2
different squares.

366
00:24:07 --> 00:24:09
I still have a square, and
want to integrate this.

367
00:24:09 --> 00:24:12
I'm still not in the easy case.
 

368
00:24:12 --> 00:24:16
I got myself back to
an easier hard case.

369
00:24:16 --> 00:24:18
But we do know what
to do about this.

370
00:24:18 --> 00:24:21
Because I just did it up there.
 

371
00:24:21 --> 00:24:24
And I could play into this
formula that we got.

372
00:24:24 --> 00:24:29
But I think it's just as easy
to continue to calculate here.

373
00:24:29 --> 00:24:33
Use the half angle formula
again for this, and

374
00:24:33 --> 00:24:34
continue on your way.
 

375
00:24:34 --> 00:24:37
So I get a 1/4 from this bit.
 

376
00:24:37 --> 00:24:45
And then - 1/4 of the cos^2 2x.
 

377
00:24:45 --> 00:24:50
And when I plug in 2x in for
theta, there in the top

378
00:24:50 --> 00:24:59
board, I'm going to get a
4x on the right-hand side.

379
00:24:59 --> 00:25:02
So it comes out like that.
 

380
00:25:02 --> 00:25:04
And I guess I could simplify
that a little bit more.

381
00:25:04 --> 00:25:05
This is a 1/4.
 

382
00:25:05 --> 00:25:07
Oh, but then there's a 2 here.
 

383
00:25:07 --> 00:25:10
It's half that.
 

384
00:25:10 --> 00:25:12
So then I can simplify
a little more.

385
00:25:12 --> 00:25:16
It's 1/4 - 1/8, which is 1/8.
 

386
00:25:16 --> 00:25:25
And then I have 1/8 cos 4x.
 

387
00:25:25 --> 00:25:27
OK, that's my side work.
 

388
00:25:27 --> 00:25:30
I just did some trig
identities over here.

389
00:25:30 --> 00:25:34
And rewrote sine squared times
cosine squared as something

390
00:25:34 --> 00:25:37
which involves just no powers
of trig, just cosine by itself.

391
00:25:37 --> 00:25:41
And a constant.
 

392
00:25:41 --> 00:25:45
So I can take that and
substitute it in here.

393
00:25:45 --> 00:25:53
And now the integration is
pretty easy. (1/8 - cos 4x /

394
00:25:53 --> 00:26:01
8) dx, which is, OK the 1/8
is going to give me x / 8.

395
00:26:01 --> 00:26:06
The integral or cosine is
plus or minus the sine.

396
00:26:06 --> 00:26:08
The derivative of the
sine is plus the cosine.

397
00:26:08 --> 00:26:11
So it's going to be plus the,
only there's a minus here.

398
00:26:11 --> 00:26:18
So it's going to be the sin -
the sin 4x / 8, but then I

399
00:26:18 --> 00:26:20
have an additional factor
in the denominator.

400
00:26:20 --> 00:26:21
And what's it going to be?
 

401
00:26:21 --> 00:26:28
I have to put a 4 there.
 

402
00:26:28 --> 00:26:32
So we've done that
calculation, too.

403
00:26:32 --> 00:26:38
So any of these, if you keep
doing this kind of process,

404
00:26:38 --> 00:26:48
these two kinds of procedures,
you can now integrate any

405
00:26:48 --> 00:26:50
expression that has a power of
a sine times a power of

406
00:26:50 --> 00:26:56
a cosine in it, by
using these ideas.

407
00:26:56 --> 00:27:01
Now, let's see.
 

408
00:27:01 --> 00:27:04
Oh, let me give you an
alternate method for

409
00:27:04 --> 00:27:16
this last one here.
 

410
00:27:16 --> 00:27:26
I know what I'll do.
 

411
00:27:26 --> 00:27:29
Let me give an alternate method
for doing, really doing the

412
00:27:29 --> 00:27:33
side work over there.
i'm trying to deal

413
00:27:33 --> 00:27:35
with sin^2 cos^2.
 

414
00:27:35 --> 00:27:50
Well that's the square
of the sin x cos x.

415
00:27:50 --> 00:27:54
And the sin x cos x
shows up right here.

416
00:27:54 --> 00:27:55
In another trig identity.
 

417
00:27:55 --> 00:27:58
So we can make use
of that, too.

418
00:27:58 --> 00:28:00
That reduces the number
of factors of sines

419
00:28:00 --> 00:28:01
and cosines by 1.
 

420
00:28:01 --> 00:28:04
So it's going in the
right direction.

421
00:28:04 --> 00:28:11
This is equal to 1/2
sine 2x, squared.

422
00:28:11 --> 00:28:17
Sin cos = 1/2, say this right.
 

423
00:28:17 --> 00:28:21
It's sin 2x / 2, and then
I want to square that.

424
00:28:21 --> 00:28:31
So what I get is
the sin^2 2x / 4.

425
00:28:31 --> 00:28:34
Which is, well, I'm not
too happy yet, because I

426
00:28:34 --> 00:28:35
still have an even power.
 

427
00:28:35 --> 00:28:37
Remember I'm trying to
integrate this thing in the

428
00:28:37 --> 00:28:39
end, even powers are bad.
 

429
00:28:39 --> 00:28:40
I try to get rid of them.
 

430
00:28:40 --> 00:28:47
By using that formula, the half
angle formula, so I can apply

431
00:28:47 --> 00:28:48
that to the sin x here again.
 

432
00:28:48 --> 00:28:57
I get 1/4 of 1 - cos of 4x / 2.
 

433
00:28:57 --> 00:29:02
That's what the half angle
formula says for the sin^2 2x.

434
00:29:02 --> 00:29:04
And that's exactly the same
as the expression that

435
00:29:04 --> 00:29:08
I got up here, as well.
 

436
00:29:08 --> 00:29:11
It's the same expression
that I have there.

437
00:29:11 --> 00:29:16
So it's the same expression
as I have here.

438
00:29:16 --> 00:29:20
So this is just an alternate
way to play this game of using

439
00:29:20 --> 00:29:24
the half angle formula.
 

440
00:29:24 --> 00:29:28
OK, let's do a little
application of these things and

441
00:29:28 --> 00:29:46
change the topic a little bit.
 

442
00:29:46 --> 00:29:48
So here's the problem.
 

443
00:29:48 --> 00:29:57
So this is an application
and example of a real

444
00:29:57 --> 00:30:07
trig substitution.
 

445
00:30:07 --> 00:30:22
So here's the problem
I want to look at.

446
00:30:22 --> 00:30:26
OK, so I have a circle
whose radius is a.

447
00:30:26 --> 00:30:31
And I cut out from it
a sort of tab, here.

448
00:30:31 --> 00:30:36
This tab here.
 

449
00:30:36 --> 00:30:38
And the height of
this thing is b.

450
00:30:38 --> 00:30:42
So this length is a number b.
 

451
00:30:42 --> 00:30:44
And what I want to do
is compute the area

452
00:30:44 --> 00:30:47
of that little tab.
 

453
00:30:47 --> 00:30:48
That's the problem.
 

454
00:30:48 --> 00:30:50
So there's an arc over here.
 

455
00:30:50 --> 00:30:55
And I want to find the area
of this, for a and b,

456
00:30:55 --> 00:30:57
in terms of a and b.
 

457
00:30:57 --> 00:31:07
So the area, well, I guess
one way to compete the

458
00:31:07 --> 00:31:12
area would be to take
the integral of y dx.

459
00:31:12 --> 00:31:16
You've seen the idea of
splitting this up into vertical

460
00:31:16 --> 00:31:21
strips whose height is
given by a function, y (x).

461
00:31:21 --> 00:31:21
And then you integrate that.
 

462
00:31:21 --> 00:31:24
That's an interpretation
for the integral.

463
00:31:24 --> 00:31:27
The area is given by y dx.
 

464
00:31:27 --> 00:31:29
But that's a little bit
awkward, because my formula

465
00:31:29 --> 00:31:31
for y is going to be
a little strange.

466
00:31:31 --> 00:31:34
Its constant value of b along
here, and then at this

467
00:31:34 --> 00:31:37
point it becomes this
arc, of the circle.

468
00:31:37 --> 00:31:40
So working this out, I could do
it but it's a little awkward

469
00:31:40 --> 00:31:44
because expressing y as a
function of x, the top edge of

470
00:31:44 --> 00:31:48
this shape, it's a little
awkward and takes two

471
00:31:48 --> 00:31:51
different regions to express.
 

472
00:31:51 --> 00:31:58
So, a different way to
say it is to say x dy.

473
00:31:58 --> 00:32:00
Maybe that'll work a
little bit better.

474
00:32:00 --> 00:32:02
Or maybe it won't, but
it's worth trying.

475
00:32:02 --> 00:32:06
I could just as well split
this region up into

476
00:32:06 --> 00:32:08
horizontal strips.
 

477
00:32:08 --> 00:32:13
Whose width is dy, and
whose length is x.

478
00:32:13 --> 00:32:17
Now I'm thinking of this
as a function of y.

479
00:32:17 --> 00:32:20
This is the graph of
a function of y.

480
00:32:20 --> 00:32:24
And that's much better, because
the function of y is, well,

481
00:32:24 --> 00:32:28
it's the square root of
a^2 - y ^2, isn't it.

482
00:32:28 --> 00:32:34
That's x ^2 + y ^2 = a ^2.
 

483
00:32:34 --> 00:32:38
So that's what x is.
 

484
00:32:38 --> 00:32:41
And that's what I'm asked
to integrate, then.

485
00:32:41 --> 00:32:45
Square root of (a
^2 - y ^2) dy.

486
00:32:45 --> 00:32:47
And I can even put in
limits of integration.

487
00:32:47 --> 00:32:49
Maybe I should do that,
because this is supposed

488
00:32:49 --> 00:32:50
to be an actual number.
 

489
00:32:50 --> 00:32:55
I guess I'm integrating it
from y = 0, that's here.

490
00:32:55 --> 00:33:00
To y = b, dy.
 

491
00:33:00 --> 00:33:01
So this is what I want to find.
 

492
00:33:01 --> 00:33:07
This is a integral formula
for the area of that region.

493
00:33:07 --> 00:33:08
And this is a new form.
 

494
00:33:08 --> 00:33:17
I don't think that you've
thought about integrating

495
00:33:17 --> 00:33:20
expressions like this
in this class before.

496
00:33:20 --> 00:33:23
So, it's a new form and I want
to show you how to do it, how

497
00:33:23 --> 00:33:30
it's related to trigonometry.
 

498
00:33:30 --> 00:33:33
It's related to trigonometry
through that exact picture that

499
00:33:33 --> 00:33:36
we have on the blackboard.
 

500
00:33:36 --> 00:33:42
After all, this a^2 - y ^2 is
the formula for this arc.

501
00:33:42 --> 00:33:47
And so, what I propose is
that we try to exploit the

502
00:33:47 --> 00:33:52
connection with the circle and
introduce polar coordinates.

503
00:33:52 --> 00:34:03
So, here if I measure this
angle then there are various

504
00:34:03 --> 00:34:04
things you can say.
 

505
00:34:04 --> 00:34:07
Like the coordinates of
this point here are

506
00:34:07 --> 00:34:10
(a, cosine theta, a.
 

507
00:34:10 --> 00:34:17
Well, I'm sorry, it's not.
 

508
00:34:17 --> 00:34:20
That's an angle, but I
want to call it theta 0.

509
00:34:20 --> 00:34:25
And, in general you know that
the coordinates of this point

510
00:34:25 --> 00:34:31
are (a, cosine theta,
a, sine theta).

511
00:34:31 --> 00:34:39
If the radius is a, then
the angle here is theta.

512
00:34:39 --> 00:34:45
So x is a cosine theta, and y
is a sine theta, just from

513
00:34:45 --> 00:34:49
looking at the geometry
of the circle.

514
00:34:49 --> 00:34:56
So let's make that
substitution. y = a sine theta.

515
00:34:56 --> 00:35:00
I'm using the picture to
suggest that maybe making

516
00:35:00 --> 00:35:02
the substitution is
a good thing to do.

517
00:35:02 --> 00:35:06
Let's follow along and
see what happens.

518
00:35:06 --> 00:35:08
If that's true, what
we're interested in is

519
00:35:08 --> 00:35:13
integrating a^2 - y ^2.
 

520
00:35:13 --> 00:35:18
Which is a ^2, we're interested
in integrating the square

521
00:35:18 --> 00:35:20
root of (a ^2 - y^2).
 

522
00:35:20 --> 00:35:23
Which is the square root
of a ^2 minus this.

523
00:35:23 --> 00:35:27
a ^2 sin ^2 theta.
 

524
00:35:27 --> 00:35:35
And, well, that's
= a cos theta.

525
00:35:35 --> 00:35:41
That's just sin^2 + cos^2
= 1, all over again.

526
00:35:41 --> 00:35:42
It's also x.
 

527
00:35:42 --> 00:35:44
This is x.
 

528
00:35:44 --> 00:35:46
And this was x.
 

529
00:35:46 --> 00:35:48
So there are a lot of different
ways to think of this.

530
00:35:48 --> 00:35:51
But no matter how you say it,
the thing we're trying to

531
00:35:51 --> 00:35:57
integrate, a squared, a ^2 - y
^2 is, under this substitution

532
00:35:57 --> 00:36:02
it is a cos theta.
 

533
00:36:02 --> 00:36:04
So I'm interested in
integrating the square

534
00:36:04 --> 00:36:09
root of (a ^2 - y ^2) dy.
 

535
00:36:09 --> 00:36:17
And I'm going to make this
substitution y = a sin theta.

536
00:36:17 --> 00:36:22
And so under that substitution,
I've decided that the

537
00:36:22 --> 00:36:31
square root of a ^2 -
y ^2 = a cos theta.

538
00:36:31 --> 00:36:33
That's this.
 

539
00:36:33 --> 00:36:34
What about the dy?
 

540
00:36:34 --> 00:36:38
Well, I'd better
compute the dy.

541
00:36:38 --> 00:36:41
So dy just differentiating
this expression, is

542
00:36:41 --> 00:36:44
a cos theta d theta.
 

543
00:36:44 --> 00:36:57
So let's put that in. dy
= a cos theta d theta.

544
00:36:57 --> 00:36:58
OK.
 

545
00:36:58 --> 00:37:04
Making that trig substitution,
y = a sin theta has replaced

546
00:37:04 --> 00:37:06
this integral that has
a square root in it.

547
00:37:06 --> 00:37:08
And no trig functions.
 

548
00:37:08 --> 00:37:12
With an integral that
involves no square roots

549
00:37:12 --> 00:37:15
and only trig functions.
 

550
00:37:15 --> 00:37:17
In fact, it's not too hard to
integrate this now, because

551
00:37:17 --> 00:37:19
of the work that we've done.
 

552
00:37:19 --> 00:37:20
The a ^2 comes out.
 

553
00:37:20 --> 00:37:26
This is cos^2 theta. d theta.
 

554
00:37:26 --> 00:37:28
And maybe we've done that
example already today.

555
00:37:28 --> 00:37:35
I think we have.
 

556
00:37:35 --> 00:37:38
Maybe we can think it through,
but maybe the easiest thing is

557
00:37:38 --> 00:37:42
to look back at notes and
see what we got before.

558
00:37:42 --> 00:37:46
That was the first example in
the hard case that I did.

559
00:37:46 --> 00:38:03
And what it came out to, I used
x instead of theta at the time.

560
00:38:03 --> 00:38:05
So this is a good step forward.
 

561
00:38:05 --> 00:38:08
I started with this integral
that I really didn't know

562
00:38:08 --> 00:38:12
how to do by any means
that we've had so far.

563
00:38:12 --> 00:38:15
And I've replaced it by
a trig integral that

564
00:38:15 --> 00:38:16
we do know how to do.
 

565
00:38:16 --> 00:38:19
And now I've done
that trig integral.

566
00:38:19 --> 00:38:22
But we're still not quite
done, because of the problem

567
00:38:22 --> 00:38:23
of back substituting.
 

568
00:38:23 --> 00:38:29
I'd like to go back and
rewrite this in terms of

569
00:38:29 --> 00:38:32
the original variable, y.
 

570
00:38:32 --> 00:38:34
Or, I'd like to go back and
rewrite it in terms of the

571
00:38:34 --> 00:38:37
original limits of integration
that we had in the

572
00:38:37 --> 00:38:40
original problem.
 

573
00:38:40 --> 00:38:43
In doing that, it's going to
be useful to rewrite this

574
00:38:43 --> 00:38:47
expression and get rid
of the sin 2 theta.

575
00:38:47 --> 00:38:52
After all, the original y was
expressed in terms of the sin

576
00:38:52 --> 00:38:54
theta, not the sin 2 theta.
 

577
00:38:54 --> 00:39:04
So let me just do that here,
and say that this, in turn,

578
00:39:04 --> 00:39:14
is equal to a^2 theta / 2 +,
well, the sin 2 theta =

579
00:39:14 --> 00:39:18
2 sin theta cos theta.
 

580
00:39:18 --> 00:39:20
And so, when there's a 4 in the
denominator, what I'll get

581
00:39:20 --> 00:39:32
is sin theta cos theta / 2.
 

582
00:39:32 --> 00:39:36
I did that because I'm getting
closer to the original terms

583
00:39:36 --> 00:39:39
that the problem started with.
 

584
00:39:39 --> 00:40:05
Which was sin theta.
 

585
00:40:05 --> 00:40:08
So let me write down the
integral that we have now.

586
00:40:08 --> 00:40:15
The square root of a ^2 - y ^2
dy is, so far, what we know

587
00:40:15 --> 00:40:26
is a ^2 (theta / 2 + sin
theta cos theta / 2) + c.

588
00:40:26 --> 00:40:28
But I want to go back and
rewrite this in terms

589
00:40:28 --> 00:40:30
of the original value.
 

590
00:40:30 --> 00:40:32
The original variable, y.
 

591
00:40:32 --> 00:40:37
Well, what is theta
in terms of y?

592
00:40:37 --> 00:40:40
Let's see. y in terms of
theta was given like this.

593
00:40:40 --> 00:40:44
So what is theta in terms of y?
 

594
00:40:44 --> 00:40:44
Ah.
 

595
00:40:44 --> 00:40:48
So here the fearsome arc
sin rears its head, right?

596
00:40:48 --> 00:40:53
Theta is the angle so
that y = a sin theta.

597
00:40:53 --> 00:40:56
So that means that theta
is the arc sign, or

598
00:40:56 --> 00:41:07
sine inverse, of y / a.
 

599
00:41:07 --> 00:41:12
So that's the first thing
that shows up here.

600
00:41:12 --> 00:41:17
Arc sin (y / a).
 

601
00:41:17 --> 00:41:18
All over 2.
 

602
00:41:18 --> 00:41:19
That's this term.
 

603
00:41:19 --> 00:41:24
Theta is the arc
sin ( y /a) / 2.

604
00:41:24 --> 00:41:26
What about the
other side, here?

605
00:41:26 --> 00:41:30
Well sine and cosine, we knew
what they were in terms of y

606
00:41:30 --> 00:41:37
and in terms of x, if you like.
 

607
00:41:37 --> 00:41:40
Maybe I'll put the
a ^2 inside here.

608
00:41:40 --> 00:41:42
That makes it a
little bit nicer.

609
00:41:42 --> 00:41:49
Plus, and the other term is
a ^2 (sin theta cos theta).

610
00:41:49 --> 00:41:52
So the a sin theta is just y.
 

611
00:41:52 --> 00:42:02
Maybe I'll write this (a sin
theta)( a cos theta) / 2 + c.

612
00:42:02 --> 00:42:03
And so I get the same thing.
 

613
00:42:03 --> 00:42:06
And now here a sin
theta, that's y.

614
00:42:06 --> 00:42:12
And what's the a cos theta?
 

615
00:42:12 --> 00:42:22
It's x, or, if you like, it's
the square root of a ^2 - y ^2.

616
00:42:22 --> 00:42:28
And so there I've rewritten
everything, back in terms of

617
00:42:28 --> 00:42:31
the original variable, y.
 

618
00:42:31 --> 00:42:36
And there's an answer.
 

619
00:42:36 --> 00:42:41
So I've done this indefinite
integration of a form of this

620
00:42:41 --> 00:42:44
quadratic, this square root of
something which is

621
00:42:44 --> 00:42:46
a constant - y ^2.
 

622
00:42:46 --> 00:42:49
Whenever you see that,
the thing to think

623
00:42:49 --> 00:42:50
of is trigonometry.
 

624
00:42:50 --> 00:42:54
That's going to play into the
sin^ 2 + cos^2 identity.

625
00:42:54 --> 00:42:56
And the way to exploit it is
to make the substitution

626
00:42:56 --> 00:43:01
y = a sin theta.
 

627
00:43:01 --> 00:43:03
You could also make a
substitution y = a cos

628
00:43:03 --> 00:43:05
theta, if you wanted to.
 

629
00:43:05 --> 00:43:12
And the result would come out
to exactly the same in the end.

630
00:43:12 --> 00:43:14
I'm still not quite done with
the original problem that I

631
00:43:14 --> 00:43:24
had, because the original
problem asked for a

632
00:43:24 --> 00:43:25
definite integral.
 

633
00:43:25 --> 00:43:33
So let's just go back and
finish that as well.

634
00:43:33 --> 00:43:38
So the area was the
integral from 0 to b

635
00:43:38 --> 00:43:45
of this square root.
 

636
00:43:45 --> 00:43:48
So I just want to evaluate
the right-hand side here.

637
00:43:48 --> 00:43:50
The answer that we came up
with, this indefinite integral.

638
00:43:50 --> 00:43:53
I want to evaluate
it at 0 and at b.

639
00:43:53 --> 00:43:54
Well, let's see.
 

640
00:43:54 --> 00:44:13
When I evaluate it at b, I get
a ^2 ( arc sin (b / a) / 2 + y,

641
00:44:13 --> 00:44:19
which is b times the square
root of a ^2 - b ^2, putting

642
00:44:19 --> 00:44:23
y = b, divided by 2.
 

643
00:44:23 --> 00:44:26
So I've plugged in y =
b into that formula,

644
00:44:26 --> 00:44:27
this is what I get.
 

645
00:44:27 --> 00:44:31
Then when I plug in y = 0,
well the, sine of 0 is 0,

646
00:44:31 --> 00:44:34
so the arc sine of 0 is 0.
 

647
00:44:34 --> 00:44:35
So this term goes away.
 

648
00:44:35 --> 00:44:38
And when y = 0, this
term is 0 also.

649
00:44:38 --> 00:44:43
And so I don't get any
subtracted terms at all.

650
00:44:43 --> 00:44:45
So there's an
expression for this.

651
00:44:45 --> 00:44:52
Notice that this arc sin ( b /
a), that's exactly this angle.

652
00:44:52 --> 00:45:00
The arc sin ( b / a), it's the
angle that you get when y = b.

653
00:45:00 --> 00:45:09
So this theta is the
arc sin (b / a).

654
00:45:09 --> 00:45:15
Put this over here.
 

655
00:45:15 --> 00:45:17
That is theta 0.
 

656
00:45:17 --> 00:45:21
That is the angle that
the corner makes.

657
00:45:21 --> 00:45:28
So I could rewrite this as a
^2 theta 0 / 2 + b times the

658
00:45:28 --> 00:45:34
square root of a ^2 - b ^2 / 2.
 

659
00:45:34 --> 00:45:36
Let's just think about
this for a minute.

660
00:45:36 --> 00:45:40
I have these two terms in the
sum, is that reasonable?

661
00:45:40 --> 00:45:44
The first term is a ^2.
 

662
00:45:44 --> 00:45:50
That's the radius squared
times this angle, times 1/2.

663
00:45:50 --> 00:45:54
Well, I think that is exactly
the area of this sector. a ^2

664
00:45:54 --> 00:46:03
theta / 2 is the formula for
the area of the sector.

665
00:46:03 --> 00:46:07
And this one, this is
the vertical elevation.

666
00:46:07 --> 00:46:14
This is the horizontal. a ^2
- b ^2 is this distance.

667
00:46:14 --> 00:46:16
Square root of a ^2 - b ^2.
 

668
00:46:16 --> 00:46:20
So the right-hand term is b
times the square root of a

669
00:46:20 --> 00:46:31
^2 - b ^2 / 2, that's the
area of that triangle.

670
00:46:31 --> 00:46:35
So using a little bit of
geometry gives you the same

671
00:46:35 --> 00:46:39
answer as all of this
elaborate calculus.

672
00:46:39 --> 00:46:41
Maybe that's enough cause
for celebration for

673
00:46:41 --> 00:46:43
us to quit for today.
 

674
00:46:43 --> 00:46:44