WEBVTT

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Welcome back to recitation.

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In this video, I'd
like us to find

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an antiderivative
of the function 1

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over x squared minus 8x plus 1.

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So I'll give you a while to work
on it, and then I'll come back,

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and I'll show you how I started.

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So welcome back.

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Well, what we'd like to do is,
find an antiderivative to 1

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over x squared minus 8x plus 1.

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And how we're going
to do that, is

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we're going to use the
technique completing the square.

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And I'm going to
set up the problem,

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I'm going to get it
to a certain place,

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and then I'm going
to let you finish it.

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And how do you know if
you got the right answer?

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Well, you actually take a
derivative of your answer,

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and see if it gives you back 1
over x squared minus 8x plus 1.

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That's how you can check.

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So let's start off.

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If I want to
complete the square,

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let's just remind
ourselves how to complete

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the square on this quadratic.

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So I'd like something
right here that

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makes this a perfect square.

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Right?

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Well, the 8 here in the middle,
if I want a perfect square,

00:01:18.570 --> 00:01:20.020
if you think about
this, I'm going

00:01:20.020 --> 00:01:22.020
to have an x minus--
I need a number here

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that when I multiply it by 2,
that's where this 8 comes from,

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it gives me 8.

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So obviously I need
this number to be a 4.

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Right?

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Which puts what here?

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Puts a 16 here, right?

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So just to double check, what
I'm looking for is a number

00:01:37.750 --> 00:01:41.110
here-- I need a number right
here that when I multiply by 2,

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gives me negative 8, and
then I need to figure out

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what it squares to.

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So that number is negative
4, and it squares to 16.

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But obviously this isn't
what I have, right?

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I have plus 1.

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So what have I had to do
to get from here to here?

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Well, I had plus 1 and
now I have plus 16,

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so obviously I've added 15.

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So I have to subtract
15 to keep this,

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to keep these three lines
all equal to each other.

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So to understand
where that comes from,

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let me just remind
you, my denominator

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was looking like this.

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I'd like it to have
a perfect square,

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and then subtract a
constant, or add a constant.

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Right now I have
something that-- I don't

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have a perfect square in here.

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I can't make this
into a perfect square

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unless I add a certain amount
to the constant right here.

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So I had to add a 15
to the constant here.

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Notice 16 minus 15 is 1.

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That's my check, also,
that the lines are equal.

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And so what I've done, is I've
added 15 and subtracted 15,

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and then I put this
plus 16 into here. x

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minus 4 quantity squared is
exactly these first three

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terms.

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And then I keep the minus 15.

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Now, you might say,
why did you do this?

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So let's make sure we
understand why we're

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completing the square on this.

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If we come back, I'm going
to put this line in place

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of what's in the
denominator there,

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because these three
things are all equal.

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So this is actually
the integral of dx

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over x minus 4 quantity
squared minus 15.

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Now you might say, Christine,
this looks no easier.

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I don't know why you did this.

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But it actually is one of our
favorite, or least favorite,

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depending on how you feel about
it, types of tricks we use now,

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which is the trig substitution.

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So some people love
this because they just

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have to memorize
a little formula,

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and some people love it because
they can draw a triangle

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and understand what they choose.

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I'm going to show you, remind
you what the formula was

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you saw in class.

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I believe Professor Jerison
said something like this.

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If the denominator is in the
form u squared minus a squared,

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this implies that you make
u equal to a secant theta.

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Now, he probably wrote it
as x, but I wrote it as u

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for a very specific reason.

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Because here I have x minus 4.

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I have x minus 4
quantity squared.

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Now, this is where it gets
a little rough, right?

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This is not a perfect
square, but it

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is the perfect square--
it is the square

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of the square root of 15.

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So I can write the denominator
in the form, something squared

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minus something else squared.

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And again, you might
say, why is this good?

00:04:24.860 --> 00:04:26.600
Well, what we're going
to be able to do,

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is we're going to be able
to rewrite this in terms

00:04:28.880 --> 00:04:30.540
of trigonometric
functions, which will

00:04:30.540 --> 00:04:32.660
make it much simpler to solve.

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So let's use what
Professor Jerison gave us.

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And so what we see, is that
this is u and this is a.

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Right?

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So I get x minus 4 is equal to
square root of 15 secant theta.

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Now, you might not like
this square root of 15,

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but it's just hanging out.

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It's not causing any problems.

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It's just a number
there, so we'll

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keep it a square root of 15.

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So you don't have
to worry about it.

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Now what's the point again?

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Let me just remind
you, the object

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is to get this in terms
of the trig functions.

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So we should anticipate
that probably we'll

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have some tangent
functions to go with this.

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And there are two
reasons to think that.

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The first reason to think
that is at some point,

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I have to find dx.

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Well, the derivative of secant
involves secant and tangent,

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right?

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So that's going to pull in a
tangent function somewhere.

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I'm also going to have a
tangent function show up

00:05:25.951 --> 00:05:26.725
somewhere else.

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And where that's
going to be, is coming

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from this denominator, this
expression in the denominator.

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Because there's a certain
trig identity that we should

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have memorized, but
I'll just remind you.

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I'll write it here and
put a star next to it.

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It's 1 plus tangent
squared theta is

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equal to secant squared theta.

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So this is a-- I'll even put
a star on the other side.

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So we should really
remember this.

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Now, where does it come from?

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It comes from the cosine squared
theta plus sine squared theta

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equals 1 identity.

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You can divide everything
by cosine squared theta

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and get this one.

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So we have this identity,
and so if you notice,

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we're going to be able to
manipulate the expression right

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here, and get the
denominator to look

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like tangent squared theta.

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So let's do some of that
work off to the right here.

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So what did I say we needed?

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We have this expression.

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We need dx, so let's
find-- actually, no.

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Let's find the
denominator first,

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because I was just
talking about it.

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So if I look at what
x minus 4 squared is,

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I'm going to substitute
in this expression.

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So x minus 4 squared
minus 15 is the same

00:06:35.850 --> 00:06:42.670
as, based on this substitution,
square root 15 secant theta

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squared minus 15, which
is 15 secant squared

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theta minus 15, which, just
to hammer home the point,

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is 15 times the quantity
secant squared theta minus 1.

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OK?

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Everybody follows, hopefully.

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All I've done is the
substitution I made,

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and then I started expanding,
or I squared this term,

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and I factored out the 15.

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And now let's go back
to my start expression.

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What is secant
squared theta minus 1?

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It's tangent squared theta.

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So we get 15 tangent
squared theta.

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So that is actually what the
denominator of our integral

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is going to be over there.

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So I'm going to come in and
put that part-- actually,

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let me even put this here, too.

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So right now, our denominator
is 15 tangent squared theta.

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So far, so good.

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But of course, if I put a
dx up here, I'm in trouble.

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Because I have, it's a
function of theta now.

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So I need to write this-- I
shouldn't write in terms of x.

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I need to figure out what
it is in terms of theta.

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And to do that, we again use
the substitution that we made.

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Which is just above
the starred expression.

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It was that x minus 4 equals
square root 15 secant theta.

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This is going to
allow us to find

00:08:12.670 --> 00:08:15.950
what d theta is in terms of dx.

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OK?

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So let's do that.

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So I'm not done, by
the way, over here.

00:08:21.130 --> 00:08:23.160
I'm not done I've
got a little gap I've

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got to fill in the numerator.

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So let's come back over here.

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So now we have x
minus 4-- let me just

00:08:30.549 --> 00:08:31.590
write that one more time.

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So we get dx is equal to
the square root of 15.

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Well, what's the
derivative of secant theta?

00:08:42.630 --> 00:08:49.120
It's secant theta
tangent theta d theta.

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So now I have all
the pieces I need.

00:08:51.480 --> 00:08:52.870
And I'm actually
going to rewrite

00:08:52.870 --> 00:08:55.640
the whole thing over
here underneath,

00:08:55.640 --> 00:08:57.990
so that I can work with
it a little bit more.

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So the dx is in the numerator.

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Square root 15 secant theta tan
theta d theta, all over 15 tan

00:09:11.720 --> 00:09:13.550
squared theta.

00:09:13.550 --> 00:09:15.300
Now, this might still
look a little messy,

00:09:15.300 --> 00:09:18.590
but we can simplify
it some more.

00:09:18.590 --> 00:09:22.920
We divide out by one tangent,
we'll pull this out in front.

00:09:22.920 --> 00:09:24.620
And notice, what's secant?

00:09:24.620 --> 00:09:28.060
Secant theta is 1
over cosine theta,

00:09:28.060 --> 00:09:30.620
and tangent theta is sine
theta over cosine theta.

00:09:30.620 --> 00:09:32.460
So let's write that down.

00:09:32.460 --> 00:09:36.510
So this becomes square
root 15 over 15.

00:09:36.510 --> 00:09:38.790
We'll just leave it out there.

00:09:38.790 --> 00:09:41.130
It's not hurting anyone.

00:09:41.130 --> 00:09:45.630
So we get a 1 over cosine theta
times-- well, tangent theta,

00:09:45.630 --> 00:09:48.090
1 over tangent theta is
cotangent theta also.

00:09:48.090 --> 00:09:49.680
There's another way
to think about it.

00:09:49.680 --> 00:09:56.820
So it's cosine theta
over sine theta d theta.

00:09:56.820 --> 00:09:58.200
So these divide out.

00:09:58.200 --> 00:10:01.170
And I'm left with, I'm taking
now an antiderivative of 1

00:10:01.170 --> 00:10:05.000
over sine theta, which
is cosecant theta.

00:10:05.000 --> 00:10:06.550
So I have to find
an antiderivative

00:10:06.550 --> 00:10:07.980
of cosecant theta.

00:10:07.980 --> 00:10:10.620
Well, you can find that
with the exact same strategy

00:10:10.620 --> 00:10:14.140
you found, or I should say,
that Professor Jerison used

00:10:14.140 --> 00:10:16.080
in class-- or maybe
it was actually

00:10:16.080 --> 00:10:18.130
Professor Miller
in that lecture--

00:10:18.130 --> 00:10:21.840
to find an antiderivative
of secant theta.

00:10:21.840 --> 00:10:24.250
So you can do the same kind
of thing with cosecant theta,

00:10:24.250 --> 00:10:26.208
because they have the
same kind of derivatives.

00:10:26.208 --> 00:10:29.610
Cosecant and cotangent have
very similar-looking derivatives

00:10:29.610 --> 00:10:31.090
to tangent and secant.

00:10:31.090 --> 00:10:33.090
Same kinds of relationships.

00:10:33.090 --> 00:10:35.190
So you can actually find
that antiderivative.

00:10:35.190 --> 00:10:37.190
So this is some constant
we don't care about.

00:10:43.150 --> 00:10:46.810
And once you find that, this
will be in terms of theta.

00:10:46.810 --> 00:10:49.400
Your final answer needs
to be in terms of x,

00:10:49.400 --> 00:10:51.760
but you saw how to
do that, actually.

00:10:51.760 --> 00:10:54.040
You just need to
make a triangle that

00:10:54.040 --> 00:10:58.690
represents the relationship
between x and theta.

00:10:58.690 --> 00:11:00.832
So I'll draw a picture
of that triangle,

00:11:00.832 --> 00:11:02.790
then I'll give a little
summary of what we did,

00:11:02.790 --> 00:11:03.975
and then we'll stop.

00:11:03.975 --> 00:11:06.087
So let me draw a picture
of that triangle.

00:11:06.087 --> 00:11:08.045
So from here, all you
would do is actually find

00:11:08.045 --> 00:11:09.959
this antiderivative,
and then you

00:11:09.959 --> 00:11:12.000
would have to make the
right kind of substitution

00:11:12.000 --> 00:11:12.850
in terms of theta.

00:11:12.850 --> 00:11:16.330
We want to know, how do we find
that, do that substitution.

00:11:16.330 --> 00:11:19.690
So the triangle is going to
come from the following thing.

00:11:19.690 --> 00:11:26.630
We know x minus 4, again, is
square root of 15 secant theta.

00:11:26.630 --> 00:11:28.840
So I'm going to make this theta.

00:11:28.840 --> 00:11:31.700
Secant theta, well, it's
1 over cosine theta,

00:11:31.700 --> 00:11:33.930
cosine is adjacent
over hypotenuse,

00:11:33.930 --> 00:11:37.430
so secant is hypotenuse
over adjacent.

00:11:37.430 --> 00:11:38.010
Right?

00:11:38.010 --> 00:11:39.700
That's the relationship.

00:11:39.700 --> 00:11:43.740
So x minus 4 over 15 is
equal to the hypotenuse

00:11:43.740 --> 00:11:46.030
over the adjacent.

00:11:46.030 --> 00:11:47.130
Did I square root?

00:11:47.130 --> 00:11:48.230
Sorry.

00:11:48.230 --> 00:11:49.450
Square root.

00:11:49.450 --> 00:11:53.510
So the hypotenuse is x
minus 4, the adjacent

00:11:53.510 --> 00:11:57.050
is square root of 15, and then
now I can fill in the opposite

00:11:57.050 --> 00:11:58.270
by Pythagorean theorem.

00:11:58.270 --> 00:11:58.770
Right?

00:11:58.770 --> 00:12:02.830
I just take this squared,
and I subtract this squared,

00:12:02.830 --> 00:12:04.440
and then I take the square root.

00:12:04.440 --> 00:12:11.870
So I get the square root of
x minus 4 squared minus 15.

00:12:11.870 --> 00:12:13.660
So whatever I have
in terms of theta,

00:12:13.660 --> 00:12:15.460
I just look at this triangle.

00:12:15.460 --> 00:12:18.910
If I had in my
answer sine theta,

00:12:18.910 --> 00:12:23.250
I would replace sine theta
by this square root divided

00:12:23.250 --> 00:12:24.300
by x minus 4.

00:12:24.300 --> 00:12:25.799
Because that's
what sine theta is.

00:12:25.799 --> 00:12:28.090
And so from-- that's how I
finish this type of problem,

00:12:28.090 --> 00:12:28.589
always.

00:12:28.589 --> 00:12:31.590
I want to have a picture of
this triangle, label a theta,

00:12:31.590 --> 00:12:34.800
use my substitution to give
me what two of the sides are,

00:12:34.800 --> 00:12:37.200
use the Pythagorean theorem
to get the third side.

00:12:37.200 --> 00:12:38.237
So that's the strategy.

00:12:38.237 --> 00:12:40.820
So let's go back and just remind
ourselves where we came from.

00:12:40.820 --> 00:12:42.819
We're going to go all the
way to the other side.

00:12:42.819 --> 00:12:45.480
This was a long, long problem.

00:12:45.480 --> 00:12:47.260
So what did we do
in this problem?

00:12:47.260 --> 00:12:50.420
I wanted us to find an
antiderivative of something.

00:12:50.420 --> 00:12:54.720
And right away, we can't
use partial fractions,

00:12:54.720 --> 00:12:57.110
because we can't
factor out an x here.

00:12:57.110 --> 00:13:00.540
So I'm forced to use
completing the square.

00:13:00.540 --> 00:13:02.560
So I completed the square first.

00:13:02.560 --> 00:13:05.660
That was the little algebra
that I had to do first.

00:13:05.660 --> 00:13:07.500
Then once I have that
little bit of algebra,

00:13:07.500 --> 00:13:11.910
I get into a situation where I'm
set up for a trig substitution.

00:13:11.910 --> 00:13:15.070
So then I had to start off
and do some trig substituting.

00:13:15.070 --> 00:13:18.200
And the things you have to do
to make a trig substitution work

00:13:18.200 --> 00:13:21.442
are, pick the right substitution
that makes sense, which

00:13:21.442 --> 00:13:22.650
you were given that in class.

00:13:22.650 --> 00:13:25.210
You can also figure it out
from a triangle picture,

00:13:25.210 --> 00:13:26.530
if you wanted to.

00:13:26.530 --> 00:13:28.780
And then you have to make
sure you substitute not just

00:13:28.780 --> 00:13:32.130
for the expression, the
function of x, but also the dx.

00:13:32.130 --> 00:13:35.690
So we did all that, and then
we came over, further, further,

00:13:35.690 --> 00:13:37.350
further, further, further, here.

00:13:37.350 --> 00:13:39.920
And we had everything
in terms of theta.

00:13:39.920 --> 00:13:42.080
So then we had to
look at-- we had

00:13:42.080 --> 00:13:44.320
all these trigonometric
functions of theta.

00:13:44.320 --> 00:13:46.730
We simplified that
as far as we could.

00:13:46.730 --> 00:13:48.520
We got one we could find.

00:13:48.520 --> 00:13:50.864
Then we finally, we take
the antiderivative there,

00:13:50.864 --> 00:13:52.405
and then in the very
end, we're going

00:13:52.405 --> 00:13:55.500
to substitute in for theta,
using the triangle we've

00:13:55.500 --> 00:13:57.810
drawn up here.

00:13:57.810 --> 00:13:58.800
So!

00:13:58.800 --> 00:14:00.950
I think that's where I'm
going to stop this one.

00:14:00.950 --> 00:14:03.811
Also, I ran out of board
space, so I have to stop.