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Welcome back to recitation.

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In this video, I'd like us to
find an antiderivative of the

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function 1 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

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back, 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 we're going to use

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the technique, completing
the square.

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And I'm going to set up the
problem, I'm going to get it

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to a certain place, and
then I'm going to

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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, and

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see if it gives you back
1 over x squared

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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, let's just remind

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ourselves how to complete the
square on this quadratic.

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So I'd like something right
here that makes

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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

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square, if you think about
this, I'm going

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to have an x minus--

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I need a number here that when
I multiply it by 2, that's

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where this 8 comes from,
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

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for is a number here--

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I need a number right here that
when I multiply by 2,

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

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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, so obviously

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I've added 15.

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So I have to subtract 15 to keep
this, to keep these three

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lines all equal to each other.

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So to understand where that
comes from, let me just remind

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you, my denominator was
looking like this.

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I'd like it to have a perfect
square, and then subtract a

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constant, or add a constant.

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Right now I have something
that--

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I don't have a perfect
square in here.

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I can't make this into a perfect
square unless I add a

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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 minus 4 quantity

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squared is exactly these first
three terms. And then I keep

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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 of

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what's in the denominator there,
because these three

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things are all equal.

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So this is actually the integral
of dx over x minus 4

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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, types of

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tricks we use now, which is
the trig substitution.

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So some people love this because
they just have to

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memorize a little formula, and
some people love it because

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they can draw a triangle and
understand what they choose.

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

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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

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squared, this implies
that you make u

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equal to a secant theta.

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

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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 is the perfect square,

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it is the square 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?

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Well, what we're going to be
able to do, is we're going to

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be able to rewrite this in
terms of trigonometric

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functions, which will 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

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of 15 secant theta.

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Now, you might not like this
square root of 15, but it's

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just hanging out.

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It's not causing any problems.
It's just a number there, so

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we'll 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 is to get this in terms

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of the trig functions.

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So we should anticipate they
probably we'll have some

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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, I

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have to find dx.

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Well, the derivative of
secant involves secant

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and tangent, 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

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somewhere else.

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And where that's going to
be, is coming from this

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denominator, this expression
in the denominator.

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Because there's a certain trig
identity that we should have

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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 equal to secant

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squared theta.

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So this is a--

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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

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squared theta 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, we're going

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to be able to manipulate the
expression right here, and get

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the denominator to look 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--

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actually, no.

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Let's find the denominator
first, because I was just

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talking about it.

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So if I look at what x minus
4 squared is, I'm going to

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substitute in this expression.

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So x minus 4 squared minus 15 is
the same as, based on this

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substitution, square root 15
secant theta squared minus 15,

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which is 15 secant squared theta
minus 15, which, just to

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hammer home the point, is 15
times the quantity secant

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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, and then

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

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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 is

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going to be over there.

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

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actually, let me even
put this here, too.

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So right now, our denominator
is tangent 15

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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--

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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 what d theta

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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, I'm not done.

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I've got a little gap I've 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 write

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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?

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It's secant theta tangent
theta d theta.

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So now I have all the
pieces I need.

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And I'm actually going to
rewrite the whole thing over

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here underneath, so that
I can work with it

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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

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tan squared theta.

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Now, this might still look a
little messy, but we can

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simplify it some more.

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We divide out by 1 tangent,
we'll pull this out in front.

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And notice, what's secant?

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Secant theta is 1 over cosine
theta, and tangent theta is

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sine theta over cosine theta.

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So let's write that down.

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So this becomes square
root 15 over 15.

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We'll just leave it out there.

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It's not hurting anyone.

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So we get a 1 over cosine
theta times--

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well, tangent theta, 1 over
tangent theta is cotangent

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theta also.

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There's another way
to think about it.

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So it's cosine theta over
sine theta d theta.

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So these divide out.

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And I'm left with, I'm taking
now an antiderivative of 1

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over sine theta, which
is cosecant theta.

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So I have to find an

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antiderivative of cosecant theta.

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Well, you can find that with
the exact same strategy you

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found, or I should say,
that Professor

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Jerison used in class--

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or maybe it was actually
Professor

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Miller in that lecture--

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to find an antiderivative
of secant theta.

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So you can do the same kind of
thing with cosecant theta,

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because they have the same
kind of derivatives.

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Cosecant and cotangent have
very similar-looking

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derivatives to tangent
and secant.

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Same kinds of relationships.

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So you can actually find
that antiderivative.

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So this is some constant
we don't care about.

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And once you find that, this
will be in terms of theta.

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Your final answer needs to be in
terms of x, but you saw how

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to do that, actually.

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You just need to make a triangle
that represents the

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relationship between
x and theta.

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So I'll draw a picture of that
triangle, then I'll give a

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little summary of what we did,
and then we'll stop.

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So let me draw a picture
of that triangle.

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So from here, all you would
do is actually find this

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antiderivative, and then you
would have to make the right

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kind of substitution
in terms of theta.

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We want to know, how do
we find that, do that

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substitution.

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So the triangle is going to come
from the following thing.

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We know x minus 4, again,
is square root

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00:11:22,065 --> 00:11:26,630
of 15 secant theta.

249
00:11:26,630 --> 00:11:28,840
So I'm going to make
this theta.

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00:11:28,840 --> 00:11:32,420
Secant theta, well, it's 1 over
cosine theta, cosine is

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00:11:32,420 --> 00:11:35,000
adjacent over hypotenuse,
so secant is

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00:11:35,000 --> 00:11:37,430
hypotenuse over adjacent.

253
00:11:37,430 --> 00:11:38,010
Right?

254
00:11:38,010 --> 00:11:39,700
That's the relationship.

255
00:11:39,700 --> 00:11:42,850
So x minus 4 over 15
is equal to the

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00:11:42,850 --> 00:11:46,030
hypotenuse over the adjacent.

257
00:11:46,030 --> 00:11:47,130
Did I square root?

258
00:11:47,130 --> 00:11:48,230
Sorry.

259
00:11:48,230 --> 00:11:49,450
Square root.

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00:11:49,450 --> 00:11:54,080
So the hypotenuse is x minus 4,
the adjacent is square root

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00:11:54,080 --> 00:11:57,050
of 15, and then now I can
fill in the opposite

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00:11:57,050 --> 00:11:58,560
by Pythagorean theorem.

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00:11:58,560 --> 00:11:58,770
Right?

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00:11:58,770 --> 00:12:02,990
I just take this squared, and
I subtract this squared, and

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00:12:02,990 --> 00:12:04,440
then I take the square root.

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00:12:04,440 --> 00:12:11,870
So I get the square root of x
minus 4 squared minus 15.

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00:12:11,870 --> 00:12:14,030
So whatever I have in terms
of theta, I just

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00:12:14,030 --> 00:12:15,460
look at this triangle.

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00:12:15,460 --> 00:12:20,130
If I had in my answer, sine
theta, I would replace sine

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00:12:20,130 --> 00:12:24,300
theta by this square root
divided by x minus 4.

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00:12:24,300 --> 00:12:26,020
Because that's what
sine theta is.

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00:12:26,020 --> 00:12:27,400
And so from, that's
how I finish this

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00:12:27,400 --> 00:12:28,370
type of problem, always.

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00:12:28,370 --> 00:12:31,590
I want to have a picture of this
triangle, label a theta,

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00:12:31,590 --> 00:12:34,960
use my substitution to give me
what two of the sides are, use

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00:12:34,960 --> 00:12:37,200
the Pythagorean theorem
to get the third side.

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00:12:37,200 --> 00:12:38,406
So that's the strategy.

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00:12:38,406 --> 00:12:40,060
So let's go back and just
remind ourselves

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00:12:40,060 --> 00:12:40,820
where we came from.

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00:12:40,820 --> 00:12:42,620
We're going to go all the
way to the other side.

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00:12:42,620 --> 00:12:45,480
This was a long, long problem.

282
00:12:45,480 --> 00:12:47,260
So what did we do
in this problem?

283
00:12:47,260 --> 00:12:50,420
I wanted us to find an
antiderivative of something.

284
00:12:50,420 --> 00:12:54,940
And right away, we can't use
partial fractions, because we

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00:12:54,940 --> 00:12:57,110
can't factor out an x here.

286
00:12:57,110 --> 00:13:00,540
So I'm forced to use completing
the square.

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00:13:00,540 --> 00:13:03,050
So I completed the square first.
That was the little

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00:13:03,050 --> 00:13:06,560
algebra that I had to do first.
Then once I have that

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00:13:06,560 --> 00:13:09,700
little bit of algebra, I get
into a situation where I'm set

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00:13:09,700 --> 00:13:11,910
up for a trig substitution.

291
00:13:11,910 --> 00:13:15,070
So then I had to start off and
do some trig substituting.

292
00:13:15,070 --> 00:13:17,570
And the things you have to do
to make a trig substitution

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00:13:17,570 --> 00:13:21,330
work are, pick the right
substitution that makes sense,

294
00:13:21,330 --> 00:13:22,650
which you were given
that in class.

295
00:13:22,650 --> 00:13:25,320
You can also figure it out from
a triangle picture, if

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00:13:25,320 --> 00:13:26,530
you wanted to.

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00:13:26,530 --> 00:13:28,990
And then you have to make sure
you substitute not just for

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00:13:28,990 --> 00:13:32,130
the expression, the function
of x, but also the dx.

299
00:13:32,130 --> 00:13:35,310
So we did all that, and then
we came over, further,

300
00:13:35,310 --> 00:13:37,350
further, further, further,
further, here.

301
00:13:37,350 --> 00:13:39,920
And we had everything
in terms of theta.

302
00:13:39,920 --> 00:13:41,790
So then we had to look at--

303
00:13:41,790 --> 00:13:44,320
we had all these trigonometric
functions of theta.

304
00:13:44,320 --> 00:13:46,730
We simplified that as
far as we could.

305
00:13:46,730 --> 00:13:48,520
We got one we could find.

306
00:13:48,520 --> 00:13:51,470
Then we finally, we take the
antiderivative there, and then

307
00:13:51,470 --> 00:13:54,320
in the very end, we're going
to substitute in for theta,

308
00:13:54,320 --> 00:13:57,810
using the triangle we've
drawn up here.

309
00:13:57,810 --> 00:13:58,800
So!

310
00:13:58,800 --> 00:14:00,950
I think that's where I'm
going to stop this one.

311
00:14:00,950 --> 00:14:04,110
Also, I ran out of board space,
so I have to stop.

312
00:14:04,110 --> 00:14:04,311