Welcome back to recitation.
In this video, I'd
like us to find
an antiderivative
of the function 1
over x squared minus 8x plus 1.
So I'll give you a while to work
on it, and then I'll come back,
and I'll show you how I started.
So welcome back.
Well, what we'd like to do is,
find an antiderivative to 1
over x squared minus 8x plus 1.
And how we're going
to do that, is
we're going to use the
technique completing the square.
And I'm going to
set up the problem,
I'm going to get it
to a certain place,
and then I'm going
to let you finish it.
And how do you know if
you got the right answer?
Well, you actually take a
derivative of your answer,
and see if it gives you back 1
over x squared minus 8x plus 1.
That's how you can check.
So let's start off.
If I want to
complete the square,
let's just remind
ourselves how to complete
the square on this quadratic.
So I'd like something
right here that
makes this a perfect square.
Right?
Well, the 8 here in the middle,
if I want a perfect square,
if you think about
this, I'm going
to have an x minus--
I need a number here
that when I multiply it by 2,
that's where this 8 comes from,
it gives me 8.
So obviously I need
this number to be a 4.
Right?
Which puts what here?
Puts a 16 here, right?
So just to double check, what
I'm looking for is a number
here-- I need a number right
here that when I multiply by 2,
gives me negative 8, and
then I need to figure out
what it squares to.
So that number is negative
4, and it squares to 16.
But obviously this isn't
what I have, right?
I have plus 1.
So what have I had to do
to get from here to here?
Well, I had plus 1 and
now I have plus 16,
so obviously I've added 15.
So I have to subtract
15 to keep this,
to keep these three lines
all equal to each other.
So to understand
where that comes from,
let me just remind
you, my denominator
was looking like this.
I'd like it to have
a perfect square,
and then subtract a
constant, or add a constant.
Right now I have
something that-- I don't
have a perfect square in here.
I can't make this
into a perfect square
unless I add a certain amount
to the constant right here.
So I had to add a 15
to the constant here.
Notice 16 minus 15 is 1.
That's my check, also,
that the lines are equal.
And so what I've done, is I've
added 15 and subtracted 15,
and then I put this
plus 16 into here. x
minus 4 quantity squared is
exactly these first three
terms.
And then I keep the minus 15.
Now, you might say,
why did you do this?
So let's make sure we
understand why we're
completing the square on this.
If we come back, I'm going
to put this line in place
of what's in the
denominator there,
because these three
things are all equal.
So this is actually
the integral of dx
over x minus 4 quantity
squared minus 15.
Now you might say, Christine,
this looks no easier.
I don't know why you did this.
But it actually is one of our
favorite, or least favorite,
depending on how you feel about
it, types of tricks we use now,
which is the trig substitution.
So some people love
this because they just
have to memorize
a little formula,
and some people love it because
they can draw a triangle
and understand what they choose.
I'm going to show you, remind
you what the formula was
you saw in class.
I believe Professor Jerison
said something like this.
If the denominator is in the
form u squared minus a squared,
this implies that you make
u equal to a secant theta.
Now, he probably wrote it
as x, but I wrote it as u
for a very specific reason.
Because here I have x minus 4.
I have x minus 4
quantity squared.
Now, this is where it gets
a little rough, right?
This is not a perfect
square, but it
is the perfect square--
it is the square
of the square root of 15.
So I can write the denominator
in the form, something squared
minus something else squared.
And again, you might
say, why is this good?
Well, what we're going
to be able to do,
is we're going to be able
to rewrite this in terms
of trigonometric
functions, which will
make it much simpler to solve.
So let's use what
Professor Jerison gave us.
And so what we see, is that
this is u and this is a.
Right?
So I get x minus 4 is equal to
square root of 15 secant theta.
Now, you might not like
this square root of 15,
but it's just hanging out.
It's not causing any problems.
It's just a number
there, so we'll
keep it a square root of 15.
So you don't have
to worry about it.
Now what's the point again?
Let me just remind
you, the object
is to get this in terms
of the trig functions.
So we should anticipate
that probably we'll
have some tangent
functions to go with this.
And there are two
reasons to think that.
The first reason to think
that is at some point,
I have to find dx.
Well, the derivative of secant
involves secant and tangent,
right?
So that's going to pull in a
tangent function somewhere.
I'm also going to have a
tangent function show up
somewhere else.
And where that's
going to be, is coming
from this denominator, this
expression in the denominator.
Because there's a certain
trig identity that we should
have memorized, but
I'll just remind you.
I'll write it here and
put a star next to it.
It's 1 plus tangent
squared theta is
equal to secant squared theta.
So this is a-- I'll even put
a star on the other side.
So we should really
remember this.
Now, where does it come from?
It comes from the cosine squared
theta plus sine squared theta
equals 1 identity.
You can divide everything
by cosine squared theta
and get this one.
So we have this identity,
and so if you notice,
we're going to be able to
manipulate the expression right
here, and get the
denominator to look
like tangent squared theta.
So let's do some of that
work off to the right here.
So what did I say we needed?
We have this expression.
We need dx, so let's
find-- actually, no.
Let's find the
denominator first,
because I was just
talking about it.
So if I look at what
x minus 4 squared is,
I'm going to substitute
in this expression.
So x minus 4 squared
minus 15 is the same
as, based on this substitution,
square root 15 secant theta
squared minus 15, which
is 15 secant squared
theta minus 15, which, just
to hammer home the point,
is 15 times the quantity
secant squared theta minus 1.
OK?
Everybody follows, hopefully.
All I've done is the
substitution I made,
and then I started expanding,
or I squared this term,
and I factored out the 15.
And now let's go back
to my start expression.
What is secant
squared theta minus 1?
It's tangent squared theta.
So we get 15 tangent
squared theta.
So that is actually what the
denominator of our integral
is going to be over there.
So I'm going to come in and
put that part-- actually,
let me even put this here, too.
So right now, our denominator
is 15 tangent squared theta.
So far, so good.
But of course, if I put a
dx up here, I'm in trouble.
Because I have, it's a
function of theta now.
So I need to write this-- I
shouldn't write in terms of x.
I need to figure out what
it is in terms of theta.
And to do that, we again use
the substitution that we made.
Which is just above
the starred expression.
It was that x minus 4 equals
square root 15 secant theta.
This is going to
allow us to find
what d theta is in terms of dx.
OK?
So let's do that.
So I'm not done, by
the way, over here.
I'm not done I've
got a little gap I've
got to fill in the numerator.
So let's come back over here.
So now we have x
minus 4-- let me just
write that one more time.
So we get dx is equal to
the square root of 15.
Well, what's the
derivative of secant theta?
It's secant theta
tangent theta d theta.
So now I have all
the pieces I need.
And I'm actually
going to rewrite
the whole thing over
here underneath,
so that I can work with
it a little bit more.
So the dx is in the numerator.
Square root 15 secant theta tan
theta d theta, all over 15 tan
squared theta.
Now, this might still
look a little messy,
but we can simplify
it some more.
We divide out by one tangent,
we'll pull this out in front.
And notice, what's secant?
Secant theta is 1
over cosine theta,
and tangent theta is sine
theta over cosine theta.
So let's write that down.
So this becomes square
root 15 over 15.
We'll just leave it out there.
It's not hurting anyone.
So we get a 1 over cosine theta
times-- well, tangent theta,
1 over tangent theta is
cotangent theta also.
There's another way
to think about it.
So it's cosine theta
over sine theta d theta.
So these divide out.
And I'm left with, I'm taking
now an antiderivative of 1
over sine theta, which
is cosecant theta.
So I have to find
an antiderivative
of cosecant theta.
Well, you can find that
with the exact same strategy
you found, or I should say,
that Professor Jerison used
in class-- or maybe
it was actually
Professor Miller
in that lecture--
to find an antiderivative
of secant theta.
So you can do the same kind
of thing with cosecant theta,
because they have the
same kind of derivatives.
Cosecant and cotangent have
very similar-looking derivatives
to tangent and secant.
Same kinds of relationships.
So you can actually find
that antiderivative.
So this is some constant
we don't care about.
And once you find that, this
will be in terms of theta.
Your final answer needs
to be in terms of x,
but you saw how to
do that, actually.
You just need to
make a triangle that
represents the relationship
between x and theta.
So I'll draw a picture
of that triangle,
then I'll give a little
summary of what we did,
and then we'll stop.
So let me draw a picture
of that triangle.
So from here, all you
would do is actually find
this antiderivative,
and then you
would have to make the
right kind of substitution
in terms of theta.
We want to know, how do we find
that, do that substitution.
So the triangle is going to
come from the following thing.
We know x minus 4, again, is
square root of 15 secant theta.
So I'm going to make this theta.
Secant theta, well, it's
1 over cosine theta,
cosine is adjacent
over hypotenuse,
so secant is hypotenuse
over adjacent.
Right?
That's the relationship.
So x minus 4 over 15 is
equal to the hypotenuse
over the adjacent.
Did I square root?
Sorry.
Square root.
So the hypotenuse is x
minus 4, the adjacent
is square root of 15, and then
now I can fill in the opposite
by Pythagorean theorem.
Right?
I just take this squared,
and I subtract this squared,
and then I take the square root.
So I get the square root of
x minus 4 squared minus 15.
So whatever I have
in terms of theta,
I just look at this triangle.
If I had in my
answer sine theta,
I would replace sine theta
by this square root divided
by x minus 4.
Because that's
what sine theta is.
And so from-- that's how I
finish this type of problem,
always.
I want to have a picture of
this triangle, label a theta,
use my substitution to give
me what two of the sides are,
use the Pythagorean theorem
to get the third side.
So that's the strategy.
So let's go back and just remind
ourselves where we came from.
We're going to go all the
way to the other side.
This was a long, long problem.
So what did we do
in this problem?
I wanted us to find an
antiderivative of something.
And right away, we can't
use partial fractions,
because we can't
factor out an x here.
So I'm forced to use
completing the square.
So I completed the square first.
That was the little algebra
that I had to do first.
Then once I have that
little bit of algebra,
I get into a situation where I'm
set up for a trig substitution.
So then I had to start off
and do some trig substituting.
And the things you have to do
to make a trig substitution work
are, pick the right substitution
that makes sense, which
you were given that in class.
You can also figure it out
from a triangle picture,
if you wanted to.
And then you have to make
sure you substitute not just
for the expression, the
function of x, but also the dx.
So we did all that, and then
we came over, further, further,
further, further, further, here.
And we had everything
in terms of theta.
So then we had to
look at-- we had
all these trigonometric
functions of theta.
We simplified that
as far as we could.
We got one we could find.
Then we finally, we take
the antiderivative there,
and then in the very
end, we're going
to substitute in for theta,
using the triangle we've
drawn up here.
So!
I think that's where I'm
going to stop this one.
Also, I ran out of board
space, so I have to stop.
