Welcome back to recitation.
In this video, we're
going to be working
on establishing the best
technique for finding
an integral or finding
an antiderivative.
We'll be doing this,
as you've seen probably
in a lot of these
videos, in a row.
And so in this
one in particular,
we're going to
work on these two.
So what I'd like us to
find, is for the letter A,
I'd like us to find
an actual value
if we take the integral from
minus 1 to 0 of this fraction.
5 x squared minus 2x plus 3 over
the quantity x squared plus 1
times x minus one.
And then, the second
problem, we're
just going to be finding
an antiderivative.
So it's finding an
antiderivative of the function
1 over x plus 1 times the square
root of negative x squared
minus 2x.
Now, this does have a domain
over which this function is
well-defined, as long as
what's inside the square root
is positive.
And there are values of x
for which that's positive.
I just didn't want us to have
to compute this one exactly.
So we're just looking for
an antiderivative here.
So the goal is, figure out
what strategy you want to use,
work through that strategy,
and then I'll be back,
and I'll show you which
strategy I picked,
and the solution that I got.
OK.
Welcome back.
Well, hopefully you were
able to make some headway
in both of these.
And so what we'll do
right away, is just
we'll start with the first one.
So on the first
one, we should be
able to get an actual numerical
value at the conclusion
of the problem.
And if you look
at the first one,
it's probably pretty obvious
you want to use partial fraction
decomposition.
You already have a
denominator that's factored,
and so this is going to
be fairly easy to do.
Now, what's one thing
you want to check
with partial
fractions, is you want
to make sure that the
degree of the numerator
is smaller than the
degree of the denominator.
Notice that the
numerator degree is 2
and the denominator
degree is 3, because we
have an x squared times x.
So we don't have to
do any long division.
We can just start the problem.
So what I'm going to do, is I'm
going to actually decompose it
without showing
you how I did it.
And you've done that
practice enough,
so I'm just going
to show you what
I got with my decomposition,
and we'll go from there.
So when I decompose,
for letter A,
I actually get two integrals.
And the first one I get is
the integral from minus 1
to 0 of 2x over x
squared plus 1 dx.
And the second one I get is
the integral, minus 1 to 0,
of 3 over x minus 1 dx.
Let me just double check
and make sure-- yes.
That's what I got when I
did this problem earlier.
So this is not magic.
I actually did this already.
That's how I got these.
And now from here, we just have
to determine-- we just have
to integrate both of these.
Now, what would be the
strategy at this point?
Well this one-- if this
had just been a 2 and no x,
you'd be dealing with an
arctan type of problem.
Because you'd be integrating
1 over x squared plus 1.
But actually, because
we have this 2x here,
this is really a
substitution problem.
Right?
The derivative of x
squared plus 1 is 2x.
Right?
So we see that really
what we're integrating
is something like du over u.
And so if you did this
substitution problem,
you should get something
like natural log
of x squared plus 1.
Let me just double
check and make sure that
gives me the derivative here.
The derivative of natural
log of x squared plus 1
is 1 over x squared
plus 1 times 2x.
That gives me
exactly what's here.
I don't need
absolute values here,
because this is always positive.
And I know I need to
evaluate it at minus 1 and 0.
OK?
So that takes care
of the left one.
Now, the right one
is-- again, it's
a pretty straightforward
one, because it's just 3
over x minus 1.
That's a natural log again.
So this natural log
is even simpler.
It's going to be 3 times the
natural log absolute x minus 1
from minus 1 to 0.
And so now we just have
to plug in everything.
OK?
So let's just do this one step
at a time, starting over here.
So the natural log,
when I put in 0,
I get the natural log of 1.
That's 0.
And I subtract what I get when
I put in negative 1 for x.
And negative 1
squared gives me 1,
so this is minus the
natural log of 2.
And then I have plus 3
times whatever's over here.
So now let's look at this.
When I plug in 0, I
get natural log of 0
minus 1, absolute value.
That's natural log of 1.
That's zero again.
And then I get a minus.
And then I put in negative 1.
Negative 1 minus 1,
negative 2, absolute value,
so it's natural log of 2.
And so if I look it
at all the way across,
I see I have a negative
natural log of 2
and then I have 3
natural logs of 2.
So the final answer is just
negative 4 natural log of 2.
And that is where
we'll stop with (a).
OK.
So let me just remind you,
actually, before we go to (b).
What we did in (a) was we did
partial fraction decomposition.
And I gave you the numerators.
And then on the first one, we
had to use maybe a substitution
to figure it out.
I didn't write explicitly
the substitution,
but a substitution
gives us that integral,
and this one is
directly a natural log.
OK.
Now let's look at (b).
So (b)-- let me rewrite
the problem, because it's
now a little far away.
I think it's x plus 1 square
root of negative x squared
minus 2x.
OK.
So (b), the reason--
I wanted to make sure
we did a trig substitution
in a particular way,
because I haven't
demonstrated those very much.
So the denominator wound up
looking a little awkward,
to force you to
do it in that way.
But what we want to do, is
we want to actually complete
the square on what's in here.
And that might make you
a little bit nervous.
But let me just do a little
sidebar work down here,
and point out what we get.
If we factor out
a negative here,
we get an x squared plus 2x.
OK?
So we're going to complete
the square on the inside.
Now this might make
some people nervous.
They might say, you've a
negative under the square root.
But I want to point out
that I have a negative here,
but I could always
make x squared
plus 2x a negative
number, and then
I would have-- the negative
times a negative is a positive.
For instance, I think
negative 1, right,
if I put a negative 1 for
x, I get negative 2 plus 1
is a negative value.
So if I put a
negative 1 for x, I'm
taking the square root
of a positive number.
So there are values of x that
make this under the square root
positive.
OK?
So you don't have
to worry about that.
Now, if I want to complete
the square on what's in here,
what do I do?
I have x squared plus 2x.
I obviously need to add a
1 to complete the square.
Why is that?
Because you take what's
here, you divide it by 2,
and you square it.
And so this actually
equals square root
of negative x squared
plus 2x plus 1.
And I want to subtract 1 so
that I haven't changed anything,
but when I pull it out from the
negative, it's another plus 1.
OK?
Let's make sure we buy that.
I've added 1 inside here, so
if I add 1 on the outside,
this is actually a minus
1, and so this is a plus 1,
so together they add up to 0.
So I haven't changed
what's in the square root.
So if I come back and put that
in right here, what do I get?
I get the integral dx over
x plus 1 square root--
let me move this over.
I'm going to bring
this to the front-- 1
minus x plus 1 squared.
All right.
So from here, we have to
do a trig substitution.
Now, what trig
substitution we want to do,
we can do sine or cosine.
But I'm going to do cosine,
because I like secants
better than cosecants, because
I have those memorized better.
So that's why I'm
choosing cosine.
You'll see why I chose
that way in a little bit.
But it would be, you will get
the same answer if you do sine.
OK.
So I'm going to come
to the other side.
Let's see.
So I'm choosing cosine
theta equals x plus 1.
That's the substitution
I'm making.
And why am I making
that substitution?
I'm making that
substitution because when
I do 1 minus x plus 1
squared, that's actually
1 minus cosine squared theta.
So that's, this in here
is sine squared theta.
And when I take the square
root, I just get a sine theta.
So that should be
pretty familiar to you
by now, this strategy.
But the point I'm
making is that x plus 1
will be a cosine theta,
and this whole square root
is what becomes a sine theta.
So you've seen
that a fair amount,
but just to remind you that.
And then the other thing we
need is to replace the dx.
So the dx is going to
be, derivative of cosine
is negative sine, so you're
going to get negative sine
theta d theta.
So now we know all the pieces.
We said this was cosine, we
said the square root is sine,
and the dx is
negative sine d theta.
So let's rewrite that over here.
So we have-- I'm going to
put the negative in front,
so I don't have to
deal with it anymore.
Negative sine theta over cosine
theta sine theta d theta.
These divide out,
and I get negative 1
over cosine theta, which is just
equal to negative secant theta.
OK?
So I have negative secant theta.
Let me actually write that here.
Negative integral of
secant theta d theta.
And now what is that?
Well, we know how
to integrate secant.
So let me write that
in terms of theta.
It's going to be negative
natural log absolute value
secant theta plus tangent theta.
And then we have
the plus c out here.
What's the point of this?
Well, we should maybe
have this memorized.
If you have to look it up,
you have to look it up,
but you saw this one in class.
And the negative is
just dropping down here,
so don't think I added
that negative in when
I was taking antiderivative.
It was already there.
All right.
So we're done.
Oh, but we're not done.
Why are we not done?
We're not done, because we
started off with something
in terms of x, and now we have
something in terms of theta,
so we have to finish up.
And how we do that,
is we go back,
we look at the
substitution we made.
If we make a triangle
based on that substitution,
we figure out the values of
secant theta and tangent theta,
and then we can plug
those in terms of x.
So let's remind
ourselves-- I'm going
to draw the triangle
in the middle here.
Let's remind ourselves
of the relationship
we had between theta and x.
If this is theta, we said
cosine theta, right here,
cosine theta was
equal to x plus 1.
Cosine theta is adjacent
over hypotenuse.
So we want to say, this is
x plus 1, and this is 1.
And that implies by the
Pythagorean theorem,
that this is square root of
1 minus quantity x plus 1
squared.
Let me move that over.
Notice, then, this
also makes sense,
why sine theta is what it is.
Sine theta is this
value divided by 1.
So that also helps
you understand that.
All right.
So now what do we
need to read off?
We need to read off secant, and
we need to read off tangent.
So secant is 1 over
cosine, so actually, we
could have gotten that one
for free, from the cosine.
So this 1 over cosine
is 1 over x plus 1.
So this thing is equal
to negative natural log
absolute value 1 over x plus
1 plus-- now, what's tangent?
If I come back and
look at the triangle,
tangent theta is
opposite over adjacent.
Right?
So I can actually just put it
all over x plus 1 if I wanted.
But I already started
writing it separately,
so I'll leave it like this.
Square root of 1 minus x
plus 1 quantity squared.
And then close that,
and then my plus c.
So now I'm actually
finished with the problem.
Because now I have an
antiderivative in terms of x.
So let me just remind you where
this problem, where we started
the problem, kind of
take us through quickly,
and then we'll be done.
So back to the
beginning, what we had,
was we had an integral that
was a fractional problem,
but we had an x plus
1 here, and then
we had this really
messy-looking quadratic in here.
To make it easy to deal with,
I factored out a negative sign,
and then I saw I could
complete the square.
Once you complete the
square, you actually
get another x plus
1 in there, which
helps us to see immediately, it
should be a trig substitution.
So the substitution
that's natural to make,
because you have a 1 minus
something involving an x,
is going to be either
cosine or sine.
I chose cosine.
If you'd chosen
sine, you probably
would have gotten a
cosecant up there,
instead of a secant, when you
were taking an antiderivative
at the very end.
So you would have gotten
the same answer because
of the substitutions in the end.
But so I chose cosine
theta is equal to x plus 1.
You do that, you
can replace this
with cosine, this with a sine,
this becomes a negative sine,
and then you start simplifying.
So once we came over
here and simplified,
we got it into
something we recognize.
We got it into secant.
We know the
antiderivative for secant,
in terms of secant and tangent.
We know it's exactly this.
And then we went back to
the relationship we had.
We made ourselves
a triangle in terms
of the theta and the x-values.
And then we were
able to substitute in
for secant and tangent.
All right.
So hopefully that was
successful for you.
And that's where I'll stop.
