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PROFESSOR: So we're
going to continue
to talk about trig integrals
and trig substitutions.
This is maybe the
most technical part
of this course, which maybe is
why professor Jerison decided
to just take a leave, go AWOL
just now and let me take over
for him.
But I'll do my best to help
you learn this technique
and it'll be useful for you.
So we've talked about
trig integrals involving
sines and cosines yesterday.
There's another
whole world out there
that involves these other trig
polynomials-- trig functions,
secant and tangent.
Let me just make a little table
to remind you what they are.
Because I have trouble
remembering myself,
so I enjoy the
opportunity to go back
to remind myself of this stuff.
Let's see.
The secant is one over one of
those things, which one is it?
It's weird, it's
1 over the cosine.
And the cosecant
is 1 over the sin.
Of course the tangent, we know.
It's the sine over the
cosine and the cotangent
is the other way around.
So when you put a
"co" in front of it,
it exchanges sine and cosine.
Well, I have a few identities
involving tangent and secant
up there, in that little
prepared blackboard up above.
Maybe I'll just go
through and check that out
to make sure that we're all
on the same page with them.
So I'm going to
claim that there's
this trig identity at the top.
Secant squared is
1 plus the tangent.
So let's just check that out.
So the secant is
1 over the cosine,
so secant squared is
1 over cosine squared.
And then whenever you
see a 1 in trigonometry,
you'll always have the option
of writing as cos^2 + sin^2.
And if I do that, then I can
divide the cos^2 into that
first term.
And I get 1 + sin^2 / cos^2.
/ ^2.
Which is the tangent squared.
So there you go.
That checks the first one.
That's the main
trig identity that's
going to be behind what
I talk about today.
That's the trigonometry
identity part.
How about this
piece of calculus.
Can we calculate what the
derivative of tan x is?
Actually, I'm going to
do that on this board.
So tan x = sin x / cos x.
So I think I was with
you when we learned
about the quotient rule.
Computing the derivative
of a quotient.
And the rule is, you
take the numerator
and you-- sorry, you take the
derivative of the numerator,
which is cosine.
And you multiply it
by the denominator,
so that gives you cos^2.
And then you take the numerator,
take minus the numerator,
and multiply that by the
derivative of the denominator,
which is -sin x.
And you put all that over the
square of the denominator.
And now I look at that and
before my eyes I see the same
trig identity, cos^2 +
sin^2 = 1, appearing there.
This is 1 / cos^2(x)
which is secant squared.
And, good.
So that's what the claim was.
The derivative of the tangent
is the secant squared.
That immediately
gives you an integral.
Namely, the integral of
secant squared is the tangent.
That's the fundamental
theorem of calculus.
So we verified the
first integral there.
Well, let's just do
the second one as well.
So if I want to
differentiate the secant,
derivative of the secant.
So that's d/dx of
1 over the cosine.
And again, I have a quotient.
This one's a little easier
because the numerator's
so simple.
So I take the derivative of
the numerator, which is 0.
And then I take the
numerator, I take
minus the numerator times the
derivative of the denominator.
Which is -sin x,
and put all that
over the square of
the same denominator.
So one minus sign came
from the quotient rule,
and the other one
came because that's
the derivative of the cosine.
But they cancel, and
so I get sin / cos^2,
which is sin / cos times 1 /
cos and so that's the secant,
that's 1 / cos, times tan x.
So, not hard.
That verifies that the
derivative of secant
is secant tangent.
And it tells you that the
integral of that weird thing
in case you ever want
to know, the integral
of the secant tangent
is the secant.
Well, there are a
couple more integrals
that I want to do for you.
Where I can't sort of
work backwards like that.
Let's calculate the
integral of the tangent.
Just do this straight out.
So the tangent is the sine
divided by the cosine.
And now there's a habit of
mind, that I hope you get into.
When you see the cosine and
you're calculating an integral
like this, it's
useful to remember
what the derivative
of the cosine is.
Because maybe it shows up
somewhere else in the integral.
And that happens here.
So that suggests we make
a substitution. u = cos x.
Which means du = -sin x dx.
That's the numerator,
except for the minus sign.
And so I can rewrite this
as, under the substitution,
I can rewrite this as
-du, that's the numerator,
sin x dx is -du, divided by u.
Well, I know how to
do that integral too.
That gives me the
natural log, doesn't it.
So this is -ln(u)
plus a constant.
I'm not quite done.
I have to back-substitute
and replace
this new variable that I've made
up, called u, with what it is.
And what you get is -ln(cos x).
So the integral of the
tangent is minus log cosine.
Now, you find these
tables of integrals
in the back of the book.
Things like that.
I'm not sure how much
memorization Professor Jerison
is going to ask
of you, but there
is a certain amount
of memorization
that goes on in calculus.
And this is one of
the kinds of things
that you probably want to know.
Let me do one more integral.
I think I'm making my way
through a prepared board here,
let's see.
Good.
So the integral of the
tangent is minus log cosine.
I'd also like to know what the
integral of the secant of x is.
And I don't know a way to
kind of go straight at this,
but let me show you a way to
think your way through to it.
If I take these two facts,
tangent prime is what it is,
and secant prime is what it
is, and add them together,
I get this fact.
That the derivative of
the sec x + tan x is,
well, it's the sum
of these two things.
Secant squared plus
secant tangent.
And there's a secant that
occurs in both of those terms.
So I'll factor it out.
And that gives me,
I'll put it over here.
There's the secant of x
that occurs in both terms.
And then in one term,
there's another secant.
And in the other term,
there's a tangent.
So that's interesting somehow,
because this same term appears
on both sides of this equation.
Let's write u, for
that sec x + tan x.
And so the equation that
I get is u' = u sec x.
I've just made a
direct substitution.
Just decide that
I'm going to write u
for that single
thing that occurs
on both sides of the equation.
So u' is on the left, and
u * sec x is on the right.
Well, there's my secant.
That I was trying to integrate.
And what it tells
you is that = u' / u.
Just divide both sides by u,
and I get this equation. u' / u,
that has a name.
Not sure that
professor Jerison's
used this in this
class, but u' / u,
we've actually used
something like that.
It's on the board right now.
It's a logarithmic derivative.
It is the derivative of the
national logarithm of u.
Maybe it's easier to read
this from right to left,
if I want to calculate the
derivative of the logarithm,
well, the chain rule says I
get the derivative of u times
the derivative of the log
function, which is 1 / u.
So often u' / u is called
the logarithmic derivative.
But it's done what I wanted.
Because it's expressed the
secant as a derivative.
And I guess I should
put in what u is.
It's the secant
plus the tangent.
And so that implies
that the integral--
Integrate both sides.
That says that the integral of
sec x dx, is ln(sec x + tan x).
So that's the last line in this
little memo that I created.
That we can use now for
the rest of the class.
Any questions about that trick?
It's a trick, I have nothing
more to say about it.
OK.
So, the next thing
I-- oh yes, so now
I want to make the point
that using these rules
and some thought,
you can now integrate
most trigonometric polynomials.
Most things that involve
powers of sines and cosines
and tangents and secants
and everything else.
For example, let's try to
integrate the integral of sec^4
x.
Big power of the
secant function.
Well, there are too many
secants there for me.
So let's take some away.
And I can take them away by
using that trig identity,
sec^2 = 1 + tan^2.
So I'm going to replace two
of those secants by 1 + tan^2.
That leaves me
with two left over.
Now there was method
to my madness.
Because I've got a secant
squared left over there.
And secant squared is the
derivative of tangent.
So that suggests a substitution.
Namely, let's say, let's let u =
tan x, so that du = sec^2 x dx.
And I have both terms
that occur in my integral
sitting there very nicely.
So this is the possibility
of making this substitution
and seeing a secant
squared up here as part
of the differential here.
That's why it was a good
idea for me to take two
of the secants and
write them as 1 + tan^2.
So now I can continue this.
Under that
substitution, I get 1.
Oh yeah, and I should add
the other fact, that--
Well I guess it's obvious
that tangent squared is u^2.
So I get 1 + u^2.
And then du-- sec^2
sec^2 x dx, that is du.
Well that's pretty
easy to integrate.
So I get u + u^3 / 3.
Plus a constant.
And then I just have
to back-substitute.
Put things back in terms
of the original variables.
And that gives me tan x
plus tangent cubed over 3.
And there's the answer.
So we could spend
a lot more time
doing more examples of this
kind of polynomial trig thing.
It's probably best for you to
do some practice on your own.
Because I want to talk
about other things, also.
And what I want to
talk about is the use
of these trig identities in
making really trig substitution
integration.
So we did a little
bit of this yesterday,
and I'll show you some
more examples today.
Let's start with a pretty hard
example right off the bat.
So this is going to be the
integral of dx over x^2 times
the square root of 1+x^2.
It's a pretty bad
looking integral.
So how can we approach this?
Well, the square root is the
ugliest part of the integral,
I think.
What we should try to do
is write this square root
in some nicer way.
That is, figure out a way
to write 1+x^2 as a square.
That'll get rid of
the square root.
So there is an example of a
way to write 1 plus something
squared in a different way.
And it's right up there.
sec^2 = 1 + tan^2.
So I want to use that idea.
And when I see this
form, that suggests
that we make a trig
substitution and write
x as the tangent of
some new variable.
Which you might as
well call theta,
to because it's like an angle.
Then 1+x^2 is the
secant squared.
According to that trig identity.
And so the square root
of 1+x^2 is sec(theta).
Right?
So this identity is the
reason that the substitution
is going to help us.
Because it gets rid
of the square root
and replaces it by some
other trig function.
I'd better be able to
get rid of the dx, too.
That's part of the
substitution process.
But we can do that,
because I know
what the derivative
of the tangent is.
It's secant squared.
So dx / d theta is sec^2(theta).
So dx is sec^2(theta) d theta.
So let's just substitute
all of that stuff
in, and rewrite the
entire integral in terms
of our new variable, theta.
So dx is in the numerator.
That's sec^2(theta) d theta.
And then the denominator,
well, it has an x^2.
That's tan^2(theta).
And then there's
this square root.
And we know what that
is in terms of theta.
It's sec(theta).
OK, now. we've done
the trig substitution.
I've gotten rid of
the square root,
I've got everything in
terms of trig functions
of the new variable.
Pretty complicated
trig function.
This often happens, you wind
up with a complete scattering
of different trig functions in
the numerator and denominator
and everything.
A systematic thing to do
here is to put everything
in terms of sines and cosines.
Unless you can see right away,
how it's going to simplify,
the systematic thing
to do is to rewrite
in terms of sines and cosines.
So let's do that.
So let's see.
The secant squared,
secant is 1 over cosine.
So I'm going to put a cosine
squared in the denominator.
Oh, I guess the first
thing I can do is cancel.
Let's do that.
That's clever.
You were all thinking that too.
Cancel those.
So now I just get one
cosine denominator
from the secant there
in the numerator.
It's still pretty complicated,
secant over tangent squared,
who knows.
Well, we'll find out.
Because the tangent
is sine over cosine.
So I should put a sine
squared where the tangent was,
and a cosine squared up there.
And I still have d theta.
And now you see some
more cancellation occurs.
That's the virtue of writing
things out in this way.
So now, the square here
cancels with this cosine.
And I'm left with cos(theta)
d theta / sin^2(theta).
That's a little simpler.
And it puts me in a position to
use the same idea I just used.
I see the sine here.
I might look around
in this integral
to see if its derivative
occurs anywhere.
The differential of
the sine is the cosine.
And so I'm very much inclined
to make another substitution.
Say, u, direct
substitution this time.
And say u is the
cosine of theta.
Because then du-- Oh, I'm sorry.
Say, u is the sine of theta.
Because then du is
cos(theta) d theta.
And then this integral becomes,
well, the numerator just is du.
The denominator is u^2.
And I think we can
break out the champagne,
because we can
integrate that one.
Finally get rid of
the integral sign.
Yes sir.
STUDENT: [INAUDIBLE]
PROFESSOR: OK, how do I know
to make u equal to sine rather
than cosine.
Because I want to see
du appear up here.
If I'd had a sine up here,
that would be a signal to me
that maybe I should say
let u be the cosine.
OK?
Also, because this thing in
the denominator is something
I want to get rid of.
It's in the denominator.
So I'll get rid of it
by wishful thinking
and just call it something else.
It works pretty
well in this case.
Wishful thinking doesn't
always work so well.
So I integrate u^(-2) du, and
I get - -1/u plus a constant,
and I'm done with the
calculus part of this problem.
I've done the integral now.
Gotten rid of the integral sign.
But I'm not quite
done with the problem
yet, because I have
to work my way back
through two substitutions.
First, this one.
And then this one.
So this first substitution isn't
so bad to get rid of, to undo,
to back-substitute.
Because u is just sin(theta).
And so 1/u is, I
guess a fancy way
to write it is the
cosecant of theta.
1 over the sine is the cosecant.
So I get -csc(theta)
plus a constant.
Is there a question in the back?
Yes sir?
STUDENT: [INAUDIBLE]
PROFESSOR: I'm sorry,
my hearing is so bad.
STUDENT: [INAUDIBLE]
PROFESSOR: How did I
know this substitution
in the first place.
It's because of the 1 + x^2.
And I want to make use
of the trig identity
in the upper left-hand corner.
I'll make you a table in a
few minutes that will put
all this in a bigger context.
And I think it'll help you then.
OK, I'll promise.
So, what I want to try
to talk about right
now is how to rewrite
a term like this.
A trig term like this,
back in terms of x.
So I want to undo this
trick substitution.
This is a trig sub.
And what I want to do now is
try to undo that trig sub.
And I'll show you
a general method
for undoing trig substitutions.
This happens quite often.
I don't know what the
cosecant of theta is.
But I do know what the
tangent of theta is.
So I want to make a
relation between them.
OK, so undoing.
Trig subs.
So let's go back to where
trigonometry always comes from,
this right angled triangle.
The theta in the corner,
and then these three sides.
This one's called
the hypotenuse.
This one is called
the adjacent side,
and that one's called
the opposite side.
And now, let's find out where
x lies in this triangle.
Let's try to write the sides
of this triangle in terms of x.
And what I know is, x
is the tangent of theta.
So the tangent of theta,
tangent of this angle,
is opposite divided by adjacent.
Did you learn SOH CAH TOA?
OK, so it's opposite
divided by adjacent.
Is the tangent.
So there are different
ways to do that,
but why not just do
it in the simplest way
and suppose that the adjacent
is 1, and the opposite is x.
This is correct now, isn't it?
I get the correct value
for the tangent of theta
by saying that the lengths
of those are 1 and x.
And that means that the
hypotenuse has length 1 + x^2.
Well, here's a triangle.
I'm interested in computing
the cosecant of theta.
Where's that appear
in the triangle?
Well, let's see.
The cosecant of theta
is 1 over the sine.
And the sine is opposite
over hypotenuse.
So the cosecant is
hypotenuse over opposite.
And the hypotenuse is the
square root of 1 + x^2,
and the opposite is x.
And so I've done it.
I've undone the
trig substitution.
I've figured out what
this cosecant of theta
is, in terms of x.
And so the final answer is
minus the square root of 1+x^2,
over x, plus a constant,
and there's an answer
to the original problem.
This took two boards
to go through this.
I illustrated several things.
Actually, this
three half boards.
I illustrated this use
of trig substitution,
and I'll come back
to that in a second.
I illustrated patience.
I illustrated rewriting things
in terms of sines and cosines,
and then making a
direct substitution
to evaluate an
integral like this.
And then there's this undoing
all of those substitutions.
And it culminated with
undoing the trig sub.
So let's play a game here.
Why don't we play
the game where you
give me-- So, there's a step in
here that I should have done.
I should've said this is
-cos(arctan(theta)) plus
a constant.
The most straightforward
thing you can do
is to say since x is
the tangent of theta,
that means that-- sorry,
if x, that means that theta
is the arctangent of x.
And so let's just put in
theta as the arctangent of x,
and that's what you get.
So really, what I
just did for you
was to show you a way to compute
some trig function applied
to the inverse of
another trig function.
I computed cosecant of the
arctangent by this trick.
So now, let's play the
game where you give me
a trig function and an
inverse trig function,
and I try to compute
what the composite is.
OK.
So who can give me
a trig function.
Has to be one of
these standard ones.
STUDENT: Tan.
PROFESSOR: Tangent.
Alright.
How about another one?
STUDENT: Sine.
PROFESSOR: Sine.
Do we have agreement on sine.
STUDENT: [INAUDIBLE]
PROFESSOR: Secant?
STUDENT: [INAUDIBLE]
PROFESSOR: Right, csc
has the best cheer.
So that's the game.
We have to compute, try to
compute, that composite.
Something wrong with this?
STUDENT: [INAUDIBLE]
PROFESSOR: What does
acceptable mean?
Don't you think--
so the question is,
isn't this a perfectly
acceptable final answer.
It's a correct final answer.
But this is much
more insightful.
And after all the original
thing was involving square roots
and things, this is
the kind of thing
you might hope for is an answer.
This is just a nicer
answer for sure.
And likely to be more
useful to you when you go on
and use that answer
for something else.
OK, so let's try
to do this this.
Undo a trig substitution
that involved a cosecant.
And I manipulate
around, and I find
myself trying to find out
what's the tangent of theta.
So here's how we go about it.
I draw this triangle.
Theta is the angle here.
This is the adjacent,
opposite, hypotenuse.
So, the first thing is how can
I make the cosecant appear here,
csc x.
What dimensions should I
give to the sides in order
for the cosecant of x,
sorry, in order for theta
to be the cosecant of x.
This thing is theta.
So, that means that
the cosecant of x--
that means the cosecant
of theta should be x.
Theta is the arccosecant, so
x is the cosecant of theta.
So, what'll I take the sides
to be, to get the cosecant?
The cosecant is 1 over the sine.
And the sine is the opposite
over the hypotenuse.
So I get hypotenuse
over opposite.
And that's supposed
to be what x is.
So I could make the
opposite anything
I want, but the simplest
thing is to make it 1.
Let's do that.
And then what does that mean
about the rest of the sides?
Hypotenuse had better be x.
And then I've recovered this.
So here's a triangle that
exhibits the correct angle.
This remaining side is
going to be useful to us.
And it is the square
root of x^2 - 1.
So I've got a triangle of
the correct angle theta,
and now I want to compute
the tangent of that angle.
Well, that's easy.
That's opposite
divided by adjacent.
So I get 1 over the
square root of x^2 - 1.
Very flexible tool
that'll be useful to you
in many different times.
Whenever you have to
undo a trig substitution,
this is likely to be useful.
OK, that was a good game.
No winners in this game.
We're all winners.
No losers, we're all winners.
OK.
So, good.
So let me make this table of the
different trig substitutions,
and how they can be useful.
Summary of trig substitutions.
So over here, we
have, if you see,
so if your integrand contains,
make a substitution to get.
So if your integrand
contains, I'll
write these things
out as square roots.
If it contains the
square root a^2 - x^2,
this is what we talked
about on Thursday.
When I was trying to find the
area of that piece of a circle.
There, I suggested that we
should make the substitution
x = a cos(theta).
Or, x = a sin(theta).
Either one works just as well.
And there's no way to
prefer one over the other.
And when you make the
substitution, x = a cos(theta),
you get a^2 - a^2 cos^2(theta).
theta.
1 - cos^2 is sin^2.
So you get a sin(theta).
So this expression becomes
equal to this expression
under that substitution.
And then you go on.
Then you've gotten rid
of the square root,
and you've got a
trigonometric integral
that you have to try to do.
If you made the substitution a
sin(theta), you'd get a a^2 -
a^2 sin^2, which
is a cos(theta).
And then you can
go ahead as well.
We just saw another example.
Namely, if you have a^2 + x^2.
That's like the example we had
up here. a = 1 in this example.
What did we do?
We tried the substitution
x = a tan(theta).
And the reason is that I can
plug into the trig identity
up here in the upper left.
And replace a^2 +
x^2 by a sec(theta).
Square root of the
secant squared.
There's one more
thing in this table.
Sort of, the only
remaining sum or difference
of terms like this.
And that's what happens
if you have x^2 - a^2.
So there, I think we can make
a substitution a sec(theta).
Because, after all,
sec^2(theta)-- so x^2 - a^2--
Sorry.
Let's see what happens when
I make that substitution.
x^2 - a^2 = a^2
sec^2(theta) - a^2.
Under this substitution.
That's sec^2 - 1.
Well, put the 1
on the other side.
And you find tan^2, coming out.
So this is a a^2 tan^2(theta).
And so that's what you
get, a times tan(theta).
After I take the square
root, I get a tan(theta).
So these are the three basic
trig substitution forms.
Where trig substitutions
are useful to get
rid of expressions like
this, and replace them
by trigonometric expressions.
And then you use this trick,
you do the integral if you can
and then you use this trick
to get rid of the theta
at the end.
So now, the last thing I
want to talk about today
is called completing the square.
And that comes in
because unfortunately,
not every square
root of a quadratic
has such a simple form.
You will often encounter
things that are not just
the square root of
something simple.
Like one of these forms.
Like there might be a
middle term in there.
I don't actually
have time to show you
an example of how this comes out
in a sort of practical example.
But it does happen
quite frequently.
And so I want to show you
how to deal with things
like the following example.
Let's try to integrate
dx over x^2 + 4x,
the square root of x^2 + 4x.
So there's a square root of
some square, some quadratic.
It's very much
like this business.
But it isn't of
any of these forms.
And so what I want
to do is show you
how to rewrite it in one of
those forms using substitution,
again.
All this is about substitution.
So the game is to rewrite
quadratic as something
like x plus something or other.
Plus some other constant.
So write it, try to write
it, in the form of a square
plus or minus another constant.
And then we'll go on from there.
So let's do that in this
case. x^2, x^2 + 4x.
Well, if you square
this form out,
then the middle term
is going to be 2ax.
So that, since I have
a middle term here,
I pretty much know
what a has to be.
The only choice in order to
get something like x^2 + 4x out
of this, is to take a to be 2.
Because then, this
is what you get.
This isn't quite right yet, but
let's compute what I have here.
x^2 + 4x, so far
so good, plus 4,
and I don't have a plus 4 here.
So I have to fix that
by subtracting 4.
So that's what I mean.
I've completed the square.
The word for this process of
eliminating the middle term
by using the square of
an expression like that.
That's called
completing the square.
And we can use that process
to compute this integral.
So let's do that.
So I can rewrite this integral,
rewrite this denominator
like this.
And then I'm going to try to use
one of these forms over here.
So in order to get a
single variable there,
instead of something
complicated like x + 2,
I'm inclined to come up
with another variable name
and write it equal, write x +
2 as that other variable name.
So here's another little
direct substitution. u = x + 2.
Figure out what du is.
That's pretty easy.
And then rewrite the
integral in those terms.
So dx = du.
And then in the
denominator I have,
well, I have the
square root of that.
Oh yeah, so I think as part of
this I'll write out what x^2 +
4x is.
The point is, it's
equal to u^2 - 4.
x^2 + 4x = u^2 - 4.
There's the data box containing
the substitution data.
And so now I can put that in.
I have x^2 + 4x there.
In terms of u, that's u^2 - 4.
Well, now I'm in a happier
position because I can look
for u^2 - 4 for something
like that in my table here.
And it actually sits down here.
So except for the
use of the letter x
here instead of u over there.
That tells me what I want.
So to handle this, what I should
use is a trig substitution.
And the trig substitution
that's suggested is,
according to the bottom
line with a = 2, so a^2 = 4.
The suggestion is,
I should take x--
But I'd better not use
the letter x any more.
But I don't have a letter
x, I have the letter u.
I should take u
equal to 2 secant.
And then some letter
I haven't used before.
And theta is available.
This is a look-up table process.
I see the square
root of u^2 - 4,
I see that that's of this form.
I'm instructed to make
this substitution.
And that's what I just did.
Let's see how it works out.
So that means the du is 2, OK.
What's the derivative
of the secant?
Secant tangent.
So du = 2 sec(theta) tan(theta).
And u^2 - 4 is,
here's the payoff.
I'm supposed to
be able to rewrite
that in terms of the tangent.
According to this. u u^2 - 4
is 4 secant squared minus 4.
And secant squared minus
1 is tangent squared.
So this is 4 tan^2(theta).
Right, yeah?
STUDENT: [INAUDIBLE]
PROFESSOR: But I squared it.
And now I'll square root it.
And I'll get a 2 and this
tangent will go away.
So there's my data box
for this substitution.
And let's go on
to another board.
So where I'm at is the integral
of du over the square root
of u^2 - 4.
And I have all the
data I need here
to rewrite that
in terms of theta.
So du = 2 sec(theta)
tan(theta) d theta.
And the denominator
is 2 tan(theta).
Ha.
Well, so some very nice
simplification happens here.
The 2's cancel.
And the tangents cancel.
And I'm left with trying to work
with the integral, sec(theta) d
theta.
And luckily enough at the
very beginning of the hour,
I worked out how to
compute the integral
of the secant of theta.
And there it is.
So this is ln(sec(theta) +
tan(theta)) plus a constant.
And we're done with
the calculus part.
There's no more integral there.
But I still am not quite
done with the problem,
because again I have these two
substitutions to try to undo.
So let's undo them one by one.
Let's see.
I have this trig
substitution here.
And I could use my triangle
trick, if I need to.
But maybe I don't need to.
Let's see, do I know
what the secant of theta
is in terms of u?
Well, I do.
So I get ln(u/2).
Do I know what the
tangent is in terms of u?
Well, I do.
It's here.
So I lucked out, in this case.
And I don't have to go through
and use that triangle trick.
So the tangent of theta is the
square root of u^2 - 4, over 2.
Good.
So I've undone this
trig substitution.
I'm not quite done yet because
my answer is involved with u.
And what I wanted
originally was x.
But this direct substitution
that I started with
is really easy to deal with.
I can just put x + 2
every time I see a u.
So this is the natural logarithm
of (x+2)/2 plus the square
root...
What's going to happen when I
put x + 2 in place for u here?
You know what you get.
You get exactly what
we started with.
Right?
I put x + 2 in
place of the u here.
I get x^2 + 4x.
So I've gotten back
to a function purely
in terms of x OK, that's
a good place to quit.
Have a great little
one-day break.
I guess this class doesn't
meet on Monday anyway.
Bye.
