Hi.
Welcome back to recitation.
I, today I wanted to teach you a variation on trig
substitution.
So this is called hyperbolic trigonometric substitution.
And I'm going to teach you it just by going through a nice
example of a question where it turns out to be useful.
So the question is the following.
Compute the area of the region below.
So what's the region?
Well, I have here the hyperbola, x squared minus y
squared equals 1.
And I've chosen a point on the hyperbola whose coordinates
are cosh t sinh t.
So remember that cosh t is a hyperbolic cosine, and sinh t
is the hyperbolic sine.
And they're given by the formulas cosh t equals e to
the t plus e to the minus t over 2, and sinh t equals e to
the t minus e to the minus t over 2.
So and we saw in an earlier recitation video that this
point, cosh t, sinh t, is a point on the right branch of
this hyperbola.
So I've got a region.
So I've got the hyperbola, I've got that point.
I've drawn a straight line between the
origin and that point.
So the region that I want you to find the area of is this
region here.
So it's the region below that line segment, above the
x-axis, and to the left of this branch of the hyperbola.
Now, I'm not going to, I'm not going to just ask you to do
that alone, because there's this technique that
I want you to use.
So let's start setting up the integral together, and then
I'll describe the technique and give you a chance to work
it out yourself, to work out the problem yourself.
So from looking at this region--so let's think about
computing the area of this region.
There are two ways we could split it up, right?
We could cut it up into vertical rectangles and
integrate with respect to x, or we could cut it up into
horizontal rectangles and integrate with respect to y.
Now, if we cut it up into vertical rectangles, our life
is a little complicated, because we have to cut the
region into 2 pieces here.
Right?
There's the, so this is the point 1, 0, where the
hyperbola crosses x-axis.
So over here, you know, the top part is the line segment
and the bottom part is the x-axis, and then over here,
the top part is the line segment and the bottom part is
the hyperbola.
So life is complicated if we use vertical rectangles.
It's a little bit simpler if we use horizontal rectangles,
so let's go with that.
You know, the amount of work will be similar either way,
but I like this way, lets you right down in one integral.
So the area--
OK.
So if we use horizontal rectangles to compute the area
of a region, then the area is, we need to integrate from the
bottom to the top, whatever the, you know,
the bounds on y are.
And then the thing we have to integrate has to be the area
of one of those little rectangles.
So the area of the little rectangle, its height is dy,
and its length, or width, I guess, is the x-coordinate of
its rightmost point, minus the x coordinate of
its leftmost point.
So OK.
So we need to integrate from the lowest value of y to the
highest value.
So we start at the bottom at y equals 0, and we need to go
all the way up to the top here, which is
y equals sinh t.
So it's an integral from 0 to sinh t.
OK.
And so we need the x-coordinate on the right
here, minus the x-coordinate on the left.
So what's the x-coordinate on the right?
Well, we need to solve this equation for x in terms of y.
Right?
This is going to be an integral with respect to y.
So that if we solve for this point, we get x is equal to
the square root of y squared plus 1.
So the right coordinate is the square root of y
squared plus 1.
And the left coordinate, this is just a straight line
passing through the origin, so its equation is y equals sinh
t over cosh t times x, or x equals cosh t over
sinh t times y.
So this is minus cosh t over sinh t times y.
And we're integrating with respect to y.
OK.
So this is the integral that we're interested in.
This integral gives us the area.
And just a couple of things to notice about it.
So t is a constant.
It's just fixed.
So cosh t over sinh t, which we could also, if we wanted
to, this is the hyperbolic cotangent.
But that's really not important at all.
But we could call it that, if we wanted to.
So this is just a constant.
So we have a minus a constant times y dy.
So this part's easy to integrate.
So the hard part is going to be integrating this y
squared plus 1.
Now, one thing you've seen is that when you have a y
squared, a square-root of, when you have a y squared plus
1, one substitution that sometimes works is a tangent
substitution.
And the reason a tangent substitution works, is that
you have a trig identity, tan squared plus 1
equals secant squared.
In this case, I'd like to suggest a different
substitution.
All right?
So this integral is the integral that you want.
And I'd like to suggest a substitution, which is that
you use a hyperbolic trig function as the thing that you
substitute.
So in particular, instead of using, instead of relying on
the trig identity, tan squared plus 1 equals secant squared,
you can use the hyperbolic trig identity, which is that
sinh squared u plus 1 equals cosh squared u.
So this identity--
so here we have a something squared plus 1 equals
something squared.
So the identity this suggests is to try the substitution y
equals sinh u.
All right?
So this is a hyperbolic trig substitution.
So why don't you take that hint, try it out on this
integral, see how it goes.
Take some time, pause the video, work it out, come back
and we can work it out together.
Welcome back.
Hopefully you had some luck solving this integral using a
hyperbolic trig substitution.
Let's work it out together, see if my answer matches the
one that you came up with.
So as I said before, this integral comes in two parts.
There's the hard part, the square root of y squared plus
1 part, and there's the easy part, this y part.
So before I make the substitution, let me just deal
with the easy part.
So I'll do that over here.
So we have the one part of the area, or one part of the
integral, really, is the integral from 0 to sinh t of
minus cosh t over sinh t times y dy.
OK.
So this is just a constant, so it's a constant times y.
So this is equal to--
well, it's the same constant comes along, minus cosh t over
sinh t, times y squared over 2, for y between 0 and this
upper bound, sinh t.
So that's equal--OK.
So when y is 0, this is just 0.
So this equals minus cosh t sinh t over 2.
So that's the easy part of integral.
So in order to compute the total area, we need to add
this expression that we just computed to the integral of
this first part.
So that's what we need to compete next.
And that's what we're going to use the hyperbolic trig
substitution on.
So we're going to complete the integral from 0 to sinh t of
the square root of y squared plus 1 dy.
And we're going to use the substitution sinh u equals y,
or y equals sinh u.
OK.
So we need, what do I need?
I need what dy is, and I need to change the bounds.
So dy--
I'm sorry, I'm going to flip this around to take the-- so
dy is, I need the differential of sine u, sorry of sinh u.
And so we saw in the earlier hyperbolic trig function
recitation that that's cosh u du, or if you like, you could
just differentiate using the formulas that we know
for sinh and cosh.
And we need bounds.
So when y is 0, we need sinh of something is 0.
And so it happens that that value is 0.
So if you remember the graph of the function, or you can
just check in the formula, when sinh is 0, when t is 0,
that's when you get sinh is 0.
It's the only time e to the t equals e to the minus t.
OK.
So when y is 0, then u is 0, and when y is sinh
t, then u is t.
Right?
Because sinh u is sinh t.
So under the substitution, this becomes the integral from
0 to t now, from u equals 0 to t, of--
well, OK.
So this becomes the square root of sinh squared u plus 1,
and then dy is times cosh u du.
OK.
Now the reason we made this substitution in the first
place is that this, we can use a hyperbolic
trig identity here.
So sinh squared u plus 1 is just cosh squared u, and
square root of cosh squared u is cosh u.
Remember that cosh u is positive, so we don't have to
worry about an absolute value here.
So this is the integral from 0 to t of cosh squared u du.
OK.
So at this point, there are a couple of different
things you can do.
One is that you can, just like when we have certain trig
identities, we have corresponding hyperbolic trig
identities that we could try out here.
So we could try something like that.
Another thing you can do, is you can just go back to the
formula, right?
Cosh t has a simple formula in terms of exponentials, so you
can go back to this formula and you can plug in.
So let's just try that quickly, because that's a sort
of easy way to handle this.
So this is cosh squared u du.
So I'm going to write--
OK.
Carry that all the way up here.
So this is the integral from 0 to t.
Well, if you take the formula for hyperbolic cosine and
square it, what you get, I'm going to do this all in one
step, is you e to the 2u plus 2 plus e to the
minus 2u over 4 du.
OK.
And so now this is, once you've replaced everything
with exponentials, this is easy to integrate.
This is so e to the 2u, the integral is e to the 2u over
2, so that comes over 8.
2 over 4, you integrate that, and that's just 2u over 4,
which is u over 2.
And now the last one is minus e to the minus 2u over 8
between 0 and t.
OK.
So now we take the difference here.
At t, we get e to the 2t over 8 plus t over 2 minus e to the
minus 2t over 8.
Minus--
OK.
And when u is equal, so that was at u equals t.
At u equals 0, we get 1/8 plus 0 minus 1/8.
So that's just 0.
OK.
So this is what we got for that part of the integral.
So OK, so we've now split the integral into two pieces.
We computed one piece, because it was just easy, we're
integrating a polynomial.
We computed the other piece, which was more complicated,
using a hyperbolic trig substitution.
The whole integral is the sum of those two pieces.
So now the whole integral, I have to take this piece, and I
have to add it to the thing that I computed for the other
piece before, which was somewhere--
where did it go?
Here it is, right here.
Which was minus cosh t sinh t over 2.
OK.
So I'm going to save you a little arithmetic, and I'm
going to observe that minus cosh t sinh t over 2 is
exactly equal to the minus e to the 2t over 8, plus e to
the minus 2t over 8.
So adding these two expressions
together gives us--
so the first expression, minus cosh t sinh t over 2, is minus
e to the 2t over 8 plus e to the minus 2t over 8, plus--
OK.
Plus what we've got right here, which is e to the 2t
over 8 minus e to the minus 2t over 8 plus t over 2.
And that's exactly equal to t over 2.
OK.
So this is the area of that sort of hyperbolic triangle
thing that we started out with at the beginning.
So let me just walk back over there for a second.
So we used this hyperbolic trig substitution in order to
compute that the area of this triangle is t over 2.
And I just want to--
first of all, I want to observe that that's a really
nice answer.
So that's kind of cool.
The other thing that I want to observe is that this is a very
close analogy with something that happens in the case of
regular circle trigonometric functions.
Which is, if you look at a regular circle, and you take
the point cosine theta comma sine theta, then the area of
this little triangle here is theta over 2.
So in this case, u doesn't measure an angle, but it does
measure an area in exactly the same way that
theta measures an area.
So there's a really cool relationship there between the
hyperbolic trig function and the regular trig function.
So that's just a kind of cool fact.
The useful piece of knowledge that you can extract from what
we've just done, though, is that you can use this
hyperbolic trig substitution in integrals of certain forms.
So in the same way that trig substitutions are suggested by
certain forms of the integrand, hyperbolic trig
substitutions are also suggested by certain forms of
the integrand, and often you have a choice about
which one to use.
And in this particular instance, a hyperbolic trig
substitution worked out quite nicely.
Much more nicely than a trig substitution
would have worked out.
So it's just another tool for your toolbox.
I'll end with that.