WEBVTT
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Hi.
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Welcome back to recitation.
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I, today I wanted to teach you a
variation on trig substitution.
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So this is called hyperbolic
trigonometric substitution.
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And I'm going to
teach you it just
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by going through a nice
example of a question
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where it turns out to be useful.
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So the question
is the following.
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Compute the area of
the region below.
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So what's the region?
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Well, I have here the hyperbola
x squared minus y squared
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equals 1.
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And I've chosen a point on the
hyperbola whose coordinates are
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(cosh t, sinh t).
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So remember that cosh t
is a hyperbolic cosine,
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and sinh t is the
hyperbolic sine.
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And they're given by the
formulas cosh t equals
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e to the t plus e to
the minus t over 2,
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and sinh t equals e to the t
minus e to the minus t over 2.
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So and we saw in an
earlier recitation video
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that this point,
(cosh t, sinh t),
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is a point on the right
branch of this hyperbola.
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So I've got a region.
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So I've got the hyperbola,
I've got that point.
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I've drawn a straight
line between the origin
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and that point.
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So the region that I want
you to find the area of
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is this region here.
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So it's the region below that
line segment, above the x-axis,
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and to the left of this
branch of the hyperbola.
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Now, I'm not going
to, I'm not going
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to just ask you
to do that alone,
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because there's this technique
that I want you to use.
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So let's start setting
up the integral together,
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and then I'll
describe the technique
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and give you a chance to work
it out yourself, to work out
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the problem yourself.
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So from looking at
this region-- so let's
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think about computing
the area of this region.
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There are two ways we
could split it up, right?
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We could cut it up into
vertical rectangles
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and integrate with
respect to x, or we
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could cut it up into
horizontal rectangles
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and integrate with respect to y.
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Now, if we cut it up
into vertical rectangles,
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our life is a
little complicated,
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because we have to cut the
region into 2 pieces here.
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Right?
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There's the-- so this
is the point (1, 0),
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where the hyperbola
crosses x-axis.
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So over here, you know, the
top part is the line segment
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and the bottom
part is the x-axis,
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and then over here, the top
part is the line segment
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and the bottom part
is the hyperbola.
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So life is complicated if
we use vertical rectangles.
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It's a little bit simpler if
we use horizontal rectangles,
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so let's go with that.
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You know, the
amount of work will
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be similar either way,
but I like this way,
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lets you right down
in one integral.
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So the area-- OK.
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So if we use horizontal
rectangles to compute
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the area of a
region, then the area
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is, we need to integrate
from the bottom to the top,
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whatever the, you know,
the bounds on y are.
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And then the thing
we have to integrate
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has to be the area of one
of those little rectangles.
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So the area of the little
rectangle, its height is dy,
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and its length,
or width, I guess,
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is the x-coordinate of
its rightmost point,
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minus the x coordinate
of its leftmost point.
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So OK.
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So we need to integrate
from the lowest value of y
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to the highest value.
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So we start at the
bottom at y equals 0,
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and we need to go all
the way up to the top
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here, which is y equals sinh t.
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So it's an integral
from 0 to sinh t.
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OK.
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And so we need the
x-coordinate on the right here,
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minus the x-coordinate
on the left.
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So what's the
x-coordinate on the right?
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Well, we need to solve this
equation for x in terms of y.
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Right?
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This is going to be an
integral with respect to y.
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So that if we solve
for this point,
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we get x is equal to the square
root of y squared plus 1.
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So the right coordinate
is the square root
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of y squared plus 1.
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And the left
coordinate, this is just
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a straight line passing
through the origin,
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so its equation is y equals
sinh t over cosh t times x,
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or x equals cosh t
over sinh t times y.
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So this is minus cosh
t over sinh t times y.
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And we're integrating
with respect to y.
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OK.
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So this is the integral
that we're interested in.
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This integral gives us the area.
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And just a couple of
things to notice about it.
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So t is a constant.
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It's just fixed.
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So cosh t over sinh t, which
we could also, if we wanted to,
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this is the
hyperbolic cotangent.
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But that's really
not important at all.
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But we could call it
that, if we wanted to.
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So this is just a constant.
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So we have minus a
constant times y dy.
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So this part's
easy to integrate.
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So the hard part is
going to be integrating
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this y squared plus 1.
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Now, one thing you've seen is
that when you have a y squared,
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a square root of-- when
you have y y squared
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plus 1, one substitution
that sometimes works
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is a tangent substitution.
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And the reason a tangent
substitution works,
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is that you have a trig
identity, tan squared
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plus 1 equals secant squared.
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In this case, I'd
like to suggest
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a different substitution.
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All right?
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So this integral is the
integral that you want.
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And I'd like to suggest
a substitution, which
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is that you use a hyperbolic
trig function as the thing
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that you substitute.
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So in particular, instead
of using, instead of
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relying on the trig
identity tan squared
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plus 1 equals
secant squared, you
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can use the hyperbolic
trig identity,
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which is that sinh squared u
plus 1 equals cosh squared u.
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So this identity-- so
here we have a something
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squared plus 1 equals
something squared.
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So the identity this suggests
is to try the substitution y
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equals sinh u.
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All right?
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So this is a hyperbolic
trig substitution.
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So why don't you take that hint,
try it out on this integral,
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see how it goes.
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Take some time, pause the
video, work it out, come back
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and we can work it out together.
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Welcome back.
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Hopefully you had some luck
solving this integral using
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a hyperbolic trig substitution.
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Let's work it out together,
see if my answer matches
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the one that you came up with.
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So as I said before, this
integral comes in two parts.
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There's the hard
part, the square root
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of y squared plus
1 part, and there's
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the easy part, this y part.
00:07:13.620 --> 00:07:15.550
So before I make
the substitution,
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let me just deal
with the easy part.
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So I'll do that over here.
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So we have the one
part of the area,
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or one part of the
integral, really,
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is the integral from 0 to sinh
t of minus cosh t over sinh t
00:07:39.970 --> 00:07:42.131
times y dy.
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OK.
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So this is just a constant,
so it's a constant times y.
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So this is equal to-- well, it's
the same constant comes along,
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minus cosh t over sinh t,
times y squared over 2,
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for y between 0 and this
upper bound, sinh t.
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So that's equal-- OK.
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So when y is 0, this is just 0.
00:08:08.760 --> 00:08:18.020
So this equals minus
cosh t sinh t over 2.
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So that's the easy
part of integral.
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So in order to compute
the total area,
00:08:23.500 --> 00:08:25.800
we need to add this
expression that we just
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computed to the integral
of this first part.
00:08:29.240 --> 00:08:31.070
So that's what we
need to compute next.
00:08:31.070 --> 00:08:33.319
And that's what we're going
to use the hyperbolic trig
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substitution on.
00:08:34.160 --> 00:08:38.770
So we're going to compute
the integral from 0
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to sinh t of the square
root of y squared plus 1 dy.
00:08:46.190 --> 00:08:51.245
And we're going to use
the substitution sinh
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u equals y, or y equals sinh u.
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OK.
00:08:58.260 --> 00:09:00.520
So we need, what do I need?
00:09:00.520 --> 00:09:05.070
I need what dy is, and I
need to change the bounds.
00:09:05.070 --> 00:09:08.210
So dy-- I'm sorry,
I'm going to flip
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this around to take
the-- so dy is,
00:09:11.940 --> 00:09:15.630
I need the differential of
sine u-- sorry, of sinh u.
00:09:15.630 --> 00:09:19.540
And so we saw in the earlier
hyperbolic trig function
00:09:19.540 --> 00:09:23.300
recitation that that's
cosh u du, or if you like,
00:09:23.300 --> 00:09:26.160
you could just differentiate
using the formulas
00:09:26.160 --> 00:09:28.500
that we know for sinh and cosh.
00:09:28.500 --> 00:09:32.200
And we need bounds.
00:09:32.200 --> 00:09:36.350
So when y is 0, we need
sinh of something is 0.
00:09:36.350 --> 00:09:40.220
And so it happens
that that value is 0.
00:09:40.220 --> 00:09:42.213
So if you remember the
graph of the function,
00:09:42.213 --> 00:09:43.962
or you can just check
in the formula, when
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sinh is 0, when t is 0,
that's when you get sinh is 0.
00:09:48.750 --> 00:09:51.780
It's the only time e to the
t equals e to the minus t.
00:09:51.780 --> 00:09:52.280
OK.
00:09:52.280 --> 00:09:59.870
So when y is 0, then u is 0, and
when y is sinh t, then u is t.
00:09:59.870 --> 00:10:00.370
Right?
00:10:00.370 --> 00:10:02.040
Because sinh u is sinh t.
00:10:02.040 --> 00:10:07.270
So under the substitution, this
becomes the integral from 0
00:10:07.270 --> 00:10:11.010
to t now, from u equals
0 to t, of-- well, OK.
00:10:11.010 --> 00:10:21.210
So this becomes the square
root of sinh squared u plus 1,
00:10:21.210 --> 00:10:30.161
and then dy is times cosh u du.
00:10:30.161 --> 00:10:30.660
OK.
00:10:30.660 --> 00:10:34.690
Now the reason we made this
substitution in the first place
00:10:34.690 --> 00:10:39.920
is that this, we can use a
hyperbolic trig identity here.
00:10:39.920 --> 00:10:44.990
So sinh squared u plus 1
is just cosh squared u,
00:10:44.990 --> 00:10:47.200
and square root of cosh
squared u is cosh u.
00:10:47.200 --> 00:10:48.640
Remember that cosh
u is positive,
00:10:48.640 --> 00:10:50.931
so we don't have to worry
about an absolute value here.
00:10:50.931 --> 00:11:02.980
So this is the integral from
0 to t of cosh squared u du.
00:11:02.980 --> 00:11:03.480
OK.
00:11:03.480 --> 00:11:05.920
So at this point, there are
a couple of different things
00:11:05.920 --> 00:11:07.380
you can do.
00:11:07.380 --> 00:11:11.980
One is that you can, just
like when we have certain trig
00:11:11.980 --> 00:11:15.280
identities, we have
corresponding hyperbolic trig
00:11:15.280 --> 00:11:19.030
identities that we
could try out here.
00:11:19.030 --> 00:11:21.140
So we could try
something like that.
00:11:21.140 --> 00:11:22.911
Another thing you can
do, is you can just
00:11:22.911 --> 00:11:24.160
go back to the formula, right?
00:11:24.160 --> 00:11:26.740
Cosh t has a simple formula
in terms of exponentials,
00:11:26.740 --> 00:11:29.520
so you can go back to this
formula and you can plug in.
00:11:29.520 --> 00:11:33.860
So let's just try that
quickly, because that's
00:11:33.860 --> 00:11:36.090
a sort of easy way
to handle this.
00:11:36.090 --> 00:11:39.470
So this is cosh squared u du.
00:11:39.470 --> 00:11:44.670
So I'm going to write-- OK.
00:11:44.670 --> 00:11:46.230
Carry that all the way up here.
00:11:46.230 --> 00:11:48.720
So this is the
integral from 0 to t.
00:11:48.720 --> 00:11:52.200
Well, if you take the
formula for hyperbolic cosine
00:11:52.200 --> 00:11:54.475
and square it, what
you get, I'm going
00:11:54.475 --> 00:12:03.090
to do this all in one step,
is you e to the 2u plus 2 plus
00:12:03.090 --> 00:12:14.070
e to the minus 2u over 4 du.
00:12:14.070 --> 00:12:14.570
OK.
00:12:14.570 --> 00:12:17.250
And so now this is, once
you've replaced everything
00:12:17.250 --> 00:12:20.000
with exponentials, this
is easy to integrate.
00:12:20.000 --> 00:12:26.340
This is-- so e to the 2u, the
integral is e to the 2u over 2,
00:12:26.340 --> 00:12:28.410
so that comes over 8.
00:12:28.410 --> 00:12:31.770
2 over 4, you integrate
that, and that's
00:12:31.770 --> 00:12:36.100
just 2u over 4,
which is u over 2.
00:12:36.100 --> 00:12:40.410
And now the last one
is minus e to the minus
00:12:40.410 --> 00:12:47.270
2u over 8 between 0 and t.
00:12:47.270 --> 00:12:48.490
OK.
00:12:48.490 --> 00:12:50.860
So now we take the
difference here.
00:12:50.860 --> 00:13:01.420
At t, we get e to the 2t
over 8 plus t over 2 minus
00:13:01.420 --> 00:13:05.910
e to the minus 2t over 8.
00:13:08.630 --> 00:13:09.450
Minus-- OK.
00:13:09.450 --> 00:13:13.150
And when u is equal-- so
that was at u equals t.
00:13:13.150 --> 00:13:18.360
At u equals 0, we get
1/8 plus 0 minus 1/8.
00:13:18.360 --> 00:13:20.310
So that's just 0.
00:13:20.310 --> 00:13:21.560
OK.
00:13:21.560 --> 00:13:26.230
So this is what we got for
that part of the integral.
00:13:26.230 --> 00:13:30.010
So OK, so we've now split
the integral into two pieces.
00:13:30.010 --> 00:13:32.020
We computed one piece,
because it was just easy,
00:13:32.020 --> 00:13:33.250
we're integrating a polynomial.
00:13:33.250 --> 00:13:35.110
We computed the other piece,
which was more complicated,
00:13:35.110 --> 00:13:36.651
using a hyperbolic
trig substitution.
00:13:36.651 --> 00:13:38.760
The whole integral is the
sum of those two pieces.
00:13:38.760 --> 00:13:41.310
So now the whole integral,
I have to take this piece,
00:13:41.310 --> 00:13:42.800
and I have to add
it to the thing
00:13:42.800 --> 00:13:44.540
that I computed
for the other piece
00:13:44.540 --> 00:13:47.990
before, which was
somewhere-- where did it go?
00:13:47.990 --> 00:13:49.530
Here it is, right here.
00:13:49.530 --> 00:13:54.530
Which was minus cosh
t sinh t over 2.
00:13:54.530 --> 00:13:55.660
OK.
00:13:55.660 --> 00:13:58.560
So I'm going to save
you a little arithmetic,
00:13:58.560 --> 00:14:02.080
and I'm going to observe that
minus cosh t sinh t over 2
00:14:02.080 --> 00:14:08.050
is exactly equal to the
minus e to the 2t over 8,
00:14:08.050 --> 00:14:11.310
plus e to the minus 2t over 8.
00:14:11.310 --> 00:14:18.340
So adding these two
expressions together gives us--
00:14:18.340 --> 00:14:22.020
so the first expression,
minus cosh t sinh t over 2,
00:14:22.020 --> 00:14:26.805
is minus e to the
2t over 8 plus e
00:14:26.805 --> 00:14:31.540
to the minus 2t
over 8, plus-- OK.
00:14:31.540 --> 00:14:33.960
Plus what we've got
right here, which
00:14:33.960 --> 00:14:41.320
is e to the 2t over 8 minus
e to the minus 2t over 8
00:14:41.320 --> 00:14:43.170
plus t over 2.
00:14:43.170 --> 00:14:45.120
And that's exactly
equal to t over 2.
00:14:49.711 --> 00:14:50.210
OK.
00:14:50.210 --> 00:14:53.910
So this is the area of that sort
of hyperbolic triangle thing
00:14:53.910 --> 00:14:55.660
that we started out
with at the beginning.
00:14:55.660 --> 00:14:58.250
So let me just walk back
over there for a second.
00:14:58.250 --> 00:15:02.630
So we used this hyperbolic
trig substitution
00:15:02.630 --> 00:15:04.800
in order to compute that
the area of this triangle
00:15:04.800 --> 00:15:05.855
is t over 2.
00:15:05.855 --> 00:15:07.616
And I just want
to-- first of all,
00:15:07.616 --> 00:15:09.740
I want to observe that
that's a really nice answer.
00:15:09.740 --> 00:15:11.080
So that's kind of cool.
00:15:11.080 --> 00:15:12.663
The other thing that
I want to observe
00:15:12.663 --> 00:15:15.220
is that this is a very close
analogy with something that
00:15:15.220 --> 00:15:20.460
happens in the case of regular
circle trigonometric functions.
00:15:20.460 --> 00:15:24.840
Which is, if you look
at a regular circle,
00:15:24.840 --> 00:15:31.340
and you take the point cosine
theta comma sine theta,
00:15:31.340 --> 00:15:38.360
then the area of this little
triangle here is theta over 2.
00:15:38.360 --> 00:15:41.390
So in this case, u
doesn't measure an angle,
00:15:41.390 --> 00:15:44.200
but it does measure an area
in exactly the same way
00:15:44.200 --> 00:15:45.420
that theta measures an area.
00:15:45.420 --> 00:15:46.980
So there's a really
cool relationship
00:15:46.980 --> 00:15:49.010
there between the
hyperbolic trig function
00:15:49.010 --> 00:15:50.710
and the regular trig function.
00:15:50.710 --> 00:15:52.460
So that's just a
kind of cool fact.
00:15:52.460 --> 00:15:55.150
The useful piece of
knowledge that you
00:15:55.150 --> 00:15:57.270
can extract from
what we've just done,
00:15:57.270 --> 00:16:01.020
though, is that you can use this
hyperbolic trig substitution
00:16:01.020 --> 00:16:03.040
in integrals of certain forms.
00:16:03.040 --> 00:16:05.800
So in the same way that trig
substitutions are suggested
00:16:05.800 --> 00:16:07.340
by certain forms
of the integrand,
00:16:07.340 --> 00:16:09.530
hyperbolic trig
substitutions are also
00:16:09.530 --> 00:16:11.380
suggested by certain
forms of the integrand,
00:16:11.380 --> 00:16:13.505
and often you have a choice
about which one to use.
00:16:13.505 --> 00:16:16.640
And in this particular instance,
a hyperbolic trig substitution
00:16:16.640 --> 00:16:18.600
worked out quite nicely.
00:16:18.600 --> 00:16:20.580
Much more nicely than
a trig substitution
00:16:20.580 --> 00:16:22.240
would have worked out.
00:16:22.240 --> 00:16:25.380
So it's just another
tool for your toolbox.
00:16:25.380 --> 00:16:26.620
I'll end with that.