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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

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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 by going through a nice

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example of a question 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

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squared equals 1.

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And I've chosen a point on the
hyperbola whose coordinates

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are cosh t sinh t.

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So remember that cosh t is a
hyperbolic cosine, and sinh t

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is the hyperbolic sine.

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And they're given by the
formulas cosh t equals e to

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the t plus e to the minus t over
2, and sinh t equals e to

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the t minus e to the
minus t over 2.

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So and we saw in an earlier
recitation video that this

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point, cosh t, sinh t, is a
point on the right branch of

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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

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origin and that point.

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So the region that I want you
to find the area of is this

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region here.

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So it's the region below that
line segment, above the

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x-axis, and to the left of this
branch of the hyperbola.

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Now, I'm not going to, I'm not
going to just ask you to do

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that alone, because there's
this technique that

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I want you to use.

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So let's start setting up the
integral together, and then

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I'll describe the technique and
give you a chance to work

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it out yourself, to work out
the problem yourself.

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So from looking at this
region--so let's think about

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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 and

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integrate with respect to x,
or we could cut it up into

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horizontal rectangles and
integrate with respect to y.

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Now, if we cut it up into
vertical rectangles, our life

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is a little complicated, because
we have to cut the

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region into 2 pieces here.

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Right?

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There's the, so this is the
point 1, 0, where the

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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, and then over here,

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the top part is the line segment
and the bottom part is

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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 be similar either way,

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but I like this way, lets you
right down in one integral.

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So the area--

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OK.

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So if we use horizontal
rectangles to compute the area

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of a region, then the area is,
we need to integrate from the

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bottom to the top, whatever
the, you know,

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the bounds on y are.

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And then the thing we have to
integrate has to be the area

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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, is the x-coordinate of

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its rightmost point, minus
the x coordinate of

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its leftmost point.

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So OK.

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So we need to integrate from the
lowest value of y to the

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highest value.

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So we start at the bottom at y
equals 0, and we need to go

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all the way up to the
top here, which is

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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

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here, 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, we get x is equal to

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the square root of
y squared plus 1.

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So the right coordinate is
the square root of y

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squared plus 1.

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And the left coordinate, this
is just a straight line

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passing through the origin, so
its equation is y equals sinh

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t over cosh t times x, or
x equals cosh t over

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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

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to, 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 a 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 this y

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squared plus 1.

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Now, one thing you've seen
is that when you have a y

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squared, a square-root of, when
you have a y squared plus

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1, one substitution that
sometimes works is a tangent

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substitution.

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And the reason a tangent
substitution works, is that

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you have a trig identity,
tan squared plus 1

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equals secant squared.

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In this case, I'd like to
suggest a different

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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 is that

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you use a hyperbolic trig
function as the thing that you

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substitute.

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So in particular, instead of
using, instead of relying on

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the trig identity, tan squared
plus 1 equals secant squared,

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you can use the hyperbolic trig
identity, which is that

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sinh squared u plus 1 equals
cosh squared u.

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So this identity--

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so here we have a something
squared plus 1 equals

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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

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integral, 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 a

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hyperbolic trig substitution.

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Let's work it out together, see
if my answer matches the

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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 of y squared plus

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1 part, and there's the easy
part, this y part.

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So before I make the
substitution, let me just deal

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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, or one part of the

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integral, really, is the
integral from 0 to sinh t of

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minus cosh t over sinh
t 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--

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well, it's the same constant
comes along, minus cosh t over

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sinh t, times y squared over
2, for y between 0 and this

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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.

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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, we need to add

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this expression that we just
computed to the integral of

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this first part.

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So that's what we need
to compete next.

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And that's what we're going
to use the hyperbolic trig

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substitution on.

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So we're going to complete the
integral from 0 to sinh t of

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the square root of y
squared plus 1 dy.

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And we're going to use the
substitution sinh u equals y,

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or y equals sinh u.

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OK.

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So we need, what do I need?

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I need what dy is, and I need
to change the bounds.

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So dy--

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I'm sorry, I'm going to flip
this around to take the-- so

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dy is, I need the differential
of sine u, sorry of sinh u.

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And so we saw in the earlier
hyperbolic trig function

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recitation that that's cosh u
du, or if you like, you could

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just differentiate using the
formulas that we know

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for sinh and cosh.

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And we need bounds.

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So when y is 0, we need sinh
of something is 0.

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And so it happens that
that value is 0.

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So if you remember the graph
of the function, or you can

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just check in the formula, when
sinh is 0, when t is 0,

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that's when you get sinh is 0.

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It's the only time e to the
t equals e to the minus t.

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OK.

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So when y is 0, then u is
0, and when y is sinh

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t, then u is t.

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Right?

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Because sinh u is sinh t.

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00:10:02,040 --> 00:10:06,930
So under the substitution, this
becomes the integral from

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0 to t now, from u equals
0 to t, of--

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well, OK.

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So this becomes the square root
of sinh squared u plus 1,

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and then dy is times
cosh u du.

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OK.

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00:10:30,660 --> 00:10:34,180
Now the reason we made this
substitution in the first

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place is that this, we
can use a hyperbolic

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trig identity here.

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So sinh squared u plus 1 is
just cosh squared u, and

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00:10:45,180 --> 00:10:47,200
square root of cosh squared
u is cosh u.

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00:10:47,200 --> 00:10:49,140
Remember that cosh u is
positive, so we don't have to

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00:10:49,140 --> 00:10:50,640
worry about an absolute
value here.

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00:10:50,640 --> 00:11:03,040
So this is the integral from 0
to t of cosh squared u du.

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00:11:03,040 --> 00:11:03,480
OK.

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00:11:03,480 --> 00:11:05,690
So at this point, there are
a couple of different

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00:11:05,690 --> 00:11:07,380
things you can do.

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00:11:07,380 --> 00:11:11,980
One is that you can, just like
when we have certain trig

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identities, we have
corresponding hyperbolic trig

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00:11:15,280 --> 00:11:19,030
identities that we could
try out here.

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00:11:19,030 --> 00:11:21,140
So we could try something
like that.

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00:11:21,140 --> 00:11:23,530
Another thing you can do, is
you can just go back to the

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00:11:23,530 --> 00:11:24,160
formula, right?

219
00:11:24,160 --> 00:11:27,120
Cosh t has a simple formula in
terms of exponentials, so you

220
00:11:27,120 --> 00:11:29,520
can go back to this formula
and you can plug in.

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00:11:29,520 --> 00:11:34,180
So let's just try that quickly,
because that's a sort

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00:11:34,180 --> 00:11:36,090
of easy way to handle this.

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00:11:36,090 --> 00:11:39,470
So this is cosh squared u du.

224
00:11:39,470 --> 00:11:42,590
So I'm going to write--

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OK.

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00:11:44,670 --> 00:11:46,230
Carry that all the
way up here.

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00:11:46,230 --> 00:11:48,720
So this is the integral
from 0 to t.

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00:11:48,720 --> 00:11:52,360
Well, if you take the formula
for hyperbolic cosine and

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square it, what you get, I'm
going to do this all in one

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step, is you e to the 2u
plus 2 plus e to the

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00:12:03,510 --> 00:12:14,160
minus 2u over 4 du.

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00:12:14,160 --> 00:12:14,570
OK.

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00:12:14,570 --> 00:12:17,250
And so now this is, once you've
replaced everything

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00:12:17,250 --> 00:12:21,100
with exponentials, this
is easy to integrate.

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00:12:21,100 --> 00:12:25,400
This is so e to the 2u, the
integral is e to the 2u over

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00:12:25,400 --> 00:12:28,410
2, so that comes over 8.

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00:12:28,410 --> 00:12:34,290
2 over 4, you integrate that,
and that's just 2u over 4,

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00:12:34,290 --> 00:12:36,100
which is u over 2.

239
00:12:36,100 --> 00:12:43,510
And now the last one is minus
e to the minus 2u over 8

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00:12:43,510 --> 00:12:47,270
between 0 and t.

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00:12:47,270 --> 00:12:48,490
OK.

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00:12:48,490 --> 00:12:50,860
So now we take the
difference here.

243
00:12:50,860 --> 00:13:02,120
At t, we get e to the 2t over 8
plus t over 2 minus e to the

244
00:13:02,120 --> 00:13:05,910
minus 2t over 8.

245
00:13:05,910 --> 00:13:08,630

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00:13:08,630 --> 00:13:09,200
Minus--

247
00:13:09,200 --> 00:13:09,450
OK.

248
00:13:09,450 --> 00:13:13,150
And when u is equal, so that
was at u equals t.

249
00:13:13,150 --> 00:13:18,360
At u equals 0, we get 1/8
plus 0 minus 1/8.

250
00:13:18,360 --> 00:13:20,310
So that's just 0.

251
00:13:20,310 --> 00:13:21,560
OK.

252
00:13:21,560 --> 00:13:26,230
So this is what we got for that
part of the integral.

253
00:13:26,230 --> 00:13:30,010
So OK, so we've now split the
integral into two pieces.

254
00:13:30,010 --> 00:13:32,190
We computed one piece, because
it was just easy, we're

255
00:13:32,190 --> 00:13:33,250
integrating a polynomial.

256
00:13:33,250 --> 00:13:35,110
We computed the other piece,
which was more complicated,

257
00:13:35,110 --> 00:13:36,600
using a hyperbolic trig
substitution.

258
00:13:36,600 --> 00:13:38,760
The whole integral is the
sum of those two pieces.

259
00:13:38,760 --> 00:13:41,720
So now the whole integral, I
have to take this piece, and I

260
00:13:41,720 --> 00:13:44,290
have to add it to the thing that
I computed for the other

261
00:13:44,290 --> 00:13:47,420
piece before, which
was somewhere--

262
00:13:47,420 --> 00:13:47,990
where did it go?

263
00:13:47,990 --> 00:13:49,530
Here it is, right here.

264
00:13:49,530 --> 00:13:54,530
Which was minus cosh
t sinh t over 2.

265
00:13:54,530 --> 00:13:55,660
OK.

266
00:13:55,660 --> 00:13:58,790
So I'm going to save you a
little arithmetic, and I'm

267
00:13:58,790 --> 00:14:02,270
going to observe that minus
cosh t sinh t over 2 is

268
00:14:02,270 --> 00:14:08,830
exactly equal to the minus e
to the 2t over 8, plus e to

269
00:14:08,830 --> 00:14:11,310
the minus 2t over 8.

270
00:14:11,310 --> 00:14:12,730
So adding these two expressions

271
00:14:12,730 --> 00:14:18,340
together gives us--

272
00:14:18,340 --> 00:14:23,270
so the first expression, minus
cosh t sinh t over 2, is minus

273
00:14:23,270 --> 00:14:31,270
e to the 2t over 8 plus e to
the minus 2t over 8, plus--

274
00:14:31,270 --> 00:14:31,540
OK.

275
00:14:31,540 --> 00:14:36,260
Plus what we've got right here,
which is e to the 2t

276
00:14:36,260 --> 00:14:43,170
over 8 minus e to the minus
2t over 8 plus t over 2.

277
00:14:43,170 --> 00:14:45,120
And that's exactly equal
to t over 2.

278
00:14:45,120 --> 00:14:50,050

279
00:14:50,050 --> 00:14:50,210
OK.

280
00:14:50,210 --> 00:14:53,800
So this is the area of that sort
of hyperbolic triangle

281
00:14:53,800 --> 00:14:55,660
thing that we started out
with at the beginning.

282
00:14:55,660 --> 00:14:58,250
So let me just walk back over
there for a second.

283
00:14:58,250 --> 00:15:03,090
So we used this hyperbolic trig
substitution in order to

284
00:15:03,090 --> 00:15:05,855
compute that the area of this
triangle is t over 2.

285
00:15:05,855 --> 00:15:07,510
And I just want to--

286
00:15:07,510 --> 00:15:09,070
first of all, I want to observe
that that's a really

287
00:15:09,070 --> 00:15:09,740
nice answer.

288
00:15:09,740 --> 00:15:11,080
So that's kind of cool.

289
00:15:11,080 --> 00:15:13,210
The other thing that I want to
observe is that this is a very

290
00:15:13,210 --> 00:15:17,120
close analogy with something
that happens in the case of

291
00:15:17,120 --> 00:15:20,460
regular circle trigonometric
functions.

292
00:15:20,460 --> 00:15:26,120
Which is, if you look at a
regular circle, and you take

293
00:15:26,120 --> 00:15:32,110
the point cosine theta comma
sine theta, then the area of

294
00:15:32,110 --> 00:15:38,360
this little triangle here
is theta over 2.

295
00:15:38,360 --> 00:15:42,090
So in this case, u doesn't
measure an angle, but it does

296
00:15:42,090 --> 00:15:44,330
measure an area in exactly
the same way that

297
00:15:44,330 --> 00:15:45,420
theta measures an area.

298
00:15:45,420 --> 00:15:47,730
So there's a really cool
relationship there between the

299
00:15:47,730 --> 00:15:50,710
hyperbolic trig function and
the regular trig function.

300
00:15:50,710 --> 00:15:52,460
So that's just a kind
of cool fact.

301
00:15:52,460 --> 00:15:56,400
The useful piece of knowledge
that you can extract from what

302
00:15:56,400 --> 00:15:58,810
we've just done, though, is
that you can use this

303
00:15:58,810 --> 00:16:03,040
hyperbolic trig substitution in
integrals of certain forms.

304
00:16:03,040 --> 00:16:05,940
So in the same way that trig
substitutions are suggested by

305
00:16:05,940 --> 00:16:08,290
certain forms of the integrand,
hyperbolic trig

306
00:16:08,290 --> 00:16:10,925
substitutions are also suggested
by certain forms of

307
00:16:10,925 --> 00:16:12,710
the integrand, and often
you have a choice about

308
00:16:12,710 --> 00:16:13,450
which one to use.

309
00:16:13,450 --> 00:16:16,170
And in this particular instance,
a hyperbolic trig

310
00:16:16,170 --> 00:16:18,600
substitution worked
out quite nicely.

311
00:16:18,600 --> 00:16:20,580
Much more nicely than
a trig substitution

312
00:16:20,580 --> 00:16:22,240
would have worked out.

313
00:16:22,240 --> 00:16:25,380
So it's just another tool
for your toolbox.

314
00:16:25,380 --> 00:16:27,080
I'll end with that.

315
00:16:27,080 --> 00:16:27,120