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PROFESSOR: Hi.

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Today we will discuss the
inverse hyperbolic functions

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and, with this lecture, finish
our block on the logarithmic

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00:00:40,840 --> 00:00:43,640
exponential and hyperbolic
functions.

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00:00:43,640 --> 00:00:47,380
And what we're going to find is
that much of what we have

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to say today is simply a
specific application to a

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special function of the same
theory that we were talking

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00:00:55,470 --> 00:00:57,200
about in general before.

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00:00:57,200 --> 00:01:00,530
Recall that we can always talk
about an inverse function if

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00:01:00,530 --> 00:01:03,690
the original function is
a one-to-one function.

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00:01:03,690 --> 00:01:07,880
For example, to introduce
today's topic, suppose we take

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00:01:07,880 --> 00:01:11,385
the function 'y' equals
hyperbolic sine 'x'.

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00:01:11,385 --> 00:01:13,080
'y' equals 'sinh x'.

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00:01:13,080 --> 00:01:17,500
Well, as we saw last time,
the graph of 'y'

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00:01:17,500 --> 00:01:20,480
equals 'sinh x' is this.

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00:01:20,480 --> 00:01:23,390
And we can see in a glance
that this clearly is a

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00:01:23,390 --> 00:01:25,160
one-to-one function.

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00:01:25,160 --> 00:01:27,850
In fact, the derivative
of sinh is cosh.

26
00:01:27,850 --> 00:01:30,480
And as we've seen also
last time, the cosh

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00:01:30,480 --> 00:01:31,970
can never be negative.

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00:01:31,970 --> 00:01:35,170
In fact, the cosh can't
be less than one.

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00:01:35,170 --> 00:01:38,050
So here is a curve that's
always rising--

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00:01:38,050 --> 00:01:41,380
and, in fact, in general
rising quite steeply.

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00:01:41,380 --> 00:01:45,770
In any event, we can therefore
talk about the inverse

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00:01:45,770 --> 00:01:47,320
hyperbolic sine.

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00:01:47,320 --> 00:01:47,990
All right?

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00:01:47,990 --> 00:01:50,420
And how do we get the inverse
function in general?

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00:01:50,420 --> 00:01:53,820
Well we've always seen that to
invert a function, all we have

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00:01:53,820 --> 00:01:57,350
to do is to reflect the original
graph with respect to

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00:01:57,350 --> 00:01:59,090
the line, 'y' equals 'x'.

38
00:01:59,090 --> 00:02:03,400
Or, again, in more slow motion,
we rotate through 90

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00:02:03,400 --> 00:02:06,750
degrees and then flip
the graph over.

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00:02:06,750 --> 00:02:09,310
And, in any event, whichever way
you want to look at this

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00:02:09,310 --> 00:02:13,240
thing, we obtain that the
graph 'y' equals inverse

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00:02:13,240 --> 00:02:17,300
hyperbolic sine 'x' is this
particular curve over here.

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00:02:17,300 --> 00:02:21,140
This is the graph of the inverse
hyperbolic sine.

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00:02:21,140 --> 00:02:25,850
Now, again, what was the main
issue or the main property of

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00:02:25,850 --> 00:02:27,120
inverse functions?

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00:02:27,120 --> 00:02:30,870
The idea was that once we knew
the original function, we

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00:02:30,870 --> 00:02:33,000
could, by a change in emphasis,

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00:02:33,000 --> 00:02:34,990
study the inverse function.

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00:02:34,990 --> 00:02:39,620
Well, by way of illustration,
let's suppose we take the

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00:02:39,620 --> 00:02:43,020
functional relationship 'y'
equals inverse hyperbolic sine

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00:02:43,020 --> 00:02:46,970
'x' and we say, let's
find the 'ydx'.

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00:02:46,970 --> 00:02:50,870
Recall that what we know how to
do, is how to differentiate

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00:02:50,870 --> 00:02:52,370
the hyperbolic sine.

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00:02:52,370 --> 00:02:55,820
Consequently, given that y
equals the inverse hyperbolic

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00:02:55,820 --> 00:03:00,010
sine 'x', we simply switch the
emphasis and say, this is the

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00:03:00,010 --> 00:03:03,820
same as saying that 'x'
equals 'sinh y'.

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00:03:03,820 --> 00:03:07,240
And if 'x' equal 'sinh y',
we already know that the

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00:03:07,240 --> 00:03:11,190
derivative of 'x' with respect
to 'y' is 'cosh y'.

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00:03:11,190 --> 00:03:15,160
And since the derivative of 'y'
with respect to 'x' is the

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00:03:15,160 --> 00:03:18,950
reciprocal of the derivative of
'x' with respect to 'y', we

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00:03:18,950 --> 00:03:24,060
can conclude from this that the
'ydx' is '1 over cosh y'.

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00:03:24,060 --> 00:03:26,460
In a certain manner
of speaking, we're

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00:03:26,460 --> 00:03:27,520
all finished now.

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00:03:27,520 --> 00:03:31,810
We have found that the
derivative off inverse 'sinh

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00:03:31,810 --> 00:03:35,070
x' is '1 over cosh y'.

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00:03:35,070 --> 00:03:38,000
The only problem is, is as we've
said many times also

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00:03:38,000 --> 00:03:41,670
before that if 'y' is given as
a function of 'x', we would

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00:03:41,670 --> 00:03:44,760
like the 'ydx' expressed
in terms of 'x'.

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00:03:44,760 --> 00:03:47,370
Or, to put it in still different
words, usually when

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00:03:47,370 --> 00:03:50,510
you're given an expression like
the 'ydx', you're asked

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00:03:50,510 --> 00:03:55,210
to evaluate this when 'x' is
some particular value, not

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when 'y' is some particular
value.

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00:03:57,340 --> 00:04:00,390
At any rate, we do much the same
here as we did with the

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inverse circular
trig functions.

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00:04:02,910 --> 00:04:08,670
Namely, having arrived at '1
over cosh y' and remembering

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00:04:08,670 --> 00:04:12,810
that the relationship that ties
in 'x' and 'y' is at 'x'

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00:04:12,810 --> 00:04:17,769
equals 'sinh y', we invoke the
fundamental identity that

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00:04:17,769 --> 00:04:21,810
'cosh squared y' minus 'sinh
squared y' is 1.

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00:04:21,810 --> 00:04:25,050
From which we can conclude
that 'cosh y'--

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00:04:25,050 --> 00:04:27,270
and here we have to be a
little bit careful--

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00:04:27,270 --> 00:04:31,700
if we solve this algebraically
we find that 'cosh y' is plus

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00:04:31,700 --> 00:04:36,180
or minus the square root of
'1 plus 'sinh squared y''.

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00:04:36,180 --> 00:04:39,670
But recalling the cosh could
never be negative, it means

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00:04:39,670 --> 00:04:42,030
that for this particular
problem, the minus

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00:04:42,030 --> 00:04:43,740
sign does not apply.

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00:04:43,740 --> 00:04:48,770
In other words, since cosh has
to be at least as big one, you

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00:04:48,770 --> 00:04:53,180
see, that 'cosh y' is plus the
square root of '1 plus 'sinh

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00:04:53,180 --> 00:04:54,370
squared y''.

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00:04:54,370 --> 00:04:57,150
Now keep in mind why we
went to this identity

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00:04:57,150 --> 00:04:58,220
in the first place.

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00:04:58,220 --> 00:05:00,840
The reason we went to this
identity in the first place,

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00:05:00,840 --> 00:05:03,390
is if we come back to the
beginning of our problem we

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00:05:03,390 --> 00:05:08,020
see that we had that 'x'
is equal to 'sinh y'.

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00:05:08,020 --> 00:05:12,460
In other words, down here now,
all we do is replace 'sinh y'

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00:05:12,460 --> 00:05:14,490
by a synonym, namely 'x'.

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00:05:14,490 --> 00:05:18,820
And we arrive at the fact that
'cosh y' is the square root of

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00:05:18,820 --> 00:05:21,340
'1 plus 'x squared''.

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00:05:21,340 --> 00:05:26,200
Therefore, since 'dy/dx' is the
reciprocal of 'cosh y',

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00:05:26,200 --> 00:05:29,550
the derivative of 'y' with
respect to 'x' is '1 over the

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00:05:29,550 --> 00:05:32,000
'square root of '1 plus
'x squared'''.

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00:05:32,000 --> 00:05:34,900
And this is summarized in
our last step over here.

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00:05:34,900 --> 00:05:38,270
In other words, the derivative
of the inverse hyperbolic sine

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00:05:38,270 --> 00:05:41,730
of 'x' with respect to 'x' is '1
over the 'square root of '1

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00:05:41,730 --> 00:05:43,160
plus 'x squared'''.

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00:05:43,160 --> 00:05:46,290
Now, you see, the first thing I
want to point out here is to

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00:05:46,290 --> 00:05:48,890
observe that without worrying
about what's important about

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00:05:48,890 --> 00:05:53,100
this result, is to observe
again that we obtain this

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00:05:53,100 --> 00:05:57,540
information virtually free of
charge by knowing the calculus

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00:05:57,540 --> 00:05:59,680
of the regular hyperbolic
functions.

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00:05:59,680 --> 00:06:01,980
You see, this result
was obtained

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00:06:01,980 --> 00:06:04,610
without any new knowledge.

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00:06:04,610 --> 00:06:07,640
The other thing I'd like to
point out here is somewhat

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00:06:07,640 --> 00:06:08,780
more subtle.

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00:06:08,780 --> 00:06:11,990
And we also mentioned this in
the same context, but from a

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00:06:11,990 --> 00:06:14,550
different point of view, when
we dealt with the inverse

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00:06:14,550 --> 00:06:15,920
circular functions.

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00:06:15,920 --> 00:06:18,970
Many people will say things
like, who needs the inverse

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00:06:18,970 --> 00:06:20,230
hyperbolic functions?

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00:06:20,230 --> 00:06:23,120
How many times am I going to be
confronted with having to

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00:06:23,120 --> 00:06:25,750
work with inverse hyperbolic
functions?

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00:06:25,750 --> 00:06:29,930
And the interesting point that's
typified by this result

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00:06:29,930 --> 00:06:32,360
is that if we now invert
this emphasis--

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00:06:32,360 --> 00:06:35,090
in other words, if we now
read this equation

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00:06:35,090 --> 00:06:36,420
from right to left--

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00:06:36,420 --> 00:06:40,000
observe that if you start with
the function '1 over the

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00:06:40,000 --> 00:06:41,970
'square root of '1 plus
'x squared''''--

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00:06:41,970 --> 00:06:45,790
which is hardly a hyperbolic
function, this is a fairly

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00:06:45,790 --> 00:06:48,070
straightforward algebraic
function--

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00:06:48,070 --> 00:06:52,000
notice that the inverse
derivative leads to an inverse

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00:06:52,000 --> 00:06:53,260
hyperbolic sine.

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00:06:53,260 --> 00:06:57,370
In other words, stated from a
different perspective and

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00:06:57,370 --> 00:07:00,500
using our language of the
indefinite integral, notice

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00:07:00,500 --> 00:07:04,320
that what we have here is that
the indefinite integral 'dx'

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00:07:04,320 --> 00:07:07,760
over the 'square root of '1 plus
'x squared''' is inverse

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00:07:07,760 --> 00:07:10,860
hyperbolic sine 'x'
plus a constant.

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00:07:10,860 --> 00:07:13,970
Now what does this mean,
say, geometrically?

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00:07:13,970 --> 00:07:17,270
Suppose we take the curve 'y'
equals '1 over the 'square

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00:07:17,270 --> 00:07:19,060
root of '1 plus 'x squared'''.

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00:07:19,060 --> 00:07:21,590
Without beating this thing to
death, it should be fairly

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00:07:21,590 --> 00:07:26,240
straightforward at this stage of
the game that the graph of

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00:07:26,240 --> 00:07:28,510
this function can be obtained,
and it looks

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00:07:28,510 --> 00:07:29,810
something like this.

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00:07:29,810 --> 00:07:33,590
In fact, intuitively notice that
'y' will be maximum when

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00:07:33,590 --> 00:07:35,330
my denominator is smallest.

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00:07:35,330 --> 00:07:38,000
My denominator is smallest
when 'x' is zero.

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00:07:38,000 --> 00:07:43,190
So the maximum value of 'y'
occurs when 'x' is 0, at which

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00:07:43,190 --> 00:07:44,740
case, 'y' is 1.

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00:07:44,740 --> 00:07:47,340
Also, if I replace 'x'
by minus 'x', I

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00:07:47,340 --> 00:07:49,710
don't change the function.

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00:07:49,710 --> 00:07:51,270
And therefore the graph--

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00:07:51,270 --> 00:07:52,520
it's an even function--

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00:07:52,520 --> 00:07:54,180
the graph is symmetric
with respect to

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00:07:54,180 --> 00:07:55,910
the y-axis, et cetera.

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00:07:55,910 --> 00:07:58,300
At any rate, I have a
picture like this.

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00:07:58,300 --> 00:08:02,330
And now suppose I want to find
the area of the region 'R',

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00:08:02,330 --> 00:08:05,970
where 'R' is bounded above by
this curve, below by the

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00:08:05,970 --> 00:08:10,230
x-axis, on the left by the
y-axis, and on the right by

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00:08:10,230 --> 00:08:11,970
the line 'x' equals 't'.

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00:08:11,970 --> 00:08:15,810
The area of the region 'R',
which is a function of 't', is

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00:08:15,810 --> 00:08:17,290
given by what?

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00:08:17,290 --> 00:08:21,710
The definite integral from 0 to
't', 'dx' over the 'square

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00:08:21,710 --> 00:08:23,940
root of '1 plus 'x squared'''.

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00:08:23,940 --> 00:08:27,730
The point is I could, as we
talked about in the previous

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00:08:27,730 --> 00:08:32,100
block, try to evaluate this as
the limit of a sum-- in other

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00:08:32,100 --> 00:08:33,590
words, an infinite sum--

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00:08:33,590 --> 00:08:36,620
and go through all sorts of work
to try to do this thing.

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00:08:36,620 --> 00:08:39,799
But the first fundamental
theorem tells us in this

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00:08:39,799 --> 00:08:43,490
particular case that this
particular area just turns out

169
00:08:43,490 --> 00:08:47,110
to be the inverse hyperbolic
sine of 't'.

170
00:08:47,110 --> 00:08:51,170
Notice that a non-trigonometric
region has

171
00:08:51,170 --> 00:08:55,060
as its answer an inverse
hyperbolic

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00:08:55,060 --> 00:08:56,380
trigonometric function.

173
00:08:56,380 --> 00:08:58,520
Or, if you want this thing
more specifically, for

174
00:08:58,520 --> 00:09:02,010
example, notice that if you want
the area of this region

175
00:09:02,010 --> 00:09:07,096
from 0 to 1, the answer to this
problem would have just

176
00:09:07,096 --> 00:09:10,470
been the inverse hyperbolic
sine of 1.

177
00:09:10,470 --> 00:09:12,750
In other words, 'e to the
1' minus 'e to the

178
00:09:12,750 --> 00:09:14,870
minus 1' over 2.

179
00:09:14,870 --> 00:09:18,490
Notice how 'e' sneaks into
a problem which basically

180
00:09:18,490 --> 00:09:21,240
doesn't seem to have any
relationship to 'e'.

181
00:09:21,240 --> 00:09:23,590
By the way-- as a very
brief aside--

182
00:09:23,590 --> 00:09:26,930
for what it's worth, it's rather
interesting to observe

183
00:09:26,930 --> 00:09:31,830
that this last equation that
we've written down gives a

184
00:09:31,830 --> 00:09:37,150
rather interesting definition of
the inverse hyperbolic sine

185
00:09:37,150 --> 00:09:40,420
without having to refer to
a hyperbola or anything

186
00:09:40,420 --> 00:09:41,420
trigonometric.

187
00:09:41,420 --> 00:09:44,320
In other words, notice that the
inverse hyperbolic sine

188
00:09:44,320 --> 00:09:48,540
can be defined as an integral,
which is what we've really

189
00:09:48,540 --> 00:09:49,330
done over here.

190
00:09:49,330 --> 00:09:50,990
But again, that's
just an aside.

191
00:09:50,990 --> 00:09:54,440
The main point that I wanted us
to get a hold of over here

192
00:09:54,440 --> 00:09:58,850
was the fact that you solve
non-hyperbolic functions

193
00:09:58,850 --> 00:10:01,140
conveniently if we
have mastered

194
00:10:01,140 --> 00:10:02,830
the hyperbolic functions.

195
00:10:02,830 --> 00:10:05,380
Well, at any rate, here's
another interesting question

196
00:10:05,380 --> 00:10:06,270
that comes up.

197
00:10:06,270 --> 00:10:08,770
And I thought that we should
mention this, also.

198
00:10:08,770 --> 00:10:12,420
Notice that we arrived at
this result by doing

199
00:10:12,420 --> 00:10:13,510
the thing in reverse.

200
00:10:13,510 --> 00:10:16,750
You'll recall that we started
with 'y' equals inverse

201
00:10:16,750 --> 00:10:19,840
hyperbolic sine 'x' and show
the derivative of that

202
00:10:19,840 --> 00:10:22,930
function was '1 over the 'square
root of '1 plus 'x

203
00:10:22,930 --> 00:10:23,920
squared'''.

204
00:10:23,920 --> 00:10:26,340
An interesting question might
have been, what if we had

205
00:10:26,340 --> 00:10:31,910
started with the integral being
given and we hadn't have

206
00:10:31,910 --> 00:10:34,890
differentiated the inverse
hyperbolic sine?

207
00:10:34,890 --> 00:10:37,910
How could we have got
from here to the

208
00:10:37,910 --> 00:10:39,630
inverse hyperbolic sine?

209
00:10:39,630 --> 00:10:41,940
And I thought I would mention
this because there may be some

210
00:10:41,940 --> 00:10:45,370
confusion, especially if you've
taken to heart certain

211
00:10:45,370 --> 00:10:49,290
advice that I gave you earlier
when we dealt with the inverse

212
00:10:49,290 --> 00:10:51,070
circular functions.

213
00:10:51,070 --> 00:10:53,490
Remember I told you that
whenever you see something

214
00:10:53,490 --> 00:10:57,810
like the sum of two squares, to
think of a right triangle?

215
00:10:57,810 --> 00:11:01,750
In other words, if you have the
square root of '1 plus 'x

216
00:11:01,750 --> 00:11:04,920
squared'', it seems to me that
the triangle that suggests

217
00:11:04,920 --> 00:11:06,610
itself is something like this.

218
00:11:06,610 --> 00:11:10,010
In other words, if I call this
side 'x', I call this side 1,

219
00:11:10,010 --> 00:11:12,740
and this the square root of '1
plus 'x squared'', it would

220
00:11:12,740 --> 00:11:15,800
seem to me that I could make
a circular trigonometric

221
00:11:15,800 --> 00:11:17,160
substitution over here.

222
00:11:17,160 --> 00:11:20,510
In fact, what seems to dictate
itself over here, is to make

223
00:11:20,510 --> 00:11:24,750
the substitution, let
tan theta equal 'x'.

224
00:11:24,750 --> 00:11:28,420
Now if I let tan theta equal
'x', watch what happens here.

225
00:11:28,420 --> 00:11:32,080
I get 'secant squared theta 'd
theta'' equals 'dx', taking

226
00:11:32,080 --> 00:11:33,850
the differential
of both sides.

227
00:11:33,850 --> 00:11:37,020
I also get, looking at my
reference triangle, that the

228
00:11:37,020 --> 00:11:41,120
square root of '1 plus 'x
squared'' is secant theta.

229
00:11:41,120 --> 00:11:44,910
See, this over this
is secant theta.

230
00:11:44,910 --> 00:11:48,830
At any rate, then, making the
substitution in here,

231
00:11:48,830 --> 00:11:52,800
replacing 'dx' by 'secant
squared theta 'd theta'', and

232
00:11:52,800 --> 00:11:56,350
the square root of '1 plus 'x
squared'' by secant theta, I

233
00:11:56,350 --> 00:11:59,970
wind up with integral of 'secant
theta 'd theta''.

234
00:11:59,970 --> 00:12:02,970
Now you see what I've done here,
is I have successfully

235
00:12:02,970 --> 00:12:06,130
transformed an integral
in terms of 'x' into

236
00:12:06,130 --> 00:12:07,640
one in terms of theta.

237
00:12:07,640 --> 00:12:11,190
But without belaboring this
point, it turns out that at

238
00:12:11,190 --> 00:12:16,270
this stage of the game, we do
not know how to exhibit a

239
00:12:16,270 --> 00:12:18,480
function whose derivative
with respect to

240
00:12:18,480 --> 00:12:20,430
theta is secant theta.

241
00:12:20,430 --> 00:12:23,150
In other words, we've made the
substitution but we wind up

242
00:12:23,150 --> 00:12:26,570
with an integral that's just as
tough to handle as the one

243
00:12:26,570 --> 00:12:27,580
that we started with.

244
00:12:27,580 --> 00:12:30,580
You see, in this case, trying
to make a circular

245
00:12:30,580 --> 00:12:32,590
trigonometric substitution
wouldn't have

246
00:12:32,590 --> 00:12:34,200
helped us very much.

247
00:12:34,200 --> 00:12:36,400
What I'd like to show you is,
again, an interesting

248
00:12:36,400 --> 00:12:39,270
connection between the circular
functions and the

249
00:12:39,270 --> 00:12:40,740
hyperbolic functions.

250
00:12:40,740 --> 00:12:45,380
Namely, when we did this
particular thing over here

251
00:12:45,380 --> 00:12:48,350
using our reference triangle,
what was the reference

252
00:12:48,350 --> 00:12:51,100
triangle really taking
the place of?

253
00:12:51,100 --> 00:12:55,190
Notice that if we let 'x' equal
tan theta, certainly '1

254
00:12:55,190 --> 00:12:58,910
plus 'x squared'' is 1 plus
tan squared theta.

255
00:12:58,910 --> 00:13:02,740
But there is a trigonometric
identity that says 1 plus tan

256
00:13:02,740 --> 00:13:05,390
squared theta is secant
squared theta.

257
00:13:05,390 --> 00:13:08,400
I think the usual way that's
given is that secant squared

258
00:13:08,400 --> 00:13:11,810
theta minus tan squared
theta is 1.

259
00:13:11,810 --> 00:13:13,290
This is the result
that we used.

260
00:13:13,290 --> 00:13:16,090
We didn't really use the
triangle other than to get

261
00:13:16,090 --> 00:13:18,380
this result more visually.

262
00:13:18,380 --> 00:13:22,860
The point is, is there a
hyperbolic function that has

263
00:13:22,860 --> 00:13:24,370
the same format?

264
00:13:24,370 --> 00:13:27,500
Is there a hyperbolic identity
which says that the difference

265
00:13:27,500 --> 00:13:29,340
of two squares is one?

266
00:13:29,340 --> 00:13:33,260
The answer is, well remember our
basic hyperbolic identity

267
00:13:33,260 --> 00:13:38,280
is the cosh squared theta minus
sinh squared theta is 1.

268
00:13:38,280 --> 00:13:43,130
Structurally, notice that sinh
theta does for the hyperbolic

269
00:13:43,130 --> 00:13:47,620
functions what tan theta does
for the circular functions.

270
00:13:47,620 --> 00:13:49,150
See the common structure,
here?

271
00:13:49,150 --> 00:13:52,380
We're going to reinforce this
in the next block by doing

272
00:13:52,380 --> 00:13:53,520
problems like this again.

273
00:13:53,520 --> 00:13:56,060
But for the time being, I
thought I would like to point

274
00:13:56,060 --> 00:13:57,920
this thing out to you.

275
00:13:57,920 --> 00:14:01,060
What the approach is, is that
when you try a circular

276
00:14:01,060 --> 00:14:04,460
function substitution and it
doesn't give you the answer

277
00:14:04,460 --> 00:14:07,490
that you want-- meaning that you
wind up with an integral

278
00:14:07,490 --> 00:14:09,930
that's just as tough to handle
as the original one--

279
00:14:09,930 --> 00:14:13,820
you look for the corresponding
hyperbolic function.

280
00:14:13,820 --> 00:14:17,940
What hyperbolic function plays
to the hyperbolic identity the

281
00:14:17,940 --> 00:14:20,900
same role that this
trigonometric function play to

282
00:14:20,900 --> 00:14:22,400
the circular identity?

283
00:14:22,400 --> 00:14:26,770
In this case, we replace tan
theta by sinh theta.

284
00:14:26,770 --> 00:14:29,010
Instead of making the
substitution 'x' equals

285
00:14:29,010 --> 00:14:32,130
tangent theta, we make
the substitution

286
00:14:32,130 --> 00:14:34,300
'x' equals sinh theta.

287
00:14:34,300 --> 00:14:37,150
And now watch what happens as
we work this thing quite

288
00:14:37,150 --> 00:14:38,200
mechanically.

289
00:14:38,200 --> 00:14:42,520
The differential of sinh theta
is 'cosh theta 'd theta''.

290
00:14:42,520 --> 00:14:45,780
And the square root of '1 plus
'x squared'' is the square

291
00:14:45,780 --> 00:14:48,440
root of 1 plus sin
squared theta.

292
00:14:48,440 --> 00:14:51,010
But notice that because
of the relationship

293
00:14:51,010 --> 00:14:52,470
between sinh and cosh--

294
00:14:52,470 --> 00:14:55,540
that's how we rig this thing,
that's why we chose 'x' to be

295
00:14:55,540 --> 00:14:56,510
sinh theta--

296
00:14:56,510 --> 00:15:00,260
notice that the square root of
1 plus sinh squared theta is

297
00:15:00,260 --> 00:15:02,510
just cosh theta.

298
00:15:02,510 --> 00:15:06,410
And therefore, when we now
substitute for the 'dx' over

299
00:15:06,410 --> 00:15:09,790
the square root of '1 plus 'x
squared'', we get what?

300
00:15:09,790 --> 00:15:12,830
For 'dx' we have 'cosh
theta 'd theta''.

301
00:15:12,830 --> 00:15:14,970
For the square root of
'1 plus 'x squared''

302
00:15:14,970 --> 00:15:16,460
we have cosh theta.

303
00:15:16,460 --> 00:15:20,240
'Cosh theta 'd theta'' over cosh
theta is just 'd theta'.

304
00:15:20,240 --> 00:15:24,350
And now we see the answer is,
quite simply, theta plus 'c'.

305
00:15:24,350 --> 00:15:26,180
But what was theta?

306
00:15:26,180 --> 00:15:30,970
Since sinh theta was 'x',
theta was inverse

307
00:15:30,970 --> 00:15:33,220
hyperbolic sine 'x'.

308
00:15:33,220 --> 00:15:37,080
And, you see, this is a
technique whereby starting

309
00:15:37,080 --> 00:15:42,000
with the integral 'dx' over the
'square root of '1 plus 'x

310
00:15:42,000 --> 00:15:45,010
squared''', we can show that
we must have started with

311
00:15:45,010 --> 00:15:46,070
inverse sinh.

312
00:15:46,070 --> 00:15:48,810
At any rate, this will be
reinforced in homework

313
00:15:48,810 --> 00:15:51,400
problems, it will be reinforced
in the next block

314
00:15:51,400 --> 00:15:53,410
when we talk about techniques
of integration.

315
00:15:53,410 --> 00:15:57,050
But I just wanted to again
show the similarity, the

316
00:15:57,050 --> 00:16:00,790
things in common, between the
hyperbolic functions and the

317
00:16:00,790 --> 00:16:04,000
circular functions and how
they're intertwined.

318
00:16:04,000 --> 00:16:07,150
Let's make a few more comments
while we're at it.

319
00:16:07,150 --> 00:16:10,600
You know, we mentioned that the
hyperbolic functions were

320
00:16:10,600 --> 00:16:13,040
really combinations of
exponential functions.

321
00:16:13,040 --> 00:16:16,690
Remember, 'cosh x' was ''e to
the x' plus 'e to the minus

322
00:16:16,690 --> 00:16:19,010
x'' over 2, et cetera.

323
00:16:19,010 --> 00:16:23,390
So somehow or other, if the
hyperbolic functions can be

324
00:16:23,390 --> 00:16:26,800
expressed in terms of
exponentials, it would seem

325
00:16:26,800 --> 00:16:29,460
that the inverse hyperbolic
functions should be

326
00:16:29,460 --> 00:16:33,570
expressible in terms of the
inverse of exponentials--

327
00:16:33,570 --> 00:16:36,590
namely, in terms
of logarithms.

328
00:16:36,590 --> 00:16:39,430
And so I thought that I would
try to go through some of

329
00:16:39,430 --> 00:16:41,010
these finer points with you.

330
00:16:41,010 --> 00:16:43,890
And, for example, ask the
following question.

331
00:16:43,890 --> 00:16:49,460
Given that 'y' equals inverse
'sinh x', is there a way of

332
00:16:49,460 --> 00:16:54,200
writing this in terms of
something that uses our

333
00:16:54,200 --> 00:16:57,360
natural logarithms?

334
00:16:57,360 --> 00:16:58,450
Another reason being, what?

335
00:16:58,450 --> 00:17:01,990
That if we've already learned
natural logs and exponentials,

336
00:17:01,990 --> 00:17:06,490
it would seem that whenever we
can reduce unfamiliar names to

337
00:17:06,490 --> 00:17:09,510
more familiar ones,
psychologically we feel much

338
00:17:09,510 --> 00:17:12,069
more at home in dealing
with the concepts.

339
00:17:12,069 --> 00:17:15,010
In other words, one might feel
strange working with inverse

340
00:17:15,010 --> 00:17:18,030
hyperbolic sine because he
hasn't seen that very much.

341
00:17:18,030 --> 00:17:20,730
But if he's used to seeing
logarithms, that wouldn't seem

342
00:17:20,730 --> 00:17:21,560
quite as strange.

343
00:17:21,560 --> 00:17:25,260
At any rate, let's see how
one could proceed here.

344
00:17:25,260 --> 00:17:28,790
Starting out with 'y' equals
inverse 'sinh x', notice that

345
00:17:28,790 --> 00:17:31,100
by the property, the basic
definition of inverse

346
00:17:31,100 --> 00:17:35,020
functions, I can now write
that 'x' equals 'sinh y'.

347
00:17:35,020 --> 00:17:37,420
Now, for obvious reasons,
since I want to get the

348
00:17:37,420 --> 00:17:40,440
inverse of exponentials in here,
it would seem to me that

349
00:17:40,440 --> 00:17:43,710
I should express 'sinh y' in
terms of exponentials.

350
00:17:43,710 --> 00:17:48,510
And going again back to basic
definitions, 'sinh y' is ''e

351
00:17:48,510 --> 00:17:52,080
to the y' minus 'e to the
minus y'' over 2.

352
00:17:52,080 --> 00:17:55,430
In other words, in terms of
exponentials, 'x' is equal to

353
00:17:55,430 --> 00:17:59,570
''e to the y' minus 'e to
the minus y'' over 2.

354
00:17:59,570 --> 00:18:04,420
If I now cross multiply, I get
that '2x' is equal to 'e to

355
00:18:04,420 --> 00:18:06,360
the y' minus--

356
00:18:06,360 --> 00:18:09,770
now notice that 'e to the minus
y' is just '1 over 'e to

357
00:18:09,770 --> 00:18:13,840
the y'', so I wind up now with
this particular equation.

358
00:18:13,840 --> 00:18:17,300
And multiplying through by 'e to
the y', to clear fractions

359
00:18:17,300 --> 00:18:20,190
in my denominator, to clear
my denominators, I

360
00:18:20,190 --> 00:18:21,570
wind up with what?

361
00:18:21,570 --> 00:18:28,950
'e to the 2y' minus ''2x 'e
to the y'' - 1' equals 0.

362
00:18:28,950 --> 00:18:33,590
And if I now recall that 'e to
the 2y' is the square of 'e to

363
00:18:33,590 --> 00:18:37,810
the y', observe that what I
now have is a quadratic

364
00:18:37,810 --> 00:18:40,300
equation in 'e to the y'.

365
00:18:40,300 --> 00:18:43,880
I have a quadratic equation
in 'e to the y'.

366
00:18:43,880 --> 00:18:47,210
Now, since I have a quadratic
equation in 'e to the y', I

367
00:18:47,210 --> 00:18:51,710
can use the quadratic formula
to solve for 'e to the y'.

368
00:18:51,710 --> 00:18:53,640
If I do this I get what?

369
00:18:53,640 --> 00:18:54,500
Remember how this thing works.

370
00:18:54,500 --> 00:18:59,230
I take the coefficient of this
term minus that, that's '2x'

371
00:18:59,230 --> 00:19:05,220
plus or minus the square root of
this squared minus 4 times

372
00:19:05,220 --> 00:19:09,050
this times the coefficient
of 'e to the y' squared.

373
00:19:09,050 --> 00:19:12,910
Leaving the details as being
fairly obvious, 'e to the y'

374
00:19:12,910 --> 00:19:17,840
is '2x' plus or minus the square
root of ''4x' squared'

375
00:19:17,840 --> 00:19:23,180
plus 4 all over 2.

376
00:19:23,180 --> 00:19:26,190
And noticing now that the 4 can
be factored out here as a

377
00:19:26,190 --> 00:19:29,980
2, and that I can cancel a 2,
then from both the numerator

378
00:19:29,980 --> 00:19:35,680
and denominator, I wind up with
'e to the y' is 'x' plus

379
00:19:35,680 --> 00:19:40,350
or minus the square root
of 'x squared' plus 1.

380
00:19:40,350 --> 00:19:43,960
The point to keep in mind, now,
is remember that in terms

381
00:19:43,960 --> 00:19:47,850
of exponentials, 'e to the
y' can never be negative.

382
00:19:47,850 --> 00:19:51,960
Observe that the square root
of 'x squared plus 1' is

383
00:19:51,960 --> 00:19:56,020
bigger than 'x' in magnitude,
you see.

384
00:19:56,020 --> 00:19:57,620
See, 'x' would be just
the positive

385
00:19:57,620 --> 00:19:59,010
square root of 'x squared'.

386
00:19:59,010 --> 00:20:02,470
So the positive square root of
'x squared plus 1' is bigger

387
00:20:02,470 --> 00:20:04,450
than 'x' in magnitude.

388
00:20:04,450 --> 00:20:07,770
Consequently, if I use the minus
sign here, I'd be taking

389
00:20:07,770 --> 00:20:10,000
away more than what I had.

390
00:20:10,000 --> 00:20:12,490
That would make my answer
negative, which would be a

391
00:20:12,490 --> 00:20:16,200
contradiction, since 'e to
the y' can't be negative.

392
00:20:16,200 --> 00:20:20,110
Again, in terms of this
particular problem, the minus

393
00:20:20,110 --> 00:20:23,870
root, the minus sign
here is extraneous.

394
00:20:23,870 --> 00:20:25,920
And we therefore wind
up with what?

395
00:20:25,920 --> 00:20:29,050
'e to the y' is 'x' plus
the square root of 'x

396
00:20:29,050 --> 00:20:30,290
squared plus 1'.

397
00:20:30,290 --> 00:20:34,870
Therefore 'y' itself is the
'log of x' plus the square

398
00:20:34,870 --> 00:20:38,030
root of 'x squared plus 1' to
the base 'e', which we've

399
00:20:38,030 --> 00:20:42,890
already seen is called
the natural log.

400
00:20:42,890 --> 00:20:46,860
Going back now, say, from the
first step to the last, I

401
00:20:46,860 --> 00:20:52,040
guess we can now fill in what's
really happened here.

402
00:20:52,040 --> 00:20:56,930
A synonym for the inverse 'sinh
x' is the natural log of

403
00:20:56,930 --> 00:21:01,020
'x' plus the square root
of 'x squared plus 1'.

404
00:21:01,020 --> 00:21:05,980
So notice that we can study the
inverse sinh, for example,

405
00:21:05,980 --> 00:21:09,040
in terms of a suitably
chosen natural log.

406
00:21:09,040 --> 00:21:11,260
And of course there are many
more examples that we could

407
00:21:11,260 --> 00:21:12,670
use along these lines.

408
00:21:12,670 --> 00:21:17,390
But again, I think that with
the previous explanation,

409
00:21:17,390 --> 00:21:21,150
coupled with the fact that there
will be ample exercises

410
00:21:21,150 --> 00:21:23,790
in the like, I think the message
has become clear as

411
00:21:23,790 --> 00:21:26,410
far as the inverse hyperbolic
sine is concerned.

412
00:21:26,410 --> 00:21:29,580
What I would like to do now is
to turn to another facet of

413
00:21:29,580 --> 00:21:33,050
inverse functions, something
that involves principal values

414
00:21:33,050 --> 00:21:35,100
the same as it did with the
circular functions.

415
00:21:35,100 --> 00:21:39,320
We wind up with the same problem
as before when we come

416
00:21:39,320 --> 00:21:42,780
to the idea that, technically
speaking, you cannot talk

417
00:21:42,780 --> 00:21:45,620
about an inverse function unless
the original function

418
00:21:45,620 --> 00:21:47,140
is one-to-one.

419
00:21:47,140 --> 00:21:49,700
And so therefore, when one
talks about the inverse

420
00:21:49,700 --> 00:21:54,300
hyperbolic cosine, one is in a
way looking for trouble if one

421
00:21:54,300 --> 00:21:56,370
doesn't keep his eye
on exactly what's

422
00:21:56,370 --> 00:21:57,690
going on around here.

423
00:21:57,690 --> 00:22:00,950
Namely, if we look at the graph
'y' equals hyperbolic

424
00:22:00,950 --> 00:22:05,750
cosine 'x', observe that whereas
the function is single

425
00:22:05,750 --> 00:22:08,310
valued, it is not one-to-one.

426
00:22:08,310 --> 00:22:12,420
In fact, there is a zero
derivative at this point here,

427
00:22:12,420 --> 00:22:15,720
which leads us to believe that
maybe what we should have done

428
00:22:15,720 --> 00:22:20,280
was to have broken this curve
down into the union of two

429
00:22:20,280 --> 00:22:21,830
one-to-one functions.

430
00:22:21,830 --> 00:22:26,510
Let me call this curve 'y'
equals 'c1 of x' and let me

431
00:22:26,510 --> 00:22:31,430
call this branch here 'y'
equals 'c2 of x'.

432
00:22:31,430 --> 00:22:35,300
Notice that both 'c1
of x' and 'c2 of x'

433
00:22:35,300 --> 00:22:38,050
are one-to-one functions.

434
00:22:38,050 --> 00:22:40,960
In fact, let's write this
more formally using the

435
00:22:40,960 --> 00:22:43,140
picture as a guide.

436
00:22:43,140 --> 00:22:45,180
Let's do the following
analytically.

437
00:22:45,180 --> 00:22:46,230
Let's say this.

438
00:22:46,230 --> 00:22:51,690
Define 'c sub 1 of x' to be
'cosh x', provided that 'x' is

439
00:22:51,690 --> 00:22:53,130
at least as big as 0.

440
00:22:53,130 --> 00:22:55,770
Again, I mentioned this with the
circular functions, let me

441
00:22:55,770 --> 00:22:57,290
reinforce this again.

442
00:22:57,290 --> 00:23:00,860
To define a function, it's not
enough to tell the rule.

443
00:23:00,860 --> 00:23:03,410
You must also tell the domain.

444
00:23:03,410 --> 00:23:07,410
Notice that 'c1' is not the
same as cosh, because the

445
00:23:07,410 --> 00:23:10,010
domain of cosh is all
real numbers.

446
00:23:10,010 --> 00:23:14,370
The domain of 'c1' is just
the non-negative reals.

447
00:23:14,370 --> 00:23:18,710
At any rate, I define 'c1' to
be 'cosh x', where 'x' is at

448
00:23:18,710 --> 00:23:20,420
least as big as 0.

449
00:23:20,420 --> 00:23:24,710
I define 'c2 of x' to be
'cosh x', where 'x' is

450
00:23:24,710 --> 00:23:26,460
no bigger than 0.

451
00:23:26,460 --> 00:23:28,530
In other words, these two
functions are different,

452
00:23:28,530 --> 00:23:31,510
because even though the
functional relations are the

453
00:23:31,510 --> 00:23:33,960
same, the domains
are different.

454
00:23:33,960 --> 00:23:35,610
The interesting point is what?

455
00:23:35,610 --> 00:23:40,710
That 'cosh x' is the union
of 'c1' and 'c2'.

456
00:23:40,710 --> 00:23:45,050
But the important point is that
both 'c1' and 'c2' are

457
00:23:45,050 --> 00:23:46,910
one-to-one.

458
00:23:46,910 --> 00:23:48,540
And because they
are one-to-one,

459
00:23:48,540 --> 00:23:49,780
their inverses exist.

460
00:23:49,780 --> 00:23:53,710
In other words, I can talk
meaningfully about 'c1

461
00:23:53,710 --> 00:23:56,060
inverse' and 'c2 inverse'.

462
00:23:56,060 --> 00:24:02,090
In fact, pictorially,
what I have is this.

463
00:24:02,090 --> 00:24:08,530
See, if I take the curve 'y'
equals 'cosh x' and reflect it

464
00:24:08,530 --> 00:24:11,680
about the 45 degree line, this
is the curve that I get.

465
00:24:11,680 --> 00:24:14,080
You see, it's a double
value curve.

466
00:24:14,080 --> 00:24:19,100
All I'm saying is if we look at
'y' equals 'c1x', which is

467
00:24:19,100 --> 00:24:23,500
a one-to-one function, its
inverse is 'c1 inverse x',

468
00:24:23,500 --> 00:24:25,630
which is this piece over here.

469
00:24:25,630 --> 00:24:30,730
And if, on the other hand, we
look at 'y' equals 'c2x',

470
00:24:30,730 --> 00:24:34,000
that's this branch over here,
its inverse is this.

471
00:24:34,000 --> 00:24:37,400
You see, notice that these two
pieces are symmetric with

472
00:24:37,400 --> 00:24:40,540
respect to the line 'y' equals
'x', and these two pieces are

473
00:24:40,540 --> 00:24:43,530
symmetric with respect to
the line 'y' equals 'x'.

474
00:24:43,530 --> 00:24:47,200
As long as we break this down
into the union of two pieces,

475
00:24:47,200 --> 00:24:49,710
we can talk about inverse
functions.

476
00:24:49,710 --> 00:24:52,740
Now you see, the interesting
point is that what most

477
00:24:52,740 --> 00:24:57,790
authors traditionally refer to
as the inverse hyperbolic

478
00:24:57,790 --> 00:25:03,210
cosine of 'x' is really what
we call 'c1 inverse of x'.

479
00:25:03,210 --> 00:25:07,610
In other words, the definition
'y' equals inverse hyperbolic

480
00:25:07,610 --> 00:25:11,750
cosine 'x' is 'x'
equals cosh 'y'.

481
00:25:11,750 --> 00:25:15,330
And this is very important,
and 'y' is at

482
00:25:15,330 --> 00:25:18,000
least as big as 0.

483
00:25:18,000 --> 00:25:23,750
Notice that the domain of 'cosh
inverse x' is really 'x'

484
00:25:23,750 --> 00:25:26,110
has to be at least
as big as one.

485
00:25:26,110 --> 00:25:28,920
But that's not the important
point here.

486
00:25:28,920 --> 00:25:32,140
What I do want to see over here
is that when you put this

487
00:25:32,140 --> 00:25:35,660
restriction on, instead if you
left this restriction out,

488
00:25:35,660 --> 00:25:38,350
there would be no inverse
function here.

489
00:25:38,350 --> 00:25:39,950
I'll come back to that
in a moment.

490
00:25:39,950 --> 00:25:42,770
Let me just reinforce what we've
talked about before, and

491
00:25:42,770 --> 00:25:45,420
let's find the derivative
of 'inverse cosh x'.

492
00:25:45,420 --> 00:25:48,640
In other words, let's find
'dy/dx', if 'y' equals

493
00:25:48,640 --> 00:25:49,910
'inverse cosh x'.

494
00:25:49,910 --> 00:25:53,000
Well again, what is the
definition, 'y' equals

495
00:25:53,000 --> 00:25:54,330
'inverse cosh x'?

496
00:25:54,330 --> 00:26:00,230
It means 'x' equals cosh y',
where 'y' is positive.

497
00:26:00,230 --> 00:26:01,350
OK.

498
00:26:01,350 --> 00:26:05,200
If 'x' equals 'cosh y',
'dx/dy' is 'sinh y'.

499
00:26:05,200 --> 00:26:07,450
And we'll keep track of the
fact that 'y' is positive.

500
00:26:07,450 --> 00:26:09,820
Actually, 'y' is non-negative.

501
00:26:09,820 --> 00:26:14,970
Therefore the reciprocal of
'dx/dy' will be 'dy/dx'.

502
00:26:14,970 --> 00:26:19,370
In other words, 'dy/dx'
is '1 over sinh y'.

503
00:26:19,370 --> 00:26:23,010
And this would be a correct
answer, except, as usual, we

504
00:26:23,010 --> 00:26:26,360
would like to be able to express
'dy/dx' for a given

505
00:26:26,360 --> 00:26:27,350
value of 'x'.

506
00:26:27,350 --> 00:26:31,520
What we do now is, remembering
that 'x' is 'cosh y', we

507
00:26:31,520 --> 00:26:33,410
invoke the identity again.

508
00:26:33,410 --> 00:26:37,500
'Cosh squared y' minus 'sinh
squared y' is one.

509
00:26:37,500 --> 00:26:41,530
From which we can solve and find
that 'sinh y' is plus or

510
00:26:41,530 --> 00:26:44,550
minus the 'square root of
'x squared minus 1'.

511
00:26:44,550 --> 00:26:47,680
And by the way, I'm not
going to remove the

512
00:26:47,680 --> 00:26:49,270
extraneous sign here.

513
00:26:49,270 --> 00:26:52,130
Because in a certain manner
of speaking, it is only

514
00:26:52,130 --> 00:26:55,810
extraneous because we are
imposing the condition that

515
00:26:55,810 --> 00:26:59,340
'y' is positive.

516
00:26:59,340 --> 00:27:03,070
See, in other words, once we
assume that y' is positive--

517
00:27:03,070 --> 00:27:06,190
remember that 'sinh y' is
positive for positive values

518
00:27:06,190 --> 00:27:09,220
of 'y', and negative for
negative values of 'y'--

519
00:27:09,220 --> 00:27:12,850
consequently, the assumption
that 'y' is positive forces us

520
00:27:12,850 --> 00:27:15,520
to accept the fact that
'sinh y' is positive.

521
00:27:15,520 --> 00:27:19,030
And that's what forces us, in
terms of the restriction that

522
00:27:19,030 --> 00:27:23,530
we imposed the fact that y has
to be at least as big as 0,

523
00:27:23,530 --> 00:27:26,260
why we can get rid of
the minus sign here.

524
00:27:26,260 --> 00:27:31,300
And so we wind up with what?
'Sinh y' is positive 'square

525
00:27:31,300 --> 00:27:34,320
root of 'cosh squared
y' minus 1''.

526
00:27:34,320 --> 00:27:37,000
But 'cosh y' is 'x'
in this problem.

527
00:27:37,000 --> 00:27:40,130
In other words, 'sinh y', in
this problem, is the positive

528
00:27:40,130 --> 00:27:42,360
square root of 'x squared
minus 1'.

529
00:27:42,360 --> 00:27:44,560
By the way, don't
be nervous here.

530
00:27:44,560 --> 00:27:46,580
You might say, couldn't
this be imaginary?

531
00:27:46,580 --> 00:27:49,370
In other words, what happens if
'x squared' is less than 1?

532
00:27:49,370 --> 00:27:52,190
Remember, 'x' is at
least as big as 1.

533
00:27:52,190 --> 00:27:55,740
So this thing here in
the square root sign

534
00:27:55,740 --> 00:27:56,870
can never be negative.

535
00:27:56,870 --> 00:27:59,960
But at any rate, what we now
wind up with is that the

536
00:27:59,960 --> 00:28:04,120
derivative of 'inverse cosh x',
with respect to 'x', is '1

537
00:28:04,120 --> 00:28:07,520
over the 'square root of ''x
squared' minus 1'''.

538
00:28:07,520 --> 00:28:10,810
Now again, there's no law that
says that a person couldn't

539
00:28:10,810 --> 00:28:13,830
have been on the negative
branch of this curve.

540
00:28:13,830 --> 00:28:17,200
In other words, if all you mean
by inverse cosh is the

541
00:28:17,200 --> 00:28:21,070
inverse of cosh with no
restriction to branch, what

542
00:28:21,070 --> 00:28:22,590
we've really proven is this.

543
00:28:22,590 --> 00:28:24,720
And let me summarize on
this particular point.

544
00:28:24,720 --> 00:28:27,130
What we've really
proven is this.

545
00:28:27,130 --> 00:28:32,160
That the derivative of 'c1'
inverse is '1 over the 'square

546
00:28:32,160 --> 00:28:34,280
root of ''x squared'
minus 1'''.

547
00:28:34,280 --> 00:28:39,430
The derivative of 'c2' inverse
is '1 over minus the 'square

548
00:28:39,430 --> 00:28:41,360
root of ''x squared'
minus 1'''.

549
00:28:41,360 --> 00:28:44,090
I've taken the liberty of
putting the minus down with

550
00:28:44,090 --> 00:28:47,370
the square root sign, rather
than with the fraction itself,

551
00:28:47,370 --> 00:28:50,370
to emphasize the fact that
which of the two signs we

552
00:28:50,370 --> 00:28:53,860
choose depends on whether we're
looking at the branch

553
00:28:53,860 --> 00:28:57,030
for which 'y' is above the
x-axis, or the branch which

554
00:28:57,030 --> 00:28:58,830
'y' is below the x-axis.

555
00:28:58,830 --> 00:29:02,280
In other words, what we don't
want to happen here is for

556
00:29:02,280 --> 00:29:05,420
people to lose track of the fact
that all we have done is

557
00:29:05,420 --> 00:29:08,430
made a convention so we can
talk about one-to-one

558
00:29:08,430 --> 00:29:11,320
functions and inverse functions
more meaningfully.

559
00:29:11,320 --> 00:29:14,150
But you can be on either of
these particular branches.

560
00:29:14,150 --> 00:29:19,160
In any event, this does complete
our discussion of the

561
00:29:19,160 --> 00:29:21,010
hyperbolic functions.

562
00:29:21,010 --> 00:29:24,510
And we will now turn
our attention to

563
00:29:24,510 --> 00:29:26,240
utilizing these results.

564
00:29:26,240 --> 00:29:28,610
We will learn some techniques of
integration, and the like.

565
00:29:28,610 --> 00:29:31,150
But at any rate, until
next time, goodbye.

566
00:29:33,820 --> 00:29:36,840
MALE SPEAKER: Funding for the
publication of this video was

567
00:29:36,840 --> 00:29:41,560
provided by the Gabriella and
Paul Rosenbaum Foundation.

568
00:29:41,560 --> 00:29:45,730
Help OCW continue to provide
free and open access to MIT

569
00:29:45,730 --> 00:29:49,930
courses by making a donation
at ocw.mit.edu/donate.