1
00:00:07,041 --> 00:00:07,540
Hi.

2
00:00:07,540 --> 00:00:10,730
Welcome back to recitation.

3
00:00:10,730 --> 00:00:15,280
I, today I wanted to teach you a
variation on trig substitution.

4
00:00:15,280 --> 00:00:17,984
So this is called hyperbolic
trigonometric substitution.

5
00:00:17,984 --> 00:00:19,400
And I'm going to
teach you it just

6
00:00:19,400 --> 00:00:22,820
by going through a nice
example of a question

7
00:00:22,820 --> 00:00:24,510
where it turns out to be useful.

8
00:00:24,510 --> 00:00:26,360
So the question
is the following.

9
00:00:26,360 --> 00:00:27,960
Compute the area of
the region below.

10
00:00:27,960 --> 00:00:29,000
So what's the region?

11
00:00:29,000 --> 00:00:33,730
Well, I have here the hyperbola
x squared minus y squared

12
00:00:33,730 --> 00:00:34,880
equals 1.

13
00:00:34,880 --> 00:00:38,670
And I've chosen a point on the
hyperbola whose coordinates are

14
00:00:38,670 --> 00:00:40,215
(cosh t, sinh t).

15
00:00:40,215 --> 00:00:43,490
So remember that cosh t
is a hyperbolic cosine,

16
00:00:43,490 --> 00:00:45,430
and sinh t is the
hyperbolic sine.

17
00:00:45,430 --> 00:00:51,570
And they're given by the
formulas cosh t equals

18
00:00:51,570 --> 00:00:56,620
e to the t plus e to
the minus t over 2,

19
00:00:56,620 --> 00:01:08,120
and sinh t equals e to the t
minus e to the minus t over 2.

20
00:01:08,120 --> 00:01:10,140
So and we saw in an
earlier recitation video

21
00:01:10,140 --> 00:01:12,620
that this point,
(cosh t, sinh t),

22
00:01:12,620 --> 00:01:15,855
is a point on the right
branch of this hyperbola.

23
00:01:15,855 --> 00:01:16,730
So I've got a region.

24
00:01:16,730 --> 00:01:20,180
So I've got the hyperbola,
I've got that point.

25
00:01:20,180 --> 00:01:23,480
I've drawn a straight
line between the origin

26
00:01:23,480 --> 00:01:25,040
and that point.

27
00:01:25,040 --> 00:01:28,270
So the region that I want
you to find the area of

28
00:01:28,270 --> 00:01:29,360
is this region here.

29
00:01:29,360 --> 00:01:37,680
So it's the region below that
line segment, above the x-axis,

30
00:01:37,680 --> 00:01:39,956
and to the left of this
branch of the hyperbola.

31
00:01:39,956 --> 00:01:41,455
Now, I'm not going
to, I'm not going

32
00:01:41,455 --> 00:01:43,980
to just ask you
to do that alone,

33
00:01:43,980 --> 00:01:46,640
because there's this technique
that I want you to use.

34
00:01:46,640 --> 00:01:49,740
So let's start setting
up the integral together,

35
00:01:49,740 --> 00:01:51,470
and then I'll
describe the technique

36
00:01:51,470 --> 00:01:54,765
and give you a chance to work
it out yourself, to work out

37
00:01:54,765 --> 00:01:55,640
the problem yourself.

38
00:01:55,640 --> 00:01:58,900
So from looking at
this region-- so let's

39
00:01:58,900 --> 00:02:01,930
think about computing
the area of this region.

40
00:02:01,930 --> 00:02:04,460
There are two ways we
could split it up, right?

41
00:02:04,460 --> 00:02:08,530
We could cut it up into
vertical rectangles

42
00:02:08,530 --> 00:02:10,330
and integrate with
respect to x, or we

43
00:02:10,330 --> 00:02:12,390
could cut it up into
horizontal rectangles

44
00:02:12,390 --> 00:02:14,040
and integrate with respect to y.

45
00:02:14,040 --> 00:02:16,120
Now, if we cut it up
into vertical rectangles,

46
00:02:16,120 --> 00:02:18,020
our life is a
little complicated,

47
00:02:18,020 --> 00:02:21,850
because we have to cut the
region into 2 pieces here.

48
00:02:21,850 --> 00:02:22,350
Right?

49
00:02:22,350 --> 00:02:27,340
There's the-- so this
is the point (1, 0),

50
00:02:27,340 --> 00:02:31,120
where the hyperbola
crosses x-axis.

51
00:02:31,120 --> 00:02:34,464
So over here, you know, the
top part is the line segment

52
00:02:34,464 --> 00:02:35,880
and the bottom
part is the x-axis,

53
00:02:35,880 --> 00:02:38,046
and then over here, the top
part is the line segment

54
00:02:38,046 --> 00:02:39,610
and the bottom part
is the hyperbola.

55
00:02:39,610 --> 00:02:41,940
So life is complicated if
we use vertical rectangles.

56
00:02:41,940 --> 00:02:44,440
It's a little bit simpler if
we use horizontal rectangles,

57
00:02:44,440 --> 00:02:45,510
so let's go with that.

58
00:02:45,510 --> 00:02:47,810
You know, the
amount of work will

59
00:02:47,810 --> 00:02:51,040
be similar either way,
but I like this way,

60
00:02:51,040 --> 00:02:52,810
lets you right down
in one integral.

61
00:02:52,810 --> 00:02:55,790
So the area-- OK.

62
00:02:55,790 --> 00:02:58,462
So if we use horizontal
rectangles to compute

63
00:02:58,462 --> 00:02:59,920
the area of a
region, then the area

64
00:02:59,920 --> 00:03:04,190
is, we need to integrate
from the bottom to the top,

65
00:03:04,190 --> 00:03:07,056
whatever the, you know,
the bounds on y are.

66
00:03:07,056 --> 00:03:08,680
And then the thing
we have to integrate

67
00:03:08,680 --> 00:03:11,620
has to be the area of one
of those little rectangles.

68
00:03:11,620 --> 00:03:15,130
So the area of the little
rectangle, its height is dy,

69
00:03:15,130 --> 00:03:18,970
and its length,
or width, I guess,

70
00:03:18,970 --> 00:03:23,950
is the x-coordinate of
its rightmost point,

71
00:03:23,950 --> 00:03:28,291
minus the x coordinate
of its leftmost point.

72
00:03:28,291 --> 00:03:28,790
So OK.

73
00:03:28,790 --> 00:03:33,032
So we need to integrate
from the lowest value of y

74
00:03:33,032 --> 00:03:33,910
to the highest value.

75
00:03:33,910 --> 00:03:36,619
So we start at the
bottom at y equals 0,

76
00:03:36,619 --> 00:03:38,410
and we need to go all
the way up to the top

77
00:03:38,410 --> 00:03:41,440
here, which is y equals sinh t.

78
00:03:41,440 --> 00:03:47,100
So it's an integral
from 0 to sinh t.

79
00:03:47,100 --> 00:03:47,600
OK.

80
00:03:47,600 --> 00:03:51,750
And so we need the
x-coordinate on the right here,

81
00:03:51,750 --> 00:03:54,240
minus the x-coordinate
on the left.

82
00:03:54,240 --> 00:03:56,370
So what's the
x-coordinate on the right?

83
00:03:56,370 --> 00:03:59,540
Well, we need to solve this
equation for x in terms of y.

84
00:03:59,540 --> 00:04:00,040
Right?

85
00:04:00,040 --> 00:04:02,123
This is going to be an
integral with respect to y.

86
00:04:02,123 --> 00:04:04,280
So that if we solve
for this point,

87
00:04:04,280 --> 00:04:07,940
we get x is equal to the square
root of y squared plus 1.

88
00:04:07,940 --> 00:04:11,130
So the right coordinate
is the square root

89
00:04:11,130 --> 00:04:13,840
of y squared plus 1.

90
00:04:13,840 --> 00:04:15,410
And the left
coordinate, this is just

91
00:04:15,410 --> 00:04:17,650
a straight line passing
through the origin,

92
00:04:17,650 --> 00:04:24,260
so its equation is y equals
sinh t over cosh t times x,

93
00:04:24,260 --> 00:04:28,290
or x equals cosh t
over sinh t times y.

94
00:04:28,290 --> 00:04:40,270
So this is minus cosh
t over sinh t times y.

95
00:04:40,270 --> 00:04:42,400
And we're integrating
with respect to y.

96
00:04:42,400 --> 00:04:42,900
OK.

97
00:04:42,900 --> 00:04:46,200
So this is the integral
that we're interested in.

98
00:04:46,200 --> 00:04:48,830
This integral gives us the area.

99
00:04:48,830 --> 00:04:51,720
And just a couple of
things to notice about it.

100
00:04:51,720 --> 00:04:53,900
So t is a constant.

101
00:04:53,900 --> 00:04:54,970
It's just fixed.

102
00:04:54,970 --> 00:04:58,130
So cosh t over sinh t, which
we could also, if we wanted to,

103
00:04:58,130 --> 00:05:00,490
this is the
hyperbolic cotangent.

104
00:05:00,490 --> 00:05:03,589
But that's really
not important at all.

105
00:05:03,589 --> 00:05:05,380
But we could call it
that, if we wanted to.

106
00:05:05,380 --> 00:05:07,970
So this is just a constant.

107
00:05:07,970 --> 00:05:11,170
So we have minus a
constant times y dy.

108
00:05:11,170 --> 00:05:13,430
So this part's
easy to integrate.

109
00:05:13,430 --> 00:05:15,500
So the hard part is
going to be integrating

110
00:05:15,500 --> 00:05:17,800
this y squared plus 1.

111
00:05:17,800 --> 00:05:20,697
Now, one thing you've seen is
that when you have a y squared,

112
00:05:20,697 --> 00:05:22,530
a square root of-- when
you have y y squared

113
00:05:22,530 --> 00:05:25,700
plus 1, one substitution
that sometimes works

114
00:05:25,700 --> 00:05:27,910
is a tangent substitution.

115
00:05:27,910 --> 00:05:30,180
And the reason a tangent
substitution works,

116
00:05:30,180 --> 00:05:32,510
is that you have a trig
identity, tan squared

117
00:05:32,510 --> 00:05:36,150
plus 1 equals secant squared.

118
00:05:36,150 --> 00:05:38,260
In this case, I'd
like to suggest

119
00:05:38,260 --> 00:05:39,810
a different substitution.

120
00:05:39,810 --> 00:05:40,310
All right?

121
00:05:40,310 --> 00:05:43,900
So this integral is the
integral that you want.

122
00:05:43,900 --> 00:05:46,400
And I'd like to suggest
a substitution, which

123
00:05:46,400 --> 00:05:51,277
is that you use a hyperbolic
trig function as the thing

124
00:05:51,277 --> 00:05:52,110
that you substitute.

125
00:05:52,110 --> 00:05:55,150
So in particular, instead
of using, instead of

126
00:05:55,150 --> 00:05:57,610
relying on the trig
identity tan squared

127
00:05:57,610 --> 00:05:59,250
plus 1 equals
secant squared, you

128
00:05:59,250 --> 00:06:01,390
can use the hyperbolic
trig identity,

129
00:06:01,390 --> 00:06:12,720
which is that sinh squared u
plus 1 equals cosh squared u.

130
00:06:12,720 --> 00:06:15,720
So this identity-- so
here we have a something

131
00:06:15,720 --> 00:06:18,230
squared plus 1 equals
something squared.

132
00:06:18,230 --> 00:06:22,720
So the identity this suggests
is to try the substitution y

133
00:06:22,720 --> 00:06:27,671
equals sinh u.

134
00:06:27,671 --> 00:06:28,170
All right?

135
00:06:28,170 --> 00:06:31,330
So this is a hyperbolic
trig substitution.

136
00:06:31,330 --> 00:06:36,070
So why don't you take that hint,
try it out on this integral,

137
00:06:36,070 --> 00:06:38,250
see how it goes.

138
00:06:38,250 --> 00:06:40,987
Take some time, pause the
video, work it out, come back

139
00:06:40,987 --> 00:06:42,320
and we can work it out together.

140
00:06:52,770 --> 00:06:53,450
Welcome back.

141
00:06:53,450 --> 00:06:56,165
Hopefully you had some luck
solving this integral using

142
00:06:56,165 --> 00:06:58,200
a hyperbolic trig substitution.

143
00:06:58,200 --> 00:07:01,640
Let's work it out together,
see if my answer matches

144
00:07:01,640 --> 00:07:03,720
the one that you came up with.

145
00:07:03,720 --> 00:07:07,400
So as I said before, this
integral comes in two parts.

146
00:07:07,400 --> 00:07:09,450
There's the hard
part, the square root

147
00:07:09,450 --> 00:07:11,320
of y squared plus
1 part, and there's

148
00:07:11,320 --> 00:07:13,620
the easy part, this y part.

149
00:07:13,620 --> 00:07:15,550
So before I make
the substitution,

150
00:07:15,550 --> 00:07:18,680
let me just deal
with the easy part.

151
00:07:18,680 --> 00:07:20,290
So I'll do that over here.

152
00:07:20,290 --> 00:07:23,320
So we have the one
part of the area,

153
00:07:23,320 --> 00:07:25,280
or one part of the
integral, really,

154
00:07:25,280 --> 00:07:39,970
is the integral from 0 to sinh
t of minus cosh t over sinh t

155
00:07:39,970 --> 00:07:42,131
times y dy.

156
00:07:42,131 --> 00:07:42,630
OK.

157
00:07:42,630 --> 00:07:46,510
So this is just a constant,
so it's a constant times y.

158
00:07:46,510 --> 00:07:52,540
So this is equal to-- well, it's
the same constant comes along,

159
00:07:52,540 --> 00:08:00,160
minus cosh t over sinh t,
times y squared over 2,

160
00:08:00,160 --> 00:08:06,170
for y between 0 and this
upper bound, sinh t.

161
00:08:06,170 --> 00:08:07,070
So that's equal-- OK.

162
00:08:07,070 --> 00:08:08,760
So when y is 0, this is just 0.

163
00:08:08,760 --> 00:08:18,020
So this equals minus
cosh t sinh t over 2.

164
00:08:18,020 --> 00:08:20,780
So that's the easy
part of integral.

165
00:08:20,780 --> 00:08:23,500
So in order to compute
the total area,

166
00:08:23,500 --> 00:08:25,800
we need to add this
expression that we just

167
00:08:25,800 --> 00:08:29,240
computed to the integral
of this first part.

168
00:08:29,240 --> 00:08:31,070
So that's what we
need to compute next.

169
00:08:31,070 --> 00:08:33,319
And that's what we're going
to use the hyperbolic trig

170
00:08:33,319 --> 00:08:34,160
substitution on.

171
00:08:34,160 --> 00:08:38,770
So we're going to compute
the integral from 0

172
00:08:38,770 --> 00:08:46,190
to sinh t of the square
root of y squared plus 1 dy.

173
00:08:46,190 --> 00:08:51,245
And we're going to use
the substitution sinh

174
00:08:51,245 --> 00:08:57,740
u equals y, or y equals sinh u.

175
00:08:57,740 --> 00:08:58,260
OK.

176
00:08:58,260 --> 00:09:00,520
So we need, what do I need?

177
00:09:00,520 --> 00:09:05,070
I need what dy is, and I
need to change the bounds.

178
00:09:05,070 --> 00:09:08,210
So dy-- I'm sorry,
I'm going to flip

179
00:09:08,210 --> 00:09:11,940
this around to take
the-- so dy is,

180
00:09:11,940 --> 00:09:15,630
I need the differential of
sine u-- sorry, of sinh u.

181
00:09:15,630 --> 00:09:19,540
And so we saw in the earlier
hyperbolic trig function

182
00:09:19,540 --> 00:09:23,300
recitation that that's
cosh u du, or if you like,

183
00:09:23,300 --> 00:09:26,160
you could just differentiate
using the formulas

184
00:09:26,160 --> 00:09:28,500
that we know for sinh and cosh.

185
00:09:28,500 --> 00:09:32,200
And we need bounds.

186
00:09:32,200 --> 00:09:36,350
So when y is 0, we need
sinh of something is 0.

187
00:09:36,350 --> 00:09:40,220
And so it happens
that that value is 0.

188
00:09:40,220 --> 00:09:42,213
So if you remember the
graph of the function,

189
00:09:42,213 --> 00:09:43,962
or you can just check
in the formula, when

190
00:09:43,962 --> 00:09:48,750
sinh is 0, when t is 0,
that's when you get sinh is 0.

191
00:09:48,750 --> 00:09:51,780
It's the only time e to the
t equals e to the minus t.

192
00:09:51,780 --> 00:09:52,280
OK.

193
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.

194
00:09:59,870 --> 00:10:00,370
Right?

195
00:10:00,370 --> 00:10:02,040
Because sinh u is sinh t.

196
00:10:02,040 --> 00:10:07,270
So under the substitution, this
becomes the integral from 0

197
00:10:07,270 --> 00:10:11,010
to t now, from u equals
0 to t, of-- well, OK.

198
00:10:11,010 --> 00:10:21,210
So this becomes the square
root of sinh squared u plus 1,

199
00:10:21,210 --> 00:10:30,161
and then dy is times cosh u du.

200
00:10:30,161 --> 00:10:30,660
OK.

201
00:10:30,660 --> 00:10:34,690
Now the reason we made this
substitution in the first place

202
00:10:34,690 --> 00:10:39,920
is that this, we can use a
hyperbolic trig identity here.

203
00:10:39,920 --> 00:10:44,990
So sinh squared u plus 1
is just cosh squared u,

204
00:10:44,990 --> 00:10:47,200
and square root of cosh
squared u is cosh u.

205
00:10:47,200 --> 00:10:48,640
Remember that cosh
u is positive,

206
00:10:48,640 --> 00:10:50,931
so we don't have to worry
about an absolute value here.

207
00:10:50,931 --> 00:11:02,980
So this is the integral from
0 to t of cosh squared u du.

208
00:11:02,980 --> 00:11:03,480
OK.

209
00:11:03,480 --> 00:11:05,920
So at this point, there are
a couple of different things

210
00:11:05,920 --> 00:11:07,380
you can do.

211
00:11:07,380 --> 00:11:11,980
One is that you can, just
like when we have certain trig

212
00:11:11,980 --> 00:11:15,280
identities, we have
corresponding hyperbolic trig

213
00:11:15,280 --> 00:11:19,030
identities that we
could try out here.

214
00:11:19,030 --> 00:11:21,140
So we could try
something like that.

215
00:11:21,140 --> 00:11:22,911
Another thing you can
do, is you can just

216
00:11:22,911 --> 00:11:24,160
go back to the formula, right?

217
00:11:24,160 --> 00:11:26,740
Cosh t has a simple formula
in terms of exponentials,

218
00:11:26,740 --> 00:11:29,520
so you can go back to this
formula and you can plug in.

219
00:11:29,520 --> 00:11:33,860
So let's just try that
quickly, because that's

220
00:11:33,860 --> 00:11:36,090
a sort of easy way
to handle this.

221
00:11:36,090 --> 00:11:39,470
So this is cosh squared u du.

222
00:11:39,470 --> 00:11:44,670
So I'm going to write-- OK.

223
00:11:44,670 --> 00:11:46,230
Carry that all the way up here.

224
00:11:46,230 --> 00:11:48,720
So this is the
integral from 0 to t.

225
00:11:48,720 --> 00:11:52,200
Well, if you take the
formula for hyperbolic cosine

226
00:11:52,200 --> 00:11:54,475
and square it, what
you get, I'm going

227
00:11:54,475 --> 00:12:03,090
to do this all in one step,
is you e to the 2u plus 2 plus

228
00:12:03,090 --> 00:12:14,070
e to the minus 2u over 4 du.

229
00:12:14,070 --> 00:12:14,570
OK.

230
00:12:14,570 --> 00:12:17,250
And so now this is, once
you've replaced everything

231
00:12:17,250 --> 00:12:20,000
with exponentials, this
is easy to integrate.

232
00:12:20,000 --> 00:12:26,340
This is-- so e to the 2u, the
integral is e to the 2u over 2,

233
00:12:26,340 --> 00:12:28,410
so that comes over 8.

234
00:12:28,410 --> 00:12:31,770
2 over 4, you integrate
that, and that's

235
00:12:31,770 --> 00:12:36,100
just 2u over 4,
which is u over 2.

236
00:12:36,100 --> 00:12:40,410
And now the last one
is minus e to the minus

237
00:12:40,410 --> 00:12:47,270
2u over 8 between 0 and t.

238
00:12:47,270 --> 00:12:48,490
OK.

239
00:12:48,490 --> 00:12:50,860
So now we take the
difference here.

240
00:12:50,860 --> 00:13:01,420
At t, we get e to the 2t
over 8 plus t over 2 minus

241
00:13:01,420 --> 00:13:05,910
e to the minus 2t over 8.

242
00:13:08,630 --> 00:13:09,450
Minus-- OK.

243
00:13:09,450 --> 00:13:13,150
And when u is equal-- so
that was at u equals t.

244
00:13:13,150 --> 00:13:18,360
At u equals 0, we get
1/8 plus 0 minus 1/8.

245
00:13:18,360 --> 00:13:20,310
So that's just 0.

246
00:13:20,310 --> 00:13:21,560
OK.

247
00:13:21,560 --> 00:13:26,230
So this is what we got for
that part of the integral.

248
00:13:26,230 --> 00:13:30,010
So OK, so we've now split
the integral into two pieces.

249
00:13:30,010 --> 00:13:32,020
We computed one piece,
because it was just easy,

250
00:13:32,020 --> 00:13:33,250
we're integrating a polynomial.

251
00:13:33,250 --> 00:13:35,110
We computed the other piece,
which was more complicated,

252
00:13:35,110 --> 00:13:36,651
using a hyperbolic
trig substitution.

253
00:13:36,651 --> 00:13:38,760
The whole integral is the
sum of those two pieces.

254
00:13:38,760 --> 00:13:41,310
So now the whole integral,
I have to take this piece,

255
00:13:41,310 --> 00:13:42,800
and I have to add
it to the thing

256
00:13:42,800 --> 00:13:44,540
that I computed
for the other piece

257
00:13:44,540 --> 00:13:47,990
before, which was
somewhere-- where did it go?

258
00:13:47,990 --> 00:13:49,530
Here it is, right here.

259
00:13:49,530 --> 00:13:54,530
Which was minus cosh
t sinh t over 2.

260
00:13:54,530 --> 00:13:55,660
OK.

261
00:13:55,660 --> 00:13:58,560
So I'm going to save
you a little arithmetic,

262
00:13:58,560 --> 00:14:02,080
and I'm going to observe that
minus cosh t sinh t over 2

263
00:14:02,080 --> 00:14:08,050
is exactly equal to the
minus e to the 2t over 8,

264
00:14:08,050 --> 00:14:11,310
plus e to the minus 2t over 8.

265
00:14:11,310 --> 00:14:18,340
So adding these two
expressions together gives us--

266
00:14:18,340 --> 00:14:22,020
so the first expression,
minus cosh t sinh t over 2,

267
00:14:22,020 --> 00:14:26,805
is minus e to the
2t over 8 plus e

268
00:14:26,805 --> 00:14:31,540
to the minus 2t
over 8, plus-- OK.

269
00:14:31,540 --> 00:14:33,960
Plus what we've got
right here, which

270
00:14:33,960 --> 00:14:41,320
is e to the 2t over 8 minus
e to the minus 2t over 8

271
00:14:41,320 --> 00:14:43,170
plus t over 2.

272
00:14:43,170 --> 00:14:45,120
And that's exactly
equal to t over 2.

273
00:14:49,711 --> 00:14:50,210
OK.

274
00:14:50,210 --> 00:14:53,910
So this is the area of that sort
of hyperbolic triangle thing

275
00:14:53,910 --> 00:14:55,660
that we started out
with at the beginning.

276
00:14:55,660 --> 00:14:58,250
So let me just walk back
over there for a second.

277
00:14:58,250 --> 00:15:02,630
So we used this hyperbolic
trig substitution

278
00:15:02,630 --> 00:15:04,800
in order to compute that
the area of this triangle

279
00:15:04,800 --> 00:15:05,855
is t over 2.

280
00:15:05,855 --> 00:15:07,616
And I just want
to-- first of all,

281
00:15:07,616 --> 00:15:09,740
I want to observe that
that's a really nice answer.

282
00:15:09,740 --> 00:15:11,080
So that's kind of cool.

283
00:15:11,080 --> 00:15:12,663
The other thing that
I want to observe

284
00:15:12,663 --> 00:15:15,220
is that this is a very close
analogy with something that

285
00:15:15,220 --> 00:15:20,460
happens in the case of regular
circle trigonometric functions.

286
00:15:20,460 --> 00:15:24,840
Which is, if you look
at a regular circle,

287
00:15:24,840 --> 00:15:31,340
and you take the point cosine
theta comma sine theta,

288
00:15:31,340 --> 00:15:38,360
then the area of this little
triangle here is theta over 2.

289
00:15:38,360 --> 00:15:41,390
So in this case, u
doesn't measure an angle,

290
00:15:41,390 --> 00:15:44,200
but it does measure an area
in exactly the same way

291
00:15:44,200 --> 00:15:45,420
that theta measures an area.

292
00:15:45,420 --> 00:15:46,980
So there's a really
cool relationship

293
00:15:46,980 --> 00:15:49,010
there between the
hyperbolic trig function

294
00:15:49,010 --> 00:15:50,710
and the regular trig function.

295
00:15:50,710 --> 00:15:52,460
So that's just a
kind of cool fact.

296
00:15:52,460 --> 00:15:55,150
The useful piece of
knowledge that you

297
00:15:55,150 --> 00:15:57,270
can extract from
what we've just done,

298
00:15:57,270 --> 00:16:01,020
though, is that you can use this
hyperbolic trig substitution

299
00:16:01,020 --> 00:16:03,040
in integrals of certain forms.

300
00:16:03,040 --> 00:16:05,800
So in the same way that trig
substitutions are suggested

301
00:16:05,800 --> 00:16:07,340
by certain forms
of the integrand,

302
00:16:07,340 --> 00:16:09,530
hyperbolic trig
substitutions are also

303
00:16:09,530 --> 00:16:11,380
suggested by certain
forms of the integrand,

304
00:16:11,380 --> 00:16:13,505
and often you have a choice
about which one to use.

305
00:16:13,505 --> 00:16:16,640
And in this particular instance,
a hyperbolic trig substitution

306
00:16:16,640 --> 00:16:18,600
worked out quite nicely.

307
00:16:18,600 --> 00:16:20,580
Much more nicely than
a trig substitution

308
00:16:20,580 --> 00:16:22,240
would have worked out.

309
00:16:22,240 --> 00:16:25,380
So it's just another
tool for your toolbox.

310
00:16:25,380 --> 00:16:26,620
I'll end with that.