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LongleatLongleat mazemaze, Somerset, Somerset, England, England
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ExampleExample①
Given
Consider the labeled triangle
= tan θ
x = 3 tan θ
Then x2
+ 9 = 9 tan2
θ
+ 9
And dx = 3 sec2
θ dθ
Then we have
29
dx
x 3
x
2 23 xθ
2
23sec9 tan 9
d
Use identity
tan2x + 1 = sec2x
Use identity
tan2x + 1 = sec2x
23sec
3 sec ln sec tan3sec
d
d C
3
x
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Back-Substitution
Our results are in terms of θ
We must back -substitute for a solution in xx
ln sec tan C
29ln
3 3
x xC
;
3tan
x
;3
9
cos
1sec2 x
3
x
2 23 xθ
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Try It!!
For each problem, identify which substitutionand which triangle should be used
3 29 x x dx
2
2
1 xdx
x
22 5 x x dx
2
4 1 x dx
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Try It!! ➁
3 29 x x dx
xx
33
92 x
;tan39;tan3
9 2
2
x
x
sec3;sec3
x x
d dx x x tansec3tan3sec393323
d dx sectan3;
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xx
33
92 x
d
d
dx x x
245
33
23
tansec3
tansec3tan3sec3
9
d 2225
tantan1sec3
C
uuduuu
duuu
53
33
1353
5425
225
d du
u
2sec
;tan
Try It!! ➁
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C
C
uu
duuu
5
tan
3
tan
3
5333
53
5
535425
xx
33
92 x
C C
x
5
tan
3
tantan3
5
tan
3
tan3
;3
9
tan
42
5
53
5
2
Try It!! ➁
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xx
33
92 x
C x x
53
9
3
93
5
252
4
2325
C
x x
5
993
252232
Try It!! ➁
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Setups for the rest
For each problem, identify which substitutionand which triangle should be used
24 1 x dx
x- 1x- 1
22
214 x
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Setups for the rest
For each problem, identify which substitutionand which triangle should be used
xx
11
21 x
2
2
1 xdx
x
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Setups for the rest
For each problem, identify which substitutionand which triangle should be used
2 2 5 x x dx xx
52 x
522
x x
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The Empire Builder, 1957The Empire Builder, 1957
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2
5 3
2 3
xdx
x x
This would be a lot easier if we could
re-write it as two separate terms.
5 3
3 1
x
x x
3 1
A B
x x
1
These are called non-
repeating linear factors.
You may already know a
short-cut for this type of problem. We will get to
that in a few minutes.
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2
5 3
2 3
x
dx x x
This would be a lot easier if we couldre-write it as two separate terms.
5 3
3 1
x
x x
3 1
A B
x x
Multiply by the common
denominator.
5 3 1 3 x A x B x
5 3 3 x Ax A Bx B Set like-terms equal to
each other.
5 x Ax Bx 3 3 A B
5 A B 3 3 A B Solve two equations with
two unknowns.
1
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2
5 3
2 3
x
dx x x
5 3
3 1
x
x x
3 1
A B
x x
5 3 1 3 x A x B x
5 3 3 x Ax A Bx B
5 x Ax Bx 3 3 A B
5 A B 3 3 A B Solve two equations withtwo unknowns.
5 A B 3 3 A B
3 3 A B
8 4 B
2 B 5 2 A
3 A
3 2
3 1dx
x x
3ln 3 2 ln 1 x x C
This technique is calledPartial Fractions
1
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2
5 3
2 3
x
dx x x
The short-cut for this type of problem is
called the Heaviside Method, after
English engineer Oliver Heaviside.
5 3
3 1
x
x x
3 1
A B
x x Multiply by the commondenominator.
5 3 1 3 x A x B x
8 0 4 A B
1
Let x = - 1
2 B
12 4 0 A B
Let x = 3
3 A
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2
5 3
2 3
x
dx x x
The short-cut for this type of problem
is called the Heaviside Method, after English engineer Oliver Heaviside.
5 3
3 1
x
x x
3 1
A B
x x
5 3 1 3 x A x B x
8 0 4 A B
1
2 B
12 4 0 A B
3 A
3 2
3 1
dx
x x
3ln 3 2 ln 1 x x C
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Why Heaviside’s Method Works: Example and Discussion
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Why Heaviside’s Method Works: Example and Discussion
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Why Heaviside’s Method Works: Example and Discussion
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26 7
2
x
x
Repeated roots: we mustuse two terms for partial
fractions.
22 2
A B
x x
6 7 2 x A x B
6 7 2 x Ax A B
6 x Ax 7 2 A B
6 A 7 2 6 B
7 12 B
5 B
2
6 5
2 2 x x
2 Moving on …
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3 2
2
2 4 3
2 3
x x x
x x
Is the degree of the numerator
higher than the degree of thedenominator?
Use long division first.
2 3 22 3 2 4 3 x x x x x
2 x
3 22 4 6 x x x
5 3 x
2
5 32
2 3
x x
x x
5 32
3 1
x x
x x
3 22
3 1 x
x x
3
(from example one)
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22
2 4
1 1
x
x x
irreducible
quadratic
factor
repeated root
22
1 1 1
Ax B C D
x x x
first degree numerator
2 2 2
2 4 1 1 1 1 x Ax B x C x x D x
2 3 2 22 4 2 1 1 x Ax B x x C x x x Dx D
3 2 2 3 2 22 4 2 2 x Ax Ax Ax Bx Bx B Cx Cx Cx C Dx D
A challenging
example:
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22
2 4
1 1
x
x x
irreducible
quadratic
factor
repeated root
22
1 1 1
Ax B C D
x x x
first degree numerator A challenging
example:
3 2 2 3 2 22 4 2 2 x Ax Ax Ax Bx Bx B Cx Cx Cx C Dx D
0 A C 0 2 A B C D 2 2 A B C 4 B C D
First solved using the above system, per student
request.
First solved using the above system, per student
request.
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0 A C 0 2 A B C D 2 2 A B C 4 B C D
DC B
C B A
DC B A
C A
4
22
20
0
DC B
C B A
DC B DC BC
C A
4
22
020
:
1
2
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0 A C 0 2 A B C D 2 2 A B C 4 B C D
D B
C B A
D B
DC B
DC B
C A
2
22
224
4
0
:
C B A
D B
C A
22
24
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0 A C 0 2 A B C D 2 2 A B C 4 B C D
122
2
BC BC
D B
C A
112 D D
C A
5
6
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0 A C 0 2 A B C D 2 2 A B C 4 B C D
1 D B
2211422
2211200
C C C
C A
C AC A
C A
7
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3 2 2 3 2 22 4 2 2 x Ax Ax Ax Bx Bx B Cx Cx Cx C Dx D
0 A C 0 2 A B C D 2 2 A B C 4 B C D
1 0 1 0 0
2 1 1 1 0
1 2 1 0 2
0 1 1 1 4
2 r 3 r 1
1 0 1 0 0
0 3 1 1 4
0 2 0 0 2
0 1 1 1 4
2
1 0 1 0 0
0 1 0 0 1
0 3 1 1 4
0 1 1 1 4
3 r 2
r 2
1 0 1 0 0
0 1 0 0 1
0 0 1 1 1
0 0 1 1 3
r 3
A B C D A B C D
x3
x2
x
1
x3
x2
x
1
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1 0 1 0 0
0 1 0 0 1
0 3 1 1 4
0 1 1 1 4
3 r 2
r 2
1 0 1 0 0
0 1 0 0 1
0 0 1 1 10 0 1 1 3
r 3
1 0 1 0 00 1 0 0 1
0 0 1 1 1
0 0 0 2 2
2
1 0 1 0 0
0 1 0 0 1
0 0 1 1 1
0 0 0 1 1
r 4
1 0 1 0 0
0 1 0 0 1
0 0 1 0 20 0 0 1 1
r 3
1 0 0 0 20 1 0 0 1
0 0 1 0 2
0 0 0 1 1
A B C D A B C D
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22
2 4
1 1
x
x x
22
1 1 1
Ax B C D
x xx
22
2 1 2 1
1 1 1
x
x x x
1 0 0 0 2
0 1 0 0 1
0 0 1 0 2
0 0 0 1 1
A B C D A B C D
12 steps / manipulations to get to this solution.
This method is 4 steps longer, but arguably lesserror-prone, since it does not involve variables.
12 steps / manipulations to get to this solution.
This method is 4 steps longer, but arguably lesslesserror error --proneprone, since it does notnot involve variables.