26. More Trigonometric Substitution

7/28/2019 26. More Trigonometric Substitution

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LongleatLongleat mazemaze, Somerset, Somerset, England, England



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



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





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



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;



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



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





xx

33

92 x

C x x

53

9

3

93

5

252

4

2325

C

x x

5

993

252232

Try It!! ➁





Setups for the rest


24 1 x dx

x- 1x- 1

22

214 x



Setups for the rest


xx

11

21 x

2

2

1 xdx

x



Setups for the rest


2 2 5 x x dx xx

52 x

522

x x



The Empire Builder, 1957The Empire Builder, 1957



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.



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



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



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



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



Why Heaviside’s Method Works: Example and Discussion









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 …



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)



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:



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:


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.



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



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



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



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






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



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



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.

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26. More Trigonometric Substitution