Chapter 9 Calculus Techniques for the Elementary Functions.

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Chapter 9 Calculus Techniques for the Elementary Functions

Transcript of Chapter 9 Calculus Techniques for the Elementary Functions.

Chapter 9 Calculus Techniques for the Elementary Functions

9.2 Integration by parts – A way to integrate Products

Integration by parts is a rule that transforms the integral of products of functions into other, hopefully simpler, integrals.

xdxdxexdxe xx coscos

.sderivativeforruleproducttheatlookmustWe

Integration by parts – A way to integrate Products

udvvdudy

vdudyudv

vduyudv

If where u and v are differentiable functions of x then:

uvy

vduuvudv Formula for integration by parts.

dvuvdudy

Integration by parts – A way to integrate Products

xdxxxxx sinsincos

xvdxdu sin1 xdvxu cos

Suppose we want to find the volume of the solid formed by rotating the region under around the y-axis.

From aside:

xy cos

xdxxV

xdxxdV

ydxxdV

cos2

cos2

2

2

0cossin2

xxxV

Aside

Let

vduuvudv

xxxxx cossincos

Integration by parts – A way to integrate Products

Examples:

1.

2.

xdxx sin2

Let

vduuvudv2xu

xdxdu 2xdxdv sinxv cos xdxxxxxdxx 2coscossin 22

dxxex

9.3 Rapid Repeated Integration by Parts

Example:

xdxx 2cos5

U dV

0

120

120

60

20

5

2

3

4

5

x

x

x

x

x

x

x

x

x

x

x

x

2cos64

1

2sin32

1

2cos16

1

2sin8

1

2cos4

1

2sin2

1

2cos

vduuvudv

+

-

+

-

+

-

+

Reason for the sign change.

vduuvuvuv

vduuvuv

vdu

dvu mult

integrate

U dV

0

120

120

60

20

5

2

3

4

5

x

x

x

x

x

x

x

x

x

x

x

x

2cos64

1

2sin32

1

2cos16

1

2sin8

1

2cos4

1

2sin2

1

2cos

+

-

+

-

+

-

+

Cxxxxxxxxxxxxdxx 2cos8

152sin

4

152cos

4

152sin

4

102cos

4

52sin

2

12cos 23455

Using Trig properties to make original reappear.

dxx 4.0sin2

xx

xx

4.0cos4.0

14.0cos4.0

4.0sin4.0sin

+

-

xxxdxx 4.0cos4.0sin4.0cos4.0

14.0sin 22

xxxdxx 4.0sin14.0sin4.0cos4.0

14.0sin 22

xdxdxxxdxx 4.0sin14.0sin4.0cos4.0

14.0sin 22

xxxdxx 4.0sin4.0cos4.0

14.0sin2 2

Cxxxdxx 5.04.0sin4.0cos8.0

14.0sin2

Reassociation between Steps

dxex x35

3

3

3

13 2

23

x

x

ex

exx

3

3

9

10

3

13 2

x

x

e

ex

reassociate

+

-

+

Integrating with substitution method

Ceexdxex xxx 333

3

1

3

1 35

9.4 Reduction Formulas Objective is to find a generic formula for

xdxncos

xdx6cosxxx

xx

sinsincos5

coscos4

5

xdxxxx 245 sincos5sincos

dxxxxx 245 cos1cos5sincos

xdxxdxxx 645 cos5cos5sincos

xdxxx 45 cos6

5sincos

6

1

xdxn

nxx

nxdx nnn 21 cos

1sincos

1cos

9 – 6 Integration by Trig Substitution9 – 6 Integration by Trig Substitution

249 x

7

Reminder:

1.

xxxxx

xxxxx

222

2

sincos2cos2cos12

1sin

cossin22sin2cos12

1cos

Since this reminds us of Pythagoras, let’s label a triangle and attempt to change into integration of trig.

Key to success is deciding whether the variable x is the leg or the hypotenuse. If we choose x as a leg we can choose which leg. Vertical or horizontal will affect which trig function we will use later on.

dxx249

x

7

49cos

7sin

2xx

d

ddx

xx

cos7cos7

cos7

49cos7sin7 2

dd 2cos12

149cos49 2

C

d

2sin2

1

2

49

2cos12

49

C cossin24

49

2

49

Cxxx

7

49

72

49

7sin

2

49 21

9 – 6 Integration by Trig Substitution9 – 6 Integration by Trig Substitution

2.When changing from must also change the bounds.

dtodx

2

024 x

dx

9 – 7 Integration of Rational Functions by Partial 9 – 7 Integration of Rational Functions by Partial FractionsFractions

101

x

B

x

A

101

1510

xx

x

10

1

1

1

101

15101

x

xB

x

xA

xx

xx

linear

ntconsta

linear

ntconsta

linear

ntconsta

Polynomial

Polynomial ...

1011

15102

xx

x

10

9115

1

925

1011

15102

xxxx

x

We are wanting to ‘undo’ the common denominator

Let x=1 so that we get rid of B and just have A left

10

1

10

1510

x

xBA

x

x

A

101

15110

10

10

1

10

101

151010

x

xB

x

xA

xx

xx

B

1

10

1

1510

x

xA

x

x

B

110

151010

SHORTCUT!!

101

1510

xx

x

101

x

B

x

A

Simply place finger over (x-1) when letting x=1 and solve to get A. Then place finger over (x-10) when solving for B.

Integration of Rational Functions by Partial Fractions

Examples:

1.

This is an example of a rational function in proper form. That is lower degree than .

Improper form

2.

3.

dxxxx

x

6

6223

xqxp xp

xq xqofegreedxpofegreed

dx

x

xxx

1

1116 23

dxxxx

x

11

14

With improper form we need to perform long division first. Partial fractions will work only with proper form.

Integration of Rational Functions by Partial Fractions

B

BAxx

x

Bx

x

Ax

x

xx

6

193

1

1

1

1

1

9312

22

2

2

A

xAx

xAx

Axx

3

113

133

6193

One problem that we may encounter is that when we factor the denominator, we find some factors repeated, that is occurring with multiplicities greater than 1.

Repeated linear factors

4. Trick to partial fractions:

dxx

x21

93

22 111

93

x

B

x

A

x

x

Cx

x

xxdx

x

x

1

61ln3

1

6

1

3

1

9322

Integration of Rational Functions by Partial Fractions

214 22 xCxBAxx

CCxBBxAxAxx 24 222

CBxBAxCAx 24 22

Unfactorable Quadratics:5. Trick:

dxxx

x

12

42

2 1212

422

2

x

C

x

BAx

xx

x

CB

BA

CA

24

0

1

1

1

2

22

12

422

2

xx

xdx

xx

x

dxx

dxx

dxx

x

1

1

2

2

2

222

9 – 8 Integrals of the Inverse Trig Functions9 – 8 Integrals of the Inverse Trig Functions

“Like climbing mount Everest, they are interesting more from the standpoint that they can be done, rather than because they are of great practical use.”

In 4 – 5 we found derivatives of inverse trig by implicit differentiation.

1

1csc

1

1sec

1

1cot

1

1tan

1

1cos

1

1sin

2

1

2

1

21

21

2

1

2

1

xxx

dx

d

xxx

dx

dx

xdx

d

xx

dx

dx

xdx

d

xx

dx

d

9 – 8 Integrals of the Inverse Trig Functions9 – 8 Integrals of the Inverse Trig Functions

Integrate using parts.

dx

x

xxxx

211

1cotcot

xx

x

2

1

1

1

1cot

x1cot+

-

Cxxxx 211 1ln2

1cotcot

Aside: our calc and the back of the book use the notion sgnx

Treat this as a constant even though we do not know the sign.

x

xx sgn

HW 1 – 9 odd

9 – 10 Improper Integrals

Improper Integrals:•Upper or lower limit of integration is infinite.•Integrand is discontinuous for at least one value of x at or between the limits of integration.

An improper integral converges to a certain number if each applicable limit shown below is finite. Otherwise, the integral diverges.

fdxxfdxxfdxxf

dxxfdxxf

dxxfdxxf

k

ack

b

kck

b

a

b

aa

b

a

b

ab

,)(lim)(lim)(

)(lim)(

)(lim)(

Is discontinuous at x=c in ba,

Improper Integrals

** If the integrand seems to approach zero as x gets very large or small, then the integral might converge. Conversely, if the integrand grows without bound as grows without bound, then the integral definitely diverges.

Examples:

1. For the improper integral

a. Graph the integrand, and tell whether or not the integral might converge.

b. If the integral might converge,find out whether or not it really does,and if so, to what limit it converges.

x

,11

07.0 dx

x

Graph looks to be approaching 0 and therefore might converge.

17.0

0

1

07.0

lim1

aa

dxxdxx

Improper Integrals

2. For the improper integral

a. Graph the integrand, and tell whether or not the integral might converge.

b. If the integral might converge, find out whether or not it really does, and if so, to what limit it converges.

3. For the improper integral

a. Graph the integrand, and tell whether or not the integral might

converge.

b. If the integral might converge, find out whether or not it really does, and if so, to what limit it converges.

,13

32

dxx

,3

4

x

dx

9.11 Miscellaneous Integrals and DerivativesDifferentiation Sum Product Quotient

Composite Implicit Power Function Exponential Function Logarithmic function

'vu '' vu 'vu

2

''

v

uvvu '

v

u

'' uvvu

'uf '' uuf

xgyf xgyyf '''

'nx 1 nnx

'xn nn x ln

'log xbbx ln

1

Differentiation Con’t

Logarithmic

Trigonometric Function

Inverse Trig Function – differentiate Implicitly

xfy

xfy lnln 'ln'1

xfyy

'ln' xfyy

xx

xx

xx

2sectan'

sincos'

cossin'

Integration Sum

Product

Reciprocal Function

Power Function

Power of a Function

dxvu vdxudx

udv vduuv

duu 1 Cu ln

duun 11

1 1

nCun

n

dxxf n formulareduction

Integration Continued

Square root of quadratic: Trig Sub Rational Algebraic Expression:

Convert to sum by long division, partial fractions

Inverse Function: Integrate by parts