508 Chapter 8 Integration Techniques, L’Hôpital’s Rule ... · Two of the most commonly overlooked integration rules are the Log Rule and the Power Rule. Notice in the next two

508 Chapter 8 Integration Techniques, L’Hôpital’s Rule, and Improper Integrals

8.1 Basic Integration Rules

Review procedures for fitting an integrand to one of the basic integration rules.

Fitting Integrands to Basic Integration RulesIn this chapter, you will study several integration techniques that greatly expand the setof integrals to which the basic integration rules can be applied. These rules are reviewedat the left. A major step in solving any integration problem is recognizing which basicintegration rule to use.

A Comparison of Three Similar Integrals

See LarsonCalculus.com for an interactive version of this type of example.

Find each integral.

a. b. c.

Solution

a. Use the Arctangent Rule and let and

Constant Multiple Rule

Arctangent Rule

Simplify.

b. The Arctangent Rule does not apply because the numerator contains a factor of Consider the Log Rule and let Then and you have

Constant Multiple Rule

Substitution:

Log Rule

Rewrite as a function of

c. Because the degree of the numerator is equal to the degree of the denominator, youshould first use division to rewrite the improper rational function as the sum of apolynomial and a proper rational function.

Rewrite using long division.

Write as two integrals.

Integrate.

Simplify.

Note in Example 1(c) that some algebra is required before applying any integrationrules, and more than one rule is needed to evaluate the resulting integral.

� 4x � 12 arctan x3

� C

� 4x � 36�13

arctan x3� � C

� � 4 dx � 36� 1

x2 � 9 dx

� 4x2

x2 � 9 dx � � �4 �

�36x2 � 9� dx

x. � 2 ln�x2 � 9� � C.

� 2 ln�u� � C

u � x2 � 9 � 2�duu

� 4x

x2 � 9 dx � 2�

2x dxx2 � 9

du � 2x dx,u � x2 � 9.x.

�43

arctan x3

� C

� 4�13

arctan x3� � C

� 4

x2 � 9 dx � 4�

1x2 � 32 dx

a � 3.u � x

� 4x2

x2 � 9 dx�

4xx2 � 9

dx� 4

x2 � 9 dx

REVIEW OF BASICINTEGRATION RULES

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20. � du

u�u2�a2�

1a

arcsec �u�a

�C

� dua2 � u2 �

1a

arctan ua

� C

� du�a2 � u2

� arcsin ua

� C

�csc u cot u du � �csc u � C

�sec u tan u du � sec u � C

�csc2 u du � �cot u � C

�sec2 u du � tan u � C

�ln�csc u � cot u� � C

�csc u du �

ln�sec u � tan u� � C

�sec u du �

�cot u du � ln�sin u� � C

� tan u du � �ln�cos u� � C

�cos u du � sin u � C

�sin u du � �cos u � C

�au du � � 1ln a�au � C

�eu du � eu � C

�duu

� ln�u� � C

n � �1

�un du �un�1

n � 1� C,

�du � u � C

� f �u� du ± �g�u� du

� f �u� ± g�u� du �

�kf �u� du � k� f �u� du

�a > 0�

9781285057095_0801.qxp 9/17/12 3:19 PM Page 508

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Using Two Basic Rules to Solve a Single Integral

Evaluate

Solution Begin by writing the integral as the sum of two integrals. Then apply thePower Rule and the Arcsine Rule.

See Figure 8.1.

Rules 18, 19, and 20 of the basic integration rules on the preceding page all haveexpressions involving the sum or difference of two squares:

and

These expressions are often apparent after a -substitution, as shown in Example 3.

A Substitution Involving

Find

Solution Because the radical in the denominator can be written in the form

you can try the substitution Then and you have

Rewrite integral.

Substitution:

Arcsine Rule

Rewrite as a function of x. �13

arcsin x3

4� C.

�13

arcsin u4

� C

u � x3 �13

� du

�42 � u2

� x2

�16 � x6 dx �

13�

3x2 dx�16 � �x3�2

du � 3x2 dx,u � x3.

�a2 � u2 � �42 � �x3�2

� x2

�16 � x6 dx.

a2 � u2

u

u2 � a2.a2 � u2,a2 � u2,

1.839

� ��3 ��

2� � ��2 � 0�

� ��4 � x2�1�2 � 3 arcsin x2�

1

0

� �12

�1

0 �4 � x2��1�2��2x� dx � 3 �1

0

1�22 � x2

dx

�1

0

x � 3�4 � x2

dx � �1

0

x�4 � x2

dx � �1

0

3�4 � x2

dx

�1

0

x � 3�4 � x2

dx.

8.1 Basic Integration Rules 509

TECHNOLOGY Simpson’s Rule can be used to give a good approximation ofthe value of the integral in Example 2 (for the approximation is 1.839).When using numerical integration, however, you should be aware that Simpson’sRule does not always give good approximations when one or both of the limits ofintegration are near a vertical asymptote. For instance, using the FundamentalTheorem of Calculus, you can obtain

For Simpson’s Rule gives an approximation of 6.889.n � 10,

�1.99

0

x � 3�4 � x2

dx 6.213.

n � 10,

Exploration

A Comparison of ThreeSimilar Integrals Which,if any, of the integrals listedbelow can be evaluated usingthe 20 basic integration rules?For any that can be evaluated,do so. For any that cannot,explain why not.

a.

b.

c. � 3x2

�1 � x2 dx

� 3x

�1 � x2 dx

� 3

�1 � x2 dx

2

1

1−1x

4 − x2y =

x + 3

y

The area of the region is approximately1.839.Figure 8.1

9781285057095_0801.qxp 9/17/12 3:19 PM Page 509



Two of the most commonly overlooked integration rules are the Log Rule and thePower Rule. Notice in the next two examples how these two integration rules can bedisguised.

A Disguised Form of the Log Rule

Find

Solution The integral does not appear to fit any of the basic rules. The quotient form,however, suggests the Log Rule. If you let then You can obtainthe required by adding and subtracting in the numerator.

Add and subtract in numerator.

Rewrite as two fractions.

Rewrite as two integrals.

Integrate.

There is usually more than one way to solve an integration problem. For instance,in Example 4, try integrating by multiplying the numerator and denominator by toobtain an integral of the form See if you can get the same answer by this procedure. (Be careful: the answer will appear in a different form.)

A Disguised Form of the Power Rule

Find

Solution Again, the integral does not appear to fit any of the basic rules. However,considering the two primary choices for

or

you can see that the second choice is the appropriate one because

and

So,

Substitution:

Integrate.

Rewrite as a function of

In Example 5, try checking that the derivative of

is the integrand of the original integral.

12

ln�sin x�2 � C

x. �12

ln�sin x�2 � C.

�u2

2� C

u � ln�sin x� � �cot x�ln�sin x� dx � � u du

du �cos xsin x

dx � cot x dx.u � ln�sin x�

u � ln�sin x�u � cot x

u

� �cot x�ln�sin x� dx.

�� du�u.e�x

� x � ln�1 � ex� � C

� � dx � � ex dx

1 � ex

� � �1 � ex

1 � ex �ex

1 � ex� dx

e x � 1

1 � ex dx � � 1 � ex � ex

1 � ex dx

exdudu � ex dx.u � 1 � ex,

� 1

1 � ex dx.


REMARK Remember thatyou can separate numerators butnot denominators. Watch out forthis common error when fittingintegrands to basic rules. Forinstance, you cannot separatedenominators in Example 4.

11 � ex �

11

�1ex

9781285057095_0801.qxp 9/17/12 3:19 PM Page 510



Trigonometric identities can often be used to fit integrals to one of the basicintegration rules.

Using Trigonometric Identities

Find

Solution Note that is not in the list of basic integration rules. However,is in the list. This suggests the trigonometric identity If you let

then and

Substitution:

Trigonometric identity

Rewrite as two integrals.

Integrate.

Rewrite as a function of x.

This section concludes with a summary of the common procedures for fitting integrands to the basic integration rules.

�12

tan 2x � x � C.

�12

tan u �u2

� C

�12� sec2 u du �

12� du

�12� �sec2 u � 1� du

u � 2x � tan2 2x dx �12� tan2 u du

du � 2 dxu � 2x,tan2 u � sec2 u � 1.

sec2 utan2 u

� tan2 2x dx.


PROCEDURES FOR FITTING INTEGRANDS TO BASIC INTEGRATION RULES

Technique Example

Expand (numerator).

Separate numerator.

Complete the square.

Divide improper rational function.

Add and subtract terms in numerator.

Use trigonometric identities.

Multiply and divide by Pythagorean conjugate.

� sec2 x �sin xcos2 x

�1 � sin x

cos2 x

�1 � sin x1 � sin2 x

1

1 � sin x� � 1

1 � sin x��1 � sin x1 � sin x�

cot2 x � csc2 x � 1

�2x � 2

x2 � 2x � 1�

2�x � 1�2

2x

x2 � 2x � 1�

2x � 2 � 2x2 � 2x � 1

x2

x2 � 1� 1 �

1x2 � 1

1�2x � x2

�1

�1 � �x � 1�2

1 � xx2 � 1

�1

x2 � 1�

xx2 � 1

�1 � ex�2 � 1 � 2ex � e2x

TECHNOLOGY If you haveaccess to a computer algebrasystem, try using it to evaluatethe integrals in this section.Compare the forms of the antiderivatives given by the software with the formsobtained by hand. Sometimesthe forms will be the same, butoften they will differ. Forinstance, why is the antiderivative

equivalent to the antiderivative ln x � C?ln 2x � C

9781285057095_0801.qxp 9/17/12 3:19 PM Page 511




8.1 Exercises See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.

Choosing an Antiderivative In Exercises 1–4, select thecorrect antiderivative.

1.

(a) (b)

(c) (d)

2.

(a) (b)

(c) (d)

3.

(a) (b)

(c) (d)

4.

(a) (b)

(c) (d)

Choosing a Formula In Exercises 5–14, select the basicintegration formula you can use to find the integral, and identify and when appropriate.

5. 6.

7. 8.

9. 10.

11. 12.

13. 14.

Finding an Indefinite Integral In Exercises 15–46, findthe indefinite integral.

15. 16.

17. 18.

19. 20.

21. 22.

23. 24.

25. 26.

27. 28.

29. 30.

31. 32.

33. 34.

35. 36.

37. 38.

39. 40.

41. 42.

43.

44.

45. 46.

Slope Field In Exercises 47 and 48, a differential equation,a point, and a slope field are given. (a) Sketch two approximatesolutions of the differential equation on the slope field, one ofwhich passes through the given point. (b) Use integration tofind the particular solution of the differential equation and usea graphing utility to graph the solution. Compare the resultwith the sketches in part (a). To print an enlarged copy of thegraph, go to MathGraphs.com.

47. 48.

4

−1

−2

1

2

x

y

t

s

1−1

1

−1

�2, 12��0, �

12�

dydx

�1

�4x � x2

dsdt

�t

�1 � t4

� 1

x2 � 4x � 9 dx�

44x2 � 4x � 65

dx

� 1

�x � 1��4x2 � 8x � 3 dx

� 6

�10x � x2 dx

� e1�t

t 2 dt� tan�2�t�

t 2 dt

� 1

25 � 4x2 dx� �1

�1 � �4t � 1�2 dt

� 1

cos � � 1 d��

1 � cos �sin �

d�

� �tan x�ln�cos x� dx� ln x2

x dx

� 2

7ex � 4 dx�

2e�x � 1

dx

� csc2 xecot x dx� sin x

�cos x dx

� csc �x cot �x dx� x cos 2�x2 dx

� x�3 �2x�

2

dx� �5 � 4x2�2 dx

� � 12x � 5

�1

2x � 5� dx� ex

1 � ex dx

� 3x

x � 4 dx�

x2

x � 1 dx

� x � 1

�3x2 � 6x dx�

t 2 � 3�t3 � 9t � 1

dt

� �4x �2

�2x � 3�2� dx� �v �1

�3v � 1�3� dv

� t 3�t 4 � 1 dt� 7

�z � 10�7 dz

� 5

�t � 6�3 dt� 14�x � 5�6 dx

� 1

x�x2 � 4 dx� �cos x�esin x dx

� sec 5x tan 5x dx� t sin t 2 dt

� �2x

�x2 � 4 dx�

3�1 � t 2

dt

� 2

�2t � 1�2 � 4 dt�

1�x �1 � 2�x � dx

� 2t � 1

t 2 � t � 4 dt� �5x � 3�4 dx

au

�2x sin�x2 � 1� � C12 sin�x2 � 1� � C

�12 sin�x2 � 1� � C2x sin�x2 � 1� � C

dydx

� x cos�x2 � 1�

ln�x2 � 1� � Carctan x � C

2x�x2 � 1�2 � Cln�x2 � 1 � C

dydx

�1

x2 � 1

ln�x2 � 1� � Carctan x � C

2x�x2 � 1�2 � Cln�x2 � 1 � C

dydx

�x

x2 � 1

ln�x2 � 1� � C12�x2 � 1 � C

�x2 � 1 � C2�x2 � 1 � C

dydx

�x

�x2 � 1

9781285057095_0801.qxp 9/17/12 3:19 PM Page 512




Slope Field In Exercises 49 and 50, use a computer algebrasystem to graph the slope field for the differential equation andgraph the solution through the specified initial condition.

49.

50.

Differential Equation In Exercises 51–56, solve the differ-ential equation.

51. 52.

53. 54.

55. 56.

Evaluating a Definite Integral In Exercises 57–64,evaluate the definite integral. Use the integration capabilities ofa graphing utility to verify your result.

57. 58.

59. 60.

61. 62.

63. 64.

Area In Exercises 65–68, find the area of the region.

65. 66.

67. 68.

Finding an Integral Using Technology In Exercises69–72, use a computer algebra system to find the integral. Usethe computer algebra system to graph two antiderivatives.Describe the relationship between the graphs of the two antiderivatives.

69. 70.

71. 72.

77. Finding Constants Determine the constants and suchthat

Use this result to integrate

78. Deriving a Rule Show that

Then use this identity to derive the basic integration rule

79. Area The graphs of and intersect at thepoints and Find such that the areaof the region bounded by the graphs of these two functions is

80. Think About It When evaluating

is it appropriate to substitute

and

to obtain

Explain.

12

�1

1

�u du � 0?

dx �du

2�ux � �u,u � x2,

�1

�1 x2 dx

23.

a �a > 0��1�a, 1�a�.�0, 0�g�x� � ax2f �x� � x

� sec x dx � ln�sec x � tan x� � C.

sec x �sin xcos x

�cos x

1 � sin x.

� dx

sin x � cos x.

sin x � cos x � a sin�x � b�.

ba

� �ex � e�x

2 �3

dx� 1

1 � sin � d�

� x � 2

x2 � 4x � 13 dx�

1x2 � 4x � 13

dx

4π

0.5

1.0

x

y

−2 2

−1

−2

1

2

y

x

y � sin 2xy2 � x2�1 � x2�

y

x1 2 3 4 5

0.2

0.4

0.6

0.8

y

x(1.5, 0)

−1 1 2

5

10

15

y �3x � 2x2 � 9

y � ��4x � 6�3�2

�7

0

1�100 � x2

dx�2��3

0

14 � 9x2 dx

�3

1 2x2

� 3x � 2x

dx�8

0

2x�x2 � 36

dx

�e

1 1 � ln x

x dx�1

0 xe�x2

dx

��

0 sin2 t cos t dt��4

0 cos 2x dx

y �1

x�4x2 � 9�4 � tan2 x�y � sec2 x

drdt

��1 � et�2

e3t

drdt

�10et

�1 � e2t

dydx

� �4 � e2x�2dydx

� �ex � 5�2

y�0� � 1dydx

� 5 � y,

y�0� � 4dydx

� 0.8y,

WRITING ABOUT CONCEPTSChoosing a Formula In Exercises 73–76, state the integration formula you would use to perform the integration.Explain why you chose that formula. Do not integrate.

73.

74.

75. 76. � 1

x2 � 1 dx�

xx2 � 1

dx

� x sec�x2 � 1� tan�x2 � 1� dx

� x�x2 � 1�3 dx

9781285057095_0801.qxp 9/17/12 3:19 PM Page 513




81. Comparing Antiderivatives

(a) Explain why the antiderivative is equivalent tothe antiderivative

(b) Explain why the antiderivative is equivalent to the antiderivative

Approximation In Exercises 83 and 84, determine whichvalue best approximates the area of the region between the -axis and the function over the given interval. (Make your

selection on the basis of a sketch of the region and not by integrating.)

83.

(a) 3 (b) 1 (c) (d) 8 (e) 10

84.

(a) 3 (b) 1 (c) (d) 4 (e) 10

Interpreting Integrals In Exercises 85 and 86, (a) sketchthe region whose area is given by the integral, (b) sketch thesolid whose volume is given by the integral when the diskmethod is used, and (c) sketch the solid whose volume is givenby the integral when the shell method is used. (There is morethan one correct answer for each part.)

85. 86.

87. Volume The region bounded by and is revolved about the -axis.

(a) Find the volume of the solid generated when

(b) Find such that the volume of the generated solid is cubic units.

88. Volume Consider the region bounded by the graphs ofand Find the

volume of the solid generated by revolving the region aboutthe -axis.

89. Arc Length Find the arc length of the graph offrom to

90. Arc Length Find the arc length of the graph offrom to

91. Surface Area Find the area of the surface formed byrevolving the graph of on the interval aboutthe axis.

92. Centroid Find the coordinate of the centroid of theregion bounded by the graphs of

and

Average Value of a Function In Exercises 93 and 94, findthe average value of the function over the given interval.

93.

94. is a positive integer.

Arc Length In Exercises 95 and 96, use the integrationcapabilities of a graphing utility to approximate the arc lengthof the curve over the given interval.

95. 96.

97. Finding a Pattern

(a) Find

(b) Find

(c) Find

(d) Explain how to find without actually integrating.

98. Finding a Pattern

(a) Write in terms of Then find

(b) Write in terms of

(c) Write where is a positive integer, interms of

(d) Explain how to find without actuallyintegrating.

99. Methods of Integration Show that the followingresults are equivalent.

Integration by tables:

Integration by computer algebra system:

� �x2 � 1 dx �12

x�x2 � 1 � arcsinh�x� � C

� �x2 � 1 dx �12

�x�x2 � 1 � ln�x � �x2 � 1�� C

� tan15 x dx

� tan2k�1 x dx.k� tan2k�1 x dx,

� tan3 x dx.� tan5 x dx

� tan3 x dx.� tan x dx.� tan3 x dx

� cos15 x dx

� cos7 x dx.

� cos5 x dx.

� cos3 x dx.

1, 8y � x 2�3,0, 14y � tan �x,

n0 x ��n,f �x� � sin nx,

�3 x 3f �x� �1

1 � x2,

x � 4.x � 0,y � 0,y �5

�25 � x2,

x-

x-0, 9y � 2�x

x � ��3.x � 0y � ln�cos x�

x � ��2.x � ��4y � ln�sin x�

y

x � �� 2.y � sin x2,y � cos x2,x � 0,

43b

b � 1.

yx � b �b > 0�x � 0,y � 0,y � e�x 2,

�4

0 �y dy�2

0 2�x2 dx

�4

0, 2f �x� �4

x2 � 1,

�8

0, 2f �x� �4x

x2 � 1,

x

y2 � tan2 x � C.y1 � sec2 x � C1

y2 � Ce x.y1 � e x�C1

82. HOW DO YOU SEE IT? Using the graph, is

positive or negative? Explain.

x

y

1 2 3 4 6

1

2

−3

3

f(x) = (x3 − 7x2 + 10x)15

�5

0 f �x� dx

PUTNAM EXAM CHALLENGE

100. Evaluate

This problem was composed by the Committee on the Putnam Prize Competition.© The Mathematical Association of America. All rights reserved.

�4

2

�ln�9 � x� dx�ln�9 � x� � �ln�x � 3�.

9781285057095_0801.qxp 9/17/12 3:19 PM Page 514



8.2 Integration by Parts 515

8.2 Integration by Parts

Find an antiderivative using integration by parts.

Integration by PartsIn this section, you will study an important integration technique called integration byparts. This technique can be applied to a wide variety of functions and is particularlyuseful for integrands involving products of algebraic and transcendental functions. Forinstance, integration by parts works well with integrals such as

and

Integration by parts is based on the formula for the derivative of a product

where both and are differentiable functions of When and are continuous, youcan integrate both sides of this equation to obtain

By rewriting this equation, you obtain the next theorem.

This formula expresses the original integral in terms of another integral. Depending onthe choices of and it may be easier to evaluate the second integral than the original one. Because the choices of and are critical in the integration by partsprocess, the guidelines below are provided.

When using integration by parts, note that you can first choose or first chooseAfter you choose, however, the choice of the other factor is determined—it must be

the remaining portion of the integrand. Also note that must contain the differentialof the original integral.dx

dvu.

dv

dvudv,u

� � u dv � � v du.

uv � � uv� dx � � vu� dx

v�u�x.vu

� uv� � vu�

ddx

�uv� � u dvdx

� v dudx

� ex sin x dx.� x2 ex dx,� x ln x dx,

THEOREM 8.1 Integration by Parts

If and are functions of and have continuous derivatives, then

� u dv � uv � � v du.

xvu

GUIDELINES FOR INTEGRATION BY PARTS

1. Try letting be the most complicated portion of the integrand that fits a basicintegration rule. Then will be the remaining factor(s) of the integrand.

2. Try letting be the portion of the integrand whose derivative is a functionsimpler than Then will be the remaining factor(s) of the integrand.

Note that always includes the of the original integrand.dxdv

dvu.u

udv

Exploration

Proof Without Words Here is a different approach to proving the formula for integration by parts. Thisapproach is from “ProofWithout Words: Integration by Parts” by Roger B. Nelsen,Mathematics Magazine, 64,No. 2, April 1991, p. 130, bypermission of the author.

Explain how this graph proves the theorem. Whichnotation in this proof is unfamiliar? What do you think it means?

u

s = g(b)

r = g(a)

u = f(x) v = g(x)

p = f(a) q = f(b)

v

�s

r u dv � �uv�

�q, s�

�p, r��p

q

v du

�s

r

u dv ��p

q

v du � �uv��q, s�

�p, r�

Area ��Area �� qs � pr

9781285057095_0802.qxp 9/17/12 3:20 PM Page 515



Integration by Parts

Find

Solution To apply integration by parts, you need to write the integral in the formThere are several ways to do this.

u dv u dv u dv u dv

The guidelines on the preceding page suggest the first option because the derivative ofis simpler than and is the most complicated portion of the integrand

that fits a basic integration formula.

Now, integration by parts produces

Integration by parts formula

Substitute.

Integrate.

To check this, differentiate to see that you obtain the original integrand.


Find

Solution In this case, is more easily integrated than Furthermore, thederivative of is simpler than So, you should let

Integration by parts produces

Integration by parts formula

Substitute.

Simplify.

Integrate.

You can check this result by differentiating.

ddx�

x3

3 ln x �

x3

9� C� �

x3

3 1x � �ln x��x2� �

x2

3� x2 ln x

�x3

3 ln x �

x3

9� C.

�x3

3 ln x �

13

� x2 dx

� x2 ln x dx �

x3

3 ln x � � x3

3 1x dx

� u dv � uv � � v du

du �1x dx u � ln x

v � � x2 dx �

x3

3dv � x2 dx

dv � x2 dx.ln x.ln xln x.x2

� x2 ln x dx.

xex � ex � C

� xex � ex � C.

� xex dx � xex � � ex dx


du � dx u � x

v � � dv � � ex dx � exdv � ex dx

dv � ex dxx,u � x

� �xex��dx�� 1� �xex dx�,� �ex��x dx�,� �x� �ex dx�,

� u dv.

� xex dx.


REMARK In Example 1,note that it is not necessary toinclude a constant of integrationwhen solving

To illustrate this, replace by and apply integration by parts to see thatyou obtain the same result.

v � ex � C1

v � ex

v � � ex dx � ex � C1.

TECHNOLOGY Try graphing

and

on your graphing utility. Do youget the same graph? (This maytake a while, so be patient.)

x3

3 ln x �

x3

9� x2 ln x dx

9781285057095_0802.qxp 9/17/12 3:20 PM Page 516



One surprising application of integration by parts involves integrands consisting ofsingle terms, such as

or

In these cases, try letting as shown in the next example.

An Integrand with a Single Term

Evaluate

Solution Let


Substitute.

Rewrite.

Integrate.

Using this antiderivative, you can evaluate the definite integral as shown.

The area represented by this definite integral is shown in Figure 8.2.

� 0.571

��

2� 1

�1

0 arcsin x dx � �x arcsin x � 1 � x2 �

1

0

� x arcsin x � 1 � x2 � C.

� x arcsin x �12

� �1 � x2��1�2 ��2x� dx

� arcsin x dx � x arcsin x � � x

1 � x2 dx

Integration by partsformula


du �1

1 � x2 dx u � arcsin x

v � �dx � xdv � dx

dv � dx.

�1

0 arcsin x dx.

dv � dx,

� arcsin x dx.� ln x dx


TECHNOLOGY Remember that there are two ways to use technology to evaluate a definite integral: (1) you can use a numerical approximation such as theTrapezoidal Rule or Simpson’s Rule, or (2) you can use a computer algebra systemto find the antiderivative and then apply the Fundamental Theorem of Calculus. Bothmethods have shortcomings. To find the possible error when using a numericalmethod, the integrand must have a second derivative (Trapezoidal Rule) or a fourthderivative (Simpson’s Rule) in the interval of integration; the integrand in Example 3fails to meet either of these requirements. To apply the Fundamental Theorem ofCalculus, the symbolic integration utility must be able to find the antiderivative.

FOR FURTHER INFORMATION To see how integration by parts is used to proveStirling’s approximation

see the article “The Validity of Stirling’s Approximation: A Physical Chemistry Project”by A. S. Wallner and K. A. Brandt in Journal of Chemical Education.

ln�n!� � n ln n � n

y = arcsin x

x

2 ))1,

1

y

2π

π

The area of the region is approximately0.571.Figure 8.2

9781285057095_0802.qxp 9/17/12 3:20 PM Page 517



Some integrals require repeated use of the integration by parts formula.

Repeated Use of Integration by Parts

Find

Solution The factors and sin are equally easy to integrate. However, thederivative of becomes simpler, whereas the derivative of sin does not. So, youshould let


First use of integration by parts

This first use of integration by parts has succeeded in simplifying the original integral,but the integral on the right still doesn’t fit a basic integration rule. To evaluate that integral, you can apply integration by parts again. This time, let


Second use of integration by parts

Combining these two results, you can write

When making repeated applications of integration by parts, you need to be carefulnot to interchange the substitutions in successive applications. For instance, in Example 4,the first substitution was and If, in the second application, youhad switched the substitution to and you would have obtained

thereby undoing the previous integration and returning to the original integral. Whenmaking repeated applications of integration by parts, you should also watch for theappearance of a constant multiple of the original integral. For instance, this occurs whenyou use integration by parts to evaluate and it also occurs in Example 5on the next page.

The integral in Example 5 is an important one. In Section 8.4 (Example 5), you willsee that it is used to find the arc length of a parabolic segment.

� ex cos 2x dx,

� � x2 sin x dx

� �x2 cos x � x2 cos x � � x2 sin x dx

� x2 sin x dx � �x2 cos x � � 2x cos x dx

dv � 2x,u � cos xdv � sin x dx.u � x2

� x2 sin x dx � �x2 cos x � 2x sin x � 2 cos x � C.

� 2x sin x � 2 cos x � C.

� 2x cos x dx � 2x sin x � � 2 sin x dx

du � 2 dx u � 2x

v � � cos x dx � sin xdv � cos x dx

u � 2x.

� x2 sin x dx � �x2 cos x � � 2x cos x dx.

du � 2x dx u � x2

v � � sin x dx � �cos xdv � sin x dx

u � x2.xx2

xx2

� x2 sin x dx.


9781285057095_0802.qxp 9/17/12 3:20 PM Page 518




Find

Solution The most complicated portion of the integrand that can be easily integrated isso you should let and


Substitute.

Trigonometric identity

Rewrite.

Collect like integrals.

Integrate.

Divide by 2.

Finding a Centroid

A machine part is modeled by the region bounded by the graph of and the axis, as shown in Figure 8.3. Find the centroid of this region.

Solution Begin by finding the area of the region.

Now, you can find the coordinates of the centroid. To evaluate the integral for firstrewrite the integrand using the trigonometric identity

You can evaluate the integral for with integration by parts. Todo this, let and This produces and and youcan write

Finally, you can determine to be

So, the centroid of the region is �1, ��8�.

x �1A

��2

0 x sin x dx � ��x cos x � sin x�

��2

0� 1.

x

� �x cos x � sin x � C. � x sin x dx � �x cos x � � cos x dx

du � dx,v � �cos xu � x.dv � sin x dxx, �1�A� ��2

0 x sin x dx,

y �1A

��2

0 sin x

2�sin x� dx �

14

��2

0 �1 � cos 2x� dx �

14 �x �

sin 2x2 �

��2

0�

�

8

sin2 x � �1 � cos 2x��2.y,

A � ��2

0 sin x dx � ��cos x�

��2

0� 1

0 � x � ��2,x-y � sin x

� sec3 x dx �12

sec x tan x �12

ln�sec x � tan x� � C.

2 � sec3 x dx � sec x tan x � ln�sec x � tan x� � C

2 � sec3 x dx � sec x tan x � � sec x dx

� sec3 x dx � sec x tan x � � sec3 x dx � � sec x dx

� sec3 x dx � sec x tan x � � sec x�sec2 x � 1� dx

� sec3 x dx � sec x tan x � � sec x tan2 x dx

Integration by partsformula


du � sec x tan x dx u � sec x

v � � sec2 x dx � tan xdv � sec2 x dx

u � sec x.dv � sec2 x dxsec2 x,

� sec3 x dx.


1

x

x

Δx

sin x2

y = sin x

y

2 )) , 1

2π

π

Figure 8.3

9781285057095_0802.qxp 9/17/12 3:20 PM Page 519



As you gain experience in using integration by parts, your skill in determining and will increase. The next summary lists several common integrals with suggestionsfor the choices of and

In problems involving repeated applications of integration by parts, a tabularmethod, illustrated in Example 7, can help to organize the work. This method workswell for integrals of the form

and

Using the Tabular Method

See LarsonCalculus.com for an interactive version of this type of example.

Find

Solution Begin as usual by letting and Next, createa table consisting of three columns, as shown.

Alternate and Its and ItsSigns Derivatives Antiderivatives

2

0

The solution is obtained by adding the signed products of the diagonal entries:

� x2 sin 4x dx � �14

x2 cos 4x �18

x sin 4x �132

cos 4x � C.

Differentiate until you obtain 0 as a derivative.

164 cos 4x�

�116 sin 4x�

�14 cos 4x2x�

sin 4xx2�

v�u

dv � v� dx � sin 4x dx.u � x2

� x2 sin 4x dx.

� xn eax dx.� xn cos ax dx,� xn sin ax dx,

dv.udv

u


SUMMARY: COMMON INTEGRALS USING INTEGRATION BY PARTS

1. For integrals of the form

or

let and let or


or

let or and let


or

let or and let dv � eax dx.cos bxu � sin bx

� eax cos bx dx� eax sin bx dx

dv � xn dx.arctan axarcsin ax,u � ln x,

� xn arctan ax dx� xn arcsin ax dx,� xn ln x dx,

cos ax dx.sin ax dx,dv � eax dx,u � xn

� xn cos ax dx� xn sin ax dx,� xn eax dx,

REMARK You can use theacronym LIATE as a guidelinefor choosing in integration byparts. In order, check the integrand for the following.

Is there a Logarithmic part?

Is there an Inverse trigonometricpart?

Is there an Algebraic part?

Is there a Trigonometric part?

Is there an Exponential part?

u

FOR FURTHER INFORMATIONFor more information on the tabular method, see the article“Tabular Integration by Parts” byDavid Horowitz in The CollegeMathematics Journal, and the article “More on Tabular Integrationby Parts” by Leonard Gillman inThe College Mathematics Journal.To view these articles, go toMathArticles.com.

9781285057095_0802.qxp 9/17/12 3:20 PM Page 520




8.2 Exercises See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises.

Setting Up Integration by Parts In Exercises 1–6,identify and for finding the integral using integration byparts. (Do not evaluate the integral.)

1. 2.

3. 4.

5. 6.

Using Integration by Parts In Exercises 7–10, evaluatethe integral using integration by parts with the given choices of

and

7.

8.

9.

10.

Finding an Indefinite Integral In Exercises 11–30, findthe indefinite integral. (Note: Solve by the simplest method—not all require integration by parts.)

11. 12.

13. 14.

15. 16.

17. 18.

19. 20.

21. 22.

23. 24.

25. 26.

27. 28.

29. 30.

Differential Equation In Exercises 31–34, solve the differential equation.

31. 32.

33. 34.

Slope Field In Exercises 35 and 36, a differential equation,a point, and a slope field are given. (a) Sketch two approximatesolutions of the differential equation on the slope field, one ofwhich passes through the given point. (b) Use integration tofind the particular solution of the differential equation and usea graphing utility to graph the solution. Compare the resultwith the sketches in part (a). To print an enlarged copy of thegraph, go to MathGraphs.com.

35. 36.

Slope Field In Exercises 37 and 38, use a computer algebrasystem to graph the slope field for the differential equation andgraph the solution through the specified initial condition.

37. 38.

Evaluating a Definite Integral In Exercises 39–48,evaluate the definite integral. Use a graphing utility to confirmyour result.

39. 40.

41. 42.

43. 44.

45. 46.

47. 48.

Using the Tabular Method In Exercises 49–54, use thetabular method to find the integral.

49. 50.

51. 52. � x3 cos 2x dx� x3 sin x dx

� x3e�2x dx� x2e2x dx

��8

0 x sec2 2x dx�4

2 x arcsec x dx

�1

0 ln�4 � x2� dx�1

0 e x sin x dx

�1

0 x arcsin x2 dx�1�2

0 arccos x dx

��

0 x sin 2x dx��4

0 x cos 2x dx

�2

0 x2e�2x dx�3

0 xex�2 dx

y�0� � 4dydx

�xy sin x,y�0� � 2

dydx

�xyex�8,

x4

5

−6

−5

y

x42−2

11

y

−4

�0, �1837�dy

dx� e�x�3 sin 2x,�0, 4�dy

dx� x y cos x,

dydx

� x2 x � 3dydt

�t2

3 � 5t

y� � arctan x2

y� � ln x

� e4x cos 2x dx� e�3x sin 5x dx

� 4 arccos x dx� arctan x dx

� x2 cos x dx� x3 sin x dx

� t csc t cot t dt� x cos x dx

� x

6x � 1 dx� x x � 5 dx

� x3e x2

�x2 � 1�2 dx� xe2x

�2x � 1�2 dx

� ln xx3 dx�

�ln x�2

x dx

� x5 ln 3x dx� t ln�t � 1� dt

� e1�t

t2 dt� x3e x dx

� 5xe2x dx� xe�4x dx

� x cos 4x dx; u � x, dv � cos 4x dx

� x sin 3x dx; u � x, dv � sin 3x dx

� �4x � 7)ex dx; u � 4x � 7, dv � ex dx

� x3 ln x dx; u � ln x, dv � x3 dx

dv.u

� x2 cos x dx� x sec2 x dx

� ln 5x dx� �ln x�2 dx

� x2e2x dx� xe2x dx

dvu

9781285057095_0802.qxp 9/17/12 3:20 PM Page 521




53. 54.

Using Two Methods Together In Exercises 55–58, findthe indefinite integral by using substitution followed by integration by parts.

55. 56.

57. 58.

63. Using Two Methods Integrate

(a) by parts, letting

(b) by substitution, letting

64. Using Two Methods Integrate

(a) by parts, letting

(b) by substitution, letting

Finding a General Rule In Exercises 65 and 66, use a computer algebra system to find the integrals for 1, 2,and 3. Use the result to obtain a general rule for the integralsfor any positive integer and test your results for

65. 66.

Proof In Exercises 67–72, use integration by parts to provethe formula. (For Exercises 67–70, assume that is a positiveinteger.)

67.

68.

69.

70.

71.

72.

Using Formulas In Exercises 73–78, find the integral byusing the appropriate formula from Exercises 67–72.

73. 74.

75. 76.

77. 78.

Area In Exercises 79–82, use a graphing utility to graph theregion bounded by the graphs of the equations. Then find thearea of the region analytically.

79.

80.

81.

82.

83. Area,Volume, and Centroid Given the region boundedby the graphs of and find

(a) the area of the region.

(b) the volume of the solid generated by revolving the regionabout the axis.

(c) the volume of the solid generated by revolving the regionabout the axis.

(d) the centroid of the region.

y-

x-

x � e,y � 0,y � ln x,

x � 3x � 1,y � 0,y � x3 ln x,

x � 1y � 0,y � e�x sin �x,

x � 2x � 0,y � 0,y �1

10xe3x,

x � 3y � 0,y � 2xe�x,

� e2x cos 3x dx� e�3x sin 4x dx

� x3e2x dx� x5 ln x dx

� x2 cos x dx� x2 sin x dx

� eax cos bx dx �eax�a cos bx � b sin bx�

a2 � b2 � C

� eax sin bx dx �eax�a sin bx � b cos bx�

a2 � b2 � C

� x neax dx �x neax

a�

na

� x n�1 eax dx

� x n ln x dx �x n�1

�n � 1�2 ��1 � �n � 1� ln x� � C

� x n cos x dx � x n sin x � n � x n�1 sin x dx

� x n sin x dx � �x n cos x � n � x n�1 cos x dx

n

� x ne x dx� x n ln x dx

n � 4.n

n � 0,

u � 4 � x.

dv � 4 � x dx.

� x 4 � x dx

u � 4 � x2.

dv �x

4 � x2 dx.

� x3

4 � x2 dx

� e 2x dx� x5ex2 dx

� 2x3 cos x2 dx� sin x dx

�

x2�x � 2�3�2 dx� x sec2 x dx

WRITING ABOUT CONCEPTS59. Integration by Parts

(a) Integration by parts is based on what differentiationrule? Explain.

(b) In your own words, state how you determine whichparts of the integrand should be and

60. Integration by Parts When evaluating explain how letting and makes thesolution more difficult to find.

61. Integration by Parts State whether you would useintegration by parts to evaluate each integral. If so, identifywhat you would use for and Explain your reasoning.

(a) (b) (c)

(d) (e) (f) � x

x2 � 1 dx�

x x � 1

dx� 2xex2 dx

� x2e�3x dx� x ln x dx� ln x

x dx

dv.u

dv � x dxu � sin x�x sin x dx,

dv.u

62. HOW DO YOU SEE IT? Use the graph of shown in the figure to answer the following.

(a) Approximate the slope of at Explain.

(b) Approximate any open intervals in which the graphof is increasing and any open intervals in which itis decreasing. Explain.

f

x � 2.f

x

y

−1 1 2 3 4−1

1

2

3

4

f ′(x) = x ln x

f�

9781285057095_0802.qxp 9/17/12 3:20 PM Page 522




84. Area,Volume, and Centroid Given the region boundedby the graphs of and find

(a) the area of the region.

(b) the volume of the solid generated by revolving the regionabout the -axis.

(c) the volume of the solid generated by revolving the regionabout the -axis.

(d) the centroid of the region.

85. Centroid Find the centroid of the region bounded by thegraphs of and How is this problem related to Example 6 in this section?

86. Centroid Find the centroid of the region bounded by thegraphs of and

87. Average Displacement A damping force affects thevibration of a spring so that the displacement of the spring isgiven by

Find the average value of on the interval from to

Present Value In Exercises 89 and 90, find the present valueof a continuous income flow of dollars per year for

where is the time in years and is the annual interest ratecompounded continuously.

89.

90.

Integrals Used to Find Fourier Coefficients InExercises 91 and 92, verify the value of the definite integral,where is a positive integer.

91.

92.

93. Vibrating String A string stretched between the twopoints and is plucked by displacing the string units at its midpoint. The motion of the string is modeled by aFourier Sine Series whose coefficients are given by

Find

94. Euler’s Method Consider the differential equationwith the initial condition

(a) Use integration to solve the differential equation.

(b) Use a graphing utility to graph the solution of the differential equation.

(c) Use Euler’s Method with and the recursivecapabilities of a graphing utility, to generate the first 80 points of the graph of the approximate solution. Use thegraphing utility to plot the points. Compare the result withthe graph in part (b).

(d) Repeat part (c) using and generate the first 40 points.

(e) Why is the result in part (c) a better approximation of thesolution than the result in part (d)?

Euler’s Method In Exercises 95 and 96, consider thedifferential equation and repeat parts (a)–(d) of Exercise 94.

95. 96.

97. Think About It Give a geometric explanation of why

Verify the inequality by evaluating the integrals.

98. Finding a Pattern Find the area bounded by the graphs ofand over each interval.

(a) (b) (c)

Describe any patterns that you notice. What is the areabetween the graphs of and over the interval

where is any nonnegative integer? Explain.

99. Finding an Error Find the fallacy in the following argument that

So, 0 � 1.

� 1 � � dxx

0 � � dxx

� 1x�x� � � �

1x2�x� dx

du � �1x2 dx u �

1x

v � � dx � x dv � dx

0 � 1.

n�n�, �n � 1��,y � 0y � x sin x

�2�, 3��, 2��0, ��y � 0y � x sin x

��2

0 x sin x dx � ��2

0 x dx.

f �0� � 1f �0� � 0

f��x� � cos xf��x� � 3x sin�2x�

h � 0.1

h � 0.05,

f �0� � 0.f��x� � xe�x

bn.

bn � h �1

0 x sin

n�x2

dx � h �2

1 ��x � 2� sin

n�x2

dx.

h�2, 0��0, 0�

��

��

x2 cos nx dx ��1�n 4�

n2

��

�� x sin nx dx � �

2� n

�2� n

,

,

n is odd

n is even

n

c�t� � 30,000 � 500t, r � 7%, t1 � 5

c�t� � 100,000 � 4000t, r � 5%, t1 � 10

rt1

P � �t1

0 c�t�e�rt dt

c�t�P

t � �.t � 0y

y � e�4t �cos 2t � 5 sin 2t�.

x � 4.x � 2,g�x� � 2x,f �x� � x2,

y � ��2.x � 0,y � arcsin x,

y

x

x � �,x � 0,y � 0,y � x sin x,

A model for the ability of a child to memorize, measuredon a scale from 0 to 10, is given by

where is the child’s age in years. Find the average value of this model

(a) between the child’s first and second birthdays.

(b) between the child’s third and fourth birthdays.

t

0 < t � 4M � 1 � 1.6t ln t,

M

88. Memory Model

Juriah Mosin/Shutterstock.com

9781285057095_0802.qxp 9/17/12 3:20 PM Page 523



Documents

508 Chapter 8 Integration Techniques, L’Hôpital’s Rule ... · Two of the most commonly overlooked integration rules are the Log Rule and the Power Rule. Notice in the next two