chapter 8 techniques of integrationmrsbustamanteala.weebly.com/uploads/8/4/5/2/... · xx dx xx...

Copyright 2014 Pearson Education, Inc. 543

CHAPTER 8 TECHNIQUES OF INTEGRATION

8.1 USING BASIC INTEGRATION FORMULAS

1. 1

20

16

8 2

xdx

x

28 2 16

2 when 0, 10 when 1

u x du x dx

u x u x

1 10

10

2 220

16 1ln

8 2ln10 ln 2 ln5

xdx du u

ux

2. 2

2 1

xdx

x

Use long division to write the integrand as 2

11

1x

.

2

2 2

1

11

1 1

tan

xdx dx dx

x x

x x C

3. 2sec tanx x dx

Expand the integrand: 2 2 2

2 2

2

sec tan sec 2sec tan tan

sec 2sec tan (sec 1)

2sec 2sec tan 1

x x x x x x

x x x x

x x x

2 2sec tan 2 sec 2 sec tan 1

2 tan 2sec

x x dx x dx x x dx dx

x x x C

We have used Formulas 8 and 10 from Table 8.1.

4. /3

2/4

1

cos tandx

x x

π

π

22

1tan sec

cos

1 when 4, 3 when / 3

u x du x dx dxx

u x u xπ π

/3 33

2 11/4

1 1ln

cos tan1

ln 3 ln1 ln 32

dx du uux x

π

π

544 Chapter 8 Techniques of Integration

Copyright 2014 Pearson Education, Inc.

5. 2

1

1

xdx

x

Write as the sum of two integrals:

2 2 2

1 1

1 1 1

x xdx dx dx

x x x

For the first integral use Formula 18 in Table 8.1 with 1.a

For the second:

21 2u x du x dx

1/22

2

1 1

21

1

xdx du

ux

u x

So 1 2

2

1sin 1

1

xdx x x C

x

6. 1

dxx x

1

12

u x du dxx

1 1

2

2ln 2ln 1

dx duux x

u C x C

7. cot

2sin

zedz

z

22

1cot csc

sinu z du z dz dz

z

cot

2

cot

sin

zu

u z

edz e du

z

e C e C

8.

3ln2

16

z

dzz

3 3ln 3lnu z z du dz

z

Using Formula 5 in Table 8.1,

3

3

ln

ln

2 12

16 48

2 2

48ln 2 48ln 2

zu

u z

dz duz

C C

Section 8.1 Using Basic Integration Formulas 545


9. 1

z zdz

e e

Multiply the integrand by .z

z

e

e

2

1

1

z

z z z

edz dx

e e e

z zu e du e du

2 2

1 1

1

1 1

tan tan

z

z

z

edx du

e u

u C e C

10. 2

21

8

2 2dx

x x

1

0 when 1, 1 when 2

u x du dx

u x u x

2 1

2 21 0

11

0

8 18

2 2 1

8 tan 8 0 24

dx dux x u

uπ π

11. 0

21

4

1 (2 1)dx

x

2 1 2

1 when 1, 1 when 0

u x du dx

u x u x

0 1

2 211

11

1

4 12

1 (2 1) 1

2 tan 24 4

dx dux u

uπ π π

12. 3 2

1

4 7

2 3

xdx

x


2 32 3

xx

.

3 2 33 3

1 1 11

3 3 3 32111 1

4 7 22 3

2 3 2 3

2 3 3 8 12 4

xdx x dx dx dx

x x

x dx dx x x

For the last integral, 2 3 2

1 when 1, 9 when 3

u x du dx

u x u x



3 9

1 1

9

1

2 1

2 3

ln ln 9 ln1 2 ln 3

dx dux u

u

So 3 2

1

4 74 2ln 3

2 3

xdx

x

13. 1

1 secdt

t

Multiply the integrand by 1 sec

.1 sec

t

t

2 22 2 2

1 1 sec 1 sec cos coscot 1 csc

1 sec 1 sec tan sin sin

t t t tt t

t t t t t

22

1 cos1 csc

1 sec sincot csc

tdt dt t dt dt

t tt t t C

Here we have used Formula 9 in Table 8.1 for the second integral, and the substitution sin , cosu t du t dt

for the third integral, which gives it the form 2

1 1 1.

sindu

u tu

14. csc sin 3t t dt

Write sin 3t as sin(2 )t t and expand.

2

cos2 sin (2sin cos )coscsc sin 3

sin

cos 2 2cos 2cos 2 1

t t t t tt t

t

t t t

csc sin 3 2cos2 1

sin 2

t t dt t dt dt

t t C

15. /4

20

1 sin

cosd

π θ θθ

Split into two integrals.

/4 /4 /4

2 2 20 0 0

/4/42

20 0

/4

0

1 sin 1 sin

cos cos cos

sinsec

cos

tan sec (1 2) (0 1) 2

d d d

d d

π π π

ππ

π

θ θθ θ θθ θ θ

θθ θ θθ

θ θ

The second integral is evaluated with the substitution cos sin ,u du dθ θ θ which gives

2 2

sin 1 1 1.

coscosd du

uu

θ θθθ



16. 2

1

2dθ

θ θ

Write the integrand as 2

1

1 ( 1)θ . With 1, ,u du dθ θ

2 2

1 1

2

1 1

2 1 ( 1)

1sin sin ( 1)

1

d d

du u C Cu

θ θθ θ θ

θ

We have used Formula 18 in Table 8.1 with 1.a

17. 2

ln

4 ln

ydy

y y

Write the integrand as 2

ln 1.

1 4 ln

y

y y

2 8 ln1 4 ln

yu y du dy

y

2 2

2

ln ln 1

4ln 1 4ln

1 1 1 1ln ln(1 4 ln )

8 8 8

y ydy dy

yy y y

du u C y Cu

Note that the argument of the logarithm is positive, so we don’t need absolute value bars.

18. 2

2

y

dyy

1

2u y du dy

y

Using Formula 5 in Table 8.1,

2

22

1 12 2

ln 2 ln 2

yu

yu

dy duy

C C

19. 1

sec tandθ

θ θ

Multiply the integrand by cos

.cos

θθ

1 cos cos

sec tan cos 1 sind d

θ θθ θθ θ θ θ

1 sin cosu du dθ θ θ



cos 1

ln1 sin

ln (1 sin )

d du u Cu

C

θ θθ

θ

We can discard the absolute value because 1 sinθ is never negative.

20. 2

1

3dt

t t

Use Formula 5 in Table 7.10, with 3.a

1

2

1 1csch

3 33

tdt C

t t

21. 3 2

2

4 16

4

t t tdt

t


44 1 .

4t

t

3 2

2 2

2 1

4 16 14 1 4

4 4

2 2 tan2

t t tdt t dt dt dt

t t

tt t C

To evaluate the third integral we used Formula 19 in Table 8.1 with 2.a

22. 2 1

2 1

x xdx

x x

Split into two integrals.

2 1 1 1

2 1 2 1

1 ln

x xdx dx dx

xx x x

x x C

For the first integral we used 1

1, ,2 1

u x du dx du u Cx

23. /2

01 cos d

πθ θ

Multiply the integrand by 1 cos

.1 cos

θθ

/2 /22/2

0 00

1 cos sin1 cos .

1 cos 1 cosd d d

π ππ θ θθ θ θ θθ θ

(Note that when 20 / 2, sin 0 so sin sin .θ π θ θ θ )

1 cos sin

2 when 0, 1 when / 2

u du d

u u

θ θ θθ θ π

/2 1 2 2

10 2 1

sin 1 12 2 2 2

1 cosd du du u

u u

π θ θθ



24. 2(sec cot )t t dt

Expand the integrand:

2 2 2

2 2

sec cot sec 2sec cot cot

sec 2sec cot csc 1

t t t t t t

t t t t

2 2 2(sec cot ) sec 2 csc csc 1

tan 2ln csc cot cot

t t dt t dt t dt t dt dt

t t t t t C

We have used Formulas 8, 9 and 15 from Table 8.1.

25. 2

1

1ydy

e

Multiply the integrand by .y

y

e

e

2 2

1

1 1

y

y y y

edy dy

e e e

; y yu e du e dy

2 2

1 1

1

1 1

sec sec

y

y y

y

edy du

e e u u

u C e C

We have used Formula 20 in Table 8.1.

26. 6

1dy

y y

1

2u y du dy

y

2

1

6 112

1 1

12 tan

dy duy y u

y C

27. 2

2

1 4 lndx

x x

2

2 lnu x du dxx

2 2

1 1

2 1

1 4 ln 1

sin sin 2 ln

dx dux x u

u C x C



28. 2

1

( 2) 4 3dx

x x x

2u x du dx

2 2

1 1

1 1

( 2) 4 3 1

sec sec 2

dx dux x x u u

u C x C

We have used Formula 20 in Table 8.1 with 1.a

29. (csc sec )(sin cos )x x x x dx

Expand the integrand and separate into two integrals.

(csc sec )(sin cos ) 1 cot tan 1 cot tanx x x x x x x x

(csc sec )(sin cos ) cot tan

ln sin ln sec ln sin ln cos

x x x x dx x dx x dx

x x C x x C


30. 3sinh ln52

xdx

1

ln 52 2

xu du dx

3sinh ln5 6 sinh2

6cosh 6cosh ln52

xdx u du

xu C C

31. 3 3

22

2

1

xdx

x


22 .

1

xx

x

3 3 33 3

2 2 22 222

2 2 22 2

1 1 1

x x xdx x dx x dx dx

x x x

For the second integral we use 2 , 2 .u x du x dx

33 332 2

2 22 22

22 ln 1

1

(9 2) (ln8 ln1)

7 ln8 9.079

xx dx dx x x

x



32. 1

2

11 sinx x dx

is the integral of an odd function over an interval symmetric to 0, so its value is 0.

33. 0

1

1

1

ydy

y

Multiply the integrand by 1

1

y

y

and split the indefinite integral into a sum.

2 2 2

1 2

1 1 1

1 1 1 1

sin 1

y y ydy dy dy dy

y y y y

y y C

The first integral is Formula 18 in Section 8.1, and for the second we use the substitution 21 , 2u y du y dy . So

0 0

1 2

11

1sin 1

1

(0 1) 0 12 2

ydy y y

y

π π

34. zz ee dz

Write the integrand as zz ee e and use the substitution , .z zu e du e dz

z z

z

z e z e u

u e

e dz e e dz e du

e C e C

35. 2

7

( 1) 2 48dx

x x x

2 2 21, ; 2 48 7u x du dx x x u

We use Formula 20 in Table 8.1.

2 2 2

1 1

7 7

( 1) 2 48 7

1 17sec sec

7 7 7

dx dux x x u u

u xC C



36. 2

1

(2 1) 4 4dx

x x x

2 2 22 1, 2 ; 4 4 1u x du dx x x u

We use Formula 20 in Table 8.1.

2 2 2

1 1

1 1 1

2(2 1) 4 4 1

1 1sec sec 2 1

2 2

dx dux x x u u

u C x C

37. 3 22 7 7

2 5d

θ θ θ θθ

Use long division to write the integrand as 2 51 .

2 5θ θ

θ

3 2

2

3 2

2 7 7 51

2 5 2 5

1 1 5ln 2 5

3 2 2

d d d d d

C

θ θ θ θ θ θ θ θ θ θθ θ

θ θ θ θ

In the last integral we have used the substitution 2 5, 2 .u du dθ θ

38. 1

cos 1dθ

θ

Multiply the integrand by cos 1

.cos 1

θθ

22 2

1 cos 1 cos 1 1 coscsc csc cot

cos 1 cos 1 cos 1 sin

θ θ θ θ θ θθ θ θ θ

2 2csc csc cot csc csc cot

cot csc

d d d

C

θ θ θ θ θ θ θ θ θ

θ θ


39. 1

1 xdx

e

Use one step of long division to write the integrand as 1 .1

x

x

e

e

11 ln 1

1 1

xx

x x

edx dx dx x e C

e e

For the second integral we have used the substitution 1 , .x xu e du e dx Note that 1 xe is always positive.



40. 31

xdx

x

3/2 1/23,

2u x du x dx

3 2

1 1 3/2

2 1

31 12 2

tan tan3 3

xdx du

x u

u C x C

41. The area is

/4 /4

/4/42cos sec 2sin ln sec tan

2 ln 2 1 2 ln 2 1

2 12 2 ln 2 2 ln 3 2 2 1.066

2 1

x x dx x x xπ π

ππ

42. The volume using the washer method is /42 2

/44cos sec .x x dx

π

ππ

Split into two integrals; for the first write 24cos x as 2 1 cos 2x and for the second use Formula 8 in

Table 8.1.

/4 /4 /42 2 2 2

/4 /4 /4

/4 /42

/4 /4

/4 /4

/4 /4

2

4cos sec 4cos sec

2 1 cos2 sec

2 sin 2 tan

1 1 1 ( 1)2 2

x x dx x dx x dx

x dx x dx

x x x

π π π

π π ππ π

π ππ π

π π

π π π

π π

π π

π ππ π π

43. For ln (cos )y x , / tan .dy dx x The arc length is given by

/3 /32

0 01 tan secx dx x dx

π π since sec x is positive on the interval of integration.

/3 /3

00sec ln sec tan

ln 2 3 ln 1 0 ln 2 3

x dx x xπ π

44. For ln (sec )y x , / tan .dy dx x The arc length is given by

/4 /42

0 01 tan secx dx x dx

π π since sec x is positive on the interval of integration.

/4 /4

00 sec ln sec tan

ln 2 1 ln 1 0 ln 2 1

x dx x xπ π



45. Since secant is an even function and the domain is symmetric to 0, 0.x For the y-coordinate:

/4 /42 12/4 /4

/4 /4

/4/4

1sec tan

2

ln sec tansec

1

ln 2 1 ln 2 1

1 10.567

ln 3 2 22 1ln

2 1

x dx xy

x xx dx

π π

π ππ π

ππ

46. Since both cosecant and the domain are symmetric around / 2,π / 2.x π

5 /6 5 /62 12/6 /6

5 /6 5 /6

/6/6

12

1csc cot

2

ln csc cotcsc

3 3

ln 2 3 ln 2 3

3 30.658

ln 7 4 32 3ln

2 3

x dx xy

x xx dx

π π

π ππ π

ππ

47. 3 331 3 x xx e dx xe C

48. 2

1

1 sindx

x


2

sec.

sec

x

x

2 2

2 2 2 2

1 sec sec

1 sin sec tan 1 2 tan

x xdx dx dx

x x x x

2tan , secu x du x dx

2

2 2

sec 1

1 2 tan 1 2

xdx du

x u

2 , 2v u dv du

2 2

1

1

1 1 1

1 2 2 11

tan2

1tan 2 tan

2

du dvu v

v C

x C

Section 8.2 Integration by Parts 555


49. 7 4 1x x dx

4 3 7 11, 4 ;

4

uu x du x dx x dx du

7 4

3/2 1/2

5/2 3/2

3/23/2 4 4

11 ( 1)

41 1

4 41 1

10 61 1

3 5 1 3 230 30

x x dx u u du

u du u du

u u C

u u C x x C

50. 2/32( 1)( 1)x x dx

The easiest substitution to use is probably 2

1 2, .

1 (1 )

xu du dx

x x

The integral can be written as

2/32/3

2

1/31/3

1 1

21( 1)

1

3 3 1

2 2 1

dx u dux

xx

xu C C

x

8.2 INTEGRATION BY PARTS

1. 2 2, ; sin , 2cos ;x xu x du dx dv dx v

2 2 2 2 2sin 2 cos 2cos 2 cos 4sinx x x x xx dx x dx x C

2. 1, ; cos , sin ;u du d dv d v πθ θ πθ θ πθ

21 1cos sin sin sin cosd d Cθ θ

π π π πθ πθ θ πθ πθ θ πθ πθ

3. cos t

( )2 sint t

( )2 cost t

( )2 sin t

0 2 2cos sin 2 cos 2sint t dt t t t t t C



4. sin x

( )2 cosx x

( )2 sinx x

( )2 cos x

0 2 2sin cos 2 sin 2cosx x dx x x x x x C

5. 2

2ln , ; , ;dx xxu x du dv x dx v

2 2 22 22 23 3

2 2 4 4 41 11 1ln ln 2 ln 2 2ln 2 ln 4x x dx x

xx x dx x

6. 43

4ln , ; , ;dx xxu x du dv x dx v

4 4 4 44

34 4 4 161 11 1

3 1ln ln

16

e ee ex x dx e x

x

ex x dx x

7. , ; , ;x xu x du dx dv e dx v e

x x x x xx e dx xe e dx xe e C

8. 3 313, ; , ;x xu x du dx dv e dx v e

3 3 3 3 31 13 3 3 9

x x x x xx xx e dx e e dx e e C

9. xe

( )2 xx e

( )2 xx e

( )2 xe

0 2 2 2 2x x x xx e dx x e x e e C

10. 2xe

( )2 2122 1 xx x e

( ) 2142 2 xx e

2 ( ) 21

8xe

0

2 2 2 2 2 21 1 12 4 4

2 23 512 2 4

2 1 2 1 (2 2)x x x x

x

x x e dx x x e x e e C

x x e C

11. 21

1tan , ; , ;dy

yu y du dv dy v y

2

1 1 1 2 1 2121

tan tan tan ln 1 tan ln 1y dy

yy dy y y y y y C y y y C



12. 2

1

1sin , ; , ;dy

yu y du dv dy v y

2

1 1 1 2

1sin sin sin 1y dy

yy dy y y y y y C

13. 2, ; sec , tan ;u x du dx dv x dx v x

2sec tan tan tan ln |cos |x x dx x x x dx x x x C

14. 2 24 sec 2 ; [ 2 , 2 ] sec tan tan tan ln | sec |x x dx y x dy dx y y dy y y y dy y y y C

2 tan 2 ln |sec 2 |x x x C

15. xe

( )3 xx e

( )23 xx e

( )6 xx e

( )6 xe

0 3 3 2 3 23 6 6 3 6 6x x x x x xx e dx x e x e xe e C x x x e C

16. pe

( )4 pp e

( )34 pp e

( )212 pp e

( )24 pp e

( )24 pe

0

4 4 3 2

4 3 2

4 12 24 24

4 12 24 24

p p p p p p

p

p e dp p e p e p e pe e C

p p p p e C

17. xe

( )2 5 xx x e

( )2 5 xx e

( )2 xe

0 2 2 2

2

5 5 (2 5) 2 7 7

7 7

x x x x x x x

x

x x e dx x x e x e e C x e xe e C

x x e C



18. re

( )2 1 rr r e

( )2 1 rr e

( )2 re

0

2 2

2 2

1 1 (2 1) 2

1 (2 1) 2 2

r r r r

r r

r r e dr r r e r e e C

r r r e C r r e C

19. xe

( )5 xx e

( )45 xx e

( )320 xx e

( )260 xx e

( )120 xx e

( )120 xe

0

5 5 4 3 2

5 4 3 2

5 20 60 120 120

5 20 60 120 120

x x x x x x x

x

x e dx x e x e x e x e xe e C

x x x x x e C

20. 4te

( )2 414

tt e

( ) 41162 tt e

( ) 41642 te

0

2 2

2

2 4 4 4 4 4 4 42 2 14 16 64 4 8 32

414 8 32

t t t t t t tt t t t

tt t

t e dt e e e C e e e C

e C

21. sin ; [ sin , cos ; , ] sin cos ;I e d u du d dv e d v e I e e dθ θ θ θ θθ θ θ θ θ θ θ θ θ

[ cos , sin ; , ] sin cos sinu du d dv e d v e I e e e dθ θ θ θ θθ θ θ θ θ θ θ θ

12sin cos 2 sin cos sin cos ,e e I C I e e C I e e Cθ θ θ θ θ θθ θ θ θ θ θ where 2

CC

is another arbitrary constant

22. cos ;[ cos , sin ; , ]y y yI e y dy u y du y dy dv e dy v e

cos ( sin ) cos sin ;y y y yI e y e y dy e y e y dy

[ sin , cos ; , ] cos sin cosy y y y yu y du y dy dv e dy v e I e y e y e y dy

12cos sin 2 sin cos sin cos ,y y y y ye y e y I C I e y y C I e y e y C where

2CC is another arbitrary constant



23. 2 2 212cos 3 ; [ cos 3 ; 3 sin 3 , ; ]x x xI e x dx u x du x dx dv e dx v e

2 2 2 231 12 2 2cos 3 sin 3 ; [ sin 3 , 3cos 3 , ; ]x x x xI e x e x dx u x du x dv e dx v e

2 2 2 2 23 3 3 91 1 12 2 2 2 2 4 4cos 3 sin 3 cos 3 cos 3 sin 3x x x x xI e x e x e x dx e x e x I C

22 213 314 2 4 13cos 3 sin 3 3 sin 3 2 cos 3 ,

xx x eI e x e x C I x x C where 413C C

24. 2 12sin 2 ; [ 2 , 2 ] sin ; [ sin , cos ; , ]x y y ye x dx y x du dx e y dy I u y du y dy dv e dy v e

12 sin cos [ cos , sin ; , ]y y y yI e y e y dy u y du y dv e dy v e

1 1 12 2 2sin cos ( sin ) sin cosy y y yI e y e y e y dy e y y I C

21 12 4 42 sin cos sin cos sin 2 cos 2 ,

xy y eI e y y C I e y y C x x C where

2CC

25. 2

3 9 2 23 32

3

3 9; ; [ , ; , ];s x x x xs x

e ds e x dx xe dx u x du dx dv e dx v eds x dx

3 9 3 92 2 2 23 3 3 3 3 9x x x x x s sxe dx xe e dx xe e C s e e C

26. 323, ; 1 , (1 ) ;u x du dx dv x dx v x

11 1 13 3 5 22 2 2 2 43 3 3 5 1500 00

1 (1 ) (1 ) 0 (1 )x x dx x x x dx x

27. 2 2

2 2 22 2 sin 1 cos

cos cos cos, ; tan , tan tan ;x x dx

x x xu x du dx dv x dx v x dx dx dx dx x x

23 3 3323 3 20 00 0

tan tan (tan ) 3 ln |cos | xx x dx x x x x x dx xπ π ππ π π

2 2313 3 2 18 3 183 ln ln 2ππ π π π

28. 2

(2 1)2 2 2 2 1( 1)ln , ; , ; ln lnx dx x

x xx xu x x du dv dx v x x x dx x x x x dx

(2 1)2 2 211 1ln ln 2 ln 2 ln | 1|x dx

x xx x x x x x dx x x x x x C

29. 1

ln

sin (ln ) ; (sin ) .uxu

u x

x dx du dx u e du

dx e du

From Exercise 21, sin cos2(sin ) u uu uu e du e C

12 cos (ln ) sin (ln )x x x x C



30. 2 2 2 21

ln

(ln ) ; ;u u uzu

u z

z z dz du dz e u e du e u du

dz e du

2

( )2 212

( ) 214

( ) 218

2

2

u

u

u

u

e

u e

u e

e

0 2 22 2 2 2 2 212 2 4 4 2 2 1

uu u u uu u eu e du e e e C u u C

2 2

4 2(ln ) 2 ln 1z z z C

31. 2 2 21 1 12 2 2sec Let , 2 sec sec ln |sec tan |x x dx u x du x dx du x dx x x dx u du u u C

2 212 ln |sec tan |x x C

32. coscos 1 12

Let , 2 2 cos 2 sin 2 sinxxx x x x

dx u x du dx du dx dx u du u C x C

33. 2 2 2 21

ln

(ln ) ; ;u u uxu

u x

x x dx du dx e u e du e u du

dx e du

2

( )2 212

( ) 214

( ) 218

2

2

u

u

u

u

e

u e

u e

e

0

2 2

2 2 2 2

2 2 2 2 2 212 2 4 4

2 24 2 2 4

2 2 1

2 ln 2 ln 1 ln ln

uu u u uu u e

x x x x

u e du e e e C u u C

x x C x x C

34. 2 2 21 1 1 1 1 1

ln(ln ) (ln )Let ln , x u xx x x x u

dx u x du dx dx du C C

35. 21 1 1ln , ; , ;x xx

u x du dx dv dx v

2 2

ln ln ln1 1x x xx x xx x

dx dx C

36. 3 3(ln ) (ln ) 3 4 41 1 1

4 4Let ln , (ln )x xx x xdx u x du dx dx u du u C x C

37. 4 4 43 4 3 3 31 1 1 1

4 4 4 4Let , 4x x u u xx e dx u x du x dx du x dx x e dx e du e C e C



38. 3 33 2 2 1

3, 3 ; , ;x xu x du x dx dv x e dx v e 3 3 3 3 3 35 3 2 3 2 31 1 1 1

3 3 3 33x x x x x xx e dx x e x dx x e e x dx x e e C

39. 3 22 2 213, 2 ; 1 , 1 ;u x du x dx dv x x dx v x

3 2 3 2 3 2 5 23 2 2 2 2 2 2 21 1 1 23 3 3 151 1 1 2 1 1x x dx x x x x dx x x x C

40. 2 3 3 2 2 2 31 1 13 3 3sin Let , 3 sin sin cosx x dx u x du x dx du x dx x x dx u du u C

313 cos x C

41. 12sin 3 , 3cos 3 ; cos 2 , sin 2 ;u x du x dx dv x dx v x

312 2sin 3 cos 2 sin 3 sin 2 cos 3 sin 2x x dx x x x x dx

12cos 3 , 3sin 3 ; sin 2 , cos 2 ;u x du x dx dv x dx v x

3 31 12 2 2 2sin 3 cos 2 sin 3 sin 2 cos 3 cos 2 sin 3 cos 2x x dx x x x x x x dx

3 912 4 4sin 3 sin 2 cos 3 cos 2 sin 3 cos 2x x x x x x dx

5 314 2 4sin 3 cos 2 sin 3 sin 2 cos 3 cos 2x x dx x x x x

325 5sin 3 cos 2 sin 3 sin 2 cos 3 cos 2x x dx x x x x C

42. 14sin 2 , 2cos 2 ; cos 4 , sin 4 ;u x du x dx dv x dx v x

1 14 2sin 2 cos 4 sin 2 sin 4 cos 2 sin 4x x dx x x x x dx

14cos 2 , 2sin 2 ; sin 4 , cos 4 ;u x du x dx dv x dx v x

1 1 1 14 2 4 2sin 2 cos 4 sin 2 sin 4 cos 2 cos 4 sin 2 cos 4x x dx x x x x x x dx

1 1 14 8 4sin 2 sin 4 cos 2 cos 4 sin 2 cos 4x x x x x x dx

3 1 14 4 8sin 2 cos 4 sin 2 sin 4 cos 2 cos 4x x dx x x x x

1 13 6sin 2 cos 4 sin 2 sin 4 cos 2 cos 4x x dx x x x x C

43. 3/21 2ln Let ln , , ,

3x x dx u x du dx dv x dx v x

x

3/2

3/2 3/2 3/2

2 2ln ln

3 32 4 2

ln 3ln 23 9 9

x x dx x x x dx

x x x C x x C



44. 1 12

Let , 2 2 2 2u u xe x e xx x x x

dx u x du dx du dx dx e du e C e C

45. 12

cos ; cos 2 2 cos ;

2x

y x

x dx dy dx y y dy y y dy

dx y dy

2 , 2 ; cos , sin ;u y du dy dv y dy v y

2 cos 2 sin 2sin 2 sin 2cos 2 sin 2 cosy y dy y y y dy y y y C x x x C

46. 212

; 2 2 ;

2

x y yx

y x

x e dx dy dx y e y dy y e dy

dx y dy

( )2

( )

( )

2

4

4

y

y

y

y

e

y e

y e

e

0 2 22 2 4 4 2 4 4y y y y x x xy e dy y e y e e C x e x e e C

47. ( )2 1

2

( ) 14

( ) 18

sin 2

cos 2

2 sin 2

2 cos 2

θ

θ θ

θ θ

θ

0 2

2 2 2

2 22 1

2 2 4 00

41 1 18 4 4 4 8 2 8

sin 2 cos 2 sin 2 cos 2

( 1) 0 ( 1) 0 0 1

dπ π

θ θ

π π π π

θ θ θ θ θ θ

48. ( )3 1

2

( )2 14

( ) 18

( ) 116

cos2

sin 2

3 cos 2

6 sin 2

6 cos 2

x

x x

x x

x x

x

0

3 2

23 2 2

2 23 3 3 3

2 4 4 8 00

3 43 3 3 3 3 316 16 8 8 8 16 4 16

cos 2 sin 2 cos 2 sin 2 cos 2

0 ( 1) 0 ( 1) 0 0 0 1

x x xx x dx x x x xπ π

ππ π π π



49. 2

2

121

sec , ; , ;dt t

t tu t du dv t dt v

2 2

2 2

2 2 221 1 2

2 2 3 3 61 2 12 32 3 2 3 2 3sec sec 2 t dtt t dt

t t tt t dt t π π

22 3 3 5 3 35 5 5 51 1 4 1

9 2 9 2 3 9 2 3 9 3 92 3

1 3 1 3t ππ π π π

50. 4

21 2 2

1sin , ; 2 , ;x dx

xu x du dv x dx v x

4

1 2 1 2 1 21 2 1/221 2 2 1 2 2 4 31 12 6 2100 0 0

2 sin sin 1 4x dx

xx x dx x x x x x dxπ

1 24 6 3 123

12 12 4 120

1 1x ππ π

51. 2

1 12

1tan Let tan , , ,

21

xx x dx u x du dx dv x dx v

x

21 2 1

2

2 12

2 1 1

2 1

1 1tan tan

2 2 1

1 1 1tan 1

2 2 11 1 1

tan tan2 2 21

1 tan2 2

xx x dx x x dx

x

x x dxx

x x x x C

xx x C

52. 3

2 1 1 22

1 / 2tan Let tan , , ,

2 2 31 ( / 2)

x x xx dx u du dx dv x dx v

x

33

2 1 12

31

2

31 2

2

11 2tan tan

2 3 2 31

4

1 2tan 2

3 3 31

4

1 1 2tan

3 2 3 31

4

xx x xx dx dx

x

x x xx dx

x

x x xx dx

x

In the remaining integral, let 2

1 , .4 2

x xw dw



2

2

1 2 1 4 4 4ln ln 1

3 3 3 3 41

4

x xdx dw w

wx

Thus the original integral is equal to

3 2

1 21 4tan ln 1

3 2 3 3 4

x x xx C

53. (a) , ; sin , cos ;u x du dx dv x dx v x

1 0 00 0sin cos cos sinS x x dx x x x dx x

π ππ ππ π

(b) 2 22 2

20

sin cos cos 3 sin 3S x x dx x x x dx xπ ππ π

π πππ π

(c) 3 33 3

3 2 22 2sin cos cos 5 sin 5S x x dx x x x dx x

π ππ ππ ππ π

π π

(d) ( 1) ( 1) ( 1)1 1

1 ( 1) sin ( 1) cos sinn n nn n

n n nnS x x dx x x x

π π ππ ππ

1 1( 1) ( 1) ( 1) ( 1) 0 (2 1)n n nn n nπ π π

54. (a) , ; cos , sin ;u x du dx dv x dx v x

3 2 3 23 2 3 23

1 2 22 22 2cos sin sin cos 2S x x dx x x x dx x

π ππ ππ ππ ππ π

π

(b) 5 2 5 25 2 5 25 3

2 2 23 2 3 23 2 3 2cos sin sin cos 4S x x dx x x x dx x


π

(c) 7 2 7 27 2 7 27 5

3 2 25 2 5 25 2 5 2cos sin sin cos 6S x x dx x x x dx x


π

(d) (2 1) 2 (2 1) 2(2 1) 2

(2 1) 2(2 1) 2 (2 1) 2( 1) cos ( 1) sin sin

n nnn nn nn n

S x x dx x x x dxπ ππ

ππ π

(2 1) 2(2 1) (2 1) 1 12 2 2(2 1) 2

( 1) ( 1) ( 1) cos 2 2 2nn nn n nn

x n n nππ ππ π π π π π

55. ln 2

02 ln 2 xV x e dxπ

ln 2 ln 2

0 02 ln 2 2x xe dx xe dxπ π

ln 2ln 2 ln 2

0 0 0(2 ln 2) 2x x xe xe e dxπ π

ln 2

02 ln 2 2 2 ln 2 2 ln 2 2xeπ π π π

2 (1 ln 2)π



56. (a) 1 11

00 02 2x x xV xe dx xe e dxπ π

141 1 1

02 2 1 2x

e e e ee ππ π π

(b) 1

02 (1 ) ;xV x e dxπ

1 , ; , ;x xu x du dx dv e dx v e

11

0 02 (1 ) x xV x e e dxπ

121

02 [0 1( 1)] 2 1 1x

e ee ππ π

57. (a) 2 22

00 02 cos 2 sin sinV x x dx x x x dx

π πππ π

22 20

2 cos 2 0 1 ( 2)xππ ππ π π π

(b) 2

202 cos ;V x x dx

πππ 2 , ; cos , sin ;u x du dx dv x dx v xπ

22 2

2 00 02 sin 2 sin 0 2 cos 2 (0 1) 2V x x x dx x

ππ πππ π π π π

58. (a) 0

2 ( sin ) ;V x x x dxπ

π

( )2

( )

( )

sin

cos

2 sin

2 cos

x

x x

x x

x

2 2 2

000 2 sin 2 cos 2 sin 2 cos 2 4V x x dx x x x x x

π ππ π π π

(b) 2 2 2 300 0 0

2 ( ) sin 2 sin 2 sin 2 cos sin 2 8V x x x dx x x dx x x dx x x xπ π π ππ π π π π π π

8π

59. (a) 11 1ln ln

e eeA x dx x x dx

1( ln 1 ln 1) ( 1) 1e

e e x e e

(b) 2 2

11 1(ln ) (ln ) 2 ln

e eeV x dx x x x dxπ π

2 21 1

(ln ) 1(ln1) 2 ln 2ee

e e x x dxπ

1(2 ln 2(1) ln 1) 2 2 (2 2) ( 2)

ee e e x e e e eπ π π



(c) 21 12 211 1 1

2 ( 2) ln 2 ( 2) ln 2 2 ln 2e e ee

V x x dx x x dx x x x x dxπ π π

2 2 2 2 291 1 1 1 12 2 4 2 4 4 21

2 2 ln 2 ln 1 2 2 2 2 9e

e e e x x e e e e eππ π

(d) 1

ln 1 (from part (a));e

M x dx 21 1 11 2 211 1

ln lne ee

x x x dx x x x dx

2 2 2 2 2 2 21 1 1 1 1 1 12 2 4 2 4 4 41

ln (1) ln1 (1) 1 ;e

e e x e e e

2 2 2 21 1 1 11 2 2 2 111 1 1

(ln ) (ln ) 2 ln (ln ) 1 (ln 1) 2 ln 2e e ee e

y x dx x x x dx e e x x dx

2 1 21 1 12 2 2 4 21

(2 ln 2(1) ln 1) 2 ( 2 2 2) ( 2) ( , ) ,e e ee e e x e e e e x y is the

centroid.

60. (a) 2

1 111 1100 0

tan tan xx

A x dx x x dx

11 21

2 0tan 1 0 ln 1 x

1 14 2 4 2(ln 2 ln 1) ln 2π π

(b) 2 2

2 2

1 1 111 1 11 1 1 1

2 2 2 21 100 0 02 tan 2 tan 2 tan 1 0 1x x

x xV x x dx x dx dxπ π π

1 ( 2)1 11 1 18 2 8 2 8 2 4 20

2 tan 2 1 tan 1 (0 0) 2 1x x π ππ π π ππ π π

61. 22 sin cos1 12 20 0

2 cos( )t tt te t dt eav y

ππ

π π

(see Exercise 22) 212( ) 1av y e ππ

62. 2

12 0

( ) 4 sin costav y e t t dtπ

π

2 22 2

0 0sin cost te t dt e t dt

π π

π π

2sin cos sin cos2

2 20

t t t tt te eπ

π

22

0sin 0te t

ππ

63. 1cos ; [ , ; cos , sin ]n n nI x x dx u x du nx dx dv x dx v x

1sin sinn nI x x nx x dx



64. 1sin ; [ , ; sin , cos ]n n nI x x dx u x du nx dx dv x dx v x

1cos cosn nI x x nx x dx

65. 1 1; [ , ; , ]n ax n n ax axaI x e dx u x du nx dx dv e dx v e

1 , 0n ax n axx e na aI x e dx a

66. 1(ln )(ln ) ; [ (ln ) , ; 1 , ]

nn xn nxI x dx u x du dx dv dx v x

1(ln ) (ln )n nI x x n x dx

67. 1

1

1 1 1

(ln ) , (ln ) , ,1

1(ln ) and (ln )

1 1

mn n m

m n m n

n xu x du x dx dv x dx v

x mn

uv x x v du x x dxm m

68. First to show that /2 /2

0 0cos sinn nx dx x dx

π π note that cos x over the interval [0, / 2]π is the reflection

of sin x over the same interval around the line / 4.x π

Each iteration of the reduction formula in Example 5 for the definite integral produces an expression like

/2

1/2

2

00

cos sin 1cos

nn

x x nx dx

n n

ππ

The evaluation on the left will be 0 as long as 2,n and factors of the form 1

n

n accumulate in front of the

integral on the right. When the initial n is even, the last iteration will have 2n and the remaining integral

will be /2

0

1 3 ( 1)1 3 11 .

2 2 2 2 4

nn ndx

n n n

π π

When the initial n is odd the last iteration will

have 3n and the remaining integral will be /2

0

2 4 ( 1)1 3 2cos 1 .

2 3 3

nn nx dx

n n n

π

69. ( ) ( ) ;b

ax a f x dx , ; ( ) , ( ) ( )

x b

b xu x a du dx dv f x dx v f t dt f t dt

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )bx b x b a b b

b a b b b a xa

x a f t dt f t dt dx b a f t dt a a f t dt f t dt dx

0 ( ) ( )b b b b

a x a xf t dt dx f t dt dx

70. 2

2 2

11 ; 1 , ; ,x

xx dx u x du dx dv dx v x

2 2 2

2 2 2 2

2 2 21 1 1 1

1 1 1 11 1 1x x x

x x x xx x dx x x dx x x dx dx



2

2 2 1

11 1

xx x x dx dx

2 2

2 2 2 2 21 1

1 11 1 1 2 1 1

x xx dx x x dx x dx x dx x x dx

2

2 2 1 12 2 1

1 1x

xx dx x dx C

71. 1 1 1 1 1sin sin sin sin cos sin cos sinx dx x x y dy x x y C x x x C

72. 1 1 1 1 1tan tan tan tan ln | cos | tan ln cos tanx dx x x y dy x x y C x x x C

73. 1 1 1sec sec sec sec ln | sec tan |x dx x x y dy x x y y C

1 1 1 1 2sec ln sec sec tan sec sec ln 1x x x x C x x x x C

74. 22 2 2 2ln 2 ln 2log log 2 log logyy xx dx x x dy x x C x x C

75. Yes, 1cos x is the angle whose cosine is x which implies 1 2sin cos 1 .x x

76. Yes, 1tan x is the angle whose tangent is x which implies 1 2sec tan 1 .x x

77. (a) 1 1 1 1 1sinh sinh sinh sinh cosh sinh cosh sinh ;x dx x x y dy x x y C x x x C

check: 2 2

1 1 1 1 11

1 1sinh cosh sinh sinh sinh sinh sinhx

x xd x x x C x x dx x dx

(b) 2

1 2 1 21 1 1 2 1 21 121

sinh sinh sinh 1 2 sinh 1x

x dx x x x dx x x x x dx x x x C

check: 2 2

1 21 2 1 1

1 1sinh 1 sinh sinhx x

x xd x x x C x dx x dx

78. (a) 1 1 1tanh tanh tanh tanh ln | cosh |x dx x x y dy x x y C

1 1tanh ln cosh tanh ;x x x C

check:

1

2 21

sinh tanh1 1 1 11 1cosh tanh

tanh ln cosh tanh tanhxx

x xxd x x x C x dx

2 21 1

1 1tanh tanhx x

x xx dx x dx

(b) 2 21 1 1 1 221 1

2 21 1tanh tanh tanh tanh ln 1x x

x xx dx x x dx x dx x x x C

check: 2 21 2 1 11

2 1 1tanh ln 1 tanh tanhx x

x xd x x x C x dx x dx

Section 8.3 Trigonometric Integrals 569


8.3 TRIGONOMETRIC INTEGRALS

1. 1 12 2cos 2 cos 2 2 sin 2x dx x dx x C

2. 91 13 3 3 3 3 2 200 0

3 sin 9 sin 9 cos 9 cos cos0 9 1x x xdx dxπ π π π

3. 3 3 414cos sin cos ( sin ) cosx x dx x x dx x C

4. 4 4 51 12 10sin 2 cos 2 sin 2 cos 2 2 sin 2x x dx x x dx x C

5. 3 2 2 2 313sin sin sin 1 cos sin sin cos sin cos cosx dx x x dx x x dx x dx x x dx x x C

6. 3 2 2 21 1 14 4 4cos 4 cos 4 cos 4 1 sin 4 cos 4 4 cos 4 4 sin 4 cos 4 4x dx x x dx x x dx x dx x x dx

31 14 12sin 4 sin 4x x C

7. 2 25 2 2 2 4sin sin sin 1 cos sin 1 2cos cos sinx dx x x dx x x dx x x x dx

2 4 3 52 13 5sin 2cos sin cos sin cos cos cosx dx x x dx x x dx x x x C

8. 5 2 42 2 2 2 2 20 0 0 0

sin (using Exercise 7) sin 2cos sin cos sinx x x x x xdx dx dx dxπ π π π

3 5 164 2 4 22 3 2 5 2 3 5 150

2cos cos cos (0) 2x x xπ

9. 3 2 2 2 313cos cos cos 1 sin cos cos sin cos sin sinx dx x x dx x x dx x dx x x dx x x C

10. 6 6 62 25 2 2

0 0 03cos 3 cos 3 cos3 3 1 sin 3 cos3 3x dx x x dx x x dx

π π π

/62 4

01 2sin 3 sin 3 cos3 3x x x dx

π

3 56 6 6 62 4 sin 3 sin 3

3 5 00 0 0cos 3 3 2 sin 3 cos 3 3 sin 3 cos 3 3 sin 3 2 x xx dx x x dx x x dx x

π π π π

82 13 5 151 (0)

11. 3 3 3 2 3 2 3 5sin cos sin cos cos sin 1 sin cos sin cos sin cosx x dx x x xdx x x x dx x x dx x x dx

4 61 14 6sin sinx x C



12. 3 5 3 5 2 51 12 2cos 2 sin 2 cos 2 sin 2 2 cos 2 cos 2 sin 2 2x x dx x x dx x x x dx

2 5 5 71 1 12 2 21 sin 2 sin 2 cos 2 2 sin 2 cos 2 2 sin 2 cos 2 2x x x dx x x dx x x dx

6 81 112 16sin 2 sin 2x x C

13. 1 cos 22 1 1 1 1 12 2 2 2 2 4cos 1 cos 2 cos 2 cos 2 2xx dx dx x dx dx x dx dx x dx

1 12 4 sin 2x x C

14. 2 2 2 2 21 cos 22 1 1 1

2 2 2 20 0 0 0 0sin 1 cos 2 cos 2xx dx dx x dx dx x dx

π π π π π

2 2 21 1 1 1 1 1 1 12 4 2 4 2 2 4 2 2 400 0

cos 2 2 sin 2 sin 2 (0) sin 2(0)dx x dx x xπ π π π π

4 40 (0 0)π π

15. 2 2 2 37 6 2

0 0 0sin sin sin 1 cos siny dy y y dy y y dy

π π π

2 2 2 22 4 6

0 0 0 0sin 3 cos sin 3 cos sin cos siny dy y y dy y y dy y y dy

π π π π

3 5 7 2cos cos cos 3 161

3 5 7 5 7 350

cos 3 3 (0) 1 1y y yyπ

16. 7 2 4 67 cos (using Exercise 15) 7 cos 3 sin cos 3 sin cos sin cost dt t dt t t dt t t dt t t dt

3 5 7 3 5 7sin sin sin 213 5 7 57 sin 3 3 7sin 7sin sin sint t tt C t t t t C

17. 21 cos 24 220 0 0

8sin 8 2 1 2cos 2 cos 2xx dx dx x x dxπ π π

1 cos 42 00 0 0 0 0

2 2 cos 2 2 2 2 2sin 2 cos 4xdx x dx dx x x dx x dxπ π π π ππ

14 0

2 sin 4 2 3x xπ

π π π π

18. 21 cos 4 1 cos 84 22 28cos 2 8 2 1 2cos 4 cos 4 2 4 cos 4 2x xx dx dx x x dx dx x dx dxπ ππ π π π

1 183 4 cos 4 cos 8 3 sin 4 sin 8dx x dx x dx x x x Cπ ππ π π π

19. 1 cos 2 1 cos 2 1 cos 42 2 22 2 216 sin cos 16 4 1 cos 2 4 4x x xx x dx dx x dx dx dx

1 12 24 2 2 cos 4 4 2 sin 4 2 sin 4 2 sin 2 cos 2x dx x dx x x x C x x C x x x C

2 32 2sin cos (2cos 1) 2 4sin cos 2sin cosx x x x C x x x x x C



20. 21 cos 2 1 cos 24 2 2 32 20 0 0 0 0 0

8 sin cos 8 cos 2 cos 2 cos 2y yy y dy dy dy y dy y dy y dyπ π π π π π

1 cos 4 212 20 0 0

sin 2 1 sin 2 cos 2yy y dy y y dyπ ππ

21 12 20 0 0 0

cos 4 cos 2 sin 2 cos 2dy y dy y dy y y dyπ π π π

π

3sin 21 1 1 12 8 2 2 3 2 2

0sin 4 sin 2 yy y y

ππ ππ π

21. 43 4cos 212 48 cos 2 sin 2 8 cos 2d C Cθθ θ θ θ

22. 2 22 3 2 2

0 0sin 2 cos 2 sin 2 1 sin 2 cos 2d d

π πθ θ θ θ θ θ θ

3 52 2 22 4 sin 2 sin 21 1

2 3 2 5 00 0sin 2 cos 2 sin 2 cos 2 0d d

π π πθ θθ θ θ θ θ θ

23. 2 2 2 21 cos

2 2 2 2 00 0 0sin sin 2cos 2 2 4x x x xdx dx dx

π π π π

24. 00 0 0

1 cos 2 2 | sin | 2 sin 2cos 2 2 2 2x dx x dx x dx xπ π π π

25. 2 22

0 20 0 0 21 sin |cos | cos cos sin sin 1 0 0 1 2t dt t dt t dt t dt t t

π π π π π πππ

26. 200 0 0

1 cos |sin | sin cos 1 1 2d d dπ π π πθ θ θ θ θ θ θ

27. 2 22 2

2 2

2 2 2 2sin 1 cos sin 1 cos1 cossin sin1 cos 1 cos 1 cos 1 cos sin3 3 3 3

x x x xxx xx x x x x

dx dx dx dxπ π π π

π π π π

/2 2 3 2 3 2 3 23 2 32 2 2 2 23 3 2 3 3 3 3 23/3

sin 1 cos (1 cos ) 1 cos 1 cosx x dx xπ π π π

ππ

3 22 3

28. 2 26 6 6 6 6

1 sin 1 sin 1 sin cos cos1 1 sin 1 sin 1 sin 1 sin0 0 0 0 0

1 sin x x x x xx x x x

x dx dx dx dx dxπ π π π π

61 2 16 20

2(1 sin ) 2 1 sin 2 1 sin 0 2 2 1 2 2xπ π

29. 4 44 4

2 2

cos 1 sin1 sin cos 1 sincos cos1 sin 1 sin 1 sin 1 sin cos5 6 5 6 5 6 5 6

x xx x xx xx x x x x

dx dx dx dxπ π π π

π π π π

4cos 1 sin 3 2cos5 6 5 6 5 6

cos 1 sin cos 1 sin 1 sinx x

x dx x x dx x x x dxπ π π

π π π

2

5 6 5 6cos 1 sin cos sin 1 sin ;x x dx x x x dx

π π

π π



5 5 36 6 2Let 1 sin 1 sin cos , 1 sin , 1 sin 1u x u x du x dx x u x uπ π π π

1 13 2 2 3 2 5 2 3 22 2

3 35 6 5 63 2 3 2(1 sin ) ( 1) (1 sin ) 2x u u du x u u u du

π π

π π

13 23 2 7 2 5 2 3 252 2 2 4 23 3 6 7 5 3 3 2

(1 sin ) 1 sin u u uππ

3 2 7 2 5 2 3 2 5 2 7 23 3 3 3 3 3 182 2 2 4 2 2 4 2 4 23 3 2 7 5 3 7 2 5 2 3 2 5 2 7 2 35

30. 2 27 12 7 12 7 12 7 121 sin 2 1 sin 2 cos 21 sin 2

1 1 sin 2 1 sin 2 1 sin 22 2 2 21 sin 2

x x xxx x x

x dx dx dx dxπ π π π

π π π π

7 12 7 12cos 2 2 17 1 112 2 21 sin 2 2 222

1 sin 2 1 sin 2 1 sin 2 1 1x

xdx x

π π π πππ

31. 2 2 2 2

00 0 01 cos 2 2 | sin | 2 sin 2 cos sin 2(1) 2d d d

π π π πθ θ θ θ θ θ θ θ θ θ θ θ

32. 03 2 3 22 2 3 3 3

01 cos sin sin sin sint dt t dt t dt t dt t dt

π π π π

π π π π

02 2

01 cos sin 1 cos sint t dt t t dt

π

π

0 02 2

0 0sin cos sin sin cos sint dt t t dt t dt t t dt

π π

π π

3 30cos cos 81 1 1 1

3 3 3 3 3 3 30cos cos 1 1 1 1t tt t

π

π

33. 2 2 212sec tan tan sec tanx x dx x x dx x C

34. 2 2sec tan sec tan tan ; tan , sec , sec tan , sec ;x x dx x x x dx u x du x dx dv x x dx v x

3 2 2sec tan sec sec tan sec sec sec tan tan 1 secx x x dx x x x x dx x x x x dx

2 2sec tan tan sec sec sec tan ln | sec tan | tan secx x x x dx x dx x x x x x x dx

2 2sec tan sec tan ln | sec tan | tan secx x dx x x x x x x dx

22 tan sec sec tan ln | sec tan |x x dx x x x x

2 1 12 2tan sec sec tan ln | sec tan |x x dx x x x x C

35. 3 2 313sec tan sec sec tan secx x dx x x x dx x C

36. 3 3 2 2 2 2sec tan sec tan sec tan sec sec 1 sec tanx x dx x x x x dx x x x x dx

4 2 5 31 15 3sec sec tan sec sec tan sec secx x x dx x x x dx x x C

37. 2 2 2 2 313sec tan tan sec tanx x dx x x dx x C



38. 4 2 2 2 2 2 2 2sec tan sec tan sec tan 1 tan secx x dx x x x dx x x x dx

4 2 2 2 5 31 15 3tan sec tan sec tan tanx x dx x x dx x x C

39. 0

3

32 sec ;x dx

π 2sec , sec tan , sec , tan ;u x du x x dx dv x dx v x

0 0 003 2 233 3 3

2 sec 2sec tan 2 sec tan 2 1 0 2 2 3 2 sec sec 1x dx x x x x dx x x dxππ π π

0 0

3

3 34 3 2 sec 2 sec ;x dx x dx

π π

0 003 333 3

2 2 sec 4 3 2ln sec tan 2 2 sec 4 3 2ln |1 0 | 2 ln 2 3x dx x x x dxππ π

034 3 2ln 2 3 2 sec 2 3 ln 2 3x dx

π

40. 3 2sec ; sec , sec tan , sec , tan ;x x x x x x x x xe e dx u e du e e e dx dv e e dx v e

3 2

2

3

3

3 12

sec sec tan sec tan

sec tan sec sec 1

sec tan sec sec

2 sec sec tan ln sec tan

sec sec tan ln sec tan

x x x x x x x

x x x x x

x x x x x x

x x x x x x

x x x x x x

e e dx e e e e e dx

e e e e e dx

e e e e dx e e dx

e e dx e e e e C

e e dx e e e e C

41. 4 2 2 2 2 2 313sec 1 tan sec sec tan sec tan tand d d d Cθ θ θ θ θ θ θ θ θ θ θ θ

2 21 1 23 3 3tan tan sec 1 tan sec tanC Cθ θ θ θ θ θ

42. 4 2 2 2 2 23sec (3 ) 1 tan (3 ) sec (3 )3 sec (3 )3 tan (3 )sec (3 )3x dx x x dx x dx x x dx

313tan(3 ) tan (3 )x x C

43. 32 2 2 2 24 2 2 2 2 2 cot

3 44 4 4 4csc 1 cot csc csc cot csc cotd d d d

π π π π πθ

ππ π π πθ θ θ θ θ θ θ θ θ θ θ

1 43 3(0) 1

44. 26 4 2 2 2 4 2 2sec sec sec tan 1 sec tan 2 tan 1 secx dx x x dx x x dx x x x dx

4 2 2 2 2 5 31 25 3tan sec 2 tan sec sec tan tan tanx x dx x x dx x dx x x x C

45. 23 2 2 tan24 tan 4 sec 1 tan 4 sec tan 4 tan 4 4ln |sec |xx dx x x dx x x dx x dx x C

2 2 2 2 22 tan 4ln |sec | 2 tan 2ln sec 2 tan 2ln 1 tanx x C x x C x x C



46. 4 4 4 44 2 2 2 2 2

4 4 4 46 tan 6 sec 1 tan 6 sec tan 6 tanx dx x x dx x x dx x dx

π π π π

π π π π

34 4 4 442 2 2 2tan

3 44 4 4 46 sec tan 6 sec 1 6 6 sec 6xx x dx x dx x dx dx

π π π ππ

ππ π π π

4 4 3 32 24 4

2 1 ( 1) 6 tan 6 4 6 1 ( 1) 3 8x xπ π π π

π π π

47. 25 4 2 4 2tan tan tan sec 1 tan sec 2sec 1 tanx dx x x dx x x dx x x x dx

4 2sec tan 2 sec tan tanx x dx x x dx x dx

3 4 214sec sec tan 2 sec sec tan tan sec sec ln |sec |x x x dx x x x dx x dx x x x C

22 2 4 21 1 14 4 2tan 1 tan 1 ln |sec | tan tan ln |sec |x x x C x x x C

48. 6 4 2 4 2 4 2 4cot 2 cot 2 cot 2 cot 2 csc 2 1 cot 2 csc 2 cot 2x dx x x dx x x dx x x dx x dx

4 2 2 2 4 2 2 2cot 2 csc 2 cot 2 cot 2 cot 2 csc 2 cot 2 csc 2 1x x dx x x dx x x dx x x dx

4 2 2 2 2cot 2 csc 2 cot 2 csc 2 cot 2x x dx x x dx xdx

4 2 2 2 2cot 2 csc 2 cot 2 csc 2 csc 2 1x x dx x xdx x dx

4 2 2 2 2 5 31 1 110 6 2cot 2 csc 2 cot 2 csc 2 csc 2 cot 2 cot 2 cot 2x x dx x xdx x dx dx x x x x C

49. 23 3 3 3 33 2 2 cot

2 66 6 6 6cot csc 1 cot csc cot cot ln |csc |xx dx x x dx x x dx x dx x

π π π π π

ππ π π π

1 1 2 42 3 33

3 ln ln 2 ln 3

50. 4 2 2 2 2 2 3 2838cot 8 csc 1 cot 8 csc cot 8 cot cot 8 csc 1t dt t t dt t t dt t dt t t dt

383 cot 8 cot 8t t t C

51. 1 1 12 2 10sin 3 cos 2 (sin sin 5 ) cos cos 5x x dx x x dx x x C

52. 1 1 1 12 2 2 10sin 2 cos 3 sin( ) sin 5 sin sin 5 cos cos 5x x dx x x dx x x dx x x C

53. 1 1 1 1 12 2 2 2 12sin 3 sin 3 cos 0 cos 6 cos 6 sin 6x x dx x dx dx x dx x x

π π π π π

ππ π π π

2 2 0π π π

54. 2 2 2 21 1 1 1 1

2 2 4 4 200 0 0sin cos sin 0 sin 2 sin 2 cos 2 ( 1 1)x x dx x dx x dx x

π π π π

55. 1 1 1 12 2 2 14cos 3 cos 4 cos( ) cos 7 cos cos 7 sin sin 7x x dx x x dx x x dx x x C



56. 2 2 21 1 1 1

2 2 6 8 22 2cos 7 cos cos 6 cos 8 sin 6 sin 8 0x x dx x x dx x x

π π π

ππ π

57. 1 cos 22 1 12 2 2sin cos 3 cos 3 cos 3 cos 2 cos 3d d d dθθ θ θ θ θ θ θ θ θ θ

1 1 1 1 12 2 2 2 4cos 3 cos(2 3) cos(2 3) cos 3 cos( ) cos 5d d d dθ θ θ θ θ θ θ θ θ θ

1 1 1 1 1 12 4 4 6 4 20cos 3 cos cos 5 sin 3 sin sin 5d d d Cθ θ θ θ θ θ θ θ θ

58. 22 2 4 2cos 2 sin 2cos 1 sin 4cos 4cos 1 sind d dθ θ θ θ θ θ θ θ θ θ

4 2 5 34 45 34cos sin 4cos sin sin cos cos cosd d d Cθ θ θ θ θ θ θ θ θ θ θ

59. 3 3 4 525cos sin 2 cos 2sin cos 2 cos sin cosd d d Cθ θ θ θ θ θ θ θ θ θ θ

60. 3 2 2 2sin cos 2 sin cos 2 sin 1 cos 2cos 1 sind d dθ θ θ θ θ θ θ θ θ θ θ

4 2 4 22cos 3cos 1 sin 2 cos sin 3 cos sin sind d d dθ θ θ θ θ θ θ θ θ θ θ θ

5 325 cos cos cos Cθ θ θ

61. 1 12 2sin cos cos 3 2sin cos cos 3 sin 2 cos 3d d dθ θ θ θ θ θ θ θ θ θ θ

1 1 1 12 2 4 4sin(2 3) sin(2 3) sin( ) sin 5 sin sin 5d d dθ θ θ θ θ θ θ θ θ

1 14 20cos cos 5 Cθ θ

62. 1 12 2sin sin 2 sin 3 cos(1 2) cos(1 2) sin 3 cos( ) cos 3 sin 3d d dθ θ θ θ θ θ θ θ θ θ θ θ

1 1 1 1 12 2 2 2 4sin 3 cos sin 3 cos 3 sin(3 1) sin(3 1) 2sin 3 cos 3d d d dθ θ θ θ θ θ θ θ θ θ θ θ

1 1 1 1 14 4 8 16 24sin 2 sin 4 sin 6 cos 2 cos 4 cos 6d d Cθ θ θ θ θ θ θ θ

63. 22 23 tan 1 secsec sec tan sec secsec

tan tan tan tan tan tan sec cscx xx x x x xx

x x x x xdx dx dx dx dx x x dx x dx

sec ln |csc cot |x x x C

64. 22 23

4 4 4 4 4

1 cos sinsin sin sin cos sin 3sincos cos cos cos cos

sec tan sec tanx xx x x x xx

x x x x xdx dx dx dx dx x x dx x x dx

2 313sec sec tan sec tan sec secx x x dx x x dx x x C

65. 2

2 2 2

2 2 2 2

1 costan sin cos1csc cos cos cos cos

sin sin sin sinxx x x

x x x x xdx x dx x dx x dx x dx

sec tan sin sec cosx x dx x dx x x C

66. 2 2

cot cos 1 2 2sin 2sin cos sin 2cos cos

csc 2 2 ln |csc 2 cot 2 |x xx x x xx x

dx dx dx dx x dx x x C



67. 1 cos 22 1 12 2 2sin cos2xx x dx x dx x dx x x dx 1

2, , cos 2 , sin 2u x du dx dv x dx v x

2 21 1 1 1 1 1 14 2 2 2 4 4 8sin 2 sin 2 sin 2 cos 2x x x x dx x x x x C

68. 3 2 2 2cos cos cos 1 sin cos cos sin cos ;x x dx x x x dx x x x dx x x dx x x x dx

cos sin sin sin cos ;x x dx x x x dx x x x , , cos , sinu x du dx dv x dx v x

2 3 31 13 3sin cos sin sin ;x x x dx x x x dx 2 31

3, , sin cos , sinu x du dx dv x x dx v x

3 2 3 21 1 1 1 13 3 3 3 3sin 1 cos sin sin sin cos sinx x x x dx x x x dx x x dx

3 31 1 13 3 9sin cos cos ;x x x x

2 3 31 1 13 3 9cos sin cos sin cos sin cos cosx x dx x x x dx x x x x x x x C

3 31 2 13 3 9sin sin cos cosx x x x x x C

69. sec tan 2 2secln(sec ); tan ;( ) tan ;x x

xy x y x y x 4 4

2

0 01 tan | sec |x dx x dx

π π

4

0ln |sec tan | ln 2 1 ln(0 1) ln 2 1x x

π

70. 4 4 2 12 144

sec ln sec tan ln 2 1 ln 2 1 ln ;M x dx x xπ π

ππ

2

2 1 2 1 2 1 2 1

2 1 2 1 2 1 2 1

14 4 2 1sec1 1 1 12 4 2 1ln 2ln 2ln ln4

tan 1 ( 1) ( , ) 0, lnxy dx x x yπ π

ππ

71. 1 cos 222 2 2 2 40 00 0 0 0

sin cos 2 sin 2xV x dx dx dx x dx x xπ π π π π ππ π π ππ π

2

2 4 2( 0) (0 0)π π ππ

72. 4 3 4

0 0 0 4 3 41 cos 4 2 cos 2 2 cos 2 2 cos 2 2 cos 2A x dx x dx x dx x dx x dx

π π π π π

π π

4 3 42 2 2 2 2 22 2 2 2 2 20 4 3 4

sin 2 sin 2 sin 2 (1 0) ( 1 1) (0 1) 2 2 2 2x x xπ π π

π π

73. 2 22 2 2 21 1 12 2 200

cos sin (2 ) sin (2 ) (0) sin (0) 2 ;M x x dx x xπ π

π π π

2 2 2 2

2 2 2 22 21 1 1 1

2 2 2 20 0 0 0cos cos cosx x x x dx x x x dx x dx x xdx

π π π π

π π π π

, , cos , sinu x du dx dv x dx v x

2 2 2 2

2 22 23 31 1 1 106 2 6 20 0 0

sin sin 8 0 2 sin 2 0 sinx x x xdx xdxπ ππ π

π π π ππ π π

2 2

24 4 4 41 13 3 3 302 2

cos cos 2 cos 0 0 ;xππ π π π

π ππ 2

2 21 122 0

cosy x x dxπ

π

2 2 2 2

2 2 2 22 2 2 21 1 1 1

4 4 2 40 0 0 02 cos cos cos cosx x x x dx x dx x xdx x dx

π π π π

π π π π

Section 8.4 Trigonometric Substitutions 577


2 2 2 2 2

2 2 22 2 cos 2 13 21 1 1 1 12 3012 2 4 8 80 0 0 0

sin cos 0 cos 2xx x x x dx x dx dxπ π ππ π π

π π π π π

2

2 2

2 22 2 8 31 1 13 3 4 120 016 8

sin 2 0x xπ ππ π π

π ππ π The centroid is 24 8 3

3 12, .π ππ

74. 3 32 2 2

0 0sin sec sin 2sin sec secV x x dx x x x x dx

π ππ π

3 3 3 3 3 31 cos 22 2

2 0 00 0 0 0sin 2 tan sec 2 ln |sec | tanxx dx x dx x dx dx x x

π π π π π ππ π π π π π

3 3

2 2 3 30 0cos 2 2 ln sec ln sec 0 tan tan 0dx x dx

π ππ π π ππ π

3 32 4 2 3 4 30 0

sin 2 2 ln 2 3 0 sin 2 sin 2(0) 2 ln 2 3x xπ ππ π π π π ππ π π π

2 4 21 3 48 ln 236 8 242 ln 2 3

π πππ π π

8.4 TRIGONOMETRIC SUBSTITUTIONS

1. 2 2

3 |cos | cos2 2 2 1 12 2 3|sec | 3 3cos 9

3 tan , , , 9 9 1 tan 9 sec ;d

xx dx xθ θ θπ π

θθθ θ θ θ

because

cos 0θ when 2 2π πθ ;

2

22

cos 29cos 3 33cos9

3 ln | sec tan | ln ln 9d xdx d x

xC C x x Cθ θ θ

θθθ θ

2. 22 2

3 22 2 cos1 9 1

; [3 ,3 ] ; tan , , , 1 | sec | sec ;dx du dttx u

x u dx du u t t du u t tπ π

22

2 2

cos (sec )1sec ln | sec tan | ln 1 ln 1 9 3du dt

t tut dt t t C u u C x x C

3. 2

2 21 1 11 1 1 1 12 2 2 2 2 4 2 4 44 22

tan tan 1 tan ( 1)dx xx

π π π

4. 2 2

2 2 21 1 11 1 1 1 1 1 1 12 2 2 2 2 2 2 2 2 4 168 2 4 00 0

tan tan 1 tan 0 0dx dx xx x

π π

5. 2

3 2 3 21 1 113 2 6 6090

sin sin sin 0 0dx x

x

π π

6. 2 2

1 2 2 1 2 1/ 22 1 1 114 4201 4 10 0

; 2 , 2 sin sin sin 0 0dx dt

x tt x dt dx t π π

7. 22 25sin , , 5 cos , 25 5 cos ;t dt d tπ πθ θ θ θ θ

1 cos 2 sin 22 22 2 425 5 cos 5 cos 25 cos 25 25t dt d d d Cθ θθθ θ θ θ θ θ

2 21 125 2525 25 252 2 5 5 5 2 5 2sin cos sin sint t tt t tC C Cθ θ θ



8. 21 13 2 2 3sin , , cos , 1 9 cos ;t dt d tπ πθ θ θ θ θ

2 2 1 21 1 1 13 3 6 61 9 cos cos cos sin cos sin (3 ) 3 1 9t dt d d C t t t Cθ θ θ θ θ θ θ θ

9. 2 27 72 2 2sec , 0 , sec tan , 4 49 49 sec 49 7 tan ;x dx d xπθ θ θ θ θ θ θ

7 22

2

sec tan 4 4921 1 17 tan 2 2 2 7 74 49

sec ln |sec tan | lnd xdx x

xd C C

θ θ θθ θ θ θ θ

10. 2 23 35 2 5sec , 0 , sec tan , 25 9 9 sec 9 3 tan ;x dx d xπθ θ θ θ θ θ θ

3 25

2

5 sec tan5 25 953 tan 3 325 9

sec ln |sec tan | lnddx xx

xd C C

θ θ θθ θ θ θ θ

11. 227 sec , 0 , 7 sec tan , 49 7 tan ;y dy d yπθ θ θ θ θ θ

2 7 tan 7 sec tan49 2 2

7 sec 7 tan 7 sec 1 7 tandy

y dy d d Cθ θ θ θ

θ θ θ θ θ θ θ

2 49 17 77 sec

y y C

12. 225 sec , 0 , 5 sec tan , 25 5 tan ;y dy d yπθ θ θ θ θ θ

2

3 3

5 tan 5 sec tan25 2 2 21 1 15 5 10125 sec

tan cos sin 1 cos 2dy

ydy d d d

θ θ θ θθ


12 25

2

sec25 251 51 110 10 5 10 2

sin cos secy

y yyy y y

C C Cθ θ θ

13. 22sec , 0 , sec tan , 1 tan ;x dx d xπθ θ θ θ θ θ

2

22 2

sec tan 1secsec tan1

sind xdx dxx x

C Cθ θ θ θθθ θ

θ

14. 22sec , 0 , sec tan , 1 tan ;x dx d xπθ θ θ θ θ θ

33 2

2 2 tan sec 1 cos 222sec tan1

2 cos 2 sin cosdx d

x xd d Cθ θ θ θ

θ θθ θ θ θ θ θ

2

2

22 1 2 1 11tan cos sec 1 sec xx x

C x x C x Cθ θ θ

15. 2 129 2 ;u x du x dx du x dx

2

21 129

9x dx

uxdu u C x C



16. 2 2 22 22 tan , , 2 sec , 4 4sec ;x dx d xπ πθ θ θ θ θ

2 22

2 2

4 tan 2sec 2 2 24 4sec

2 tan 2 sec 1 2 sec 2dx dx

xd d d d

θ θ θ

θθ θ θ θ θ θ θ

122 tan 2 2 tan xC x Cθ θ

17. 2

2 2 22 2 coscos

2 tan , , , 4 ;dx dx xθπ πθθ

θ θ

3 233

2 4 42

8 tan cos cos 1 sinsin

cos cos cos48 8 ;

d ddx dx

x

θ θ θ θ θ θθ θθ θ θ

[ cos , sin ]t dt dθ θ θ

3 2222 3

4 2 4 3

441 sec1 1 1 13 2 8 33

8 8 8 8 sec 8xxt

tt t t tdt dt C C Cθθ

3 22 2 2 21 13 34 4 4 8 4x x C x x C

18. 2 22 2tan , , sec , 1 sec ;x dx d xπ πθ θ θ θ θ

2 2

2 22 2

sec cos 11sintan sec sin1

d d xdxxx x

C Cθ θ θ θθθ θ θ

19. 22 22 sin , , 2 cos , 4 2 cos ;w dw d wπ πθ θ θ θ θ

2

2 22 2

8 8 2 cos 2 44 sin 2 cos sin4

2 2 cotdw d wdww w

C Cθ θ θθ θ θ

θ

20. 22 23 sin , , 3 cos , 9 3 cos ;w dw d wπ πθ θ θ θ θ

2 2

2 2 2

3cos 3cos 2 29 1 sin9 sin sin

cot csc 1dww

dw d d dθ θ θ θθ θ

θ θ θ θ θ

2 193cot sinw w

wC Cθ θ

21. 1

1

xdx

x


.1

x

x

2

1 1where 1 1

1 1

x xdx dx x

x x

2sin , cos , so that cos 0 and 1 cos .2 2

x dx d xπ πθ θ θ θ θ θ

2

1 2

1 sin 1cos

cos1

sin 1 cos

sin 1

xdx d

x

d C

x x C

θ θ θθ

θ θ θ θ



22. 2 124 2 ;u x du x dx du x dx

3 22 3 2 21 1 12 3 34 4x x dx u du u C x C

23. 3 22 33sin , 0 , cos , 1 cos ;x dx d xπθ θ θ θ θ

22 2

3 2 3 22

3 2 3 3 3 34sin cos 24 1 cos 430cos cos0 0 0 01

4 4 sec 1 4 tan 4 3dx dx

xd d

π π π πθ θ θ θ πθ θ

θ θ θ θ θ

24. 3 22 362sin , 0 , 2cos , 4 8 cos ;x dx d xπθ θ θ θ θ

3 2 3 22

1 6 6 62 cos 31 1 14 4 120 4 38 cos cos0 0 04

tanddx d

x

π π πθ θ θθ θ

θ

25. 3 22 32sec , 0 , sec tan , 1 tan ;x dx d xπθ θ θ θ θ θ

3 2 3 2 22

sec tan cos 1sintan sin 11

d ddx x

xxC Cθ θ θ θ θ

θθ θ

26. 5 22 52sec , 0 , sec tan , 1 tan ;x dx d xπθ θ θ θ θ θ

22 3

5 2 5 4 3 3 22 2

sec sec tan cos 1tan sin 3 sin1 3 1

dx dx x

x xd C Cθ θ θ θ θ

θ θ θθ

27. 3 22 32 2sin , , cos , 1 cos ;x dx d xπ πθ θ θ θ θ

3 22 3 25

6 6

51 cos cos 4 2 1cot 15 5sin

cot cscx dx d x

xxd C Cθ θ θ θ

θθ θ θ

28. 1 222 2sin , , cos , 1 cos ;x dx d xπ πθ θ θ θ θ

1 2223

4 4


cot cscx dx d x

xxd C Cθ θ θ θ

θθ θ θ

29. 22 2 41 12 2 2 2tan , , sec , 4 1 sec ;x dx d xπ πθ θ θ θ θ

212

2 4 22

8 sec 2 18 4sec 4 14 1

4 cos 2 sin cos 2 tan 2ddx x

xxd C x C

θ θ

θθ θ θ θ θ

30. 2 2 21 13 2 2 3tan , , sec , 9 1 sec ;t dt d tπ πθ θ θ θ θ

21

32 4 22

6 sec6 2 1 3sec 9 19 1

2 cos sin cos tan 3ddt t

ttd C t C

θ θ

θθ θ θ θ θ



31. 2 121 2 ;u x du x dx du x dx

3

2 2 22 2 2 21 1 1 1 1 1 1

2 2 2 2 2 21 1 1ln | | ln 1x x x

ux x xdx x dx x dx dx x du x u C x x C

32. 2 1825 4 8 ;u x du x dx du x dx

221 1 1 1

8 8 825 4ln | | ln 25 4x

uxdx du u C x C

33. 5 22 52 2sin , , cos , 1 cos ;v dv d vπ πθ θ θ θ θ

22 3

5 2 5 22

3sin cos 2 2 tan 1

3 3cos 11tan secdv dv v

vvd C Cθ θ θ θ

θθ θ θ

34. 5/22 52 2sin , ; cos , 1 cos ;r dr d rπ πθ θ θ θ θ

5 22 5 27

8 8


cot cscr dr d r

rrd C Cθ θ θ θ

θθ θ θ

35. Let 21 1 2 2sec1 43 3 tan3tan , ln(3 tan ), tan tan , , 9 9 tan 9 3 sec ;t te t dt d eθ

θθ θ θ θ θ θ

1 1 12

12 1 1

ln 4 tan (4 3) tan (4 3) tan (4 3)3 tan sectan 3sec tan (1 3)90 tan (1 3) tan (1 3)

sec ln sec tant

t

e dt d

edθ θ θ

θ θ θ θ θ θ

105 4 13 3 3 3ln ln ln 9 ln 1 10

36. Let 21 1 2 2 23 sec44 3 tantan , ln(tan ), tan tan , , 1 1 tan sec ;t te t dt d eθ

θθ θ θ θ θ θ

2sec1 1

1tan

13 2 31 12

tanln (4 3) tan (4 3) tan (4 3) tan (4 3) 34 15 5 5tan (3 4)secln (3 4) tan (3 4) tan (3 4)1

cos sint

t

de dt

ed

θθθ θ

θθ θ θ

37. 2

1 4 12 2 2 2 216 44 11 12 1 3

; 2 , ; tan , , sec , 1 sec ;dt du

t t t t uu t du dt u du d uπ πθ θ θ θ θ

2

2 2

1 4 42 2 sec4 6 661 sec1 3 6

2 2du d

u

π πθ θ π π ππθπ

θ

38. 2tan tan 2 24, 0 , sec , 1 ln 1 tan sec ;y e dy e d yθ θπθ θ θ θ θ

tan 2

tan2

4 44sec

0sec1 0 01 lnsec ln sec tan ln 1 2

edy e

ey yd d

θ

θ

π π πθθ

θ θ θ θ θ

39. 2 22sec , 0 , sec tan , 1 sec 1 tan ;x dx d xπθ θ θ θ θ θ θ

2

sec tan 1sec tan1

secddx

x xC x Cθ θ θ

θ θ θ



40. 2 2 2tan , sec , 1 sec ;x dx d xθ θ θ θ 2

2 2

sec 11 sec

tanddxx

C x Cθ θθ

θ

41. 2 2sec , sec tan , 1 sec 1 tan ;x dx d xθ θ θ θ θ θ

2

sec sec tan 2 2tan1

sec tan 1x dx d

xd C x Cθ θ θ θ

θ θ θ θ

42. 2 22 2sin , cos , , 1 1 sin cos ;x dx d xπ πθ θ θ θ θ θ

2

cos 1cos1

sinddx

xC x Cθ θ

θ θ

43. 2 2 2 4 212 2Let tan , 0 , 2 sec sec ; 1 1 tan secx x dx d x dx d xπθ θ θ θ θ θ θ θ

2

4

4 2sec1 1 1 12 sec 2 2 21

sec ln |sec tan | ln 1x

xdx d d C x x Cθ

θ θ θ θ θ θ

44. 212 2Let ln sin , 0 or 0 , cos , 1 ln cosxx dx d xπ πθ θ θ θ θ θ

2 2 21 ln cos 1 sinln sin sin csc sin ln | csc cot | cos

x

x x dx d d d d Cθ θθ θθ θ θ θ θ θ θ θ θ

2 2

1 ln 1 1 ln2 21ln ln lnln 1 ln ln 1 ln

x x

x x xx C x C

45. 2

22 24 4Let 2 2 2 4 ;x u

x uu x x u dx u du dx u du u du

222sin , 2cos , 0 , 4 2 cosu du d uπθ θ θ θ θ

1 cos 22 222 4 2 2 cos 2 cos 8 cos 8 4 4 cos 2u du d d d d dθθ θ θ θ θ θ θ θ θ

21 142 2 2 24 2 sin 2 4 4sin cos 4 sin 4 4 sin 4u xu uC C C x x Cθ θ θ θ θ

1 224 sin 4x x x C

46. 3 2 2 3 1 323Let u x x u dx u du

2 3 1 3

3 3 1 32 22 3

1 3 1 1 3 22 2 2 1 2 23 3 3 31 31 11

sin sinx u ux uu uu

dx u du du du u C x C

47. 2 2 2 2Let 2 1 1 2 2 1 ;u x x u dx u du x xdx u u u du u u du

22 2sin , cos , , 1 cosu du d uπ πθ θ θ θ θ

1 cos 42 2 2 2 2 21 12 2 22 1 2 sin cos cos 2 sin cos sin 2u u du d d d dθθ θ θ θ θ θ θ θ θ θ

21 1 1 1 1 1 1 14 4 4 16 4 8 4 4cos 4 sin 4 sin 2 cos 2 sin cos 2cos 1d d C C Cθ θ θ θ θ θ θ θ θ θ θ θ

3 23 1 2 21 1 1 1 1 14 2 4 4 2 4sin cos sin cos sin 1 1C u u u u u Cθ θ θ θ θ

3 211 1 14 2 4sin 1 1x x x x x C



48. 22 22 1

1Let 1 1 2 2 2 1x w

wxw x w x w dw dx dx w dw w dw

22sec , sec tan , 0 , 1 tanw dx d wπθ θ θ θ θ θ

22 1 2 tan sec tan ;w dw dθ θ θ θ 2tan , sec , sec tan , secu du d dv d vθ θ θ θ θ θ θ

3 22 tan sec tan 2 sec tan 2 sec 2 sec tan 2 sec secd d dθ θ θ θ θ θ θ θ θ θ θ θ θ

2 22 sec tan 2 tan 1 sec 2 sec tan 2 tan sec secd d dθ θ θ θ θ θ θ θ θ θ θ θ

22sec tan 2 ln |sec tan | 2 tan sec dθ θ θ θ θ θ θ

2 2 22 tan sec sec tan ln |sec tan | 1 ln 1d C w w w w Cθ θ θ θ θ θ θ

1 2 ln | 1 2 |x x x x C

49. 22 2 44; 4 ; ;dy xdx

dx x xx x dy x y dx

222sec , 0 , 2sec tan , 4 2 tanx dx d xπθ θ θ θ θ θ

2 tan 2 sec tan 2 22 sec 2 tan 2 sec 1 2 tan

dy d d C

θ θ θ θθ θ θ θ θ θ θ 2 14

2 22 sec ;x x C

2x and 2 142 20 0 0 0 2 secx xy C C y

50. 2 2

2

9 99 1, ; ;dy dx dx

dx x xx dy y

2

23sec , 0 , 3sec tan , 9 3 tanx dx d xπθ θ θ θ θ θ

23 sec tan 93 tan 3 3sec ln |sec tan | ln ;d xxy d C Cθ θ θ

θ θ θ θ θ 5x and ln 3y

2 93 3ln 3 ln 3 0 ln xxC C y

51. 2 2

32 132 24 4

4 3, ; 3 tan ;dxdy dx xdx x x

x dy y C

2x and 13 32 80 0 tan 1y C C π

13 32 2 8tan xy π

52. 3 22

2 3 22 2 2 2 3

11 1, ; tan , sec , 1 sec ;dy dx

dxx

x x dy x dx d xθ θ θ θ

2

3 2

sec tansecsec 1

cos sin tan cos ;d x

xy d C C C Cθ θ θ

θθθ θ θ θ θ

0x and 1y

2 11 0 1 1x

xC C y

53. 23

2 293 20

; 3 sin , 0 , 3 cos , 9 9 9 sin 3 cos ;xA dx x dx d xπθ θ θ θ θ θ

2 2 23 cos 3 cos 2 3 3

3 2 400 03 cos sin cosdA d

π π πθ θ θ πθ θ θ θ θ



54. 22 2

2 2 21 1 ;yx xa b a

y b 2 2

2 20 0

4 1 4 1a a

x xa a

A b dx b dx

2

22 2 2sin , , cos , 1 cos , 0 sin 0, sinxa

x a dx a d x a x a aπ π πθ θ θ θ θ θ θ θ θ

2

2

2 2 2 1 cos 2220 0 0 0

4 1 4 cos cos 4 cos 4a

xa

b dx b a d ab d ab dπ π π θθ θ θ θ θ θ

2 2 2 2

20 00 02 2 cos 2 2 sin 2 2 0 sin sin 0ab d ab d ab ab ab ab ab

π π π π πθ θ θ θ θ π π

55. (a) 2

1 21 1 1

10sin sin , , ,

xA x dx u x du dx dv dx v x

2

1 21 21 21 1 2 6 3 121 12 2 120 10 0

sin sin 0 1x

xx x dx x π

(b) 1 2

1 6 3 12120

sin ;M x dx π

6 3 1212

1 2 1 21 11 12

6 3 120 0sin sinx x x dx x x dxπ π

2

1 21 121

sin , , ,x

u x du dx dv x dx v x

2

2

1 21 22 112 1 12 26 3 12 0 10

sin x

xx x dx

π

2 12 2 2 6sin , , cos , 1 cos , 0 sin 0, sinx dx d x x xπ π πθ θ θ θ θ θ θ θ θ

26 62 1 2sin12 1 1 1 1 12 12 2 2 2 cos 48 26 3 12 6 3 120 0

sin 0 cos sind dπ π

θ πθπ π

θ θ θ θ

6 6 61 cos 212 1 12 1 148 2 2 48 4 46 3 12 6 3 120 0 0

cos 2d d dπ π πθπ π

π πθ θ θ θ

6 3 312 1

48 4 86 3 12 0 4 6 3 12sin 2 ;

π ππ θπ π

θ

6 3 1212

1 2 211 120

siny x dxπ 1

2

2 2 sin1

1sin , , ,x

xu x du dx dv dx v x

1

2

1 2 1 22 2 sin166 3 12 100

sin x x

xx x dx

π

2 2

1 221

1 1sin , , , 2 1x

x xu x du dx dv dx v x

2

2

1 2 1 221 2 1 2 16 1 12 26 3 12 100

sin 0 2 1 sin x

xx x dx

π

22 22 1 21 3 12 3 726 61 1

72 2 2 72 606 3 12 6 3 12 12 6 3 122 1 sin 0 2 1x π π ππ π

π π π

56. 21 11 1

0 0tan tanV x x dx x x dxπ π 2

1 21 121

tan , , ,x

u x du dx dv x dx v x

2

2 2 2

1 1 112 1 11 1 1 1 1 1 12 2 2 2 8 21 1 10 0 0 0

tan tan 1 0 1 1xx x x

x x dx dx dxππ π π

2

1 1 1 ( 2)1 11 1 1 1 1 1 18 2 2 8 2 2 8 2 2 41 00 0

tan tan 1 0 0x

dx dx x x π ππ π ππ π π

Section 8.5 Integration of Rational Functions by Partial Fractions 585


57. (a) Integration by parts: 3 22 2 213, 2 , 1 , 1u x du x dx dv x x dx v x

3 23 2 3 2 5 23 2 2 2 2 2 2 21 1 1 23 3 3 151 1 1 2 1 1x x dx x x x x dx x x x C

(b) Substitution: 2 2 121 1 2u x x u du x dx du x dx

3 2 2 2 3/2 3/2 5/21 1 1 12 2 3 51 1 1x x dx x x x dx u u du u u du u u C

3/2 5/22 21 13 51 1x x C

(c) Trig substitution: 22 2sin , , cos , 1 cosx dx d xπ πθ θ θ θ θ

3 2 3 2 2 2 21 sin cos cos sin cos sin 1 cos cos sinx x dx d d dθ θ θ θ θ θ θ θ θ θ θ θ

3 2 5 22 4 3 5 2 21 1 1 13 5 3 5cos sin cos sin cos cos 1 1d d C x x Cθ θ θ θ θ θ θ θ

58. (a) The slope of the line tangent to ( )y f x is given by ( ).f x Consider the triangle whose hypotenuse is

the 30 ft rope, the length of the base is x and height 2900 .h x The slope of the tangent line is also 2900 ,x

x thus

2900( ) .xxf x

(b) 2900( ) x

xf x dx 2230sin , 0 , 30cos , 900 30cosx dx d xπθ θ θ θ θ

22 1 sin30cos cos

30sin sin sin30cos 30 30 30 csc 30 sind d d d dθθ θ

θ θ θθ θ θ θ θ θ θ θ

2 29003030ln csc cot 30cos 30ln 900 ;xx xC x Cθ θ θ

2 22 2900 30 90030 3030 30(30) 0 0 30ln 900 30 ( ) 30ln 900x

x xf C C f x x

8.5 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS

1. 5 13( 3)( 2) 3 2 5 13 ( 2) ( 3) ( ) (2 3 )x A Bx x x x x A x B x A B x A B

5(10 13) 3 2;

2 3 13

A BB B A

A B

thus, 5 13 32( 3)( 2) 3 2

xx x x x

2. 25 7 5 7

( 2)( 1) 2 13 25 7 ( 1) ( 2) ( ) ( 2 )x x A B

x x x xx xx A x B x A B x A B

52 3;

2 7

A BB A

A B

thus, 25 7 3 2

2 13 2x

x xx x



3. 2 24

1( 1) ( 1)

14 ( 1) ( ) 1

4x A B

xx x

Ax A x B Ax A B A

A B

and 3;B thus,

2 24 31

1( 1) ( 1)x

xx x

4. 2 2 22 2 2 2

12 1 ( 1) ( 1)

22 2 ( 1) ( )

2x x A B

xx x x x

Ax A x B Ax A B

A B

2A and 4;B thus, 2 22 2 2 4

12 1 ( 1)x

xx x x

5. 2 22 21

1( 1)

0

1 ( 1) ( 1) 1 ( ) ( ) 1

1

Cz A Bz zz z z

A C

z Az z B z Cz z A C z A B z B A B

B

1 2 2;B A C thus, 2 21 2 1 2

1( 1)z

z zz z z

6. 3 2 21 1

( 3)( 2) 3 26 6

01 ( 2) ( 3) ( ) (2 3 )

2 3 1z A B

z z z zz z z z z

A BA z B z A B z A B

A B

1 15 55 1 ;B B A thus,

1 15 5

3 2 3 26z

z zz z z

7. 2

2 28 5 2

5 6 5 61t t

t t t t

(after long division); 2

5 2 5 2( 3)( 2) 3 25 6

5 2 ( 2) ( 3)t t A Bt t t tt t

t A t B t

5( ) ( 2 3 ) (10 2) 12 12 17;

2 3 2

A BA B t A B B B A

A B

thus,

2

28 17 12

3 25 61t

t tt t

8. 4 2 2

4 2 4 2 2 29 9 9 9 9

9 9 91 1t t t

t t t t t t


2 22 29 9

99t Ct DA B

t t tt t

2 2 2 2 3 29 9 9 9 ( ) ( ) ( ) 9 9t At t B t Ct D t A C t B D t At B

0

90 0; 1 10;

9 0

9 9

A C

B DA C B D

A

B

thus, 4

4 2 2 29 101

9 91t

t t t t

9. 21 1 1

1 1 2 211 (1 ) (1 ); 1 ; 1 ;A B

x xxA x B x x A x B

21 1 12 1 2 1 21

ln 1 ln 1dx dx dxx xx

x x C



10. 21 1 1

2 2 221 ( 2) ; 0 ; 2 ;A B

x xx xA x Bx x A x B

21 1 12 2 2 22

ln ln 2dx dx dxx xx x

x x C

11. 24 5 2 2

6 1 7 7 75 64 ( 1) ( 6); 1 ; 6 ;x A B

x xx xx A x B x x B x A

22 54 5 52 2 1

7 6 7 1 7 7 75 6ln 6 ln 1 ln ( 6) ( 1)x dx dx

x xx xdx x x C x x C

12. 22 1 7 9

4 3 1 17 122 1 ( 3) ( 4); 3 7; 4 9;x A B

x xx xx A x B x x B x A

9

2 7

( 4)2 14 37 12 ( 3)

9 7 9ln 4 7ln 3 ln xx dx dxx xx x x

dx x x C C

13. 231 1

3 1 4 4 42 3( 1) ( 3); 1 ; 3 ;y A B

y yy yy A y B y y B y A

2

8 8 8 83 3 3 31 1 1 14 3 4 1 4 4 4 4 4 42 3 44 4 4

ln 3 ln 1 ln 5 ln 9 ln1 ln 5y dy dy dyy yy y

y y

ln 151 12 2 2ln 5 ln 3

14. 2

41 4 ( 1) ;y A B

y yy yy A y By

3

10 4; 1 3;y A y B

2

1 1 1 14 311 2 21 21 2 1 2 1 2

4 3 4ln 3ln 1 (4 ln1 3ln 2) 4 ln 3lny dy dyy yy y

dy y y

27 27 271 1 18 16 8 8 8 4ln ln ln ln 16 ln

15. 3 21 1 1

2 1 2 621 ( 2)( 1) ( 1) ( 2); 0 ; 2 ;CA B

t t tt t tA t t Bt t Ct t t A t B

3 21 1 1 1 1 1 13 2 6 2 3 1 2 6 32

1 ; ln | | ln | 2| ln | 1 |dt dt dt dtt t tt t t

t C t t t C

16. 33 31 1

2 2 2 8 162 8( 3) ( 2)( 2) ( 2) ( 2); 0 ; 2 ;x CA B

x x xx xx A x x Bx x Cx x x A x B

35 3 3 5 3 51 1

16 8 16 2 16 2 8 16 162 82 ; ln ln 2 ln 2x dx dx dx

x x xx xx C dx x x x C

5

6

( 2) ( 2)116 ln x x

xC

17. 3

2 23 2

2 1 ( 1)( 2)x x

x x xx

(after long division); 2 2

3 21( 1) ( 1)

3 2 ( 1) ( )x A Bxx x

x A x B Ax A B

3

2 2

1 1 1 1

12 1 ( 1)0 0 0 03, 2 3, 1; ( 2) 3x dx dx dx

xx x xA A B A B x dx

2 11 1 1

2 1 2 202 3ln 1 2 3ln 2 (1) 3ln 2 2x

xx x



18. 3

2 23 2

2 1 ( 1)( 2)x x

x x xx

(after long division); 2 2

3 21( 1) ( 1)

3 2 ( 1) ( )x A Bxx x

x A x B Ax A B

3, 2 3, 1;A A B A B 3

2 2

0 0 0 0

12 1 ( 1)1 1 1 1( 2) 3x dx dx dx

xx x xx dx

2 01 1 1 1

2 1 ( 1) 2 ( 2)12 3ln 1 0 0 3ln1 2 3ln 2 2 3ln 2x

xx x

19. 2 2 2 22 2 2 21

1 1( 1) ( 1) ( 1)1 ( 1)( 1) ( 1)( 1) ( 1) ( 1) ;CA B D

x xx x xA x x B x x C x D x

141 ;x C 1

41 ;x D coefficient of 3 0;x A B A B constant A B C D 121 ;A B C D A B thus, 1 1

4 4 ;A B

2 2 2 22

11 1 1 1 14 1 4 1 4 4 4 1( 1) ( 1) 2 11

lndx dx dx dx dx x xx x xx x xx

C

20. 2

22

2 21 1 ( 1)( 1) 2 1

( 1) ( 1)( 1) ( 1);x CA Bx x xx x x

x A x B x x C x 1

21 ;x C

141 ;x A coefficient of

2

2

2 34 ( 1) 2 1

1 ; x dxx x x

x A B A B B

3

2

ln ( 1)( 1)3 31 1 1 1 1

4 1 4 1 2 4 4 2( 1) 4 2( 1)( 1)ln 1 ln 1

x xdx dx dxx x x xx

x x C C

21. 22

21 11 21( 1) 1

1 1 ( )( 1); 1 ;Bx CAx xx x

A x Bx C x x A

coefficient of 2x A B

120 ;A B B constant 1

21 ;A C A C C 22

1 1 1( 1)1 1

2 1 2 1( 1) 10 0 0

xdx dxx xx x

dx

12 1 1 11 1 1 1 1 1 1 1 12 4 2 2 4 2 2 4 20

ln | 1| ln 1 tan ln 2 ln 2 tan 1 ln1 ln1 tan 0x x x

( 2ln 2)1 14 2 4 8ln 2 ππ

22. 2

3 22 23 4

13 4 1 ( ) ; 0 4;t t Bt CA

tt t tt t A t Bt C t t A

coefficient of 2 3t A B A B

1;B coefficient of 2

3 2

3 3 3( 1)3 4

1 11 1 11; 4 tt t dt

tt tt C C dt dt

32 1 1 11 1 1

2 2 214 ln ln 1 tan 4ln 3 ln 4 tan 3 4ln1 ln 2 tan 1t t t

91 13 2 4 2 12 122

2 ln 3 ln 2 ln 2 2ln 3 ln 2 lnπ π π π

23. 2

2 2 22 2

2 1 2 2 3 211 1

2 1 ( ) 1 ( ) ( )y y Ay B Cy D

yy yy y Ay B y Cy D Ay By A C y B D

0, 1; 2 2; 1 0;A B A C C B D D 2

2 2 22 2

2 1 111 1

2y y y

yy ydy dy dy

21 1

1tan

yy C



24. 2

2 2 22 2

2 2 3 28 8 24 14 1 4 1

8 8 2 ( ) 4 1 4 4 ( ) ( );x x Ax B Cx Dxx x

x x Ax B x Cx D Ax Bx A C x B D

0, 2; 8 8; 2 0;A B A C C B D D

2

2 2 22 2

8 8 24 14 1 4 1

2 8 x dxx x dxxx x

dx

21 1

4 1tan 2

xx C

25. 2 2 32 32 2

11 ( 1) ( 1)1 ( 1)s As B C D E

ss s ss s

3 2 2 2 22 2 ( )( 1) 1 ( 1) 1 ( 1) 1s As B s C s s D s s E s

4 3 2 4 3 2 3 2 2( 3 ) (3 3 ) ( 3 ) 2 2 2 1 1 1As A B s A B s A B s B C s s s s D s s s E s 4 3 2( ) ( 3 2 ) (3 3 2 ) ( 3 2 ) ( )A C s A B C D s A B C D E s A B C D s B C D E

0

3 2 0

3 3 2 0

3 2 2

2

A C

A B C D

A B C D E

A B C D

B C D E

summing all equations 2 4 2;E E

summing eqs (2) and (3) 2 2 0 1;B B summing eqs (3) and (4) 2 2 2 0;A A

0C from eq (1); then 1 0 2 2D from eq (5) 1;D

2 2 32 3

2 1 12 21 ( 1) ( 1)1 ( 1)

2 ( 1) ( 1) tans ds ds dss s ss s

ds s s s C

26. 4

2 2 22 2

24 2 28199 9

81 9 ( ) 9 ( )s Bs C Ds EAs ss s s

s A s Bs C s s Ds E s

4 2 4 3 2 218 81 9 9A s s Bs Cs Bs Cs Ds Es

4 3 2( ) (18 9 ) (9 ) 81 81 81A B s Cs A B D s C E s A A or 1; 1 0;A A B B 0;C

9 0 0; 18 9 0 18;C E E A B D D

4

2 2 2 22

81 9( 9) 99

18 ln | |s dss dss s ss s

ds s C

27. 2

3 22 2 22

11 12 1 ( )( 1) ( ) ( ) ( )x x Bx CA

xx x xx x A x x Bx C x A B x A B C x A C

1, 1, 2A B A B C A C adding eq(2) and eq(3) 2 1,A B add this equation to eq (1)

2 1 43 3 33 2 1 1 ;A A B A C A B 2

3 2

2 3 (1 3) 4 3211 1

xx xxx x x

dx dx

2 312 4

42 1 1 1 13 1 3 2 2

xx x

dx dx u x u x du dx

92

2 2 23 3 34 4 4

32 1 1 2 1 1 13 1 3 3 1 3 2

u ux xu u u

dx du dx du du

12

2 1 2 13 3 2 12 1 1 2 13 6 2 4 3 63 3 2 3

ln 1 ln tan ln 1 ln 1 3 tanx xx x C x x x C



28. 4 22 21

1 11 ( 1) 1 1 ( ) ( 1)Cx DA B

x xx x x xA x x x B x x x C x D x x

3 2( ) ( ) ( ) 1, 0 , 0A B C x B C D x B D x A A B D D B B C D 1 2 13 3 32 0 2 , 0 1 2 0 ;B C C B A B C B B B C D

4 2 2

1 3 ( 2 3) 1 3 2 11 1 1 1 1 11 3 1 31 1

x xx x x xx x x x x x

dx dx dx dx dx

21 13 3ln | | ln | 1| ln | 1|x x x x C

29. 2

4 22 2 2

1 11 1( 1) 1 ( 1) 1 ( )( 1)( 1)x Cx DA B

x xx xx A x x B x x Cx D x x

3 2( ) ( ) ( ) 0, 1,A B C x A B D x A B C x A B D A B C A B D

0, 0A B C A B D adding eq (1) to eq (3) gives 2 2 0,A B adding eq (2) to eq (4) gives

2 2 1,A B adding these two equations gives 144 1 ,B B using 1

42 2 0 ,A B A using

120 ,A B D D and using 2

4 2

1 4 1 4 1 21 11 1

0 0; xx xx x

A B C C dx dx

21 111 1 1 1 1 1 1 1 1 1 1

4 1 4 1 2 4 4 2 4 1 21ln 1 ln 1 tan ln tanx

x x xxdx dx dx x x x C x C

30. 2

4 2 22 2 2

2 23 4 1( 2) 1 ( 2) 1 ( )( 2)( 2)x x Cx DA B

x xx x xx x A x x B x x Cx D x x

3 2( ) (2 2 ) ( 4 ) 2 2 4 0, 2 2 1,A B C x A B D x A B C x A B D A B C A B D

4 1, 2 2 4 0A B C A B D subtracting eq (1) from eq (3) gives 155 1 ,C C subtracting

eq (2) from eq (4) gives 155 1 ,D D substituting for C in eq (1) gives 1

5 ,A B and substituting

for D in eq (4) gives 4 25 52 2 ,A B A B adding this equation to the previous equation gives

3 3 15 10 102 ;A A B 2

4 2 2

3 10 1 10 ( 1 5) 1 52 23 4 1

xx xx xx x x

dx dx

2 22 13 31 1 1 1 1 1 1 1 1

10 2 10 2 5 5 10 10 10 51 1ln 2 ln 2 ln 1 tanx

x x x xdx dx dx dx x x x x C

31. 3 2

2 2 22 2

3 2 22 5 8 42 22 2 2 2

2 5 8 4 ( ) 2 2A B C D A B C Dθ θ θ θ θθ θθ θ θ θ


3 2(2 ) (2 2 ) (2 ) 2; 2 5 1; 2 2 8 2;A A B A B C B D A A B B A B C Cθ θ θ

3 2

2 2 22 2

2 5 8 4 2 1 2 22 22 2 2 2

2 4 2;B D D d d dθ θ θ θ θθ θθ θ θ θ

θ θ θ

2 2 2 2 22

(2 2) (2 2) 2 12 2 2 2 ( 1) 1 2 22 2

ln 2 2d dd dθ θ θ θθ θθ θ θ θ θ θ θθ θ

θ θ

22 1 1

2 2ln 2 2 tan ( 1) C

θ θθ θ θ

32.

4 3 2

3 2 2 32 2 2

4 2 3 111 1 1

A B C D E Fθ θ θ θ θ θ θθθ θ θ

24 3 2 2 24 2 3 1 ( ) 1 ( ) 1A B C D E Fθ θ θ θ θ θ θ θ θ

4 2 3 2( ) 2 1A B C D C D E Fθ θ θ θ θ θ θ



5 4 3 2 3 22 2A B A B A B C D C D E Fθ θ θ θ θ θ θ θ θ

5 4 3 2(2 ) (2 ) ( ) ( ) 0; 1;A B A C B D A C E B D F A Bθ θ θ θ θ

2 4 4; 2 2 0; 3 1; 1 0;A C C B D D A C E E B D F F

4 3 2

3 2 2 32 2 2

1 21 2 24 2 3 1 1411 1 1

4 tan 2 1 1d d dd Cθ θ θ θ θ θ θ θ θθθ θ θ

θ θ θ θ

33. 3 2

2 22 2 1 1 1 1

( 1) ( 1) 12 2 ; 1 ( 1) ; 0 1; 1 1;x x A Bx x x x x xx x x x

x x A x Bx x A x B

3 2

22 22 2 1 1

12 ln ln 1 lnx x dx dx xx x xx x

x dx x x x C x C

34. 4

2 22 21 1 1

( 1)( 1) ( 1)( 1) 1 11 11 1 ; 1 ( 1) ( 1);x A B

x x x x x xx xx x A x B x

4

221 1 1 1

2 2 2 1 2 111 ; 1 ; 1x dx dx

x xxx A x B dx x dx

33 11 1 1 13 2 2 3 2 1ln 1 ln 1 lnx x

xx x x x C x C

35. 3 2

3 2 29 3 1 9 3 1

( 1)9x x x x

x x x x


2 29 3 1

1( 1)x x CA B

x xx x x

2 29 3 1 ( 1) ( 1) ; 1 7; 0 1; 9 2;x x Ax x B x Cx x C x B A C A 3

3 2 29 3 1 1

19 2 7 9 2ln 7ln 1x x dx dx dxx x xx x x

dx dx x x x C

36. 3

2 2 2 216 12 4 12 4

2 14 4 1 4 4 1 (2 1) (2 1)(4 4) ; 12 4 (2 1)x x x A B

xx x x x x xx x A x B

3

2 216

2 14 4 1 (2 1)6; 4 2; 4 ( 1) 6 2x dx dx

xx x xA A B B dx x dx

2 2 1112 12( 1) 3ln 2 1 2 4 3ln 2 1 (2 1) ,xx x C x x x x C

where 12C C

37. 4 2

3 22 2

1 2 21 111 1

; 1 1 ( ) ( )y y By CAyy y yy y y y

y A y By C y A B y Cy A

4 2 2

3 2

1 212 21

1; 0 1; 0; ln | | ln 1y dyy y dy yyy y y

A A B B C dy y dy y y C

38. 4

3 2 3 2 3 2 22

2 2 2 211 1 1 11 ( 1)

2 2 ;y By CAyy y y y y y y y y yy y

y

2 2 2 22 1 ( )( 1) ( ) ( ) ( )A y By C y Ay A By Cy By C A B y B C y A C

0, 0A B B C or , 2 1, 1, 1;C B A C A B A B C

4

3 2 2 2

2 2 2 1111 21 1 1

2 ( 1) ( 1) ln 1 ln 1 tany dy y dyyy y y y y

dy y dy dy y y y y C

2 2 1122 ln 1 ln 1 tan ,y y y y y C where 1 1C C

39. 2 2

1 11 2 23 2 3 2 2

;[ , ] ln lnt t

t t t

dy dy dy yt te dt ey y ye e y y e

e y e dt dy C C

40. 3 24 2 3

2 2 2 2 2 2

2 1 12 2 121 1 1 1 1 1

; ,t t t t t

t t

y y y y y dyt t te e e e ee e y y y y

dt e dt y e dy e dt dy y dy dy

2 2 1 2 2 11 1 12 2 2 2ln 1 tan ln 1 tany t t ty y C e e e C



41. 2 2

cos sin 221 1 1 1 15 2 3 5 3 5 sin 3sin sin 6 6

; [sin , cos ] ln lny dy ydt tt t t yy y t t

y t y dy dt dt C C

42. 2 2

2sin cos 21 1 1 13 2 3 1 3 1 3 cos 1cos cos 2 2

; [cos , sin ] ln lndy dy dy ydy y yy y

y d dy C Cθ θ θθθ θ

θ θ θ

cos 113 cos 2ln Cθ

θ

43. 2 1 3 1

2 2 2 22 2

2( 2) tan (2 ) 12 3 tan (2 ) 112 24 1 ( 2) 4 1 ( 2)4 1 ( 2)

3 tan (2 ) 3 6dxx x x x x x dx dxxx x x xx x

dx dx dx x

21 614 2tan 2 3ln | 2| xx x C

44. 2 1 3 1

2 2 2 22 2

3( 1) tan (3 ) 9 tan (3 ) 113 19 1 ( 1) 9 1 ( 1)9 1 ( 1)

tan (3 ) dxx x x x x x dx dxxx x x xx x

dx dx dx x

211 16 1tan 3 ln 1 xx x C

45. 3 2 21 1 1 1 2

( 1) 2 1; Let 2 ;

x x x xx x udx dx u x du dx du dx du

22

1 112 ( 1) ( 1) ( ) 0, 2 1 1;A B

u uuA u B u A B u A B A B A B B A

212 1 1 1 1

1 1 1 1 11ln | 1| ln | 1| ln x

u u u u xudu du du du u u C C

46. 2

2 2 21 3 2 3

6 5 5 6 6 61 11 1 11 1

; [Let 6 ] 6 6 6 ;uu u ux x u u

dx x u dx u du u du du du du du

26

1 116 ( 1) ( 1) ( ) 0, 6 3 3;A B


26 3 3 1 1

1 1 1 116 6 6 3 3 6 3ln | 1| 3ln | 1|u u u uu

du du u du u du du u u u C

1 6

1 61 6 1

16 3ln x

xx C

47. 2

2 2 2 221 2 2 2

1 1 1 1; Let 1 2 2 2 2 ;x u u

x u u u udx x u dx u du u du du du du du

22

1 112 ( 1) ( 1) ( ) 0, 2 1 1;A B


22 1 1 1 1

1 1 1 112 2 2 2 ln | 1| ln | 1|u u u uu

du du u du u du du u u u C

1 11 1

2 1 ln xx

x C

48. 2 22

21 1 2 23 39 9 99

; Let 9 2 2 ; A Bu ux x u uu u

dx x u dx u du u du du

1 13 32 ( 3) ( 3) ( ) 3 3 0, 3 3 2 ;A u B u A B u A B A B A B A B

2

1 3 1 3 9 32 1 1 1 1 1 1 13 3 3 3 3 3 3 3 3 9 39

ln | 3| ln | 3| ln xu u u u xu

du du du du u u C C



49. 3

4 4 4

4 31 1 1 14 ( 1) ( 1) 11 1

; Let 4 ;x A Bu u u u u ux x x x

dx dx u x du x dx du

1 ( 1) ( ) 1 1;A u Bu A B u A A B

4

41 1 1 1 1 1 1 1 1 1 1 14 ( 1) 4 1 4 4 1 4 4 4 1

ln | | ln | 1| ln xu u u u u u x

du du du du u u C C

50. 4

2 2 26 5 10 5

5 41 1 1 15 4( 4) ( 4)4 4

; Let 5 ;x CA Bu uu u u u ux x x x

dx dx u x du x dx du

2 2 141 ( 4) ( 4) ( ) (4 ) 4 0, 4 0, 4 1Au u B u Cu A C u A B u B A C A B B B

1 116 16 ;A C 2 2 2

1 16 1 4 1 161 1 1 1 1 1 1 1 15 5 4 80 20 80 4( 4) u u u uu u u u

du du du du du

5

5 5 55 5 41 1 1 1 1 1 1 1

80 20 80 80 80 8020 20ln | | ln | 4| ln | | ln | 4| ln x

u x x xu u C x x C C

51. 22 2 2

2 1 1 13 23 2 1; ln ; ; 3xdx dt dt dt t t

dt t t t tt tt t x C Ce t

and 0x 1

2 C

2 211 2 1ln 2 ln | 2| ln | 1| ln 2xt t

t te x t t

52. 4 2 2 213

4 2 1 13 4 1 1

3 4 1 2 3; 2 3 3 3 3tan 3 3 tan ;dx dt dt dtdt t t t t

t t x t t C

1t and

1 13 3 34 4 4 3tan 3 3 tanx C C x t tπ π π π π π

53. 22 1 1 1 1

2 1 2 2 2 2 222 2 2; ln | 1| ln | 1| ln ;dx dx dt dt dt t

dt x t t tt tt t x x x C 1t and

6 613 2 2 21 ln 2 ln ln 2 ln 3 ln 6 ln | 1| ln 6 1 1, 0t t t

t t tx C C x x x t

54. 22 1

11( 1) 1 tan ln | 1| ; 0dx dx dt

dt txt x x t C t

and 10 tan 0 ln |1|x C

1 1tan 0 0 tan ln | 1| tan ln( 1) , 1C x t x t t

55. 2

2.5 2.5 2.5 2.52 9 1 13 33 0.50.5 0.5 0.5

3 3 ln 3 ln 25xx x xx x

V y dx dx dxπ π π π π

56. 1 1 1 12 4 41 1 2 1

( 1)(2 ) 3 1 3 2 3 300 0 02 2 4 ln | 1| 2 ln |2 | (ln 2)x

x x x xV xy dx dx dx x xπ ππ π π



57. 2

3 331 1

100 0tan tan x

xA x dx x x dx

323 31

3 2 30ln 1 ln 2;xπ π

2

2

3 331 2 11 1 1 12 2 100 0

tan tan xA A x

x x x dx x x dx

31 3 321 1 1 1

2 2 2 2 6 3 20tan 1.10A A Ax xπ π π π

58. 2

3 2

5 5 5 5 54 13 9 1253 1 932 33 3 3 3

3 2 3ln ln 3 2ln 1 ln ;x x dx dx dxx x xx x x

A dx x x x

2

3 2

5 5 54 13 9 51 1 13 132 33 3 3

4 3 2 (8 11ln 2 3ln 6) 3.90x x x dx dx

A A x x Ax x xx dx x

59. (a) 1 1 1( )( ) ln ;dx dx dx dx x

dt x N x N x N N x N N xkx N x k dt k dt kt C

1250 , 1000, 0k N t and 1 2 1 1 1

1000 998 1000 1000 250 1000 4992 ln ln lnx txx C

4

44 4 4 4499 499 1000

1000 1000 499ln 4 499 (1000 ) 499 1000

t

tt t t tx x e

x x et e x e x e x e x

(b) 4

44 4 410001 1

2 4499500 500 500 499 500 1000 499 ln 499 1.55days

t

tt t te

ex N e e e t

60. ( )( )( )( )dx dxdt a x b xk a x b x k dt

(a) 21

( ): ; 0dx

a xa xa b k dt kt C t and 1 1 10 a a x ax C kt

2111 1 1

akt a a a kta x a akt akt akta x x a

(b) 1 1 1( )( ): ln ;dx dx dx b xa x b x b a a x b a b x b a a xa b k dt k dt kt C 0t and

( )

( )

1( )10 ln ln ( ) lnb a kt

b a kt

ab eb a ktb b x b b x bb a a a x a a x a a be

x C b a kt e x

8.6 INTEGRAL TABLES AND COMPUTER ALGEBRA SYSTEMS

1. 1 3233 3

tandx xx x

C

(We used FORMULA 13(a) with 1,a 3b )

2. 4 4 4 21 124 4 4 4 4 2

ln lnx xdxx x x x

C C

(We used FORMULA 13(b) with 1,a 4b )

Section 8.6 Integral Tables and Computer Algebra Systems 595


3. 1 1( 2)

2 2 22 2 2 2x dx x dx dx

x x xx dx x dx

3 12 2 2( 2)2 2

1 3 1 1 32 2 4x x xx C

(We used FORMULA 11 with 1, 2, 1a b n and 1, 2, 1a b n )

4. 3 2 3 2 3 2 3

(2 3) 3 31 12 2 2 22 3(2 3) (2 3) (2 3) 2 3

x dx x dx dx dx dxxx x x x

1 131 2 3 2 33 31 1 2 22 2 2 2 1 2 2 ( 1)2 3 2 3

x xx dx x dx C

( 3)12 2 3 2 3

(2 3 3) x

x xx C C

(We used FORMULA 11 with 2,a 3,b 1n and 2, 3, 3a b n )

5. 3 13 31 1

2 2 2 22 3 (2 3) 2 3 2 3 2 3 2 3x x dx x x dx x dx x dx x dx

5 33 2 3 22 3 2 3 (2 3) (2 3) ( 1)3 2 31 2 2

2 2 5 2 2 3 2 5 51x x x x xxC C C


6. 5 33 2 3 2 3 25 51 17 7 7 7(7 5) (7 5)(7 5) (7 5) 7 5 7 5x x dx x x dx x dx x dx x dx

7 5

5 2 5 27 5 7 5 (7 5) 2(7 5) (7 5)5 14 41 2 27 7 7 7 7 5 49 7 49 72

x x x x x xC C C


7. 2

( 4)9 4 9 42 9 4

x x dxx x xx

dx C

(We used FORMULA 14 with 4, 9a b )

9 4 9 4 919 9 4 9

2 lnx xx x

C

(We used FORMULA 13(b) with 4, 9a b )

9 4 9 4 323 9 4 3

lnx xx x

C

8. 24 9 4

( 9) 18 4 94 9xdx dx

x x xx xC


14 9 4 92 29 9 99

tanx xx C

(We used FORMULA 13(a) with 4, 9a b )

14 9 4 949 27 9tanx x

x C



9. 2 3( 2)(2 3 2) 2 22 2 1 226 2 24 2 2 sinx x x x xx x x dx x x x dx C

2 2( 2)(2 6) 4 ( 2)( 3) 41 12 26 2 3 24sin 4sinx x x x x x x xx xC C

(We used FORMULA 51 with 2a )

10. 21 12

2 212

2 2 1 2 11 1 12 2 22 sin sin (2 1)

x x xx xx xdx dx x x C x x x C


11.

2 22

2 2 2

7 7 7 71 17 77 7

ln lnx xdx dx

x xx x x xC C


12.

2 22

2 2 2

7 7 7 71 17 77 7

ln lnx xdx dx

x xx x x xC C


13. 2 2 2 2 2 22 2 24 2 2 2 2 42 2 ln 4 2lnx x x x

x x x xdx dx x C x C


14. 2 2 2 2 2 1 2 14 2

2 22 2sec 4 2secx x x xx xdx dx x C x C


15. 2 2

2 22

132 3cos3 (2 cos3 3sin 3 ) (2cos3 3sin 3 )

t tt e ee t dt t t C t t C


16. 3 3

2 23

25( 3) 4sin 4 ( 3sin 4 4 cos4 ) ( 3sin 4 4cos 4 )

t tt e ee t dt t t C t t C


17. 1 1 1 1 2 2

2 2

1 1 1 1 11 11 1 1 1 2 21 1

cos cos cos cosx x dx x x dx

x xx x dx x x dx x x

(We used FORMULA 100 with 1, 1a n )

2 21 1 2 1 1 21 1 1 1 1 12 2 2 2 2 2 4 4cos sin 1 cos sin 1x xx x x x C x x x x C


18. 1 1 1 1 2 2

2 2 21 1 1 1 11 1

1 1 1 1 2 21 (1) 1tan tan (1 ) tan (1 ) tanx x dx x x dx

x xx x dx x x dx x x


2

21 1 1

2 2 1tan 1x

xx dx

(after long division)

2 2

21 1 1 2 11 1 1 1 1 1

2 2 2 2 2 2 21tan tan tan 1 tanx x

xx dx dx x x x C x x x C



19. 2 1 2 1 3 3

2 22 1 1 11 1

2 1 2 1 3 31 1tan tan tanx x x x

x xx x dx x dx x dx


3 2 3 2

2 22 2 1 1 21 1

2 2 3 6 61 1ln 1 tan tan ln 1x dxx x x x

x xdx x dx x C x x dx x x C

20. 1 ( 2 1) ( 2 1) 1 1

2 2 22 1 1 1tan 1

( 2 1) ( 2 1) ( 1)1 1tan tan tanx x x x x

x x xdx x x dx x dx x dx


1 1

2 2 22

2 1 2tan1 1 12 21 11

ln | | ln 1 tan ln | | ln 1x dxx dx dx dx xx xx x xx x

x x C dx x x x C

21. cos5 cos10 2sin 3 cos2 x xx x dx C


22. cos5 cos10 2sin 2 cos3 x xx x dx C


23. 7 92 2

sin sin8 7 8 92 7 2 9 2 7 98sin 4 sin sin sin 8

t tt t tt dx C C

(We used FORMULA 62(b) with 124,a b )

24. 3 6 6 2sin sin 3sin sint t t tdt C

(We used FORMULA 62(b) with 1 13 6,a b )

25. 6 73 4 12 7 12cos cos 6sin sind Cθ θ θ θθ

(We used FORMULA 62(c) with 1 13 4,a b )

26. 13 152 2

sin sin13 151 12 13 2 15 2 13 15cos cos7 sin sind C C

θ θθ θ θθ θ

(We used FORMULA 62(c) with 12 , 7a b )

27.

3

2 2 2 2 2 22 2 2

2 2 11 1 1 12 2 21 1 2 11 1 1

ln 1 tanx dx x dxx x dx dx xx x xx x x

dx x x C

(For the second integral we used FORMULA 17 with 1a )



28.

2

2 2 2 2 2 2 222 2 2 2 22

2 32 2 2

6 26 33 33 3 3 3 3

1 131 13 3 33 2 32 3 3

3 3

tan 3 tan

x dx x dxx x dx dx dx dxx xx x x x x

x x xx x

dx

C

For the first integral we used FORMULA 16 with 3a = for the third integral we used FORMULA 17

with 3a = )

2 2

1 312 3 3 3 2 3

tan x xx x

C

29. 1 1 1 1 2

2 2

1 2 1 1 1 2 111 1 1 1 1 1

sin ; 2 sin 2 sin sin

2

u u u du

u u

u x

x dx x u u u du u du u u

dx u du


2 1 1 2 2 1 21 1 1 12 2 2 2sin sin 1 sin 1u u u u u C u u u u C


1 21 12 2sinx x x x C

30. 1 12 1 1 2cos cos 11; 2 2 cos 2 cos 1

2

x uux

u x

dx x u u du u du u u u C

dx u du


12 cos 1x x x C

31. 2

2 2

2 1 2 1 22 1 12 21 1 1

; 2 2 sin 1 sin 1

2

x u u ux u u

u x

dx x u du du u u u C u u u C

dx u du

(We used FORMULA 33 with 1a ) 1 1 2sin 1 sinx x x C x x x C

32. 2

2 2 2 22 2 2 12 22 2 2

; 2 2 2 2 2 sin

2

x u u uux

u x

dx x u u du u du u C

dx u du


2 1 2 122

2 2sin 2 2sinu xu u C x x C



33. 2 2 21 sin (cos )2 21 1 1sin

sin(cot ) 1 sin ; 1 ln

cost t dt u du u

t u u

u tt t dt u C

du t dt

(We used FORMULA 31 with 1a ) 22 1 1 sin

sin1 sin ln ttt C

34. 2

2 2 2

cos 2 412(tan ) 4 sin (sin ) 4 sin 4

; [ sin , cos ] lnt dt udt duut t t t u u

u t du t dt C

(We used FORMULA 34 with 2)a 22 4 sin1

2 sinln tt C

35. 2 2 2

2 2

3 (ln ) 3 3

ln

; ln 3 ln ln 3 (ln )u

u

dy u e du du

y y e u uu

u y

y e u u C y y C

dy e du


36. 2 2 2

2 21 2 1 1 2 11

2 2 1 1tan ; 2 tan 2 tan tan

2

t t tt t

t y

y dy y t t t dt t dt t t dt

dy t dt

(We used FORMULA 101 with 1, 1n a ) 2

2 22 1 2 1 1 1 11

1 1tan tan tan tan tant dt

t tt t dt t t t t C y y y y C

37. 2 2 2

21 1 1

2 5 ( 1) 4 4; [ 1, ] ln 4

x x x tdx dx t x dt dx dt t t C

(We used FORMULA 20 with 2a ) 2 2ln ( 1) ( 1) 4 ln ( 1) 2 5x x C x x x C

38. 22 2 2 2

2 2 2 2 2 2 2

( 2) 4 2 4 4

4 5 ( 2) 1 1 1 1 1 1

2; tx x t t t t

x x x t t t t t

t xdx dx dt dt dt dt dt

dt dx

(We used FORMULA 25 with 1a ) (We used FORMULA 20 with 1a ) 22 2 211

2 2ln 1 4 1 4ln 1t tt t t t t C

2( 2) ( 2) 12 2 212 2ln ( 2) ( 2) 1 4 ( 2) 1 4ln ( 2) ( 2) 1

x xx x x x x C

2( 6) 4 5272 2ln ( 2) 4 5 x x xx x x C

39. 22 2 2 2 132 2 35 4 9 ( 2) ; [ 2, ] 9 9 sint tx x dx x dx t x dt dx t dt t C


2 1 2 12 9 2 2 9 22 2 3 2 2 39 ( 2) sin 5 4 sinx x x xx C x x C



40. 2 2 2 2 2 2 2 22 1 ( 1) ; [ 1, ] ( 1) 1 2 1 1x x x dx x x dx t x dt dx t t dt t t t dt

2 2 2 21 2 1 1t t dt t t dt t dt

(We used FORMULA 30 with 1a ) (We used FORMULA 29 with 1a )

4 23 21 2 2 2 2 2 11 1 2 18 1 8 3 2 2 1sin 1 1 2 1 1 sint t tt t t t t C

3 21 2 2 2 2 2 111 1 2 18 8 3 2 2sin ( 1) ( 1) 1 ( 1) 1 2( 1) 1 ( 1) 1 ( 1) sin ( 1)xx x x x x x x C

3 21 2 2 25 128 3 8sin ( 1) 2 2 2 4 5xx x x x x x x C

41. 4 4 25 3sin 2 cos2 5 1 sin 2 cos2 sin 2 cos 2 3 14

5 2 5 10 5 3 2 3sin 2 sin 2 sin 2x x x x x xx dx x dx x dx

(We used FORMULA 60 with 2, 5a n and 2, 3a n )

4 4 22sin 2 cos2 8 sin 2 cos2 2sin 2 cos2 4cos22 110 15 15 2 10 15 15sin 2 cos 2 cos 2x x x x x x xx x x C C

42. 3cos 2 sin 24 24 14 2 48cos 2 8 cos 2t tt dt t dtπ π

ππ π

(We used FORMULA 61 with 2 , 4a nπ ) 3cos 2 sin 2 sin(2 2 )

2 4 26t t tt Cπ π ππ π

(We used FORMULA 59 with 2a π ) 3 3cos 2 sin 2 cos 2 sin 2 3cos2 sin 23sin 4

4 23 3t t t t t ttt C t Cπ π π π π πππ π π π

43. 3 22 3 2sin 2 cos 2 3 12(2 3) 3 2sin 2 cos 2 sin 2 cos2d dθ θθ θ θ θ θ θ

(We used FORMULA 69 with 2, 3, 2a m n ) 3 2 3 2 3 2 32 2sin 2 cos 2 sin 2 cos 2 sin 2 cos 2 sin 22 2 1

10 5 10 5 2 10 15sin 2 cos 2 sin 2 (cos 2 )2d d Cθ θ θ θ θ θ θθ θ θ θ θ θ

44. 32 4 2 4 4sin cos 2 12 4 2 42sin sec 2sin cos 2 cost tt t dt t t dt t dt

(We used FORMULA 68 with 1, 2, 4a n m )

23 4 3 4 3 2sec tan 4 24 1 4 1sin cos cos sin cos sec sin cos sect tt t t dt t t t dt t t t dt


23 2 2 3sec tan 2 2 2 2 23 3 3 3 3 3sin cos tan sec tan tan tan sec 1 tant tt t t C t t t C t t C t C

An easy way to find the integral using substitution:

22 4 2 2 2 3232sin cos 2 tan sec 2 tan sec tant t dt t t dt t t dt t C

45. 23 2tan 22 24 tan 2 4 tan 2 tan 2 4 tan 2xx dx x dx x x dx

(We used FORMULA 86 with 3, 2n a ) 2 24

2tan 2 ln sec 2 tan 2 2ln sec 2x x C x x C



46. 34 2cot38cot 8 cottt dt t dt


3138 cot cott t t C


47. 3 sec tan 3 2(3 1) 3 12sec 2 secx xx dx x dxπ π

ππ π

(We used FORMULA 92 with 3,n a π ) 1 1sec tan ln |sec tan |x x x x Cπ ππ π π π

(We used FORMULA 88 with a π )

48. 24 2sec 3 tan3 4 23(4 1) 4 13sec 3 3 sec 3x xx dx x dx

(We used FORMULA 92 with 4, 3n a ) 2sec 3 tan 3 2

3 3 tan 3x x x C


49. 3 35 3csc cot 5 2 csc cot 3 csc cot 3 25 1 5 1 4 4 3 1 3 1csc csc cscx x x x x xx dx x dx x dx

(We used FORMULA 93 with 5, 1n a and 3, 1n a ) 3 3 31

4 8 8csc cot csc cot ln |csc cot |x x x x x x C


50. 4 2 4 2 4(ln ) (ln ) (ln )3 2 3 32 1 1

4 4 4 2 4 416 (ln ) 16 ln 16x x x x x xx x dx x x dx x dx

(We used FORMULA 110 with 1, 3, 2a n m and 1, 3, 1a n m )

4 2 4 4 4(ln ) (ln ) 4 2 44 8 32 216 4 (ln ) 2 lnx x x x x xC x x x x C

51. 3 3 sec tan 3 23 1 3 1sec 1 ; 1, sec sect t t t x xe e dt x e dx e dt x dx x dx


sec tan 1 12 2 2ln |sec tan | sec 1 tan 1 ln sec 1 tan 1t t t tx x x x C e e e e C

52. 3 2 3csc csc cot 3 2 csc cot 1

3 1 3 1 2 2; 2 csc 2 csc 2 ln |csc cot |

2

t t t t

t

d t t dt t dt t t C

d t dt

θθ

θθ θ

θ


csc cot ln csc cot Cθ θ θ θ



53. 1 4 4

2 2 2 3

0 0 02 1 ; tan , sec 2 sec sec 2 secx dx x t dx t dt t t dt t dt

π π

44sec tan 3 23 1 3 10 0

2 sect t t dtππ

(We used FORMULA 92 with, 3, 1n a )

4

0sec tan ln sec tan 2 ln 2 1t t t t

π

54.

2

5 2 52

3 2 3 3 33cos 4 2sec tan 4 24 1 4 1cos 00 0 0 01

; [ sin , cos ] sec secx dxdy x xxy

y x dy x dx x dx x dxπ π ππ


2 3sec tan 2 4 2

3 3 3 30tan 3 3 2 3x x x

π

55. 3 22

32 3 31 4tansec1 0 0

; [ sec , sec tan ] (sec tan ) tanr

r dr r dr d d dπ π

θθθ θ θ θ θ θ θ θ θ

3 333 32 3 3tan tan

4 1 3 3 3 30 00tan tan 3d

ππ πθ θ π πθ θ θ θ

(We used FORMULA 86 with 1, 4a n and FORMULA 84 with 1a )

56.

2

7 2 72

1 3 6 6sec2 5sec0 0 01

; [ tan , sec ] cosddt

tt dt d d

π πθ θθ

θ θ θ θ θ

4 4 26 66 6 63cos sin 5 1 cos sin cos sin 3 14

5 5 5 5 3 30 0 00 0cos cosd d

π ππ π πθ θ θ θ θ θθ θ θ θ

4 62cos sin 84

5 15 15 0cos sin sin

πθ θ θ θ θ

(We used FORMULA 61 with 1, 5a n and 1, 3a n )

4

3 12 2

23 8 9 3 9 48 32 4 2034 1 1 1 4

5 15 2 2 15 2 160 10 15 480 480

57. 2 2

02 1S y y dxπ

2

2

22

202 2 1 x

xx dxπ

22

02 2 1x dxπ

22

21 12 2

02 2 ln 1x x x xπ


2 6 ln 2 3 2 3 2 ln 2 3π π π



58. 3 23 2 3 22 2 2 21 1 1 1 1

4 2 4 4 2 40 0 01 (2 ) 2 2 lnxL x dx x dx x x x


3 22 2 3 33 31 1 1 1 1 1

2 4 2 4 4 4 2 2 4 4 20

1 4 ln 1 4 1 4 ln 1 4 lnx x x x

3 3 31 1 14 4 2 4 2 4(2) ln 1 ln 2 ln 3 2

59. 33

10 02 1 2;dx

xA x

3 3 31 1 1

1 10 0 01x dx dx

A A Ax xx x dx

33 21 2 42 3 30

( 1) 1 ;x


3 31 1 1 1

2 1 4 4 200ln( 1) ln 4 ln 2 ln 2dx

A xy x

60. 3 3 3 336 2 3

2 3 2 3 2 3 00 0 018 54 18 27ln |2 3| 18 3 27ln 9 ( 27ln 3)x dx

y x x xM x dx dx x x

54 27 2ln 3 27 ln 3 54 27ln 3

61. 1

2 2

12 1 4 ;S x x dxπ

[ 2 , 2 ]u x du dx

22 2

4 21u u duπ

22 2 21

4 8 82

1 2 1 ln 1u u u u uπ


2 1 2 14 8 8 8 8(1 2 4) 1 4 ln 2 1 4 (1 2 4) 1 4 ln 2 1 4π

2 59 14 2 8 2 5

5 ln 7.62π



62. (a) The volume of the filled part equals the length of the tank times the area of the shaded region shown in the accompanying figure. Consider a layer of gasoline of thickness dy located at height y where .r y r d The width of

this layer is 2 22 .r y Therefore,

2 22r d

rA r y dy

and

2 22r d

rV L A L r y dy

(b) 2 2 22 2 12 22 2 sin

r dr d y r y yrrr r

L r y dy L

(We used FORMULA 29 with )a r

2 2 2( ) 2 1 2 12 2 2 2 2 2 22 2 sin 2 2 sind r d r d r d rr r r

r rL rd d L rd dπ π

63. The integrand 2( )f x x x is nonnegative, so the integral is maximized by integrating over the function’s

entire domain, which runs from 0x to 1x

21 1 12 2 2

12

11 1

2 2 2 11 12 2 2 20 0

0

2 2 sinx x

x x dx x x dx x x

(We used FORMULA 48 with 12 )a

12

12 11 1 1

2 8 8 2 8 2 80

sin (2 1)x

x x x π π π

64. The integrand is maximized by integrating 2( ) 2g x x x x over the largest domain on which g is

nonnegative, namely [0, 2]

222

( 1)(2 3) 22 116 20 0

2 sin ( 1)x x x xx x x dx x


1 12 2 2 2 2

π π π

CAS EXPLORATIONS

65. Example CAS commands:

Maple:

q1 : Int( x*ln(x), x ); # (a)

q1 value( q1 );

q2 : Int( x^2*ln(x), x ); # (b)

q2 value( q2 );

q3 : Int( x^3*ln(x), x ); # (c)

q3 value( q3 );

q4 : Int( x^4*ln(x), x ); # (d)



q4 value( q4 );

q5 : Int( x^n*ln(x), x ); # (e)

q6 value( q5 );

q7 : simplify(q6) assuming n::integer;

q5 collect( factor(q7), ln(x) );


Maple:

q1 : Int( ln(x)/x, x ); # (a)

q1 value( q1 );

q2 : Int( ln(x)/x^2, x ); # (b)

q2 value( q2 );

q3 : Int( ln(x)/x^3, x ); # (c)

q3 value( q3 );

q4 : Int( ln(x)/x^4, x ); # (d)

q4 value( q4 );

q5 : Int( ln(x)/x^n, x ); # (e)

q6 : value( q5 );

q7 : simplify(q6) assuming n::integer;

q5 collect( factor(q7), ln(x) );


Maple:

q : Int( sin(x)^n/sin(x)^n cos(x)^n), x 0..Pi/2 ); # (a)

q value( q );

q1 : eval( q, n 1 ): # (b)

q1 value( q1 );

for N in [1,2,3,5,7] do

q1 : eval( q, n N );

print( q1 evalf(q1) );

end do:

qq1 : PDEtools[dchange]( x Pi/2-u, q, [u] ); # (c)

qq2 : subs( u x, qq1 );

qq3 : q q q qq2;

qq4 : combine( qq3 );

qq5 : value( qq4 );

simplify( qq5/2 );



65-67. Example CAS commands:

Mathematica: (functions may vary)

In Mathematica, the natural log is denoted by Log rather than Ln, Log base 10 is Log[x, 10]

Mathematica does not include an arbitrary constant when computing an indefinite integral,

Clear[x, f, n]

n_f[x ]: Log[x] / x

Integrate[f [x], x]

For exercise 67, Mathematica cannot evaluate the integral with arbitrary n. It does evaluate the integral (value is /4π in each case) for small values of n, but for large values of n, it identifies this integral as Indeterminate

65. (e) 1 ln 1

1 1ln , 1nx xn nn nx x dx x dx n

(We used FORMULA 110 with 1, 1a m )

1 1 1

2ln 11 1 1( 1)

lnn n nx x x xn n nn

C x C

66. (e) 1 ln 1

1 ( ) 1ln , 1nn nx xn nx x dx x dx n

(We used FORMULA 110 with 1, 1,a m n n )

1 1 1ln 1 11 1 1 1 1ln

n n nx x x xn n n n nC x C

67. (a) Neither MAPLE nor MATHEMATICA can find this integral for arbitrary n. (b) MAPLE and MATHEMATICA get stuck at about 5.n

(c) Let 2 ;x u dx duπ 20 ,x u π 2 0;x uπ

2

2 2

/2 0 /2 /2sinsin cos cos

sin cos cos sin cos sinsin cos0 /2 0 0

nn n n

n n n n n nn n

u dux dx u du x dx

x x u u x xu uI

π

π π

π π π

π

/2 /2sin cos

2 4sin cos0 0

n n

n nx x

xI I dx dx I

π ππ π

Section 8.7 Numerical Integration 607


8.7 NUMERICAL INTEGRATION

1. 2

1x dx

I. (a) For 2 1 1 14 4 2 84, ;b a x

nn x ΔΔ

318 2( ) 12 (12) ;imf x T

( ) ( ) 1 0f x x f x f

0 0TM E

(b) 22 2

312 2 211

2xx dx

2

10TE x dx T

(c) True Value 100 0%TE

II. (a) For 2 1 1 14 4 3 124, ;b a x

nn x ΔΔ

3112 2( ) 18 (18) ;imf x S

(4) ( ) 0 0 0sf x M E

(b)

2 2

3 3 32 2 21 1

sx dx E x dx S (c) True Value 100 0%sE

ix ( )if x m ( )imf x

0x 1 1 1 1

1x 5/4 5/4 4 5

2x 3/2 3/2 2 3

3x 7/4 7/4 4 7

4x 2 2 1 2

2. 3

1(2 1)x dx

I. (a) For 3 1 2 14 4 24, b a

nn xΔ 1

2 4 ;xΔ

14( ) 24 (24) 6;imf x T

( ) 2 1 ( ) 2 0f x x f x f

0 0TM E

(b) 3 332

11 1(2 1) (9 3) (1 1) 6 (2 1) 6 6 0Tx dx x x E x dx T

(c) True Value 100 0%sE

II. (a) For 3 1 2 1 14 4 2 3 64, ;b a x

nn x ΔΔ

16( ) 36 (36) 6;imf x S

(4) ( ) 0 0 0sf x M E

(b) 3 3

1 1(2 1) 6 (2 1)sx dx E x dx S

6 6 0


ix ( )if x

m

( )imf x

0x 1 1 1 1

1x 3/2 2 4 8

2x 2 3 2 6

3x 5/2 4 4 16

4x 3 5 1 5


0x 1 1 1 1

1x 5/4 5/4 2 5/2

2x 3/2 3/2 2 3

3x 7/4 7/4 2 7/2

4x 2 2 1 2


0x 1 1 1 1

1x 3/2 2 2 4

2x 2 3 2 6

3x 5/2 4 2 8

4x 3 5 1 5



3. 12

11x dx

I. (a) For 1 ( 1) 2 1 14 4 2 2 44, ;b a x

nn x ΔΔ 14( ) 11 (11) 2.75;imf x T

2( ) 1 ( ) 2 ( ) 2f x x f x x f x

21 ( 1) 1 112 2 122 (2)TM E or

0.08333


0x 1 2 1 2

1x 1/2 5/4 2 5/2

2x 0 1 2 2

3x 1/2 5/4 2 5/2

4x 1 2 1 2

(b) 31 112 28 81 1 11 1

3 3 3 3 3 4 1211 11 1 1 1x

Tx dx x E x dx T

112 0.08333TE

(c) 1

1283

True Value 100 100 3%TE

II. (a) For 1 ( 1) 2 1 14 4 2 3 64, ;b a x

nn x ΔΔ

16( ) 16 (16)imf x S 8

3 2.66667;

3 ( ) 0 (4)( ) 0 0f x f x M

0sE

(b) 31 12 8

3 3111 xx dx x

12 8 8

3 311 0sE x dx S



0x 1 2 1 2

1x 1/2 5/4 4 5

2x 0 1 2 2

3x 1/2 5/4 4 5

4x 1 2 1 2

4. 02

21x dx

I. (a) For 0 ( 2) 2 14 4 24, b a

nn xΔ

12 4 ;xΔ 31

4 4( ) 3 (3) ;imf x T

2( ) 1 ( ) 2 ( ) 2f x x f x x f x

20 ( 2) 1 112 2 122 (2)TM E

0.08333

(b) 30 002 28 32 2 1 1

3 3 3 3 4 12 1222 21 0 2 1x

T Tx dx x E x dx T E

(c) 1

1223



0x 2 3 1 3

1x 3/2 5/4 2 5/2

2x 1 0 2 0

3x 1/2 3/4 2 3/2

4x 0 1 1 1



II. (a) For 0 ( 2) 2 1 14 4 2 3 64, ;b a x

nn x ΔΔ

1 26 3( ) 4 (4) ;imf x S (3) ( ) 0f x

(4) ( ) 0 0 0sf x M E

(b) 0 02 22

32 21 1sx dx E x dx S

2 23 3 0



0x 2 3 1 3

1x 3 / 2 5/4 4 5

2x 1 0 2 0

3x 1 / 2 3 / 4 4 3

4x 0 1 1 1

5. 23

0t t dt

I. (a) For 4,n 2 0 2 1 14 4 2 4 ;b a x

nx ΔΔ

2514 4( ) 25 (25) ;imf t T

3 2( ) ( ) 3 1 ( ) 6f t t t f t t f t t

22 0 1 112 2 212 (2) (12)TM f E

it ( )if t m ( )imf t

0t 0 0 1 0

1t 1/2 5/8 2 5/4

2t 1 2 2 4

3t 3/2 39/8 2 39/4

4t 2 10 1 10

(b) 4 2 4 22 223 3 252 2 1 1

4 2 4 2 4 4 400 00 6 6t t

T Tt t dt E t t dt T E

(c) 14


II. (a) For 2 0 2 1 14 4 2 3 64, ;b a x

nn x ΔΔ

16( ) 36 (36) 6;imf t S

(3) (4)( ) 6 ( ) 0 0 0sf t f t M E

(b) 2 23 3

0 06 st t dt E t t dt S

6 6 0



0t 0 0 1 0

1t 1/2 5/8 4 5/2

2t 1 2 2 4

3t 3/2 39/8 4 39/2

4t 2 10 1 10

6. 13

11t dt

I. (a) For 1 ( 1) 2 1 14 4 2 2 44, ;b a x

nn x ΔΔ

14( ) 8 (8) 2;imf t T

3 2( ) 1 ( ) 3 ( ) 6f t t f t t f t t

21 ( 1) 1 112 2 46 (1) (6)TM f E


0t 1 0 1 0

1t 1/2 7/8 2 7/4

2t 0 1 2 2

3t 1/2 9/8 2 9/4

4t 1 2 1 2

(b) 44 4

1

1 11 ( 1)3 314 4 41 1

1 1 ( 1) 2 1 2 2 0tTt dt t E t dt T

(c) True Value 100 0%TE



II. (a) For 1 ( 1) 2 1 14 4 2 3 64, ;b a x

nn x ΔΔ

16( ) 12 (12) 2;imf t S

(3) (4)( ) 6 ( ) 0 0 0sf t f t M E

(b) 1 13 3

1 11 2 1st dt E t dt S

2 2 0

(c) True Valule 100 0%sE


0t 1 0 1 0

1t 1/2 7/8 4 7/2

2t 0 1 2 2

3t 1/2 9/8 4 9/2

4t 1 2 1 2

7. 2

21

1 sds

I. (a) For 2 1 1 14 4 2 84, ;b a x

nn x ΔΔ

179,573 179,573 179,573144,100 8 44,100 352,800( )imf s T

2 31 20.50899; ( ) ( )s s

f s f s

46( ) 6 (1)s

f s M f

22 1 1 112 4 32(6) 0.03125TE

(b) 2 2

2 2 2221 1 1 1 1 1 12 1 2 211 1 1

0.50899 0.00899Tss sds s ds E ds T

0.00899TE

(c) 0.00899True Value 0.5100 100 2%TE

II. (a) For 2 1 1 14 4 3 124, ;b a x

nn x ΔΔ

264,821 264,821144,100 12 44,100( )imf s S

264,821529,200 0.50042;

5 6(3) (4) 12024( ) ( ) 120

s sf s f s M

42 1 1 1180 4 384(120) 0.00260sE

(b) 2 2

2 21 1 1 1

2 21 10.50042 0.00042 0.00042s ss s

ds E ds S E

(c) 0.0004True Value 0.5100 100 0.08%sE

8. 2

41

( 1)2 sds

I. (a) For 4 2 1 14 2 2 44, ;b a x

nn x ΔΔ

1269 1269 12691450 4 450 1800( ) 0.70500;imf s T

3 42 62

( 1) ( 1)( ) ( 1) ( ) ( )

s sf s s f s f s

24 2 1 112 2 4(6) 0.25 6TE M

is ( )if s m ( )imf s

0s 2 1 1 1

1s 5/2 4/9 2 8/9

2s 3 1/4 2 1/2

3s 7/2 4/25 2 8/25

4s 4 1/9 1 1/9

is ( )if s m ( )imf s

0s 1 1 1 1

1s 5/4 16/25 2 32/25

2s 3/2 4/9 2 8/9

3s 7/4 16/49 2 32/49

4s 2 1/4 1 1/4

is ( )if s

m ( )imf s

0s 1 1 1 1

1s 5/4 16/25 4 64/25

2s 3/2 4/9 2 8/9

3s 7/4 16/49 4 64/49

4s 2 1/4 1 1/4



(b) 2 2

4 441 1 1 1 2 1 2

( 1) 4 1 2 1 3 3( 1) ( 1)22 20.705 0.03833Tss s

ds E ds T

0.03833TE

(c) 23

0.03833True Value 100 100 6%TE

II. (a) For 4 2 1 14 2 3 64, ;b a x

nn x ΔΔ

1813 18131450 6 450( )imf s S

5(3)1813 24

2700 ( 1)0.67148; ( )

sf s

6(4) 120

( 1)( ) 120

sf s M

44 2 1 1180 2 12(120) 0.08333sE

is ( )if s

m

( )imf s

0s 2 1 1 1

1s 5/2 4/9 4 16/9

2s 3 1/4 2 1/2

3s 7/2 4/25 4 16/25

4s 4 1/9 1 1/9

(b) 2 2

4 41 2 1 2

3 3( 1) ( 1)2 20.67148 0.00481 0.00481s ss s

ds E ds S E

(c) 23

0.00481True Value 100 100 1%sE

9. 0

sin t dtπ

I. (a) For 04 4 2 84, ;b a x

nn x π π πΔΔ

( ) 2 2 2 4.8284;imf t

8 2 2 2 1.89612;T π ( ) sinf t t

( ) cos ( ) sin 1f t t f t t M

32012 4 192(1) 0.16149TE π π π

(b) 00 0sin cos ( cos ) ( cos 0) 2 sin 2 1.89612 0.10388Tt dt t E t dt T

π ππ π

(c) 0.10388True Value 2100 100 5%TE

II. (a) For 04 4 3 124, ;b a x

nn x π π πΔΔ

12( ) 2 4 2 7.6569 2 4 2imf t S π

2.00456; (3) (4)( ) cos ( ) sinf t t f t t

40180 41 (1) 0.00664sM E π π

(b)

0 0sin 2 sin 2 2.00456st dt E t dt S

π π

0.00456 0.00456sE

(c) 0.00456True Value 2100 100 0%sE


0t 0 0 1 0

1t /4π 2 /2 4 2 2

2t /2π 1 2 2

3t 3 /4π 2 /2 4 2 2

4t π 0 1 0


0t 0 0 1 0

1t /4π 2 /2 2 2

2t /2π 1 2 2

3t 3 /4π 2 /2 2 2

4t π 0 1 0



10. 1

0sin tdtπ

I. (a) For 1 0 1 14 4 2 84, ;b a x

nn x ΔΔ

( ) 2 2 2 4.828imf t

18 2 2 2 0.60355; ( ) sinT f t tπ

2( ) cos ( ) sinf t t f t tπ π π π

22 21 0 112 4 0.05140TM Eπ π


0t 0 0 1 0

1t 1/4 2 /2 2 2

2t 1/2 1 2 2

3t 3/4 2 /2 2 2

4t 1 0 1 0

(b) 1 111 1 1 200 0

sin cos cos cos0 0.63662 sinTt dt t E t dt Tπ π π ππ π π π

2 0.60355 0.03307π

(c) 20.03307


π

II. (a) For 1 0 1 14 4 3 124, ;b a x

nn x ΔΔ

( ) 2 4 2 7.65685imf t

112 2 4 2 0.63807;S

(3) 3 (4) 4( ) cos ( ) sinf t t f t tπ π π π

44 41 0 1180 4 0.00211sM Eπ π


0t 0 0 1 0

1t 1/4 2 /2 4 2 2

2t 1/2 1 2 2

3t 3/4 2 /2 4 2 2

4t 1 0 1 0

(b) 1 1

2 2

0 0sin 0.63662 sin 0.63807 0.00145 0.00145s st dt E t dt S Eπ ππ π

(c) 20.00145

True Value 100 100 0%sE

π

11. (a) 0M (see Exercise 1): Then 2 41121 1 (1) (0) 0 10Tn x EΔ

(b) 0M (see Exercise 1): Then 2n (n must be even) 4 41 1 12 180 2 (0) 0 10sx EΔ

12. (a) 0M (see Exercise 2): Then 2 42121 2 (2) (0) 0 10Tn x EΔ

(b) 0M (see Exercise 2): Then 2n (n must be even) 4 421801 (1) (0) 10sx EΔ

13. (a) 2M (see Exercise 3): Then 2

2 4 2 4 42 2 2 4 4 412 3 33

(2) 10 10 10Tn n nx E n nΔ

115.4,n so let 116n

(b) 0M (see Exercise 3): Then 2n (n must be even) 4 421801 (1) (0) 0 10sx EΔ


2 4 2 4 42 2 2 4 4 412 3 33

(2) 10 10 10Tn n nx E n nΔ

115.4,n so let 116n

(b) 0M (see Exercise 4): Then 2n (n must be even) 4 421801 (1) (0) 0 10sx EΔ




2 4 2 4 482 2 212 (12) 10 8 10 8 10Tn n n

x E n nΔ

282.8,n so let 283n

(b) 0M (see Exercise 5): Then 2n (n must be even ) 4 421801 (1) (0) 0 10sx EΔ


2 4 2 4 42 2 2 412 (6) 10 4 10 4 10Tn n n

x E n nΔ

200, so let 201n

(b) 0M (Exercise 6): Then 2n (n must be even) 4 421801 (1) (0) 0 10sx EΔ


2 4 2 4 41 1 1 1 1 112 2 22

(6) 10 10 10Tn n nx E n nΔ

70.7,n so let 71n

(b) 120M (see Exercise 7): Then 4

4 4 4 41 1 1 2 2180 33

(120) 10 10sn n nx E nΔ

4243 10 9.04,n n so let 10n (n must be even)


2 4 2 4 42 2 2 412 (6) 10 4 10 4 10

Tn n nx E n nΔ

200,n so let 201n

(b) 120M (see Exercise 8): Then 4

4 4 4 464 642 2 2180 33

(120) 10 10sn n nx E nΔ

46443 10 21.5,n n so let 22n (n must be even)

19. (a) 3 3

1/2 3/21 1 1 1 12 4 4

4 1 4 1( ) 1 ( ) ( 1) ( ) ( 1) .

xf x x f x x f x x M

Then 2

2 4 2 4 43 3 3 9 9 9112 4 16 1616

10 10 10 75,Tn n nx E n n nΔ so let

76n

(b) 7 7

(3) 5/2 (4) 7/23 15 15 15 158 16 16

16 1 16 1( ) ( 1) ( ) ( 1) .

xf x x f x x M

Then

5 4 5 45

4

3 (15) 10 3 (15) 104 3 (15) 4 4 43 3 3 15180 16 16(180) 16(180)16(180)

10 10.6,sn n nx E n n nΔ so let

12n (n must be even)

20. (a) 5 5

3/2 5/23 3 3 31 12 4 41 4 1 4 1

( ) ( ) ( 1) ( ) ( 1) .x x

f x f x x f x x M

Then

4 4 4 44

2

3 10 3 102 4 23 3 3 3 312 4 48 4848

10 129.9,Tn n nx E n n nΔ so let 130n

(b) 9 9

(3) 7/2 (4) 9/215 105 105 105 1058 16 16

16 1 16 1( ) ( 1) ( ) ( 1) .

xf x x f x x M

Then

5 4 5 45

4

3 (105) 10 3 (105) 104 3 (105) 4 4 43 3 3 105180 16 16(180) 16(180)16(180)

10 17.25,sn n nx E n n nΔ

so let 18n (n must be even)



21. (a) ( ) sin ( 1) ( ) cos ( 1) ( ) sin ( 1) 1.f x x f x x f x x M Then

4 4

2

8 10 8 102 4 282 2 212 12 1212

| | (1) 10 81.6,Tn n nx E n n nΔ so let 82n

(b) (3) (4)( ) cos ( 1) ( ) sin ( 1) 1.f x x f x x M Then 42 2 2180 (1)sn nx EΔ

4 4

4

32 10 32 1044 432180 180180

10 6.49,n

n n n so let 8n (n must be even)

22. (a) ( ) cos ( ) ( ) sin ( ) ( ) cos ( ) 1.f x x f x x f x x Mπ π π Then

4 4

2

8 10 8 102 4 282 2 212 12 1212

(1) 10 81.6,Tn n nx E n n nΔ so let 82n

(b) (3) (4)( ) sin ( ) ( ) cos ( ) 1.f x x f x x Mπ π Then 42 2 2180 (1)sn nx EΔ

4 4

4

32 10 32 1044 432180 180180

10 6.49,n n so let 8n (n must be even)

23. 352 6.0 2(8.2) 2(9.1) 2(12.7) 13.0 (30) 15,990 ft .

24. Use the conversion 30 mph 44 fps (ft per sec) since time is measured in seconds. The distance traveled as the car accelerates from, say, 40 mph 58.67 fps to 50 mph

73.33 fps in (4.5 3.2) 1.3 sec is the area of the

trapezoid (see figure) associated with that time interval: 12 (58.67 73.33)(1.3) 85.8 ft. The total distance

traveled by the Ford Mustang Cobra is the sum of all these eleven trapezoids (using 2

tΔ and the table below):

(mph)v 0 30 40 50 60 70 80 90 100 110 120 130

(fps)v 0 44 58.67 73.33 88 102.67 117.33 132 146.67 161.33 176 190.67

(sec)t 0 2.2 3.2 4.5 5.9 7.8 10.2 12.7 16 20.6 26.2 37.1

/2tΔ 0 1.1 0.5 0.65 0.7 0.95 1.2 1.25 1.65 2.3 2.8 5.45

(44)(1.1) (102.67)(0.5) (132)(0.65) (161.33)(0.7) (190.67)(0.95) (220)(1.2) (249.33)(1.25)s

(278.67)(1.65) (308)(2.3) (337.33)(2.8) (366.67)(5.45) 5166.346 ft 0.9785 mi

25. Using Simpson’s Rule, 13 31 ;xx ΔΔ

33.6imy Cross Section Area 13 (33.6)

211.2 ft . Let x be the length of the tank. Then the

Volume V (Cross Sectional Area) 11.2 .x x

Now 5000 lb gasoline at 342 lb/ft 35000

42 119.05 ftV

119.05 11.2 10.63 ftx x

ix iy m imy

0x 0 1.5 1 1.5

1x 1 1.6 4 6.4

2x 2 1.8 2 3.6

3x 3 1.9 4 7.6

4x 4 2.0 2 4.0

5x 5 2.1 4 8.4

6x 6 2.1 1 2.1



26. 242 0.019 2(0.020) 2(0.021) 2(0.031) 0.035 4.2 L

27. (a) 4180 ;b a

sE x MΔ 20

4 84 ;n xπ

πΔ 2

0 4(4)180 81 1 (1) 0.00021sf M Eπ

π

(b) 8 3 24 ;xx π πΔΔ

( ) 10.47208705imf x

24 (10.47208705) 1.37079S π

(c) 0.000211.37079 100 0.015%

ix ( )if x m 1( )imf x

0x 0 1 1 1

1x /8π 0.974495358 4 3.897981432

2x /4π 0.900316316 2 1.800632632

3x 3 /8π 0.784213303 4 3.136853212

4x /2π 0.636619772 1 0.636619772

28. (a) 1 0 0.120 1 2 3 9 1010 33

0.1 erf (1) 4 2 4 4b anx y y y y y yΔ

0 0.01 0.04 0.09 0.81 1230

4 2 4 4 0.843e e e e e eπ

(b) 4 61 0180 (0.1) (12) 6.7 10sE

29. 0 1 2 3 12 2 2 2 2xn nT y y y y y yΔ where b a

nxΔ and f is continuous on [ , ].a b So

0 1 1 2 2 1 1 0 1 11 2( ) ( ) ( ) ( )( ) ( )2 2 2 2 .n n n n ny y y y y y y y f x f x f x f xf x f xb a b a

n nT Since f is continuous

on each interval 1, ,k kx x and 1( ) ( )2

k kf x f x is always between 1( ) and ( ),k kf x f x there is

a point kc in 1,k kx x with 1( ) ( )2( ) ;k kf x f x

kf c this is a consequence of the Intermediate Value Theorem.

Thus our sum is 1

( )n

b akn

k

f c

which has the form

1

( )n

k kk

x f cΔ with b a

k nxΔ for all k. This a Riemann

Sum for f on [ , ].a b

30. 0 1 2 3 2 13 4 2 4 2 4xn n nS y y y y y y yΔ where n is even, b a

nxΔ and f is continuous on

[ , ].a b So 0 1 2 2 3 4 4 5 6 2 14 4 4 43 3 3 3

n n ny y y y y y y y y y y yb anS

0 1 2 2 3 4 4 5 6 2 1

2

( ) 4 ( ) ( ) ( ) 4 ( ) ( ) ( ) 4 ( ) ( ) ( ) 4 ( ) ( )6 6 6 6

n n nn

f x f x f x f x f x f x f x f x f x f x f x f xb a

2 2 1 2 2( ) 4 ( ) ( )6

k k kf x f x f x is the average of the six values of the continuous function on the interval 2 2 2, ,k kx x

so it is between the minimum and maximum of f on this interval. By the Extreme Value Theorem for continuous functions, f takes on its maximum and minimum in this interval, so there are ax and bx with

2 2 2,k a b kx x x x and 2 2 1 2 2( ) 4 ( ) ( )6( ) ( ).k k kf x f x f x

a bf x f x

By the Intermediate Value Theorem, there is kc in 2 2 2,k kx x with 2 2 1 2 2( ) 4 ( ) ( )6( ) .k k kf x f x f x

kf c

So our sum has the form /2

1

( )n

k kk

x f cΔ with ( /2) ,b a

k nxΔ a Riemann sum for f on [ , ].a b



31. (a) /2

21 12 40

1, Length 4 1 cosa e t dtπ

/2 /22

0 02 4 cos ( ) ;t dt f t dt

π π use the

Trapezoid Rule with 10n

20

10 20 .b ant

ππΔ

10/22

00

4 cos ( ) 37.3686183nn

t dt mf xπ

2 40(37.3686183) (37.3686183)tT πΔ

2.934924419 Length 2(2.934924419) 5.870

(b) ( ) 1 1f t M

20 22

12 12 20 1 0.0032b aTE t M

ππΔ


32. 08 8 3 24 ;xx π π πΔΔ

( ) 29.184807792imf x

24 (29.18480779) 3.82028S π


0x 0 1.414213562 1 1.414213562

1x / 8π 1.361452677 4 5.445810706

2x / 4π 1.224744871 2 2.449489743

3x 3 / 8π 1.070722471 4 4.282889883

4x / 2π 1 2 2

5x 5 / 8π 1.070722471 4 4.282889883

6x 3 / 4π 1.224744871 2 2.449489743

7x 7 / 8π 1.361452677 4 5.445810706

8x π 1.414213562 1 1.414213562

33. The length of the curve 320siny xπ from 0 to 20 is: 20 2

01 ;dy

dxL dx 3 320 20cosdy

dx xπ π

2 22022 29 3 9 3

400 20 400 200cos 1 cos .dy

dx x L x dxπ π π π Using numerical integration we find

21.07L in

34. First, we’ll find the length of the cosine curve: 25 2

251 ;dy

dxL dx

2550 50sindy x

dxπ π

2 22522 2

4 50 4 5025sin 1 sin .dy x x

dx L dxπ π π π

Using a numerical integrator we find 73.1848 ft.L

Surface area is: length width (73.1848)(300) 21,955.44 ft.A Cost 2.35 (2.35)(21,955.44)A

$51,595.28. Answers may vary slightly, depending on the numerical integration used.

Section 8.8 Improper Integrals 617


35. 2 2 2

0sin cos cos 2 (sin ) 1 cos ;dy dy

dx dxy x x x S x x dxπ

π a numerical integration

gives 14.4S

36. 2 2 2 222

4 2 4 4 402 1 ;dy dyx x x x x

dx dxy S dxπ a numerical integration gives 5.28S

37. A calculator or computer numerical integrator yields 1sin 0.6 0.643501109.

38. A calculator or computer numerical integrator yields 3.1415929.π

39. The amount of medication absorbed over a 12-hr period is given by 122

06 ln 2 3 3 .t t dt A numerical

integrator yields a value of 28.684 for this integral, so the amount of medication absorbed over a 12-hr period is approximately 28.7 milligrams.

40. The average concentration of antihistamine over a 6-hr period is given by 12

0

112.5 4ln 3 4 .

6t t dt A

numerical integrator yields a value of 6.078 for this integral, so the average concentration is approximately 6.1 grams per liter.

8.8 IMPROPER INTEGRALS

1. 2 21 1 1

2 21 1 00 0lim lim tan lim tan tan 0 0

b bdx dx

x xb b bx b π π

2. 1.001 1.001 0.0010.001 1000

11 1lim lim 1000 lim 1000 1000

b bdx dx

x x bb b bx

3. 1 1 11/2 1/2

0 0 0 0lim lim 2 lim 2 2 2 0 2dx

x bbb b bx dx x b

4. 41/2

40 04 4lim (4 ) lim 2 4 2 4 0 4 4

bdx

x b bx dx b

5. 2/3 2/3 2/3

1 0 1 11/3 1/3 1/3 1/3 1/3 1/3

11 1 0 0 0 0 0lim 3 lim 3 lim 3 3( 1) lim 3(1) 3

bdx dx dx

x x x cb c b cx x b c

(0 3) (3 0) 6

6. 1/3 1/3 1/3

1 0 1 12/3 2/33 32 288 8 0 0 0

lim limb

dx dx dxx x x cb c

x x

2/3 2/3 2/3 2/33 3 3 3 3 3 92 2 2 2 2 2 2

0 0lim ( 8) lim (1) 0 (4) 0

b cb c



7. 2

11 1 1

2 2010 1 1lim sin lim sin sin 0 0

bdx

x b bx b π π

8. 0.999

1 10.001 0.001

0 0 0lim 1000 lim 1000 1000 1000 0 1000dr

r bb br b

9. 2

2 2 2 22 22 11 1 11

lim ln 1 lim ln 1 lim lndx dx dx xx x xb bx bb b b

x x

3 1 11 1 1lim ln ln ln 3 ln lim ln 3 ln1 ln 3b b

b bb b

10. 2

2 22 1 1 1 32 2 4 2 44

lim tan lim tan 1 tandx x bx bb b

π π π

11. 2

2 1 1 2 1 12 222

lim 2ln lim 2ln 2ln 2ln(1) 2 ln 0 2ln 2 ln 4bdv v b

v bv v b b

12. 2

2 1 1 2 1 11 1 2 1 31 22

lim ln lim ln ln ln(1) ln 0 ln 3 ln 3bdt t b

t bt b b

13. 2 2 22 2 2

02 2 2

01 1 1;x dx x dx x dx

x x x

2 2

2 1 11 111

1lim lim

2

cdu duu uu u bb c

u x

du x dx

1 1lim 1 lim ( 1) ( 1 0) (0 1) 0b cb c

14. 3/2 3/2 3/22 2 2

0

04 4 4;x dx x dx x dx

x x x

3/2 3/2

2 441 1

2 24 4

4lim lim

2

cdu du

u uu u b cb

u x

du x dx

1 1 1 1 1 12 2 2 2lim lim 0 0 0

b cb c

15. 2

11

20;dθ

θ θθ

2 3 3 3

2 20 0 0 0

2lim lim lim 3 3 0 3

2( 1)du du

u u bbb b b

uu b

du d

θ θθ θ

16. 2 2 2

2 2 221 124 4 40 0 0

;s dss ds

s s sds

2

2 012 44 02

4lim

2

cdu ds

u sc

u s

du s ds

2

4 4 1 1 12 22 0400 2 0 2 0 2

lim lim lim lim sin lim 2 lim sin sin 0c c

du ds s cu bsbb c b c b c

u b

42 2(2 0) 0π π

17. (1 )0

;dxx x

2 2

2 2 du 1 1 1

1 1 00 02

lim lim 2 tan lim 2 tan 2 tan 0b bdu

dx u ub b bx

u xu b

du

22 2(0)π π



18. 2 2 2 2 2

2 2 21 1

21 1 1 1 11 1 2 21 1lim lim lim sec | | lim sec | |

c cdx dx dx dx dx

bx x x x x x x x x xc cbb bx x

1 1 1 13 2 3 2

1lim sec 2 sec lim sec sec 2 0

cbb c π π π π

19. 2 1

1 1 121 1 tan0 0

lim ln 1 tan lim ln 1 tan ln 1 tan 0 ln 1 ln(1 0)b

dvv v b b

v b π

2ln 1 π

20. 1

2

2 2 2 21 1 1 216tan210 0

lim 8 tan lim 8 tan 8 tan 0 8 8(0) 2b

xx b b

dx x b π π

21. 0 0 0 0 1 1lim lim 0 1 lim 1 limb bb b b

e ebb b b be d e e e e be eθ θ θθ θ θ

ˆ(l'Hopital's rule for form)

1 0 1

22. 1 1 00 02 sin lim 2 sin lim 2 ( sin cos )

b be

b be d e d

θθ θθ θ θ θ θ θ

(Formula 107 with 1, 1a b )

0

2(sin cos ) 2(sin0 cos0) 2(0 1)22 2

lim 0 1b

b b

e eb

23. 0 0 0lim lim 1 (1 0) 1x x x b

bb be dx e dx e e

24. 2 2 2 2 200

0 02 2 2 lim lim

cx x x x x

b cbxe dx xe dx xe dx e e

2 2

lim 1 lim ( 1) ( 1 0) (0 1) 0b c

b ce e

25. 2 2 2 2

22

1 1 ln1 1 12 4 2 4 2 4 40 0 0 0

ln lim ln lim ln1 ln lim 0b

bx x b b

bb b bx x dx x b

12

43

1 1 1 14 4 4 4 4

0 0lim lim 0b

b

b

b b

26. 1

1 12

1 1 ln

0 0 0 0 0ln lim ln lim 1 1ln1 ln 1 0 lim 1 lim b

bb

bbb b b b

x dx x x x b b b

01 lim 1 0 1

bb

27. 2

21 1 1

2 2 2 2040 2 2lim sin lim sin sin 0 0

bds s b

s b b

π π



28. 4

1 4 1 2 1 2 121 00 1 1

lim 2sin lim 2sin 2sin 0 2 0br dr

r b br b π π

29. 2

2 21 1 13 311 1 1

lim sec lim sec 2 sec 0ds

bs s b bs b π π

30. 2

4 41 1 11 1 4 1 1 12 2 2 2 2 2 2 3 2 642 2 2

lim sec lim sec sec 0dt t b

bt t b b

π π

31. 4 4 4

11 10 0 0 0lim lim lim 2 lim 2

b bdx dx dx

x xx ccb c b cx x

0 0

lim 2 2 ( 1) lim 2 4 2 0 2 2 2 0 6b c

b c

32. 2 1 2 2

1 11 00 0 1 1 1lim 2 1 lim 2 1

bdx dx dx

x xx cb cx x

1 1

lim 2 1 2 1 0 lim 2 2 1 2 1 0 2 2 0 4b c

b c

33. 22 2 1 2 13 3 1 3 25 6 11

lim ln lim ln ln 0 ln ln 2b

d bbb b

θ θθθ θ

34. 2 2

2 1 111 1 1 1 12 4 2 2 2( 1) 1 100 0

lim ln 1 ln 1 tan lim ln tanbb

dx xx x xb b

x x x x

2

1 111 1 1 1 1 1 1 1 12 2 2 2 2 2 2 2 2 411

lim ln tan ln tan 0 ln1 ln1 0b

bbb π π

35.

2 2 2

/2

00tan lim ln cos lim ln |cos | ln1 lim ln |cos | ,

b

b b bd b b

π π π

πθ θ θ

the integral diverges

36. /2 /2

0 0 0 0cot lim ln sin lim ln1 ln |sin | lim ln |sin | ,

bb b bd b b

π πθ θ θ the integral diverges

37. 1

20

ln xdx

x

1

21/3

ln xdx

x is bounded, so convergence is determined by

1/3

20

ln xdx

x .

On (0, 1 / 3], ln 1x and 2 2

ln 1.

x

x x Since

1/3

20

1dx

x

diverges to , so does 1/3

20

ln xdx

x and

hence 1

20

ln xdx

x diverges.



38. Since 1ln ln ,

lndx x

x x

2

11

1lim ln(ln 2) ln(ln ) ;

ln adx a

x x

the integral diverges. (In this case

we don’t need a comparison test.)

39. ln 2

2 1/

0;xx e dx 2

3

1/ln 21/ln 21

1/ln 21/ln 2lim lim

y by e dy y y bx y b b

y e dy e e e

1/ln 2 1/ln 20 ,e e so the integral converges.

40. 1

0;

xex

dx

12

02 2 ,y

ey x e dy so the integral converges.

41. sin0

.dtt t

π

Since for 1 1sin

0 ,0t t t

t π

and 0

dtt

π

converges, then the original integral converges as

well by the Direct Comparison Test.

42. 1

sin0;dt

t t Let 1sin( ) t tf t and 3

1( ) ,t

g t then 3 2( ) 3 6 6

( ) sin 1 cos sin cos0 0 0 0 0lim lim lim lim lim 6.f t t t t

g t t t t t tt t t t t Now,

3 2 2

1 11 1 1

22 20 0 0lim lim ,dt

t t bbb b which diverges

1

sin0

dtt t diverges by the Limit

Comparison Test.

43. 2 2 2

2 1 2

1 1 10 0 1

dx dx dxx x x and 2

11 11 1

2 1 2 11 00 1 1lim ln lim ln 0 ,

bdx x b

x bx b b

which diverges

2

2

10

dxx

diverges as well.

44. 2 1 2 1

1 1 1 1 00 0 1 0 1 1and lim ln(1 ) lim ln(1 ) 0 ,

bdx dx dx dxx x x x

b bx b

which diverges

2

10

dxx diverges as well.

45. 1 0 1

1 1 0ln ln( ) ln ;x dx x dx x dx

1 1

0 0 0ln lim ln lim 1 0 1 ln

bb bx dx x x x b b b

1 0 1; 0 1

1 1ln( ) 1 ln 2x dx x dx

converges.

46. 2 2 2 21 0 1 1 1

2 4 2 41 1 0 0 0ln | | ln( ) ln lim ln lim lnx x x x

b cb cx x dx x x dx x x dx x x

2 2 2 21 1 1 1 1 12 4 2 4 2 4 2 4 4 4

0 0lim ln1 ln lim ln1 ln 0 0 0b b c c

b cb c

the integral

converges (see Exercise 25 for the limit calculations).

47. 3 3 31 1

1 11;0dx

x x x

for 1 x and 3

1

dxx

converges 311

dxx

converges by the Direct Comparison

Test.



48. 14;dx

x

11

11

1 11 011

lim lim lim 1x

xx

xxx x x

and

44lim 2 ,

bdx

x bx

which diverges

14

dxx

diverges by the Limit Comparison Test.

49. 12

;dvv

11

11

1 11 1 01

lim lim lim 1v

vv

vvv v v

and

22lim 2 ,

bdv

v bv

which diverges

12

dvv


50. 10

;deθθ

1 11

0e eθ θ

for 0 θ and 00 0

lim lim 1 1b bd d

e eb be eθ θ

θθ θ

converges

10

deθθ

by the Direct Comparison Test.

51. 36 6 6 6

1 1

1 1 1 10 0 1 0 1

dx dx dx dx dxxx x x x

and 3 2 2

1 1 1 12 22 211

lim limb

dxx x bb b

6 10

dx

x

converges by the Direct Comparison Test.

52. 2 12

;dx

x

12 1

1 2 12

1

11lim lim lim 1;x

xx

x

xx x x

1

22lim ln ,

bx b

dx b

which diverges

2 12

dx

x


53. 21

1;x

xdx

2

112

11 1

lim lim lim 1;x

x

xx

x

xxx x x

2 3/21/2 2

11 1lim 2 lim 2

bx dxbx x b b

dx x

21

12 x

xdx

converges by the Limit Comparison Test.

54. 4 12

;x dx

x

44 1

4 14

4

1

11lim lim lim 1;

x

x

xx

x

x

xx x x

4 22 2lim ln ,

bx dx dxxx x

x

which diverges

4 12

x dx

x


55. 2 cos ;xx dx

π

2 cos10 xx x

for x π and lim ln ,bdx

x bx ππ

which diverges 2 cos x

x dxπ

diverges by the Direct Comparison Test.

56. 2

1 sin ;x

xdx

π

2 2

1 sin 20 x

x x

for x π and 2 2

22 2 2 2 2lim limb dx

x bx xb bdx π πππ π

converges

2

1 sin x

xdx

π




57. 3/2

2

14;dt

t

3/2

3/2 1lim 1t

tt and 3/2 3/2

2 21/2 4

44 4lim 4 lim 2 2

bdt dt

bt tb bt

converges 3/2

2

14

dt

t


58. ln2;dx

x

1 1ln0 x x for 2x and ln2 2

divergesdx dxx x


59. 1

;xe

x dx

10xe

x x for 1x and 1 1

divergesxdx e dx

x x


60. ln (ln ) ;ee

x dx

[ ] (ln ) ;y y

ex e y e dy

0 ln (ln ) yy y e for y e and

ln lim ln ,b

ebey dy y y y

which diverges ln y

ey e dy

diverges ln (ln )

eex dx

diverges by

the Direct Comparison Test.

61. 1

;x

dx

e x

1

1

1 11 01

lim lim lim 1;xxe x

x xxexe

e

e xx x x

/2 /2

11 1lim 2

x

bx xdx

e be dx e

/2 1/2 /22

1lim 2 2b x

ebe e e dx

converges

1 x

dx

e x

converges by the Limit Comparison

Test.

62.

1

2

21

1 11 02 2 11

; lim lim lim 1xx xe

x x x x x

exe

dx ee ex x x

and 1 1

11lim limx

bx bdxee b b

e e e

1x

dxe

converges

21x xdx

e


63. 4 41 10

2 ;dx dx

x x

24 4 4 4

1 1

1 1 1 10 0 1 0 1

dx dx dx dx dxxx x x x

and 2

111

limbdx

xx b

4

1

1lim 1 1 dx

b xb


64. 0

2 ;x x x xdx dx

e e e e

1 10 x x xe e e

for 0;x 0

xdxe

converges 0

2 x xdx

e e

converges by the Direct

Comparison Test.

65. (a) 2

(ln )1;p

dxx x 1ln 2 ln 2

1 11 11 1 10 0 0

ln lim lim (ln 2)p

pp pdt b

p p pt bb bt x t

the integral

converges for 1p and diverges for 1p

(b) (ln )2

;pdx

x x

ln 2

[ ln ] pdtt

t x

and this integral is essentially the same as in Exercise 65(a): it converges

for 1p and diverges for 1p



66. 2

2 2 2 21 00

lim ln 1 lim ln 1 0 lim ln 1bx dx

x b b bx b b

the integral 2

21

xx

dx

diverges.

But 2

2 2

2 2 2 2 11 1

lim lim ln 1 lim ln 1 ln 1 lim lnb bx dx b

x bb b b bbx b b

lim (ln1) 0

b

67. 00

limbx x

bA e dx e

0lim 0 1 1b

be e

68. 0 01

00lim lim 0 0 1 1;

bx x x b bA b b

x xe dx xe e be e e e

2 2 2 2 2 01 1 1 1 1 1 1 1 1 12 2 2 2 2 2 2 2 4 400 0

lim lim 0bx x x b

A b by e dx e dx e e e

69. 00 0

2 2 2 lim 2 lim 1 2bx x x x b b

b bV xe dx xe dx xe e be eπ π π π π

70. 2 2 2 21 1 12 2 2 200 0

lim limbx x x b

b bV e dx e dx e e ππ π π π

71.

2 2

/2tansec00

(sec tan ) lim ln |sec tan | ln |sec | lim ln 1 ln 1 0b b

bb b

A x x dx x x xπ π

π

2

lim ln 1 sin ln 2b

bπ

72. (a) /2 /2 /2 /22 2 2 2 2 2

0 0 0 0sec tan sec tan sec sec 1V x dx x dx x x dx x x dx

π π π ππ π π π

2/2

20dx

πππ

(b) 2

/2 /22 2 2

outer 00 02 sec 1 sec tan 2 sec (sec tan ) lim tan

b

bS x x x dx x x x dx x

π

π ππ π π

2 2

2 2outerlim tan 0 lim tan

b bb b S

π ππ π

diverges; /2

4inner

02 tan 1 secS x x dx

ππ

2 2 2

/22 2 2 2

inner002 tan sec lim tan lim tan 0 lim tan

b

b b bx x dx x b b S

π π π

ππ π π π

diverges



73. (a) 1

0

1

(1 )dt

t t

With 1

and2

u t du dtt

the limits of integration are unchanged.

1 1

200

1 1

0

1 2

(1 ) 1

2 lim tan 1 tan

24 2

a

dt dut t u

a

π π

(b) 0

1

(1 )dt

t t

With 1

and2

u t du dtt

the limits of integration are unchanged. We split the integral into two

integrals, the first of which was evaluated in (a).

1

2 20 10

1 1

1 2 2

(1 ) 1 1

2 lim tan tan 12

22 2 2

b

dt du dut t u u

bπ

π π π π

74. Let c be any number in (3, ).

2 2 2

3 3

1 1 1

9 9 9

c

c

dx dx dxx x x x x x

provided both integrals on the right converge.

Formula 20 in Table 8.1 gives 1

2

1 1sec .

3 39

xdx

x x

(The definition of the inverse secant is given in

Section 3.9.) Both integrals do converge:

1 1 1

2 33

1 1 1 1lim sec sec sec

3 3 3 3 3 39

c

a

c a cdx

x x

1 1 1

2

1 1 1 1lim sec sec sec

3 3 3 3 6 3 39 bc

b c cdx

x x

π

Thus 2

3

1.

69dx

x x

π



75. (a) 3 3 3 3 3 9 91 1 1 1 13 3 3 3 333

lim lim 0 0.0000411bx x b

b be dx e e e e e

0.000042.

Since 2 3x xe e for 3,x then

2

30.000042xe dx

and therefore 2

0

xe dx can be replaced by

23

0

xe dx without introducing an error greater than 0.000042.

(b) 23

00.88621xe dx

76. (a) 21 1 1 1111

lim lim (0 1)b

x x bb bV dxπ π π π π

(b) When you take the limit to , you are no longer modeling the real world which is finite. The comparison step in the modeling process discussed in Section 4.2 relating the mathematical world to the real world fails to hold.

77. (a)

(b) int((sin(t))/t, t 0.. infinity); 2answer is π

78. (a)

(b) f: 2*exp( t^2)/sqrt(Pi);

int(f , t 0..infinity); (answer is 1)

79. (a) 2 /21

2( ) xf x e

π

f is increasing on ( , 0],

f is decreasing on [0, ),

f has a local maximum at 12

0, (0) 0,fπ

(b) Maple commands: f: exp( x^2/2)(sqrt(2*pi);

int(f , x 1..1); 0.683

int(f, x 2..2); 0.954

int(f, x 3..3); 0.997



(c) Part (b) suggests that as n increases, the integral approaches 1. We can take ( )n

nf x dx

as close to 1 as

we want by choosing 1n large enough. Also, we can make ( )n

f x dx

and ( )n

f x dx

as small as

we want by choosing n large enough. This is because /20 ( ) xf x e for 1.x (Likewise,

/20 ( ) xf x e for 1.x ) Thus, /2( ) .x

n nf x dx e dx

/2 /2 /2 /2 /2 /2lim lim 2 lim 2 2 2c cx x x c n n

nc c cn ne dx e dx e e e e

As /2,2 0,nn e for large enough n, ( )n

f x dx

is as small as we want.

Likewise for large enough n, ( )n

f x dx

is as small as we want.

80. (a) The statement is true since ( ) ( ) ( ) ,b a b

af x dx f x dx f x dx

( ) ( ) ( )b

b a af x dx f x dx f x dx

and ( )

b

af x dx exists since ( )f x is integrable on every interval

[ , ].a b

(b) ( ) ( ) ( ) ( ) ( ) ( )a a b b

a a a af x dx f x dx f x dx f x dx f x dx f x dx

( ) ( ) ( ) ( ) ( )b a b

b a bf x dx f x dx f x dx f x dx f x dx


Maple:

f : (x,p) - x^p*ln(x);

domain : 0..exp(1);

_fn list : [seq( f (x,p), p -2..2 )];

_plot( fn list, x domain, y -50..10, color [red,blue,green,cyan,pink], linestyle [1,3,4,7,9],

thickness [3,4,1,2,0], legend ["p -2","p -1","p 0","p 1","p 2"], title "#81 (Section 8.8)" );

q1 : Int( f(x,p), x domain );

q2 : value( q1 );

q3 : simplify( q2 ) assuming p -1;


q5 : value( eval( q1, p -1 ) );

i1 : q1 piecewise( p -1, q4, p -1, q5, p -1, q3 );


Maple:

f : (x,p) - x^p*ln(x);

domain : exp(1)..infinity;

_fn list : [seq( f(x,p), p -2..2 )];



_plot( fn list, x exp(1)..10, y 0..100, color [red,blue,green,cyan,pink], linestyle [1,3,4,7,9],



q7 : value( q6 );



q10 : value( eval( q6, p -1 ) );

i2 : q6 piecewise( p -1, q9, p -1, q10, p -1, q8 );


Maple:

f : (x,p) - x^p*ln(x);

domain : 0..infinity;

_fn list : [seq( f(x,p), p -2..2 )];

_plot( fn list, x 0..10, y -50..50, color [red,blue,green,cyan,pink], linestyle [1,3,4,7,9],


q11 : Int( f(x,p), x domain ):

q11 lhs(i1 i2);

`` rhs(i1 i2);

`` piecewise( p -1, q4 q9, p -1, q5 q10, p -1, q3 q8 );

`` piecewise( p -1, -infinity, p -1, undefined, p -1, infinity );


Maple:

f : (x,p) - x^p*ln(abs(x));

domain : -infinity..infinity;

_fn list : [seq( f(x,p), p -2..2 )];

_plot( fn list, x 4..4, y -20..10, color [red,blue,green,cyan,pink], linestyle [1,3,4,7,9],

legend ["p -2","p -1","p 0","p 1","p 2"], title "#84 (Section 8.8)" );


q12p : Int( f(x,p), x 0..infinity );

q12n : Int( f(x,p), x -infinity..0 );

q12 q12p q12n;

`` simplify( q12p q12n );

81-84. Example CAS commands:

Mathematica: (functions and domains may vary) Clear[x, f, p]

p_f[x ]: x Log[Abs[x]]

Section 8.9 Probability 629


int Integrate[f[x], {x, e, 100)]

int / . p 2.5

In order to plot the function, a value for p must be selected.

p 3;

Plot[f[x], {x, 2.72, 10}]

85. Maple gives2/

0

1 1sin 2 Ci 0.16462,

2dx

x

π πππ

where Ci is the cosine integral function defined

by cos

Ci( ) .t

xt dx

x

86. Maple gives2/

20

1 2 1sin Si 0.10276,

2 2 4x dx

x

π π ππ

where Si is the sine integral function defined

by 0

sinSi( ) .

t xt dx

x

8.9 PROBABILITY

1. 8

4

1 41;

18 3x dx

not a probability density.

2. 2

0

1(2 ) 1;

2x dx

a probability density.

3.

ln(1 ln 2)/ln 2ln(1 ln 2)/ln 2

00

22

ln 2

1 ln 2 11

ln 2 ln 2

xxdx

This is a probability density.

4. 1x is not nonnegative on [0, 1 3], so not a probability density.

5. 2

1

11;dx

x

a probability density.

6. 1

200

8 4lim tan

24

42

2

b

b

xdx

x ππ

ππ

This is not a probability density.



7. /4

02cos 2 1;x dx

π a probability density.

8. 0

1e

dxx

diverges; not a probability density.

9. (a) The probability that a tire lasts between 25,000 and 32,000 miles

(b) The probability that a tire lasts more than 30,000 miles

(c) The probability that a tire lasts less than 20,000 miles

(d) The probability that a tire lasts less than 15,000 miles

10. (a) 3 /2

/2

1 10.5

2 2dx dx

π π

π ππ π

(b) 2

2

1 11 0.682

2dx

π

π π

11. 3 3 3 1

11( 1) 4 2 0.537x xxe dx x e e e

12. 1515

22 2

ln ln 1 ln15 1 ln 2 10.599

15 15 2 2

x xdx

xx

13. 1/21/21/2

2 2 3

0 0 0

3 3 3 1 5(2 ) 3 0.3125

2 2 2 2 16x x dx x x dx x x

14. Using software to evaluate the Sine Integral we find 2

2200

1059

sin1.00004780741

xdx

x

ππ

so the given function

is very nearly a probability density over the given interval. Again using software we find that

/6 2

2200

1059

sin0.6732.

xdx

x

π ππ

15. 9

34

2 650.0502

1296dx

x

16. /4 /4

/6/6

2 3sin cos 0.159

2 2x dx x

π πππ

17. 2

3

1 1 3

6 12 4

c

x dx c . Solving 21 3

1,12 4

c we find 21c .

18. 1 1 1

ln( 1) ln lnc

c

cdx c c

x c

. Solving

1ln 1,

c

c

we find

1ce

c

and thus 1

1c

e

.

19. 2 2

04 2 2

cx ce dx e . Solving 22 2 1,ce we find

1ln 2

2c .



20. 55 3/22 2

0 0

1 12525 25 ,

3 3cx x dx c x c

so 3

125c .

21. We will assume that the given function is to be a probability density over the whole real line.

21

cdx c

xπ

so we take

1c

π . Then

22 1 1

21 1

1 tan tan 2 10.10242

41

xdx

x π ππ

.

22. 1

13/2 5/2

00

2 2 4(1 ) ,

3 5 15c x x dx c x x c

so 15

.4

c Then 1/2

1/4

15 7 17(1 ) 2 0.353

4 16 64x x dx

23. 0

1 1lim 1 .cx bcx

be dx e

c c

Thus multiplying cxe by c produces a probability density on [0, ).

24. 2

2 2

Var( ) ( )

( ) ( 2 ) ( ) ( )

X X f X dX

X f X dX X f X dX f X dX

μ

μ μ

2( 2 ) ( ) 2 ( ) 2X f X dX X f X dXμ μ μ

2 2 2 2( ) ( ) (1)f X dX f X dXμ μ μ μ

Thus 2 2Var( ) ( )X X f X dX μ

.

25. 4

0

1 8mean

8 3x x dx

To find the median we need to solve 2

0

1 1 1

8 16 2

c

x dx c for .c Thus the median is 8 .

26. 3

2

0

1 9mean

9 4x x dx

To find the median we need to solve 2 3

0

1 1 1

9 27 2

c

x dx c for .c Thus the median is 2/33

2 2.381.2

27. 3

11

2 2mean lim 2

b

bx dx

xx

To find the median we need to solve 3 2

1

2 1 11

2

c

dxx c

for .c Thus the median is 2.

28. 1

1mean 1 1.718

e

x dx ex

To find the median we need to solve 1

1 1ln

2

c

dx cx

for .c Thus the median is 1.649.e



29. The exponential density with mean 1 is .Xe The probability that the food is digested in less than 30 minutes

is 1/2

1/2

01 0.3935.Xe dX e

30. The exponential density with mean 4 is /4(1 / 4) .Xe The probability that a flower is pollinated within 5

minutes is 5

/4 5/4

0(1 / 4) 1 0.7135.Xe dX e Out of 1000 flowers we would expect 713 or 714 to be

pollinated within 5 minutes.

31. The exponential density with mean 1200 is /1200(1 / 1200) .Xe

(a) The probability that a bulb will last less than 1000 hours is 1000

/1200 5/6

0(1 / 1200) 1 0.5654.Xe dX e

(b) By Example 9 the median lifetime is 1200 ln 2 831.8 so the expected time until half the bulbs in a batch

fail is 832 hr.

32. To find the density, solve 3

/ 3/

0

1(1 / ) 1 .

3X c cc e dX e Then

37.3989,

ln(3 / 2)c so the mean

lifetime of the components is 7.4 years. The probability of failure within 1 year is ln(3/2)1 1/3

3

0

ln(3/2) 21 0.1264.

3 3

X

e dX


/ 2/

0

2 2(1 / ) 1 ; .

5 ln(5 / 3)X c cc e dX e c The probability that a hydra dies

within 6 months, or half a year, isln(5/3)1/2 1/4

2

0

ln(5/3) 31 0.1199,

2 5

X

e dX

so we would expect

(0.12)(500) 60 hydra to die within the first six months.


/ 50/

0

3 50(1 / ) 1 ; .

10 ln(10 / 7)X c cc e dX e c The probability that a high-

risk driver is involved in an accident in the first 80 days is ln(10/7)80

50

0

ln(10/7)

50

X

e dX

8/57

1 0.4349,10

so we would expect 43 or 44 out of 100 high-risk drivers to be involved in an accident

in the first 80 days.

35. Using seconds as the time unit, the density is /30(1 / 30) .Xe

(a) 15

/30 1/2

0(1 / 30) 1 0.393Xe dX e

(b) /30 2

60(1 / 30) 0.135Xe dX e

(c) In a continuous distribution the probability of a particular number is 0.

(d) The probability than a single customer waits less than 3 minutes is 6 1 0.997521.e The probability

that at least one customer out of 200 waits longer than 3 minutes is 2001 (0.997521) 0.391 0.5, so

the most likely outcome is that all 200 are served within 3 minutes.



36. For parts (a) and (b) the density is /16(1 / 16) .Xe For parts (c) and (d) the density is /32(1 / 32) .Xe

(a) 30

/16 15/8 5/8

10(1 / 16) 0.382Xe dX e e

(b) /16 25/16

25(1 / 16) 0.210Xe dX e

(c) 50

/32 25/16 35/32

35(1 / 32) 0.125Xe dX e e

(d) 20

/32 5/8

0(1 / 32) 1 0.465Xe dX e

37. The expected payout per printer is 1 2

/2 /2

0 1200 (1 / 2) 100 (1 / 2) $102.56.X Xe dX e dX Thus the

expected refund total for 100 machines is $10,256.


/ 2/

0

1 11 ,

2X c ce dX e

c

which gives 2

.ln(2)

c The probability of failure

in the first year is ln 21

2

0

ln 2 21 0.293.

2 2

X

e dX

We expect (150)(0.293) 43.934 or about 44

copiers to fail during the first year.

For Exercises 39–52, the density function is

21

21( )

2

x

f X e

μσ

σ π

with andμ σ as given in the solution.

39. 162, 28μ σ 193

165( ) 0.323;f X dX about 323 children

167

148( ) 0.262;f X dX about 262 children

40. 20.11, 4.7μ σ

17 17

1( ) ( ) 0.74593

2f X dX f X dX

μ

41. 55, 4μ σ 60

0( ) 0.89435f X dX

42. 22,000, 4000μ σ

(a) 18,000 18,000

1( ) ( ) 0.84134;

2f X dX f X dX

μ (4000)(0.84134) 3365 tires

(b) We want to find L such that ( ) 0.9.L

f X dX

A CAS gives 16,874,L so 90% of tires will have a

lifetime of at least 16,874 miles.

43. 65.5, 2.5μ σ

(a) 68

68

1( ) ( ) 0.159,

2f X dX f X dX

μ

or 16%.

(b) 64

61( ) 0.23832f X dX



44. 81, 7μ σ 85

75( ) 0.520;f X dX one would expect 52 of the babies to live to between 75 and 85.

45. 266, 16; (36)(7) 252, (40)(7) 280μ σ 280

252( ) 0.6184;f X dX we would expect 618 of the women to have pregnancies lasting between 36 and 40

weeks.

46. 1400, 100μ σ

(a) 1450

1325( ) 0.46484f X dX

(b) 1480

1480

1( ) ( ) 0.21186

2f X dX f X dX

μ

(500)(0.21186) 106; we would expect about 106 males to have a brain weight exceeding 1480 gm.

47. 80, 12μ σ 70

0( ) 0.20233f X dX

(300)(0.20233) 61; about 61 adults.

48. 4.4, 0.2μ σ 4.45

4.3( ) 0.29017f X dX

49. 35, 9μ σ 40

40

1( ) ( ) 0.28926

2f X dX f X dX

μ

About 289 shafts would need more than 45 grams of added weight.

50. 200, 30μ σ 260

260

1( ) ( ) 0.02275

2f X dX f X dX

μ

170

0( ) 0.15866f X dX

0.02275 0.15866 0.18141 or about 18%

51. (0.8)(2500) 2000, (0.4)(50) 20μ σ

(a) 1960 1960

1( ) ( ) 0.977

2f X dX f X dX

μ

(b) 1980

0( ) 0.159f X dX

(c) 2020

1940( ) 0.840f X dX



52 . 20

(0.5)(400) 200, 102

μ σ

To improve the approximation to the binomial distribution we will modify the interval of integration. We give the true binomial values for comparison.

(a) 209.5

189.5( ) 0.68208;f X dX correct to 5 places

(b) 169.5

0( ) 0.00114;f X dX true value 0.00112

(c) 220.5

220.5

1( ) ( ) 0.02018;

2f X dX f X dX

μ

true value 0.02012

(d) The value is very close to 0, in fact about 2410 .

53. (a) and (b)

(c)

0 2 4

0.1

0.2

0.3

0.4

1 3

P

X

The probability of at least 2 heads is 1 111 4 6 .

16 16

Outcome X

HHHH 0

THHH 1

HTHH 1

HHTH 1

HHHT 1

TTHH 2

THTH 2

THHT 2

HTTH 2

HTHT 2

HHTT 2

TTTH 3

TTHT 3

THTT 3

HTTT 3

TTTT 4



54. (a) The first die is listed first in each pair, the second die second.

1 + 1 = 2 2 + 1 = 3 3 + 1 = 4 4 + 1 = 5 5 + 1 = 6 6 + 1 = 7

1 + 2 = 3 2 + 2 = 4 3 + 2 = 5 4 + 2 = 6 5 + 2 = 7 6 + 2 = 8

1 + 3 = 4 2 + 3 = 5 3 + 3 = 6 4 + 3 = 7 5 + 3 = 8 6 + 3 = 9

1 + 4 = 5 2 + 4 = 6 3 + 4 = 7 4 + 4 = 8 5 + 4 = 9 6 + 4 = 10

1 + 5 = 6 2 + 5 = 7 3 + 5 = 8 4 + 5 = 9 5 + 5 = 10 6 + 5 = 11

1 + 6 = 7 2 + 6 = 8 3 + 6 = 9 4 + 6 = 10 5 + 6 = 11 6 + 6 = 12

(b)

0.1

0.2

2 3 4 5 6 9 117 8 10 12

P

X

(c) 5

(8)36

P

(d) 1 2 3 4 5

( 5)36 18

3 2 1 1( 9)

36 6

P X

P X

55. (a) {LLL, LLD, LDL, DLL, LLU, LUL, ULL, LDD, DLD, DDL,LUU, ULU, UUL, DDU, DUD, UDD,

DUU, UDU, UUD,LUD, LDU, ULD, UDL, DLU, DUL, DDD, UUU}

(b) We will assume that the three answers are equally likely, though other assumptions might be reasonable. The plots for the X number of Ls, the number of Us and the number of Ds are identical.

0 21 3

0.1

0.2

0.3

0.4

0.5

P

X

(c) 1 3 3 7

(at least two L) 0.2627 27

P

(d) (no more than one D) 1 (at least two D)

7 201 0.74

27 27

P P

Chapter 8 Practice Exercises 637


56. The probability that both systems fail is 0.0148. Since the two systems have the same performance

distribution, the failure probability for a single system is 0.0148 0.121655. The success probability for a

single system is 1 0.0148 so the probability that both succeed is 21 0.0148 0.771489. The

probability that one fails and one succeeds is 1 0.0148 0.771489 0.213711. Since the events “main fails, backup succeeds” and “main succeeds, backup fails” have the same probability, the probability that only the main fails is 0.213711 / 2 0.106856. Thus the probability that the main fails, either along with the backup or by itself, is 0.0148 0.106856 0.121656.

CHAPTER 8 PRACTICE EXERCISES

1. 1ln( 1), ;dxxu x du , ;dv dx v x

11 1ln( 1) ln( 1) ln( 1) ln( 1) ln( 1)x dxx xx dx x x dx x x dx x x x x C

1( 1) ln( 1) ( 1) ln( 1) ( 1) ,x x x C x x x C where 1 1C C

2. ln , ;dxxu x du 2 31

3, ;dv x dx v x

3 32 3 31 1 13 3 3 9ln ln lnx x

xx x dx x x x dx x C

3. 2

311 9

tan 3 , ; , ;dx

xu x du dv dx v x

2

31 1

1 9tan 3 tan 3 ;x dx

xx dx x x

21 1 21 1

6 6

1 9tan 3 tan (3 ) ln 1 9

18dyy

y xx x x x x C

dy x dx

4. 2

12 4

cos , ;x dx

xu du

, ;dv dx v x

2

1 12 2 4

cos cos ;x dxx x

xdx x

21 1 21

2 2 2

4cos cos 4

2dyx x

y

y xx x x C

dy x dx

212 2cos 2 1x xx C

5. ( )2

( )

( )

( 1)

2( 1)

2

x

x

x

x

e

x e

x e

e

0 2 2( 1) ( 1) 2( 1) 2x xx e dx x x e C

6.

( )2

( )

( )

sin(1 )

cos(1 )

2 sin(1 )

2 cos(1 )

x

x x

x x

x

0 2 2sin(1 ) cos(1 ) 2 sin(1 ) 2cos(1 )x x dx x x x x x C



7. cos 2 , 2sin 2 ;u x du x dx , ;x xdv e dx v e

cos 2 cos2 2 sin 2 ;x x xI e x dx e x e x dx

sin 2 , 2cos 2 ;u x du x dx , ;x xdv e dx v e

cos2 2 sin 25 5cos 2 2 sin 2 2 cos 2 cos2 2 sin 2 4

x xx x x x x e x e xI e x e x e x dx e x e x I I C

8. sin cosx x x dx

21, , sin cos , sin

2u x du dx dv x x dx v x

cos3 , 3sin 3 ;u x du x dx 2 212, ;x xdv e dx v e

2 2

2

2

1sin cos sin sin

2 21

sin 1 cos 22 4

1 1sin sin 2

2 4 8

xx x x dx uv v du x x dx

xx x dx

xx x x C

9. 2

22 13 2

2 ln | 2 | ln | 1 |x dx dx dxx xx x

x x C

10. 23 31 12 3 2 1 2 24 3

ln | 3 | ln | 1 |x dx dx dxx xx x

x x C

11. 2 21 1 1 1

1 1( 1) ( 1)ln | | ln | 1 |dx

x x xx x xdx x x C

12. 2 21 12 2 1 1 1

1( 1)2 ln 2 ln | |+ 2 ln | 1 |x x

x x x x xx x xdx dx C x x C

13. 2

sin

cos cos 2;dθ θ

θ θ 2

2 cos 21 1 1 13 1 3 2 3 1 3 cos 12

[cos ] ln lndy dy dy yy y yy y

y C Cθθθ

cos 113 cos 2ln Cθ

θ

14. 2

cos

sin sin 6;dθ θ

θ θ 2sin 21 1 1

5 2 5 3 5 sin 36[sin ] lndx dx dx

x xx xx Cθ

θθ

15. 2

3 22 13 4 4 44 1

214 ln | | ln 1 4 tanx x x

xx x xdx dx dx x x x C

16. 3 2

4 4 124 4

2 tanx dx dx xx x x

C

17. 5

3 6

( 3) ( 2) ( 2)3 5 3 51 1 1 12 4 8( 2) 8( 2) 8 16 16 162 8

ln | | ln | 2 | ln | 2 | lnv dv v vv v vv v v

dv v v v C C

18. 2

(3 7) ( 2) ( 2)( 3)( 1)( 2)( 3) 1 2 3 ( 1)

lnv dv dv v vdv dvv v v v v v v

C



19. 4 2 2 21 1 1 131 1 1 1 1

2 2 2 2 62 3 3 34 3 1 3tan tan tan tandt dt dt t t

t t t tt C t C

20. 4 2 2 22 21 1 1 1

3 3 6 62 2 1ln 2 ln 1t dt t dt t dt

t t t tt t C

21. 3 2 2

2 22 2 4 4 2

3 1 3 2 2 3 32 2ln | 2 | ln | 1 |x x x dx dx x

x xx x x xdx x dx x dx x x C

22. 3

3 31 1 1

( 1) 11 1 ln | 1 | ln | |x x dx dxx x x xx x x x

dx dx dx dx x x x C

23. 3 2 2

2 24 3 3 9 9 3

2 1 2 3 2 2 24 3 4 3ln | 3 | ln | 1 |x x x dx dx x

x xx x x xdx x dx x dx x x C

24. 3 2

2 22 21 24 1 2

3 2 3 42 8 2 8(2 3) (2 3)x x x x dx dx

x xx x x xdx x dx x dx

2 2 13 33 ln | 4 | ln | 2 |x x x x C

25. 3 1;dx

x x 2

1 12 1 1 1 1 13 3 1 3 1 3 3 32 1 1 11

1

ln | 1 | ln | 1 | ln

2

u du xdx du duu ux xu u

u x

du u u C C

dx u du

26. 31;dx

x x 2 3

2/3 3 3

3

3(1 ) 1 13 (1 )

2

3 3ln 3ln

3

u du xdx du uu u u xx u u

u x

du C C

dx u du

27. 1;s

dse 1

( 1) 1 1

1

1

ln ln ln |1 |s

s

s

s sdu du du u eu u u u u e

duu

u e

du e ds C C e C

ds

28. 1;

s

ds

e 2

2

2 1 11( 1)( 1) 1 1 112 1 1 1

2

1

1

2 ln lnss

s s

s

u du ee ds du du du uu u u u uu ue e

u du

u

u e

du C C

ds

29. (a) 2 2

2 21216 16

16y dy y dy

y yy C

(b) 2

2

4 16sin cos 2cos 416

; [ 4sin ] 4 4cos 16yy dy x x dx

xyy x x C C y C

30. (a) 2 2

2 2124 4

4x dx x dx

x xx C

(b) 24;x dx

x 22 tan 2sec 2

2sec[ 2 tan ] 2 sec tan 2sec 4y y dyyx y y y dy y C x C



31. (a) 2 2

( 2 ) 21 12 24 4

ln 4x dx x dx

x xx C

(b) 24;x dx

x 2

2

2sin 2cos 424cos

[ 2sin ] tan ln | cos | lnd xx d C Cθ θ θθ

θ θ θ θ

212 ln 4 x C

32. (a) 2 2

8 21 18 44 1 4 1

4 1t dt t dt

t tt C

(b) 24 1

;t dt

t 1 1 22 2

sec tan sec 2 4 1tan1 12 tan 4 4 4sec sec

d tt d C Cθ θ θ θ θ

θθ θ θ

33. 29;x dx

x 2

21 1 1 12 2 9

9ln | | ln ln

2duu u x

u xu C C C

du x dx

34. 2

21 1 1 1 1 1 1 19 18 3 18 3 9 18 18 9 189

ln | | ln | 3 | ln | 3 | ln | | ln 9dx dx dx dxx x xx x

x x x C x x C

35. 231 1 1 1 1

6 3 6 3 6 6 6 39ln | 3 | ln | 3 | lndx dx dx x

x x xxx x C C

36. 29;dx

x 13cos3cos 3

sinsin

3cosxx

d d C Cdx d

θθ

θθ θ θ

θ θ

37. 5 73 4 4 2 4 6 cos cos5 7sin cos cos 1 cos sin cos sin cos sin x xx x dx x x x dx x x dx x x dx C

38. 25 5 5 4 5 2cos sin sin cos cos sin 1 sin cosx x dx x x x dx x x x dx

6 8 105 7 9 sin 2sin sin6 8 10sin cos 2 sin cos sin cos x x xx x dx x x dx x x dx C

39. 54 2 tan

5tan sec xx x dx C

40. 3 3 2 2 4 2tan sec sec 1 sec sec tan sec sec tan sec sec tanx x dx x x x x dx x x x dx x x x dx

5 3sec sec5 3

x x C

41. 1 1 1 1 12 2 2 2 22sin5 cos6 sin( ) sin(11 ) sin( ) sin(11 ) cos( ) cos11d d d d Cθ θ θ θ θ θ θ θ θ θ θ θ

1 12 22cos cos11 Cθ θ

42. 2

2 2

sin (1 cos )2 3 1sincos cos

sec sin sin cos ( cos ) sec cosd d d C Cθ θ θθ θ

θ θ θ θ θ θ θ θ θ θ

43. 2 4 41 cos 2 cos 4 2 sint t tdt dt C



44. 2tan 1 sec ln sec tant t t t t te e dt e e dt e e C

45. 43 1180| | ( )sE x MΔ where 3 1 2 ;n nxΔ 1 2 3 41( ) ( ) ( ) 2 ( ) 6xf x x f x x f x x f x x

(4) 5( ) 24f x x which is decreasing on [1, 3] maximum of (4) ( )f x on [1, 3] is

(4) (1) 24 24.f M Then 4

43 1 7682 1180 180| | 0.0001 (24) 0.0001 0.0001s n n

E

44180 7681

768 180(0.0001) 10,000 14.37 16n

n n n (n must be even)

46. 21 012| | ( )TE x MΔ where 1 0 1 ;n nxΔ 0 ( ) 8 8.f x M Then 23 31 1

12| | 10 (8) 10T nE 2

23 23 20002

2 3310 1000 25.82 26n

nn n n

47. 0

6 6 2 12 ;b a xnx π π πΔΔ

6

120

( ) 12 (12) ;ii

mf x T π π


0x 0 0 1 0

1x /6π 1/2 2 1

2x /3π 3/2 2 3

3x /2π 2 2 4

4x 2 /3π 3/2 2 3

5x 5 /6π 1/2 2 1

6x π 0 1 0

6

0

( ) 18ii

mf x

and 3 18x πΔ

18 (18) .S π π


0x 0 0 1 0

1x /6π 1/2 4 2

2x /3π 3/2 2 3

3x /2π 2 4 8

4x 2 /3π 3/2 2 3

5x 5 /6π 1/2 4 2

6x π 0 1 0

48. (4) ( ) 3 3;f x M 2 1 1 .n nxΔ Hence 5

4

45 5 5 4 102 1 1 1180 6060

| | 10 (3) 10 10s n nE n

6.38 8n n (n must be even)



49. 365 3652 365 21 1

365 0 365 365 2 365 0037sin ( 101) 25 37 cos ( 101) 25avy x dx x xπ π

π

365 2 365 21365 2 365 2 36537 cos (365 101) 25(365) 37 cos (0 101) 25(0)π π

π π

37 2 37 2 37 2 22 365 2 365 2 365 365cos (264) 25 cos ( 101) cos (264) cos ( 101) 25π π π ππ π π

372 (0.16705 0.16705) 25 25 Fπ

50. 5 5

675 6755 2 2 313 0.623331 1

675 20 655 10 10 2020( ) 8.27 10 26 1.87 8.27vav C T T dT T T T

1655 (5582.25 59.23125 1917.03194) (165.4 0.052 0.04987) 5.434;

26 676 4(1.87)(283,600)5 2 22(1.87)8.27 10 26 1.87 5.434 1.87 26 283,600 0 396.45T T T T T C

51. (a) Each interval is 5 min 112 hour. 291

24 12[2.5 2(2.4) 2(2.3) 2(2.4) 2.3] 2.42 gal

(b) (60 mph) 1229 hours/gal 24.83 mi/gal

52. Using the Simpson’s rule, 315 5;xx ΔΔ

( ) 1211.8imf x Area 2(1211.8)(5) 6059 ft ;

The cost is 2 2 2Area $2.10/ft 6059 ft $2.10/ft

$12,723.90 the job cannot be done for $11,000.


0x 0 0 1 0

1x 15 36 4 144

2x 30 54 2 108

3x 45 51 4 204

4x 60 49.5 2 99

5x 75 54 4 216

6x 90 64.4 2 128.8

7x 105 67.5 4 270

8x 120 42 1 42

53. 2 2

31 1 1 0

3 3 3 2 209 90 03 3 3lim lim sin lim sin sin 0

b bdx dx x b

x xb b b

π π

54. 1

1 12

1 1 ln

0 0 0 0 0ln lim ln lim (1 ln1 1) ln 1 lim 1 lim 1 0 1b

bb

bbb b b b

x dx x x x b b b

55. 2/3 2/3 2/3 2/3

2 1 2 1

( 1) ( 1) ( 1) ( 1)0 0 1 0

1/3 1/3 1/3

00 0

2

2 3 lim ( 1) 6 lim ( 1) ( 1) 6(0 1) 6

dy dy dy dy

y y y y

b

b by b



56. 3/5 3/5 3/5

0 1 0 02/5 2/55 52 2( 1) ( 1) ( 1) 22 2 1 1 1

2/5 2/5 2/5 2/55 5 5 52 2 2 2

1 1

lim ( 1) lim ( 1)

lim ( 1) ( 1) 1 lim ( 1) ( 1) (1) 0

bd d d

bb b

b bb b

θ θ θθ θ θ

θ θ

converges if each integral converges, but 3/5

3/5( 1)lim 1θ

θθ and 3/5

2

dθθ

diverges 3/5( 1)2

dθθ

diverges

57. 2

2 2 2 3 2 12 3 32 33 3 3

lim ln lim ln ln 0 ln ln 3bdu du du u b

u u u bu u b b

58. 3 2 23 1 1 1 4 1 1

4 1 4 14 11 1lim ln ln(4 1) lim ln (ln1 1 ln 3)

bv bv v v b bv v v b b

dv dv v v

314 4ln 1 ln 3 1 ln

59. 2 2 2

00lim 2 2 lim 2 2 ( 2) 0 2 2

bx x x x b b b

b bx e dx x e xe e b e be e

60. 0 03 3 3 3 31 1 1 1 13 9 9 3 9 9 9lim lim 0x x x b bx b

bb bxe dx e e e e

61. 2 2 2 94

1 1 12 21 1 2 1 2 22 2 3 3 2 3 3 34 9 4 9 00 0

2 lim tan lim tan tan (0)b

dx dx dx x bx x x b b

1 22 3 2 60π π

62. 2 21 1 14 4

4 4 216 16 002 2 lim tan 2 lim tan tan (0) 2 0

bdx dx x b

x x b b

π π

63. 2 1

lim 1θθθ

and 6

dθθ

diverges 2 16

dθθ

diverges

64. 0 00 0 0

cos lim cos sin 1 lim sin cosb bu u u u u

b bI e u du e u e u du e u e u du

121 0 2 1I I I I converges

65. 2 2 22(ln ) (ln ) (ln )ln ln ln 1 12 2 2 2 21 1 1

lim 0 lime be

z z bz z zz z z b be e

dz dz dz

diverges

66. 0t tet

e for 1t and

1

te dt converges

1

tet

dt

converges

67. 2 2 4

0 02x x x x x

dx dx dx

e e e e e

converges 2

x xdx

e e

converges



68. 2 2 2 2 2

1 0 1

1 1 1 1 11 0 1;

x x x x xdx dx dx dx dx

x e x e x e x e x e

1 22

2

12 1

1

0 0 0lim lim lim 1 2

xx

xx e

x e x

xx x xe

and

2

1

0

dxx diverges 2

1

10x

dxx e

diverges 2 1 xdx

x e

diverges

69. 1

;x dx

x 2 2 2 3 22 21 1 3

2

2 2 2 2 2ln |1 |u u duu udx

x

u xu u du u u u u C

du

3/223 2 2 ln 1x x x x C

70. 3 2

2 22 4 2 3 5 3 5

2 2 2 2 2 2 24 4ln | 2 | ln | 2 |x x dx dx x

x xx xdx x dx x dx x x C

71. 22x x dx ; 1 sin cosx dx dθ θ θ

For the integrand to be nonnegative x must be between 0 and 2 so 1x is between 1 and 1 and we can take θ between 2π and / 2,π where cosine is nonnegative. Thus

2 2 2 22 1 ( 1) 1 sin cos cos .x x x θ θ θ

2 2

1 2

12 cos cos cos 1 cos 2

21 1 1 1

sin 2 sin cos2 4 2 21 1

sin ( 1) ( 1) 22 2

x x dx d d d

C C

x x x x C


θ θ θ θ θ

72. 2 2

1

2 1 ( 1)sin ( 1)dx dx

x x xx C

73. 2 2

cos22 cos sin 1sinsin sin

2csc csc 2cot ln | csc cot |x dxx xxx x

dx x dx x dx x x x C

2cot csc ln | csc cot |x x x x C

74. 2 5sin cos dθ θ θ

22 5 2 2sin cos sin 1 sin cosd dθ θ θ θ θ θ θ ; sin cosu du dθ θ θ

22 2 2 4 6

32 4 6 5 7

35 7

sin 1 sin cos sin 2sin sin cos

2 12

3 5 7

sin 2 1sin sin

3 5 7

d d

uu u u du u u C

C

θ θ θ θ θ θ θ θ θ

θ θ θ



75. 4 2

9 131 1 1 1 12 12 3 12 3 12 3 6 381 9

ln tandv dv dv dv v vv v vv v

C

76. 21 1

1 1( 1) 22lim lim ( 1) 0 1 1

bdxx bx b b

77. ( ) 1

2( ) 1

4

cos(2 1)

sin(2 1)

1 cos(2 1)

θ

θ θ

θ

0 12 4cos(2 1) sin (2 1) cos (2 1)d Cθθ θ θ θ θ

78. 3 2

2 2 23 2 1

1 2 12 1 2 1 ( 1)2 ( 2) 3 2 3ln | 1 |x dx x dx dx x

x xx x x x xx dx x dx x x C

79. 2 2

sin 2 ( 2sin 2 ) 21 1 12 42 1 cos21 cos2 1 cos 2

secd d C Cθ θ θ θθθ θ

θ

80. /2/2 /2

2 22 2/4 /4 /4

1 cos4 2 cos 2 sin 2x dx x dx xππ π

π π π

81. 2

;x dx

x (2 ) 3/2 1/2 3/2 1/22 23 3

24 (2 ) 4(2 )y dy

y

y xy y C x x C

dy dx

32

32 2 2x

x C

82. 2

21 ;vv

dv 2

2 2

1 sincos cos 2

sin sin[ sin ] csc cot

ddv d d Cθ θθ θ θ

θ θθ θ θ θ θ θ

21 1sin vvv C

83. 2 21

2 2 ( 1) 1tan ( 1)dy dy

y y yy C

84.

2

2 4 22

2 1 11 12 2 38 2 9 1

sinx dxx dx x

x x xC

85. 2 22 2

2 11 1 1 1 1 1 1 1 14. 4 4 8 8 244

ln | | ln 4 tanz z zz zz zz z

dz dz z z C



86. 2 1/3( 1)x x dx ; 2 21 ( 1)u x du dx x u

2 1/3 2 1/3

7/3 4/3 1/3

10/3 7/3 4/3

10/3 7/3 4/3

( 1) 2 1

2

3 6 3

10 7 43 6 3

( 1) ( 1) ( 1)10 7 4

x x dx u u u du

u u u du

u u u C

x x x C

87. 2 2

8 21 18 49 4 9 4

9 4t dtt dt

t tt C

88. 21

1tan , ;dx

xu x du

2

1, ;dxxx

dv v 1

2 22

tan 1 11 111

tan tanx dx x dxdx dxx x xx xx x

x x

11 2 2tan1 12tan ln | | ln 1 ln | | ln 1x

x xx x x C x x C

89. 2 3 2;

t

t te dt

e e 1 1( 1)( 2) 1 2 2 2

[ ] ln | 1 | ln | 2 | ln lnt

tt dx dx dx x e

x x x x x ee x x x C C C

90. 2 23 2 tan tan2 2tan (tan ) sec 1 tan ln | sec |t tt dt t t dt t dt t C

91. 3

ln

1;y dy

y

32 2 21

2 4 00 0

ln

limx

x

bdy x x xx e xy e bx

x y

dx dx xe dx e e

dy e dx

2 21 1 1

4 42 4lim 0b b

be eb

92. 3/2 2(ln )y y dy ; 2 3/2 5/22 ln 2(ln ) , , ,

5

yu y du dy dv y dy v y

y

3/2 2 5/2 2 3/22 4(ln ) (ln ) ln

5 5y y dy y y y y dy

Now we compute 3/2 ln :y y dy 3/2 5/21 2ln , , ,

5u y du dy dv y dy v y

y

3/2 5/2 3/2

5/2 5/2

2 2ln ln

5 52 4

ln5 5

y y dy y y y dy

y y y



3/2 2 5/2 2 3/2

5/2 2 5/2 5/2

5/2 2

2 4(ln ) (ln ) ln

5 52 4 2 4

(ln ) ln5 5 5 5

2 8 16(ln ) ln

5 25 125

y y dy y y y y dy

y y y y y

y y y C

93. ln 3/223

xe dx x dx x C

94. 3 4 ;e e dθ θ θ 3/23/21 1 2 14 4 3 6

43 (3 ) 3 4

4

u eu du u C e C

du e d

θθ

θ θ

95. 2

sin5

1 cos5;t dt

t 21 11 1 1

5 5 51

cos5tan tan (cos5 )

5sin5duu

u tu C t C

du t dt

96. 2 1

;v

dv

e 2

1 1

1sec sec

vvdx

v x x

x ex C e C

dx e dv

97. 1

;drr 2 2

1 12

2 2 2ln |1 | 2 2 ln 1u duu udr

r

u rdu u u C r r C

du

98. 3

3

4 2 4 2

4 20 4 24 2010 9 10 9

ln 10 9x x dxx x

x x x xdx x x C

99. 3

2 2 22 221 1 1

2 2 21 1 1ln 1x x x

x x xdx x dx x dx dx x x C

100. 2 2

3 3331 1

3 31 1ln 1x x

x xdx dx x C

101. 2

311

;xx

dx 2

3 22 21

11 11 1 ( )(1 )x Bx CA

xx x xx A x x Bx C x

2 2 1 13 3 3( ) ( ) ( ) 1, 0, 1 , , ;A B x A B C x A C A B A B C A C A B C

2

3 2 2 23 14 2

(1/3) 1/31 2/3 1 12 1 1 2 1 11 3 1 3 3 1 31 1 1

;xx x xx x xx x x x x x

dx dx dx dx dx dx

322 2 23 3 3

4 4 4

12 12 31 1 1 1 1 1

3 3 2 6 4 3 3/2ln tan

u u uu u u

u xdu du du u

du dx

12

2 1 2 13 2 11 1 1 1 16 4 2 63 3/2 3 3

ln tan ln 1 tanx xx x x

22 11 2 12 1 1 2 1 1

3 1 3 3 6 3 31ln |1 | ln 1 tanx x

x x xdx dx x x x C



102. 2

31

(1 );x

xdx

2 2

3 3 2 3 2

1 ( 1) 2 2 1 2 2 2 11ln | |u u u

u uu u u u u

u xdu du du du du u C

du dx

22 1

1 (1 )ln |1 | x x

x C

103. 1 ;x x dx 2

22 12

w x w xw w dw

w dw dx

( )2 3/22

3( ) 5/24

15( ) 7/28

105

1

2 (1 )

4 (1 )

4 (1 )

w

w w

w w

w

0 2 2 3/2 5/2 7/216 3243 15 1052 1 (1 ) (1 ) (1 )w w dw w w w w w C

3/2 5/2 7/216 324

3 15 1051 1 1x x x x x C

104. 1 1 ;x dx 21 1

2 1 ;2

w x w xw w dw

w dw dx

3/2232 , 2 , 1 , (1 )u w du dw dv w dw v w

3/2 3/2 3/2 5/284 4 43 3 3 152 1 (1 ) (1 ) (1 ) (1 )w w dw w w w dw w w w C

3/2 5/284

3 151 1 1 1 1x x x C

105. 11

;x x

dx

2

22

1;

2 u

u x u xdu

u du dx

2 2

2 2tan , , sec , 1 secu du d uπ πθ θ θ θ θ

2

2

22sec2sec1

2sec 2ln sec tan 2ln 1 2ln 1u

du d d C u u C x x Cθθ θ θ θ θ θ

106. 1/2

2

01 1 ;x dx

2 12 2 2 6sin , , cos , 1 cos , 0 sin 0, sinx dx d x x xπ π πθ θ θ θ θ θ θ θ θ

2/6 /6 /6 /61 cos sin cos sin cos1 cos 1 cos 1 cos0 0 0 0

1 cos cos cos lim ;cc

d d d dπ π π πθ θ θ θ θ

θ θ θθ θ θ θ θ θ θ

1/2sin1 cos

cos , sin , , 2 1 cosu du d dv d vθθ

θ θ θ θ θ

/6 /61/2 1/2

0lim 2cos 1 cos 2 1 cos sin

cc cd

π πθ θ θ θ θ

/61/2 1/2 3/24

6 6 30

lim 2cos 1 cos 2cos 1 cos 1 cosc c

c cπ

π π θ

1/2 3/21/2 3/23 4 4

2 3 6 30

lim 3 1 2cos 1 cos 1 cos 1 cosc

c c cπ



1/2 3/21/2 3/23 34 4

2 3 2 30

lim 3 1 2cos 1 cos 1 1 cosc

c c c

1/2 3/2 1/2 4 3 2 33 3 3 4 342 3 2 2 3 3 2

3 1 1 1

107. ln ln

ln 1 ln;x x

x x x x xdx dx 1 1

1

1 lnln | |u

u ux

u xdu du du u u C

du dx

1 ln ln |1 ln | ln ln 1 lnx x C x x C

108. 1

ln ln ln;

x x xdx

1

1ln

ln lnln | | ln ln lnu

x x

u xdu u C x C

du dx

109. ln ln ;

xx xx dx ln2ln ln 2ln 2 ln 2 ln1ln ln ln

xx x x u x x xu x x xu x u x x du dx du dx dx

ln1 1 12 2 2

xdu u C x C

110. ln lnln 1ln ;xx

x xx dx ln ln lnln ln 1lnln ln ln ln ln ln ln

x xx xu x x xu x u x x x du dx

ln ln ln lnln ln1 1ln lnx xx x

x x x xdu u dx x dx du u C x C

111. 4 2 4

1

1 1;x

x x x xdx dx

2 4

2sin , 0 , 2 cos , 1 cosx x dx d xπθ θ θ θ θ

4 4

2 2 21 1 1cos1 1 1 1 1 1

2 sin cos 2 2 2 2csc ln csc cot ln lnx xx x x

d d C C Cθθ θ θ θ θ θ θ

112. 1 ;xx dx 2 2

2 2 22 2 2 2

1 1 11 1 2 2 ;u u

u u uu x u x u du dx du du du

22

1 112 ( 1) ( 1) ( ) 0, 2 1 1;A B

u uuA u B u A B u A B A B A B A B

21 12 1 1 1

1 1 2 1 112 2 2 ln | 1 | ln | 1 | 2 1 ln x

u u xudu du du u u u C x C

113. (a) 0

( ) ;a

f a x dx 0

0, 0 , 0 ( ) ( ) ,

a

au a x du dx x u a x a u f u du f u du

which is the same integral as 0

( ) .a

f x dx

(b)

2 2 2

2 2 2 2 2 2

/2 /2 /2sin sin cos cos sinsinsin cos sin cos sin cos cos sin cos cos sin sin0 0 0

x x xxx x x x x x x x

dx dx dxπ π π

π π π π π π

π π π

/2 /2 /2 /2cos sin sin cos

cos sin sin cos sin cos cos sin0 0 0 02x x x x

x x x x x x x xdx dx dx dxπ π π π

/2 /2 /2 /2/2sin cos sin sin

sin cos 2 sin cos 2 sin cos 400 0 0 02x x x x

x x x x x xdx dx x dx dxπ π π ππ π π π



114. sin sin cos cos sin sin sin cos cos sin sinsin cos sin cos sin cos sin cos sin cos

x x x x x x x x x x xx x x x x x x x x xdx dx dx dx dx

sincos sin sinsin cos sin cos sin cosln sin cos xx x x

x x x x x xdx dx dx x x x dx

sin sin 1sin cos sin cos 2 22 ln sin cos ln sin cosx x x

x x x xdx x x x dx x x C

115.

2sin2 2 2 2 2 2 2 22cos

2 2 2 2 2 2 2 2 2 2sin12 2cos cos

sin tan tan sec sec tan sec sec1 sin sec tan sec tan sec tan sec tan

x

x

x

x x

x x x x x x x xx x x x x x x x x

dx dx dx dx dx dx

2

21sec 1

21 2 tantan 2 tanx

xdx dx x x C

116. 2 2 2

2 2 2 2 2

1 cos1 cos 1 2cos cos 2cos cos11 cos 1 cos sin sin sin sin

xx x x x xx x x x x x

dx dx dx dx dx dx

2 2 2csc 2 csc cot cot cot 2csc csc 1 2cot 2cscx dx x x dx x dx x x x dx x x x C

CHAPTER 8 ADDITIONAL AND ADVANCED EXERCISES

1. 1

2

2 2sin1

1sin , ; , ;x dx

xu x du dv dx v x

1

2

2 2 2 sin1 1

1sin sin ;x x dx

xx dx x x

2 2

21 2

1 1sin , ; , 2 1 ;x dxdx

x xu x du dv v x

1

2

2 sin 1 2 1 2

12 sin 1 2 2 sin 1 2 ;x x dx

xx x dx x x x C

therefore

2 21 1 1 2sin sin 2 sin 1 2x dx x x x x x C

2. 1 1 ,x x

1 1 1( 1) 1 ,x x x x

1 1 1 1( 1)( 2) 2 1 2( 2) ,x x x x x x

1 1 1 1 1( 1)( 2)( 3) 6 2( 1) 2( 2) 6( 3) ,x x x x x x x x

1 1 1 1 1 1( 1)( 2)( 3)( 4) 24 6( 1) 4( 2) 6( 3) 24( 4)x x x x x x x x x x

the following pattern: ( 1)1( 1)( 2) ( ) ( !)( )!( )

0

;k

m

x x x x m k m k x kk

therefore ( 1)( 1)( 2) ( ) ( !)( )!

0

lnk

mdx

x x x x m k m kk

x k C

Chapter 8 Additional and Advanced Exercises 651


3. 2

2

121

sin , ; , ;dx x

xu x du dv x dx v

2 2

2

1 12 2 1

sin sin ;x x dx

xx x dx x

22 2sin cos1 1 1 212 2cos 2 2

sinsin sin sin sin

cosdx xx

x x dx x x ddx d

θ θ θθ

θθ θ

θ θ

2 12 2 21 1 1 1 sinsin 2 sin cos12 2 2 4 2 4 2 4sin sin sin x x xx x xx C x C x Cθ θ θ θ θ

4. 1sin ;y dy 1

2

2 sin ;dy

y

z yz z dz

dz

from Exercise 3, 2 12 11 1 sinsin

2 4sin z z zz zz z dz C

1 2 11 sin sin1 1 12 2 2sin sin sin

y y y y y yy dy y y C y y C

5. 21;dt

t t cossin cos tan 1

sin;

cosd dt

dt dθ θ θ

θ θ θθ

θ θ

2

2

2

( 1) 1

1

tan

sec duu u

duu

u

du d

d

θθ θ

θ

2 2 2

11 tan 11 1 1 1 1 1 12 1 2 2 2 2 2 sec 21 1 1

ln tan lnu dudu du uu u u u

u C Cθθ θ

2 11 12 2ln 1 sint t t C

6. 4 2 2 2 2 22 22 2

2 2 2 21 1 1 1 2 2164 2 2 ( 1) 1 2 2 ( 1) 12 2 2 22 4

x xx x x x x x xx x x xx x

dx dx dx dx

2

21 12 21 1

16 82 2ln tan ( 1) tan ( 1)x x

x xx x C

7. lim sin lim cos lim cos cos( ) lim cos cos lim 0 0x x

xx x x x xxt dt t x x x x

8. 2

1cos

0lim ;t

txxdt

12

2cos

2

1cos1

cos0 0 0

lim lim 1 limt

t

t

tt txt t x

dt

diverges since 2

1

0

dtt diverges; thus 2

1cos

0lim t

txxx dt

is an indeterminate 0 form and we apply l′Hôpital′s rule: coscos

2212 1

12

1cos

0 0 0 0lim lim lim lim cos 1

x xt

xt

xx

dtt

txx x x xx dt x

9. 1

1 1

01 1

lim ln 1 lim ln 1 ln 1 ;n n

knn n nn n

k k

k x dx

1 ,

0 1, 1 2

u x du dx

x u x u

2 2

11ln ln 2 ln 2 2 ln1 1 2ln 2 1 ln 4 1u du u u u

10.

2 2 2 2 2 21

1 1 1 1 111 1 1 1 1201010 0 0

lim lim lim sinn

n n nn

n nn k n k xn n n kk k k

dx x π



11. 2 2cos2 1 1 cos 2 2cos ;dy dydx dxx x x /4 /42 2

0 01 cos 2 2 cosL t dt t dt

π π

/4

02 sin 1t

π

12.

22 22 4 2

2 2 2 22 2

22 1 42 1 2 11 11 1

1 ;x xdy dyx x x x

dx dxx xx x

2

2

1/2 1/22110 0

1 dy xdx x

L dx dx

2

1/2 1/2 1/212 1 1 1 1

1 1 1 2 21 00 01 1 ln ln 3 (0 ln1) ln 3x

x x xxdx dx x

13. 1shell shellradius height

02 2

b

aV dx xy dxπ π

12

06 1 ;x x dxπ

2 21 , , (1 )u x du dx x u

0 12 1/2 3/2 5/2

1 06 (1 ) 6 2u u du u u u duπ π

13/2 5/2 7/22 4 23 5 7 0

6 u u uπ

70 84 30 16 322 4 23 5 7 105 105 356 6 6 ππ π π

14. 2 2

4 4252 51 15(5 )1 1

b dxx xx x xa

V y dx dxπ π π

45 5 1

5 4 41ln ln 4 ln 5x

x xπ π π

154 2 ln 4π π

15. shell shellradius height2

b

aV dxπ

1 1

002 2 2x x xxe dx xe eπ π π

16. ln 2

02 ln 2 1xV x e dxπ

ln 2

02 ln 2 ln 2x xe xe x dxπ

2 ln 2

2 02 ln 2 ln 2x x x xe x xe eπ

2

ln 2222 2 ln 2 ln 2 2ln 2 2 2 ln 2 1π π

2ln 222 ln 2 1π



17. (a) 21

1 lne

V x dxπ

211

ln 2 lne e

x x x x dxπ π

(FORMULA 110)

2

1ln 2 ln

ex x x x x xπ

21

ln 2 lne

x x x x xπ

2 ( 1)e e eπ π

(b) 2 2

1 11 ln 1 2ln ln

e eV x dx x x dxπ π

211

2 ln ln 2 lne e

x x x x x x x dxπ π

2

12 ln ln 2 ln

ex x x x x x x x xπ

21

5 4 ln lne

x x x x xπ

(5 4 ) (5) (2 5)e e e eπ π

18. (a) 22 21 1 12 32 12 2 2 200 0

1 1 1y ey y e eV e dy e dy y

ππ π π π

(b) 2 21 1 12 2 12 2 200 0

1 2 1 2 2 1 2yy y y ye eV e dy e e dy e y eπ π π π

22 4 552 2 22

e ee eπ

π

19. (a) 0 0

lim ln 0 lim ( ) 0 (0)x x

x x f x f f

is continuous

(b) 2 22

0ln ;V x x dxπ 32 2

3ln , 2 ln ; ,dx xxu x du x dv x dx v

3 3 3 32 2 22 28 23 3 3 3 3 900 0

lim ln 2ln ln 2 lim lnx x dx x xx bb bb

x x xπ π

28 ln 2 16 ln 2 16

3 9 27π

20. 1 12 2

0 0ln lnV x dx x dxπ π

1 12

00lim ln 2 ln

b bx x x dxπ

1

02 lim ln 2

bbx x xπ π



21. 11ln ln ( ) (0 1) 1;

e eM x dx x x x e e

2ln 12 21 1

ln lne e

xxM x dx x dx

21 12 211

ln 2 ln ( 2);e e

x x x dx e

2 ln 12 211 1

lne ee

x xyM x x dx x dx

2 22 2 21 1 1 12 2 2 2 2 41

ln 1 ;e

x ex x e e

therefore, 2 14

yM eMx and 2

2xM e

My

22. 2

1 112

0102sin ;dx

xM x π

2

11 2 2

10 02 1 2;x dx

yx

M x

therefore,

2yM

Mx π and 0y by symmetry

23. 2

211

1 11 ;

e exxx

L dx dx 1 1 22tan tan (sec ) tan 1sec sec

tan tan2 /4 /4

tan

sec

e edxL d

dx d

θ θθ θ θθ θπ π

θθ

θ θ

1 1tan tan

/4/4tan sec csc sec ln csc cot

e ed

ππθ θ θ θ θ θ θ

2 22 21 11 11 ln 2 ln 1 2 1 ln 2 ln 1 2e ee e e ee e

24. 122 2 2

0ln 1 1 2 1 2 1 ;

dy ydx

dy cy x x S x x dy S e e dyπ π

2

12 1 ;

y e

y

u eS u du

du e dyπ

1tan2

2 /4

tan2 sec sec

sec

eud

du d π

θπ θ θ θ

θ θ

1tan 2 212 /4

2 sec tan ln sec tan 1 ln 1 2 1 ln 2 1e

e e e eπ

π θ θ θ θ π π

22 12 1

1 ln 2e ee eπ

25. 2/3 2/3

1 13/2 3/22 22/3 2/31

1 12 ( ) 1 ( ) ; ( ) 1 ( ) 1 2 1 dx

x xS f x f x dx f x x f x S xπ π

1/3

1 3/22/3 1

04 1 ;

xx dxπ

1/3

2/3 1 13/2 3/232 2 0 03

4 1 6 1 ( 1)dxx

u xu du u du

duπ π

15/2 122

5 50

6 1 u ππ



26. 216 16 16 165/44 45 11 1 1 1

1 1 1 1 1 1x

dydxy t dt x L x dx x dx x dx x

5/4 5/44 4 1245 5 5(16) (1)

27. 2

2 2

121 1 1 12 2 2 2 21 1 11 1

1

lim lim ln 1 ln lim lna b

b b xax ax ax x xx xb b b

dx dx x x

2 112lim ln ln 2 ;

ab a

bb

2 122

1 2lim lim lim

aab ab

b bb b bb

if 1

2a the improper integral diverges

if 12 ;a for 1

2 :a 1/22

2

2

1 1/21 1 12lim lim 1 1 lim ln ln 2

bbb bbb b b

1 1 ln 2

2 2 4ln1 ln 2 ; if

12 :a

2 21 ( 1) 2 110 lim lim lim ( 1) 0

aab b a

b bb b bb

2 1

lim lna

b

bb

the improper integral

diverges if 12 ;a in summary, the improper integral 2

1211

axxx

dx

converges only when 1

2a and has

the value ln 24

28. 1 1 01 1

00( ) lim lim lim

xbb bxt xt ex x x xb b b

G x e dt e

if 10 ( ) 1xx xG x x if 0x

29. 1

pdxx

A

converges if 1p and diverges if 1.p Thus, 1p for infinite area. The volume of the solid of

revolution about the -axisx is 2

21

1 1p p

dxx x

V dxπ π

which converges if 2 1p and diverges if

2 1.p Thus we want 12p for finite volume. In conclusion, the curve py x gives infinite area and finite

volume for values of p satisfying 12 1.p

30 . The area is given by the integral 1

0;p

dxx

A

1

0 01 : lim ln lim ln ,

bb bp A x b

diverges;

11 1

0 01 : lim 1 lim ,p p

bb bp A x b

diverges;

11 1

0 01 : lim 1 lim 1 0,p p

bb bp A x b

converges; thus, 1p for infinite area.



The volume of the solid of revolution about the -axisx is 2

1

0p

dxx x

V π which converges if 2 1p or 12 ,p

and diverges if 12 .p Thus, xV is infinite whenever the area is infinite ( 1).p The volume of the solid of

revolution about the -axisy is 2/

2

1 1( ) p

dyy y

V R y dyπ π

which converges if 2 1 2p p (see

Exercise 29). In conclusion, the curve py x gives infinite area and finite volume for values of p satisfying

1 2,p as described above.

31. See the generalization proved in 32.

32.

0

22

0 00

0 ( )2

( ) (2 ) ( )2

a

aa a

af x x dx

af x dx x a f x dx x dx

The last integral is 3 3

0

1.

3 2 12

aa a

x

Using integration by parts with 2 , 2 , ( ), ( ),u x a du dx dv f x v f x and the fact that

( ) (0) ,f a f b the second integral is 0 0 0(2 ) ( ) 2 ( ) 2 2 ( ) .

a aax a f x f x dx ab f x dx Thus

3

2

0 0( ) 2 ( ) 2 .

12

a a af x dx f x dx ab

33.

2xe ( ) cos3x

22 xe ( ) 13 sin 3x

24 xe ( ) 19 cos3x

2 2 2 22 1343 9 9 9 9 13sin 3 cos3 3sin 3 2cos3 3sin 3 2cos3

x x x xe e e eI x x I I x x I x x C

34. 3xe ( ) sin 4x

33 xe ( ) 14 cos4x

39 xe ( ) 116 sin 4x

3 3 3 33 9 254 16 16 16 16 25cos 4 sin 4 3sin 4 4cos 4 3sin 4 4cos 4

x x x xe e e eI x x I I x x I x x C



35.

sin 3x ( ) sin x

3cos3x ( ) cos x

9sin 3x ( ) sin x

sin 3 cos 3cos3 sin8sin 3 cos 3cos3 sin 9 8 sin 3 cos 3cos3 sin x x x xI x x x x I I x x x x I C

36. cos5x ( ) sin 4x

5sin5x ( ) 14 cos4x

25cos5x ( ) 116 sin 4x

5 25 9 51 14 16 16 16 4 16cos5 cos 4 sin5 sin 4 cos5 cos 4 sin5 sin 4I x x x x I I x x x x

19 4cos5 cos 4 5sin5 sin 4I x x x x C

37. axe ( ) sin bx

axae ( ) 1 cosb bx

2 axa e ( ) 21 sinb

bx

2 2 2

2 2 2 2 2 2cos sin sin cos sin cosax ax ax axe ae a a b e eb b b b b a b

I bx bx I I a bx b bx I a bx b bx C

38.

axe ( ) cosbx

axae ( ) 1 sinb bx

2 axa e ( ) 21 cosb

bx

2 2 2

2 2 2 2 2 2sin cos cos sin cos sinax ax ax axe ae a a b e eb b b b b a b

I bx bx I I a bx b bx I a bx b bx C

39.

ln ax ( ) 1

1x ( ) x

1ln lnxI x ax x dx x ax x C

40.

ln ax ( ) 2x

1x ( ) 31

3 x

33 3 31 1 1 13 3 3 9ln lnx

xI x ax dx x ax x C



41. 2

2122

221

2 2 21 sin 1(1 ) 1 tan1

dz

zxz

z

dzdxx zz

C C

42.

2

212 222 1

2 21 1

21 sin cos 1 21 2 11

ln |1 | ln tan 1dz

z

z z

z z

dzdx dz xx x zz z z

z C C

43. 2

2122

21

/2 1 1 12 21 sin 1(1 ) 00 0 01

(1 2) 1dz

z

z

z

dzdxx zz

π

44. 2

21221

21

/2 1 1 111 cos 1/ 3/3 1/ 3 1/ 31

3 1dz

z

z

z

dx dzx zz

π

π

45. 2

212 2 221

21

1/2 1 1 12 2 1 1 32 2 12 cos 93 3 3 3 3 32 2 1 30 0 0 0 02

tan tandz

z

z

z

dz dzd zz z z

π πθ πθ

46.

2 21 2

2 22 21 13 3

22 12

2 2121

2 /3 3 3 3 32 1cos 1 1sin cos sin 2 2 42 2 2 2 1/2 1 1 1

lndzz

z z

z zz

zz

z dzd z zzz z z z

dz zπ θ θ

θ θ θπ

3 ln31 1 1 1 12 4 4 4 2 4 2ln 3 0 (ln 3 2) ln 3 1

47.

22 21

2 222 122 21 1

tan 1 21 22 2 1 1sin cos 2 1 2 22 1 ( 1) 2 tan 1 2

ln lndz tz

tz z

z z

zdt dz dzt t zz z z

C C

48.

2 21 2 2 22 21 1

2 22 2 2 2 2 2 2 22 2 2121

2 1 2 1 1cos1 cos 11 1 1 1 11 1 11

dzz

z z

z

z

z dz z dz z dzt dt dz dzt zz z z z z z zz z z

2 211

212 2 tan cotdz dz t

zz zz C t C

49.

22 21

221221

1 tan2 2cos (1 )(1 ) 1 11 1 tan

sec ln |1 | ln |1 | lndz

z

z

z

dzd dz dz dzz z z zz

d z z C Cθ

θθθθ θ

50.

221

221

sin 2csc ln | | ln tandz

z

z

z

d dzzd z C Cθ θ

θθ θ



51. (a) 1

00 0(1) lim lim lim ( 1) 0 1 1b

b bt t t

eb b be dt e dt eΓ

(b) 1, ; , ;x x t tu t du xt dt dv e dt v e x fixed positive real

1 0

00 0( 1) lim lim 0 ( ) ( )

x

b

bx t x t x t xbeb b

x t e dt t e x t e dt e x x x xΓ Γ Γ

(c) ( 1) ( ) ! :n n n nΓ Γ

0 : (0 1) (1) 0!;n Γ Γ

: Assume ( 1) !n k k kΓ for some 0;k

1: ( 1 1) ( 1) ( 1)n k k k kΓ Γ

from part (b)

( 1) !k k induction hypothesis ( 1)!k definition of factorial

Thus, ( 1) ( ) !n n n nΓ Γ for every positive integer n.

52. (a) 2( )xx

e xx πΓ and 2( ) ! ! 2n nn n

e n en n n n n nπ πΓ

(b)

n 2nn

e nπ calculator

10 3598695.619 3628800

20 182.4227868 10 182.432902 10

30 322.6451710 10 322.652528 10

40 478.1421726 10 478.1591528 10

50 643.0363446 10 643.0414093 10

60 818.3094383 10 818.3209871 10 (c)

n 2nn

e nπ 1/122n nn

e n eπ calculator

10 3598695.619 3628810.051 3628800

chapter 8 techniques of integrationmrsbustamanteala.weebly.com/uploads/8/4/5/2/... · xx dx xx...

Documents