5.1 Antiderivatives and Indefinite Integration Apply ... · Ch 05 Homework Complete Solutions V2:...

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Calculus: Early Transcendental Functions, 4e Larson

S. Stirling 2016-2017 Page 1 of 41

5.1 Antiderivatives and Indefinite Integration

Apply & Practice 5.1 Set 1:

P 291-294

#6, 8 solve differential, general

#15-42 (mult of 3), 26, 38 indefinite integral

#47 graph antiderivative family

#50-52 graphical

Simplifying is the key to most of these!

39. Need the identity: 2 2tan 1 sec to get an integrand that we can integrate.

No Quotient Property, so you have to distribute first.




No Chain Rule, so must expand.

Linear, slope = 1,

goes through (0, 0).

The antiderivatives

are a family of

functions shifted up

or down C.

Quadratric, flipped

and shifted up 1.

Volcano so 21 x

Use 0f at –1 and 1. As well

as when 0f and 0f .

Answers may vary here by the amounts that you

choose for C.




5.1 Antiderivatives and Indefinite Integration

Page 292 #57-59 Intro to slope fields. See NOTES! These slope fields are in the notes.


P 291-294

#54, 55 graphical

#57-59 Intro. slope fields. Read intro!!

Graphs are ON the NOTES page 6.

#63-72 (mult of 3) solve differential eqns.

#75, 76 writing

#81, 82 (use result from 81), 85, 88, 90,

93 applica.

#95-100 T or F?

#102




To “solve the differential” means to integrate.

Notationally, ( )dy

f xdx

( ) dy f x dx and

( )y F x C




(1) 4f , (1) 3f

Read it from the graph.

(0) 4f and (1) 4f so on [0, 2], f increases

and at 1, the rate of change is 4. After x = 1, f goes

above the x-axis so it can’t be negative when x = 2.

0f on [4, 5].

f changes sign from positive to negative.

f decreasing. Change in concavity.

f decreasing at the quickest rate at x =3.

Figure 75.

Velocity

mi

hr

Time

+f +f f

( )f x

Time

(hr)

Position

(mi)




Initial velocity = 0 “dropped”

“Hits surface” (20) 0s .




95 – 98. All True

v0 = 0

Starting point

No product property

Is a family of functions.




Use to define the function.




Apply & Practice 5.2 Set 1: P 303

#1-6 set up and sum by hand

#7-13 odds (work backwards)

2(15) 5

3 8 15 24

170 85 34 17 10 316 158

170 170 170 170 170 170 85

Note: 170 2 5 17

Note: 60 3 2 2 5 20 15 12 47

60 60 60 60




Apply & Practice 5.2 Set 2: P 304 Draw the graphs!

#23, 26

#27, 30

Note: 23. f is increasing so the upper sum S is from RRAM (an overestimate) and

the lower sum s is from LRAM (an underestimate).

Note: 26. f is decreasing so the upper sum S is from LRAM (an overestimate) and

the lower sum s is from RRAM (an underestimate).

Note: 27. f is increasing so the upper sum S is from RRAM (an overestimate) and

the lower sum s is from LRAM (an underestimate).

Note: 30. f is decreasing so the upper sum S is from LRAM (an overestimate) and

the lower sum s is from RRAM (an underestimate).




47. 1

1 3 1 22

A 49. 2 2y x , [0, 1]

1 0 1x

n n

Using right endpoints…

x-coord: 1

0a i x in

1

1 1lim 0

n

ni

Area f in n

2

1

1lim 2

n

ni

iArea

n n

2

1

1lim 2

n

ni

i

n n

To find limn

use n = 500.

1/500*sum(seq((I/500)^2+2, I, 1, 500))

≈2.334 (Actual answer is 7/3 ≈2.3333)

47. 2 3y x , [0, 1]

1 0 1x

n n

Using right endpoints…

x-coord: 1

0a i x in

1

1 1lim 0

n

ni

Area f in n

1

1lim 2 3

n

ni

iArea

n n

1

1 2lim 3

n

ni

i

n n

To find limn

use n = 500.

1/500*sum(seq(–2I/500+3, I, 1, 500))

≈1.998 (Actual answer is 2)

49.

51. 216y x , [1, 3]

3 1 2x

n n

Using right endpoints, RRAM…

x-coord: 2

1a i x in

2

1

2 2lim 16 1

n

ni

iArea

n n

2

1

2 2lim 16 1

n

ni

i

n n

To find limn

use n = 500. 2/500*sum(seq(16–(1+2I/500)^2, I, 1, 500)) ≈23.317

(Actual answer is 70/3 ≈23.3333)

Apply & Practice 5.2 Set 3: P 304-305

#47, 49, 51 Do these like Ex 5

#57, 59 Do these like Ex 5, but look at Ex 7 too.

#77 VIP! Show all work!




On the next page is #77…

57. ( ) 3f y y , 0 ≤ y ≤ 2

2 0 2y

n n

Using upper endpoints (RRAM)…

1

2 2lim 0

n

ni

Area f in n

1

2 2lim 3

n

ni

iArea

n n

21

12lim

n

ni

in

To find limn

use n = 500.

12/500^2*sum(seq(I, I, 1, 500))

≈ 6.012 (Actual answer is 6.)

59. 2( )f y y , 0 ≤ y ≤ 3

3 0 3y

n n

Using upper endpoints (RRAM)…

1

3 3lim 0

n

ni

Area f in n

2

1

3 3lim

n

ni

iArea

n n

2

31

27lim

n

ni

in

To find limn

use n = 500.

27/500^3*sum(seq(I^2, I, 1, 500))

≈ 9.0270 (Actual answer is 9.)

57. 12 6 6

2A


#77 VIP! Show all work!

(a) – (c) Do by hand, calculator only for crunching the sums.

(d) You should practice setting up the sums based on your work in (a) – (c), then use

your answers to help you with (e)

(e) The table is easier to create if you generate the sums in Y= and use X for n:

Y1 = 8X/(X+1)

Y2 = 4/X*sum(seq(Y1((I – 1)*(4/X)), I, 1, X)) this is LRAM.

Then use [Tblset]: Indpnt, Ask and type 4, 8, …. (f) Think about how lower and upper sums relate to the exact areas.




(d) LRAM = lower sum

( 1)ix a i x

40 ( 1)ix i

n

(d) RRAM = upper sum

ix a i x

40ix i

n

(d) MRAM = midpt. sum

1

2ix a i x

1 40

2ix i

n

4 01

4x

4 0 4x

n n

(e) Use the calculator to generate the sums in [Y=].

Use I for the index, then X can be used for n.

Y1=8X/(X+1)

Y2 = 4/X*sum(seq(Y1((I – 1)*(4/X)), I, 1, X))

Y3 = 4/X*sum(seq(Y1(I*4/X), I, 1, X))

Y4 = 4/X*sum(seq(Y1((I – .5)*(4/X)), I, 1, X))

Then use [Tblset]: Indpnt, Ask and type 4, 8, ….




Apply & Practice

Set 5.3 Set 1: P 314-317

#9-21 odds

#24, 26, 27, 30, 32

#46, 47-48

on interval [1, 5] is

5

1

31 dx

x




6 6

4 4

( ) 2 f x dx dx

Use what you know!

Three subintervals.




Apply & Practice Set 5.3 Set 2: #33-40, 42, 44 Properties of integrals

#65-70 T or F?




65. True

67. True

68. True




Think of it as a piecewise,

vertex (4, 3):

3 4 , 43 4

3 4 , 4

x xx

x x


P 327-330

#6-36 (mult. of 3) Evaluate

definite integrals

#39, 42, 44, 46, 49, 50 Areas of

regions.







Note F is the antiderivative

of the integral defined

function that goes through

(0, 0)


#79, 82, 84 Evaluate definite integrals, function

before evaluating.

#87, 90, 92, 94 Do to confirm FTC Antideriv

#95, 96, 100 FTC Antiderivative part

#85, 86 evaluate integral functions Ex 6. AP loves

these! Remember you are sweeping out area.

#65 writing

#2, 4 graphical reasoning

So FTC

Evaluation

Part.

These problems should help you confirm

the FTC Antiderivative part.

Note: by FTC Antideriv, 0

2 2

xd

t dt xdx

Note: by FTC Antideriv,

4

xd

t dt xdx

Note: by FTC Antideriv, 3

sec tan sec tan

xd

t t dt x xdx

See how easy it is to use the FTC Antiderivative

part when lower limit is a constant.

Note:

1

1 1

xd

dtdx t x




Is an area accumulation function:

0

( ) ( )

x

g x f t dt and by FTC ( ) ( )g x f x so the graph shows the derivative of g!!

No area accumulated

Use general shapes to

approximate the areas

under the curve, f x .

Justification for parts (b) & (c)

0

( ) ( ) ( )

xd

g x f t dt f xdx

by FTC Antider

( )g x is positive on (0, 4), so g is

increasing and ( )g x negative on (4, 8),

so g is decreasing.

(c) g has a max at x = 4 b/c

( )g x changes sign from pos. to neg.

No area accumulated

Use general shapes to

approximate the areas.

Areas accumulate under f(t).

Justification for parts (b) & (c)

0

( ) ( ) ( )

xd

g x f t dt f xdx

by FTC Antider

( )g x is negative on (0, 4), so g is

decreasing, and positive on (4, 8), so g is

increasing.

(c) g has a min. at x = 4 b/c

( )g x changes sign from negative to

positive.

f x g x by FTC

g x

g x f x

by FTC Antideriv.

g x





#101, 104, 106 FTC antiderivative part

Use FnInt or [CALC] f x .

Using FTC

Antiderivative

Using FTC

Evaluation

Using FTC Evaluation

Using FTC Antiderivative w/ Chain Rule

Using FTC Antiderivative w/ Chain Rule. ONLY

Note you must use this because you can’t

analytically find the antiderivative of the integrand.

It’s a composite.





P 327-330

#63, 64 velocity graphs

#107, 108, 109, 111

Using FTC Evaluation

g x f x by FTC

From the graph each square is 15 ft/sec by 2 sec. or

30 ft. You are finding the net change in distance.

Going from velocity to position, so integrate.

From the graph each square is 10 ft/sec by 0.5 sec.

or 5 ft. Again, you are finding the net change in

distance.

Is approx. ½ of a 1 x 2 rectangle. POI of g

POI of g

Max. of g

No area

Area above and

below the x-axis

are =.

By End behavior (or relative

magnitude). 44 4

Use end behavior, think 44 8

Is an area accumulation function.




Position x t Velocity x t

To find the total distance traveled, you would need to accumulate the area under the

velocity function. (mi/hr * hr = mi)

But the integral under the x-axis is negative (left direction) that’s why they put the

absolute value in. So you will need to break up the integral into three sections…

2 4 2 1

4 2 2 units

time

vel

oci

ty





#51, 54, 55 MVT for integrals

(need calculator) Isolate c!

#57, 58, 61 AV of a function

#66, 67-72 writing

#74 application

MVT for Integrals: ( )

b

a

f c b a f x dx by FTC

51.

2

0

23

2 2

0

322

2

1 4

2 3

1 42 2 0

2 3

8 22

3

x x dx

x x

MVT for integrals:

2

0

( )

2 2 0 2

8 22 4 2

3

b

a

f c b a f x dx

c c x x dx

c c

Use the intersection capabilities of your

calculator to solve.

0.4380c or 1.7908c

1

2( ) 3

3

3 3cos

1 2

3 3cos

2

0.5971

f c

c

c

c

Can use the intersection

feature to solve:

1 cosY x

2 3 3 2Y

[2nd] [CALC] 5: intersect

If you use

1cos 3 3 2c

0.5971c

Then need Quad IV

0.5971c c is defined in QI & QIV




Avg. Value: 1

b

a

AV f x dxb a

57. 2( ) 4f x x on 2,2

Avg. Value

58. 2

2

4 1( )

xf x

x

on 1,3

61. ( ) sinf x x on 0,

Best to use

trapezoids and

rectangles.

(b) Avg. Value

Area = 8

Area = 20 Just add on the

rectangle area.

The integral.




MAKE SURE THAT YOU CAN USE

THE INTEGRAL NOTATION TO

DESCRIBE YOUR SOLUTIONS!!

Below x-axis

Rectangle

Velocity

From 68.





#9 – 33 (mult. of 3); 48, 51, 54, 57, 59, 35, 38

9. 1

2 29 2 x x dx

2

312 2

32 2

9 , 2

2

3

2 9

3

u x du x dx

u du u C

x C

Opt. Check:

32 2

12 2

12 2

29

3

2 39 2 0

3 2

9 2

dx C

dx

x x

x x

12. 4

2 3 5 x x dx

3 2

4 5

53

5, 3

1 1 1

3 3 5

15

15

u x du x dx

u du u C

x C

Opt. Check:

53

43 2

42 3

15

15

15 5 3 0

15

5

dx C

dx

x x

x x

15. 2 2 t t dt

2

312 2

32 2

2, 2

1 1 2

2 2 3

12

3

u t du t dt

u du u C

t C

Opt. Check:

32 2

12 2

12 22

12

3

1 32 2 0

3 2

2 2

dt C

dx

t t

t t t t

18. 2 3 5 u u du

3 2

312 2

33 2

5, 3

1 1 2

3 3 3

25

9

x u dx x dx

x dx x C

u C

Opt. Check:

33 2

13 22

12 3 2 32

25

9

2 35 3 0

9 2

5 5

du C

dx

u u

u u u u

21.

2

23

1

xdx

x

3 2

2

22

2 2

1

3

1 , 3 , 3

1

3 3

1 1( 1)

3 3 1

duu x du x dx dx

x

x duu du

u x

u C Cx

Opt. Check:

3

23 2

2

23

1

3 1

11 1 3 0

3

1

dC

dx x

x x

x

x

24.

3

4

1

xdx

x

4 3

3

31

23

114 22

1 , 4 , 4

1

4 4

1 2 11

4 1 2

duu x du x dx dx

x

x duu du

xu

u C x C

Opt. Check:

14 2

14 32

3

4

11

2

1 11 4 0

2 2

1

dx C

dx

x x

x

x

27. 1

2

dxx

1 12 2

12

2 , 2

1 1 2

2 2 1

2

u x du dx

u du u C

x C

Opt. Check:

12

12

2

12 2 0

2

1

2

dx C

dx

x

x




30.

22

t tdt

t

31

2 2

3 52 2

3 52 2

2

2 22

3 5

2 4

3 5

t t dt

t t C

t t C

Opt. Check:

3 52 2

312 2

312 2

2 4

3 5

3 2 5 40

2 3 2 5

2

dt t C

dx

t t

t t

33.

31

2 2

3 52 2

3 52 2

9

9

2 29

3 5

26

5

y y dy

y y dy

y y C

y y C

Opt. Check:

3 52 2

312 2

312 2

26

5

3 2 56 0

2 5 2

9

dy y C

dx

y y

y y

48. 3 44 sin x x dx

4 3

4

, 4

sin cos

cos

u x du x dx

u du u C

x C

51. 2

1 1cos d

2

2

2

2

1 1, ,

1cos

cos sin

1sin

u du d du d

u du

u du u C

C

54. 3 23 xe x dx

3

3 2, 3

u u

x

u x du x dx

e du e C

e C

57. sin 2 cos2 x x dx

2

2

sin 2 , 2cos 2

1 1 1

2 2 2

1sin 2

4

u x du x dx

u du u C

x C

OR

sin 2 cos2 x x dx

2

1

2

1

cos 2 , 2sin 2

1 1 1

2 2 2

1cos 2

4

u x du x dx

u du u C

x C

59.

4 2tan sec x x dx

2

4 5

5

tan , sec

1

5

1tan

5

u x du x dx

u du u C

x C




2

1

2

1

4

14

2

2

x dx

x C

x C

2

1 1 12 2 2

12 2

8 1, 2 8 ,

2 4 , 4 2

1 2

2 2

8 1

u x x du x dx

dudu x dx x dx

duu u C u C

x x C

2

2

1 12 2

2

2

2

22

16

16 , 2

2 2 2

4 16

xdx

x

u x du x dx

u du u C

x C

2 22 4 16x x C





P 340-344

#39, 41, 43, 46 DO ON YOUR NOTES

#60 – 78 (mult. of 3)

#81, 82, 85, 90, 92

#136

39. 24y x x dx .

Let 24u x ;

2

dudx

x

1 12 2

1

2 2

duy x u u du

x

32

1

3y u C or

32 21

43

y x C

Using (2, 2): 3

2 212 4 (2)

3C

2C 3

2 214 2

3y x

41. 2cos y x x dx .

Let 2u x ; 2 du x dx ,

2

dux dx

1cos cos

2 2

duy u u du

21 1sin sin

2 2y u C x C

Using (0, 1): 21

1 sin 02

C

1C 21

sin 12

y x

43. 22 x

y e dx

.

Let 1

2u x ;

1

2du dx , 2 du dx

2 2 4 u uy e du e du

24 4x

uy e C e C

Using (0, 1): 01 4e C 5C

24 5x

y e

46. sin cos xy e x dx .

Let sinu x ; cos du x dx

uy e du

sinu xy e C e C

Using , 2 : sin2 e C 1C

sin 1xy e




60. 2tan sec x x dx

2

312 2

32

tan , sec

2

3

2tan

3

u x du x dx

u du u C

x C

63. 2cot x dx

Need identity: 2 21 cot cscx x

2csc 1 x dx

2csc x dx + 1 dx

1 2cot

cot

x C x C

x x C

66. 1 3 x xe e dx

2

2

1 3 , 3 , 3

1

3 3

1 1 11 3

3 2 6

x x

x

x

x

x

duu e du e dx dx

e

due u u du

e

u C e C

69. 2

5

x

x

edx

e

U-sub won’t work…yet!

25 x xe e dx

1 2

2

2

2 , 2 , 2

, 1

5 • 1 2

5

2

5

2

u v

u v

x x

duu x du dx dx

v x dv dx

due e dv

e C e C

e e C

72. tan2 2sec 2 xe x dx

2

2

2

2

tan 2

tan 2 , 2sec 2

2sec 2

sec 2 2sec 2

1

2 2

1

2

u

u u

x

u x du x dx

dudx

x

due x

x

due e du

e C

75. 23 x

dx

1, , 2 2 2

3 2 2 3

1 22 3 3

ln 3 ln 3

u u

u u

xu du dx du dx

du du

C C

78. 2

33 7

xx dx

2

2

3

3 , 2 3 , 2 3

13 7 7

2 3 2

1 1 17 7

2 ln 7 2ln 7

u u

xu

duu x du x dx dx

x

dux du

x

C C




Note:

2

2

so 1

2 1

3

u x

u x

x

u

u

Note:

2 1, 4

2 4 1

x x u

u





#99 – 108 (mult. of 3); 110

#117, 122

#137, 140, 146 Use calculator and even & odd properties.

99. 4

0

1

2 1dx

x

Use indefinite integral till you get it set up :

1

2

2 1, 3 , 2

1 1

2 2

duu x du dx dx

duu du

u

New limits:

2 0 1 1

2 4 1 9

u

u

99

1122

11

1 12

2 2

9 1 2

u du u

102. 2

1

1

xe dx

Start w/ indefinite integral:

1 , 1

u

u x du dx

e du

New limits:

1 1 0

1 2 1

u

u

1-1

00

1 0

11

u ue du e

e ee

105.

9

2

1

1

1

dx

x x


1 12 2

12

12

2

1 22

11 , ,

2

2

22

u x du x dx

x du dx

x duu du

x u

New limits:

1 1 2

1 9 4

u

u

44

2 1

22

2 2

1 1 1 12 2

4 2 4 2

u du u

108. 5

1

2 1

xdx

x


1 12 2

1 12 2

2 1, 2 , 2

1

2

1

2 2 2

1

4

duu x du dx dx

ux

du u dux u u

u u du

New limits:

2 1 1 1

2 5 1 9

u

u

99

31 1 12 2 2 2

11

3 31 12 2 2 2

1 1 22

4 4 3

1 2 29 2 9 1 2 1

4 3 3

1 218 6 2

4 3

1 22 1 2 11 1 33 1 32

4 1 4 3 2 6 6 6 6

16

3

u u du u u




110.

2

3

2

2

3

2 2

2 2 2

cos

1sin

2

1 1sin sin

2 2 2 2 3 3

3 5 2 31

8 18 2 72 2

x x dx

x x




Apply & Practice 5.6: P 350-351

On ALL trapezoidal rule only

#2, 5, 6, 10, 40

#49(a) only




Apply & Practice 5.7: P 358-360

#3 – 24 (mult of 3)

#25, 27

3 2, 3

1 1 1

3 3

u x du dx

dudu

u u

2

3 2

3 2

3 3 5

3

5

x

x x x

x x

2 5

3x

x

Integrand is easily split up by using

the distributive property.

Besides, if you go this way… 2 4

2

2

u x

du x dx

dudx

x

then 2

u du

x x UGH!

3 2 2

2

3 4, 3 6

2 3 6

2 1 1 3 2 3

u x x du x x dx

x x du

u x x

x x dudu

u x x u

3 21ln 3 4

3x x C

Numerator is degree

higher than the

denominator, so…




2

2

2

1

3, 2

3 2

1ln 3

2

u x du x dx

x x dudx

x u x

x C

18.

3 2

2

2

22

3 4 9

3

3 3

13 ln 3

2 2

x x xdx

x

xx dx

x

xx x C

3 21ln 3 4

3x x C

Even though numerator is degree higher than the denominator… 3 2 23 4 9, 3 6 4 u x x x du x x dx and there is no match in the integrand, try division!

23

3

xx

x

2 3 2

3

2

2

3

3 3 4 9

3

3 9

3 9

x

x x x x

x x

x x

x

x

24. THIS IS A CHALLENGE PROBLEM

3

2

3

2

3

2

3 3

3

2

2

1

2 1 1

1

1 1

1

1 1

1 1

1 1

1

1ln 1

2 1

x xdx

x

x xdx

x

xdx

x

xdx dx

x x

dx x dxx

x Cx

If you expand the numerator, you may see the opportunity to rewrite

as a perfect square. Then you can separate the rational expressions.

Numerator: 2 2x x complete the square

2

2

2 1 1

1 1

x x

x




25.

12

12

1

1

1

1 2

21

1

1

11

ln

1 2 ln 1 2

dxx

x du

u

udu

u

duu

u u C

x x C

Or if needed:

2 ln 1 2x x C

Where 1 1C C

12

12

12

1 2

12 2

2

2

u x

du x dx

x du dx

To get the integrand all

in terms of u:

Since 1 2u x

2 1x u

27.

1 12 2

2

2

1

2

1

3

2 2=

1

6 92

92 6

12 2 6 2 9ln

2

3 12 3 18ln 3

xdx

x

x x du xdu

u u

u udu

u

u duu

u u u C

x x x C

Or (if needed)

2

1

1

3 12 3 9ln 3

6 9 12 36 18ln 3

6 18ln 3

x x x C

x x x x C

x x x C

Where 1 27C C

12

12

12

3

1

2

2

u x

du x dx

x du dx

To get the integrand all in

terms of u:

2

2

3

3

3

6 9

u x

x u

x u

x u u




Remember!

1

2 2sin

du uC

aa u

1

2 2

1 tan

du uC

a u a a

1

2 2

1sec

uduC

a au u a

Apply & Practice 5.8: P 366

#1 – 17 (odd)

#21, 23

3a , u x

4a , u x

5.

2

2 2

2 2

1 1

1

4 1

2 1

2 2 1

1sec sec 2

1 1

dxx x

dxx x

du

u u a

uC x C

1

2

2

a

u x

du dx

Now: 2 1

2

1 1

2 2

u x

du x dx

x dudu

u x u

7.

3

2

2

2

2 2

1

1

1

1 1ln 1

2 2

xdx

x

xx dx

x

xx dx dx

x

x x C

You would need long division to get u-substitution to work.

2 1

xx

x

2 3

3

1

x

x x

x x

x

9.

2

2 2

1 1

1

1 1

sin sin 11

dxx

du

a u

uC x C

1

1

a

u x

du dx

11.

4

22

2 2

1

1 2

1

1

1

2

1sin

2 1

1sin

2

tdt

t

tdt

t

du

a u

uC

t C

2

1

2

2

a

u t

du t dt

dut dt




13.

2

4

2

2 22 2

1

2 2

21

21

4

22

1 1 1 tan

2 2

1 1tan

2 2 2

1tan

4 2

x

x

x

xx

x

x

edx

e

e du

ee

du uC

a u a a

eC

eC

2

2

2

2

2

2

x

x

x

a

u e

du e dx

dudx

e

15.

12

21 12

2 2

1

2 2

1

1

1

1 2

11

2 2sin

2sin

dxx x

x du

x x

du uC

aa u

x C

12

12

12

1

1

2

2

a

u x

du x dx

x du dx

17.

2

2 2

3

1

3

1 1

xdx

x

xdx dx

x x

2 2

2 1

1 2

2 1

1 1 3

2

1 1ln 1 3 tan

2 1

1ln 1 3tan

2

dudx

u u a

x C x C

x x C

2 1

2

2

u x

du x dx

dux dx

1a

u x

du dx

21.

16

20

16

2 20

12

2 20

11 2

0

1 1

1

1 9

1

1 3

1

3

1sin

3

1 1sin sin 023

10

3 6 18

dxx

dxx

du

a u

u

1

3

3

3

a

u x

du dx

dudx

Change limits:

3

1 136 2

3 0 0

u x

u

u

23.

32

2

0

32

2 2

0

3

2 2

0

31

0

1 1

1

1 4

1

1 2

1

2

1tan

2

1tan 3 tan 0

2

10

2 3 6

dxx

dxx

du

a u

u

1

2

2

2

a

u x

du dx

dudx

Change limits:

2

32 32

2 0 0

u x

u

u

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