Solution manual calculus 8th edition varberg, purcell, rigdon ch07

The proof fails to consider the constants when integrating .1t The symbol ∫ 1 t dtis a family of functions, all of who whom have derivative .1 t We know that any two of these function

CHAPTER 7 Techniques of Integration

1 1/ 2 0

dx u e

=+

dx

u x

=+

1ln

[tan ]2tan2

2 sin u du

= ∫

1 1/ 2

2

1– cos(ln 4 )

1tan(ln )

22 u=x2, du = 2x dx

12

−

2 –1

1

tan

x C

1 –

x x

dx

u e

41

0

2–

13sin(t 2) C

1

tan 2 2

12

u C

− ⎛ ⎞

⎝ ⎠2 1

sin

y C

3

1cosh

u C

u C

− ⎛ ⎞

3 1

1sin

t e C

−

du u

=

+

∫

1 1 0

e e

d x x

=∫

–1 – 3sin

5

x C

sec

t C

−

1

du u

= −

−

∫

11sin

b a

56 sec 1 1 sin

x x

x C x

1 cos

du u

π π

2 2

π π

−

2 2

0 0

π π

13 u = arctan x dv = dx

211

=1

t

=+

212

v= t

2 2

z C z

45 u=sin(ln )x dv = dx

1cos(ln )

2 sin(ln )∫ x dx=xsin(ln )x −xcos(ln )x +C

sin(ln ) [sin(ln ) cos(ln )]

2 cos(ln )∫ x dx=x[cos(ln ) sin(ln )]x + x +C

cos(ln ) [cos(ln ) sin(ln )]

1 1 1 1sin sin(3 ) – sin cos(3 ) cos sin(3 ) sin sin(3 )

x v

αα

+

=+ , α ≠ –1

αα

+

=+ , α ≠ –1 1

cosα− xsinx (α 1) cosα− x(1 cos x dx)

= + − ∫ − =cosα –1xsinx+( – 1) cosα ∫ α –2x dx– ( – 1) cosα ∫ αx dx

63 4cos 3 1 4sin 3 –4 3sin 3

65 First make a sketch

From the sketch, the area is given by

2 1

1(ln )2

e

x dx

∫

2(ln )

74 a We note that the nth arch extends from x=2 (π n−1)to x=π(2n−1), so the area of the nth arch is

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

π π

n n

n

G

x dx n

77 The proof fails to consider the constants when integrating 1

t

The symbol ∫ ( )1 t dtis a family of functions, all of who whom have derivative 1

t We know that any two of these

functions will differ by a constant, so it is perfectly correct (notationally) to write ∫ ( )1t dt=∫ ( )1t dt+1

78 d[e5x(C1cos 7x C2sin 7 )x C3] 5e5x(C1cos 7x C2sin 7 )x e5x(–7C1sin 7x 7C2cos 7 )x

+

=+

+ +

Assume we know the formula for n−1, and we want to show it for n

+ + +

1

( )

j n

=

0

( )( 1)

j n

j j

d P x e

3. ∫sin3x dx=∫sin (1 cosx − 2x dx)

2

sinx dx sin cosx x dx

31

8. ∫(sin 2 ) cos 23 t tdt=∫(1 – cos 2 )(cos 2 )2 t t1/ 2sin 2t dt 1 1/ 2 5 / 2

– [(cos 2 ) – (cos 2 ) ](–2 sin 2 )

1(sin ) (cos ) sin

= Making the substitution and changing the limits as necessary, we get

L L

π π

If m > N, there is no term where n = m, while if m ≤ N, then n = m occurs When n = m

t dt x

36 Since (k−sin )x2 =(sinx−k) ,2 the volume of S is 2 2 2

0ππ −(k sin )x =π 0π(k −2 sink x+sin x dx)

u x

t dt t

cos

16 –

dt t x

π π −+ π

⎛ ⎞

–1 – 3sin

5

x C

⎛ ⎞

–1 – 2sin

2

x C

2 0

/ 4

2 tan (1/ 2)

–1

/ 4 tan (1/ 2)

12

2

x dx x

=+

3 2 23

sin

t dt t

the area of the region bounded by x = 0, y = 0,

/ 2 2 sin ( / )

33 a The coordinate of C is (0, –a) The lower arc

of the lune lies on the circle given by the equation x2 + (y+a) 2 = 2a2 or

2

a a

b Without using calculus, consider the

following labels on the figure

Area of the lune = Area of the semicircle of

radius a at O + Area (ΔABC) – Area of the sector ABC

Note that since BC has length 2 , a the

measure of angle OCB is ,

34 Using reasoning similar to Problem 33 b, the area is

++

17

3

3 – 2– 1

x dx

33 x = sin t, dx = cos t dt

sin – ln sin 3 – tan (sin – 2) – ln sin – 4 sin 5

=+

C x

+

2 0

e

y t

e

=+

e

y t

e

=+

b

3 3

1

e y

e

+

12( )

C

t t

e

y t

e

=+

t

e

y t

t e

=+

b

3.6 3.6

12

5

e y

8000( )

C

t t

t

e

y t

t e

=+

b

7.2 7.2

8000

7

e y

4000( )

C

t t

e

y t

e

=+

t

e

y t

t e

=+

12

4000e

y t

e

=+

( )

L y y

kLt kLt

−

1

kLt kLt

t = 0 Further, since

0 0

is monotonic as t→ ∞ ,we conclude

that the population would decrease toward a limiting value of L

47 If y0<L, then y′(0)=ky0(L−y0)>0 and the population is increasing initially

48 The graph will be concave up for values of t that

Thus if y0<L, then y t( )<L for all positive t

(see note at the end of problem 45 solution) and

so the graph will be concave up as long as

=+(1 7)

50 a dy ky(10 – )y

dt =(10 – )

t

y e

y=(1ln8) (1 ln8)

=+(1 8)

The population will be 9 billion in 2108

51 a Separating variables, we obtain

( )1

a b kt

a b kt a b

( )

4 2

kt kt

2 / 20 ln 2 / 3 ln(2 / 3) 2

3

t t

kt t

= ⎜ ⎟⎝ ⎠

( ) ( )

/ 20 2 3 / 20 2 3

4 1( )2

t t

3 2 3 3 2 3

4 1

38

232

d If a = b, the differential equation is, after

separating variables 2

+

1( )

kt C

= −+

Since x = 0 when t = 0, C = 1,

a so

1

1( )

= −

+1

1

1

a akt

−

=+

2 1 2

Note: Throughout this section, the notation Fxxx

refers to integration formula number xxx in the

back of the book

9 2

1 1

29

2 4

4

8 8

1 cos 4cos 2

2

1

1 2 cos 4 cos 44

sinsin cos

8 Partial fractions

2

(1 )(1 )1

0

11

ln 1 ln 12

8 6

2

22

2

cos xsinx dx 0

π π

2

2 2

3 2

3 2

2

2401(2 1)(3 4 )10

F a b

3 2

1(2 cos 1)(3 4 cos )20

x

C x

b Substitution, Formula 18

18 2

2 5

2 2

510

3 2

3

2 2

3 3

55 4

3

2

6 6

u du u

3

2

33

3 5

48 3

21 a Complete the square; substitution;

Formula 45

45 2 1

29 4

2

2 2

2 2

2 2

ln5

a b

x

C x

1 sin

cos sin

1(2 cos 2) 2 cos 16

1(1 cos ) 2 cos 13

F a

4 sin

1

3

1(6 cos 1)(4 cos 1)60

29. Substitution; Formula 99, Formula 98

99 2 cos

1 sin

95 2 3

F n a

dx x

cos

2 1.14159

1 sin

x dx x

π

π

= − ≈+

3

x dx

t ≈+

∫

38. Using a CAS, we obtain:

3 / 4

4 tan

x dx x

π π

tan 0.54632

F

c c

0 0

1 2

45. There is no antiderivative that can be expressed

in terms of elementary functions; an approximation for the integral, as well as a process such as Newton’s Method, must be used

Several approaches are possible c≈0.59601

46 Integration by parts; partial fractions; Formula 17

2 3

3 ln( 1)

3

3

3 1 ,

3 2

4

2 3

2 3

3 0 3

Therefore 3

0cln(x +1)dx= ⇒ ≈1 c 1.6615

∫

47 There is no antiderivative that can be expressed

in terms of elementary functions; an approximation for the integral, as well as a process such as Newton’s Method, must be used Several approaches are possible c≈0.16668

48 There is no antiderivative that can be expressed

in terms of elementary functions; an approximation for the integral, as well as a process such as Newton’s Method, must be used Several approaches are possible c≈0.2509

49 There is no antiderivative that can be expressed

in terms of elementary functions; an approximation for the integral, as well as a process such as Newton’s Method, must be used Several approaches are possible c≈9.2365

50 There is no antiderivative that can be expressed

in terms of elementary functions; an approximation for the integral, as well as a process such as Newton’s Method, must be used Several approaches are possible c≈1.96

2 0

8

1

1 2 0

c c

3

3 0

0 0

f x g x dx e dx e

318

11

Newton’s Method to h we discover that the zeros

of h are -1, 0, and 1 Since we are dealing with

positive values, we conclude that x

c =1 or x = c.)

a

,

2 0

3 0

3 0

0

3 2

2( )

sin( )

ππ

ππ

60. From problem 58 we know that ( )S x is concave

up on (0,1) and concave down on (1, 3) so the first point of inflection occurs at x= Now 1

be integrated directly, we must use some approximation method Methods may vary but the result will be (1)S ≈0.43826 Thus the first point of inflection is (1, 0.43826)

7. True: This integral is most easily solved

with a partial fraction decomposition

8. False: This improper fraction should be

reduced first, then a partial fraction decomposition can be used

9. True: Because both exponents are even

positive integers, half-angle formulas are used

10 False: Use the substitution

14 True: The trigonometric substitution

x = 3sin t will eliminate the radical

x x

b a

22 False: Polynomials can be factored into

products of linear and quadratic polynomials with real coefficients

23 True: Polynomials with the same values for

all x will have identical coefficients

for like degree terms

24. True: Let u=2x; then du=2dxand

erf x is an increasing function

27. True: by the First Fundamental Theorem of

Sample Test Problems 1.

2

cos 2cot (2 )

1 sin 2sin 2 θ θdθ

−

=∫ =∫(csc 22 θ−1)dθ

1cot 2

sec

2 sec

2 3

t dy

dt t y

=+

3

1ln3

Note that

2 3tant= y , so

3 2 3

y t

=+

29 ∫tan3 / 2xsec4x dx=∫tan3 / 2x(1 tan+ 2x) sec2x dx =∫tan3 / 2xsec2x dx+∫tan7 / 2xsec2x dx

129

y y

t t

x x

1

3

t C a

– tan

y C

–( – 1)

–– 1 ( – 1)

– 1

t t

cos1sin

x dx x

π π

2 / 6

sin

dx x

π π

54 a First substitute u=2 ,x du=2dxto obtain

1 3

1 2

1

0.51

c

dx x

=+

1.0000 0.3287 0.5090 0.5165 0.5165 0.5165

n an

Thus c≈0.5165

Review and Preview Problems

We would conjecture lim 3 x 0

We would conjecture lim 2 x 0

1 1 2 2

2

ln 1 0.3466 0.8047 1.4166 2.0872 2.7745

2 2

21

1

1

a a

a

a

a a