Chapter 6 Differential Equations and Mathematical Modeling

(1)

Chapter

6 Differential Equations and

Mathematical Modeling

Section 6.1

Slope Fields and Euler’s Method

(pp. 321–330)

Exploration 1 Seeing the Slopes

1. Sincedy

dx=0 represents a line with a slope of 0, we should

expect to see intervals with no change in y. We see this at odd multiples of π/ .2

2. Since y is the dependent variable, I t will have no effect on the value of dy

dx=cos .x

3. The graph of dy

dxwill look the same at all values of y.

4. When x dy

dx x

=0, =cos =1. This can be seen on the graph

near the origin. At that point, the change in y and change in

x are the same.

5. When x dy

dx x

=π, =cos = −1 This can be seen in the. graph at x=π. At this point, the change in y is negative of the change in x.

6. This is true because each point on the graph has a negative of itself.

Quick Review 6.1

1. Yes. d

dxe e x₌ x

2. Yes. d

dxe e

x x

4 ₌₄ 4

3. No. d

dx x e xe x e

x x x

( 2 )=2 + 2

4. Yes. d

dxe xe x2 x2

2

=

5. No. d

dx e xe

x x

( 2+ =5) 2 2

6.Yes. d

dx 2x x x

1

2 2 2

1 2

= ( )=

7. Yes. d

dxsecx=sec tanx x

8. No. d

dxx x

−1_{= −} −2

9. y= x + x+C

= + +

= −

3 2 4 2 3(1) 4 (1)

5

2 _C

C

10. y=2sinx−3cosx+C

4 2 sin(0) 3 cos(0) 7

= − +

= −

C C

11. y=e2x+secx+C

5 0

3 2 0

= + +

=

e C

C

( ) sec( )

12. y=tan−1x+ln(2x− +1) C π

π

= + − +

=

−

tan (1) ln(2(1) 1) 3

4 1

C

Section 6.1

Exercises

1.

∫

dy=

∫

(5x4−sec2x dx) y=x5−tanx+C

2.

∫

dy=

∫

(sec tanx x−ex)dx

y=secx− +ex C

3.

∫

dy=

∫

(sinx−e−x+8x3)dx y= −cosx+e−x+2x4+C

4. dy

x x dx x x C

= ⎛ −

⎝⎜ ⎞⎠⎟ = + +

∫

1 12 ln 1

5. dy

x dx x C

x x

= +

+ ⎛

⎝⎜

⎞

⎠⎟ = + − +

∫

5 5 1

1 5

2

1

ln tan

6. dy

x x dx x x C

=

− − ⎛

⎝

⎜ ⎞

⎠

⎟ = − − +

∫

₁1 1 2

2

1 sin

7.

∫

dy=

∫

(3tcos( ))t3 dt=sin( )t3 +C

8.

∫

dy=

∫

cost esintdt =esint+C

9.

∫

dy=

∫

(sec (2 x5)(5x4))dx =tanx5+C

10.

∫

dy=

∫

4(sin ) cosu3 u du =(sin )u4+C

11.

∫

dy=

∫

3sinx dx= −3cosx+C

2= −3cos( )0 +C, C=5

y= −3cosx+5

12.

∫

dy=

∫

2ex−cosx dx=2ex−sinx+C

3 2 0 1

2 1

0

= − + =

= − +

e C C

y ex x

sin( ) , sin

13.

∫

du=

∫

(7x6−3x2+5)dx=x7−x3+5x+C

1 1 1 5 4

5 4

7 3

= − + + = −

= − + −

C C

u x x x

(2)

14.

∫

dA=

∫

(10x9+5x4−2x+4)dx=x10+x4−x2+4x+C

6 1 1 1 4 1 1

4 1

10 4 2

= + − + + =

= + − + +

( ) C, C

A x x x x

15. dy

x x dx x x x c

=

_∫

⎛_⎝⎜− − + ⎞_⎠⎟ = − + − + +

∫

1₂ 3₄ 12 1 3 12

3 1 1 12 1 11

12 1

1 3

= + + + = −

= + + −

− − − −

( ) C, C y x x x 11 (x>0)

16. dy= ⎛ x− x dx x x c

⎝⎜

⎞

⎠⎟ = − +

∫

5 3

2 5

2 3 2

sec tan /

7 5 0 0 7

5 7

3 2

= − + =

= − +

tan( ) ( ) ,

tan

/

C C

y x x

17. dy

t dt t C

t t

=

+ + ⎛

⎝⎜ ⎞⎠⎟ = + +

∫

1 −

1 2 2 2 2

1

ln tan

3 0 2 2

2

1 0

1

= + + =

= + +

− −

tan ( ) ,

tan

C C

y t t C

18. dx

t t dt t t t C

=

_∫

⎛_⎝⎜ − + ⎞_⎠⎟ = + + +

∫

1 1 ₆ − ₆

2

1 ln

0 1 1 6 1 7

6 1 1

= + + + = −

= + + −

− −

ln( ) ( ) ,

ln

C C

x t t t 77 (t>0)

19.

∫

dv=

∫

(

4sec tant t+ +et 6t dt

)

=4sect+ +et 3t2+C

5=4 sec( )0 + +e0 3 0( )2+C, C=0

V=4 t+ +et 3t − t

2 2

sec ⎛ π < <<

⎝⎜ ⎞⎠⎟ π

2

20.

∫

ds=

∫

t t(3 −2)dt= − +t3 t2 C

0 13 12 0

3 2

= − + =

= −

( ) ( ) C, C s t t

21. dy

dx d

dx f t dt t dt x

a x

=

_∫

( ) =

_∫

sin( )2 1

y=

∫

xsin ( )t2 +5

1 dt

22. du

dx d

dx f t dt t dt

x a

x

=

_∫

( ) =

_∫

2+cos 0

u=

∫

x 2+ t dt−3

0 cos

23. F x d

dx f t dt e dt t x a

x 1

2 ( )=

∫

( ) =

∫

cos

F x( )=

∫

xecostdt+9 2

24.G s′ = d

∫

=

∫

dx f t dt t dt x a

x

( ) ( ) 3tan

0

G s( )=

∫

x3tant dt+

0 4

25.Graph (b). (sin 0) (sin1) (sin (

2

=

−

0

1 2 0

>0 >

))

26.Graph (c). (sin 0) (sin1) (sin

3 3

=

−

0 0 1 3 0

> <

( ))

27.Graph (a). (cos 0) (cos1) (cos(

2 2

> >

>

0 0 12 0

− )

28.Graph (d). (cos 0) (cos1) (cos(

3

> >

0 0 2 3 0

− )) <

29.

x y

1 2

0

−1 1

−1

30.

x y

1 2

0

−1 1

−1

31.

x y

1 2

0

−1 1

−1

32.

x y

1 2

0

−1 1

(3)

33.

x y

1 2

0

−1 1

−1

34.

x y

1 2

0

−1 1

−1

35.

36.

37.

38.

39.

40.

41.

(x, y) dy

dx=x – 1 Δx Δy= dy

dxΔx (x +Δx, y +Δy)

(1, 2) 0.0 0.1 0 (1.1, 2)

(1.1, 2) 0.1 0.1 0.01 (1.2, 2.01)

(1.2, 2.01) 0.2 0.1 0.02 (1.3, 2.03)

y= 2.03

42.

(x, y) _dxdy=y – 1 Δx Δy=dy_dxΔx (x +Δx, y +Δy)

(1, 3) 2.0 0.1 0.2 (1.1, 3.2)

(1.1, 3.2) 2.2 0.1 0.22 (1.2, 3.42)

(1.2,

3.42) 2.42 0.1 0.242 (1.3, 3.662)

y= 3.662

43.

(x, y) dy_dx = 2x–y Δx Δy= dy

dx Δx (x +Δx, y +Δy)

(1, 2) 1.0 0.1 0.1 (1.1, 2.1)

(1.1, 2.1) 1.0 0.1 0.1 (1.2, 2.2)

(1.2, 2.2) 1.0 0.1 0.1 (1.3, 2.3)

y= 2.3

44.

(x, y) _dxdy = 2x – y Δx Δy= dy

(1, 0) 2.0 0.1 0.2 (1.1, 0.2)

(1.1, 0.2) 2.0 0.1 0.2 (1.2, 0.4)

(1.2, 0.4) 2.0 0.1 0.2 (1.3, 0.6)

y= 0.6

45.

(x, y) dy_dx = 2 – x Δx Δ =y dyΔ

dx x (x +Δx, y +Δy)

(2, 1) 0.0 –0.1 0.0 (1.9, 1)

(1.9, 1) 0.1 –0.1 –0.01 (1.8, 0.99)

(1.8, 0.99) 0.2 –0.1 –0.02 (1.7, 0.97)

46.

(x, y) dy_dx = 1 + y Δx Δy=dy

(2, 0) 1.0 –0.1 –0.1 (1.9, –0.1)

(1.9, –0.1) 0.9 –0.1 –0.09 (1.8, –0.19) (1.8, –0.19) 0.81 –0.1 –0.081 (1.7, –0.271)

y= –0.271

47.

(x, y) dy_dx=x – y Δx Δy=dy

(2, 2) –0.0 –0.1 0 (1.9, 2.0)

(1.9, 2) –0.1 –0.1 0.01 (1.8, 2.01)

(1.8, 2.01) –0.21 –0.1 0.021 (1.7, 2.031)

y= 2.031

48.

(x, y) dy_dx =x – 2y Δx Δy=dy

dxΔx (x + Δx, y + Δy)

(2, 1) 0.0 –0.1 0.0 (1.9, 1.0)

(1.9, 1) –0.1 –0.1 0.01 (1.8, 1.01)

(1.8, 1.01) –0.22 –0.1 0.022 (1.7, 1.032)

(4)

49. (a) Graph (b)

(b) The slope is always positive, so (a) and (c) can be ruled out.

x

␲

2

–1 0 1

y

(a)

x

p 2

–1 0 1 (b)

y

(c)

x p

2 –1 0 1

y

50. (a) Graph (b)

(b) The solution should have positive slope when x is negative, zero slope when x is zero and negative slope

when x is positive since slope = dy

dx = –x.

Graphs (a) and (c) don’t show this slope pattern.

x y

0 (–1, 1)

(a)

x y

0 (–1, 1)

(b)

x y

0 (–1, 1)

(c)

51. There are positive slopes in the second quadrent of the slope field. The graph ofy=x2_{has negative slopes in the} second quadrent.

52. The slope of y= sin x would be +1 at the origin, while the slope field shows a slope of zero at every point on the

y-axis.

53.

(x, y) _dxdy=2x+1 Δx Δy=dy

(1, 3) 3.0 0.1 0.3 (1.1, 3.3)

(1.1, 3.3) 3.2 0.1 0.32 (1.2, 3.62)

(1.2, 3.62) 3.4 0.1 0.34 (1.3, 3.96)

(1.3, 3.96) 3.6 0.1 0.36 (1.4, 4.32)

y = 4.32

Euler’s Method gives an estimate f( 1.4) ≈ 4.32. The solution to the initial value problem is

f(x) = x2+x+ 1, from which we get f( 1.4) = 4.36. The

percentage error is thus 4 36 4 32

4 36 0 9

. .

. . %.

− ₌

54.

(x, y) _dxdy = 2x – 1 Δx Δy= dy

dx Δx (x +Δx, y +Δy)

(2, 3) 3.0 –0.1 –0.3 (1.9, 2.7)

(1.9, 2.7) 2.8 –0.1 –0.28 (1.8, 2.42) (1.8, 2.42) 2.6 –0.1 –0.26 (1.7, 2.16) 1.7, 2.16) 2.4 –0.1 –0.24 (1.6, 1.92)

y= 1.92

Euler’s Method gives an estimate f( 1.6) ≈ 1.92. The solution to the initial value problem is f(x) = x2– x+ 1, from which we get f( 1.6) = 1.96. The percentage error is

thus 1 96 1 92

1 96 2

. .

. %.

− ₌

55. At every (x, y), (−e(x y− )/2)(−e(y x− )/2)= − = −e0 _{1 so the}, slopes are negative reciprocals. The slope lines are therefore perpendicular.

56. Since the slopes must be negative reciprocals, g(x) = –cos x.

57. The perpendicular slope field would be produced by

dy

dx = –sin x, so y= cos x+C for any constant C.

58. The perpendicular slope field would be produced by

dy

dx = –x, so y= 0.5x

2₊_C_{for any constant}_C_.

59. True. They are all lines of the form y= 5x+C.

60. False. For example, f(x) = x2is a solution of dy

dx = 2x,

butf−1( )x = x is not a solution of dy

dx = 2y.

61. C. m= 42 – 42 = 0

62. E. y < 0, x2 > 0, therefore dy

dx < 0.

63. B.y( )0 =e02=1 dy

dx xe xy

x =2 2=2 .

(5)

65.(a) dy

dx= −x x

1 2

dy

dxdx x x dx

y x x C x

x C = −

= + + = + +

−

∫

( 2)

2

1 2

2 2

1

Initial condition: y(1) = 2

2 1

2 1 1

2 3

2 1 2

2

= + +

= +

=

C

Solution:y x

x x

= 2+ + >

2

1 1

2, 0

(b) Again,y x x C = + +

2 1 2

.

Initial condition: y(–1) = 1

1 1

2 1

1

1 1

2 3 2

2

= − + − + = +

=

( )

( ) C

C

C −

Solution:y x

x x

= 2+ + <

2

1 3

2, 0

(c) Forx dy dx

d dx x

x C < = ⎛ + +

⎝

⎜ ⎞_⎠⎟

0 1

2 2

1 ,

= − +

= −

1

1 2

2

x x

x x .

Forx dy

dx d dx x

x C > = ⎛ + +

⎝

⎜ ⎞_⎠⎟

0 1

2 2

2 ,

= − +

= −

1

1 2

2

x x

x x .

And for x= 0,dy

dxis undefined.

(d) LetC₁be the value from part (b), and let C₂be the value from part (a). Thus, C₁=3

2and C2= 1 2.

(e) y( )2 = −1 y(− =2) 2

− = + + =

− + − +

− = + = +

1 1

2 2

1 2

2 2

1 5

2 2

3 2 2

2

2 1

2

C C

( ) ( )

1 1

2 1

7 2

1 2

− =C =C

Thus, C₁= 1

2 and C2= − 7 2.

66. (a) d

dx(lnx+C)=x x>

1

0 for

(b) d

dx x C x

d

dx x x

ln (− +) ( ) ( )

⎡⎣ _{⎤⎦ = −} − = − ⎛

⎝⎜ ⎞⎠⎟ − =

1 1

xx

for x < 0

(c) For x>0, lnx+ =C lnx+C, which is a solution to the differential equation, as we showed in part (a). For

x<0, lnx+ =C ln(− +x) C, which is a solution to the differential equation, as we showed in part (b). Thus,

d

dxln x =x

1

for all x except 0.

(d) For x < 0, we have y= ln (–x) +C₁, which is a solution to the diferential equation, as we showed in part (a). For

x > 0, we have y= ln x+ C₂, which is a solution to the differential equation, as we showed part (b). Thus,

dy dx=x

1

for all x except 0.

67. (a) y′ =

∫

12x+4dx

y x x C

y x x C dx

y x x C x C

′ = + +

= + +

=

∫

+ + +

6 4

2 2

2

1 2

1

3 2

1 2

(b) y′ =

∫

ex+sinx dx

y e x C

y e x C dx

y e x C x C

x x x ′ = − +

= − +

=

∫

− + +

cos cos sin

1

1 22

(c) y′ =

∫

x3+x−3dx y x

x C

y x

x C dx

y x

x C ′ = − +

= − +

= + +

∫

4

2 1

4

2 1

5

1 4

1 2

4 1 2

20 1

(6)

68. (a) y x dx

y x x C

C C

′ = − ′ = − +

= − +

=

∫

24 10

8 10

3 8 1 10 1 5

2

3 3

( ) ( )

yy x x dx

y x x x C

= − +

= − + +

= − +

∫

8 10 5

2 5 5

5 2 1 5 1

3

4 2

( ) ( ) 55 1 3

2 4 5 2 5 3

( )+

=

= − + +

C C

y x x x

(b) y x x dx

y x x C

C C

′ = −

′ = + +

= + +

∫

cos sin sin cos sin cos

2 0 0

==

= + +

= − + + +

= − +

∫

1

0 0

y x x dx

y x x x C

sin cos cos sin cos ssin

cos sin

0 0

1

+ + =

= − + + +

C C

y x x x

(c) y e x dx

y e x C

e C

C

y e x

x

x ′ = −

′ = − +

= − + = −

= −

∫

2

0 2

2 2

0 0

2 1

2 −−

= − − +

= − − + =

= − −

∫

1

6

1 0

6 0

0

6 3

0 3

3

dx

y e x x C

e C

C

y e x x

x

69. (a) y x

y x dx x C ′ =

=

_∫

= 2+

2

(b) y x

y x dx x C

′ = −

= −

_∫

= − 2+

2

(c) y′ =y d

dx Ce Ce y Ce

x x

x

( )=

=

(d) y′ = −y d

dx Ce Ce

y Ce

x x

x

( − ) −

−

= − =

(e) y′ =xy d

dx Ce Cxe y Ce

x x

x

2 2

2

2 2

2

/ /

/

( )

= =

70. (a) y′′ =x

′ = = +

= + = + +

∫

y x dx x C

y x C dx x C x C

2 1 2

1 3

1 2

2

2 6

(b) y′′ = −x

y x dx x C

′ = − = − +

= − + = − + +

∫

2 1 2

1

3

1 2

2

2 6

(c) y′′ = −sinx

y x dx x C

′ = − = +

=

∫

_∫

+ = + +

sin cos

cos sin

1

1 1 2

(d) y′′ =y d

dx C e C e C e C e y d

dx C e C e

x x x x

x x

1 2 1 2

1 2

+

(

)

= − = ′

−

(

− −

−

))

₌ ₊ _{= ′′}

= +

−

C e C e y

y C e C e

x x

1 2

(e) y′′ = −y d

dx C x C x C x C x y

d dx C

( sin cos ) cos sin

cos

1 2 1 2

1

+ = − = ′

xx C x C x C x

y C x C x

−

(

)

= − −

= +

2 1 2

1 2

sin sin cos

sin cos

Section 6.2

Antidifferentiation by

Substitution (pp. 331–340)

Exploration 1 Are

∫

f u du( ) and

∫

f u dx( ) the Same Thing?

1. f u du u du u

C

( )

∫

=

∫

= +

3

4

2. u x x

4 2 4 6

4 = 4 = 4

( )

3. f u u x x x dx x

( )= =( ) =

=

∫

3 2 3 6

6 7

7

(7)

1. x dx4 x

0

2 ₅

0

2 ₅ ₅

1 5

1 5 2

1 5 0

32 5

∫

= ⎤_{⎦ =} ( ) − ( ) =

2.

∫

x−1dx=

∫

x−1 dx=2 x− _⎦⎤

3 1

1

5 _{1 2}

1

5 _{3 2}

1 5

( )/ ( ) /

=2 −

3 4 2 3 0

3 2 3 2

( )/ ( )/

=2 = 3 8

16 3 ( )

3. dy

dx x =3

4. dy

dx x =3

5. dy

dx=4x −2x +3 3x −4x

3 2 3 2

( ) ( )

6. dy

dx=2sin (4x−5) cos (4x−5)i4 =8sin (4x−5)cos (4x−5)

7. dy

dx= x − x= − x

1

cos i sin tan

8. dy

dx= x x= x

1

sin icos cot

9. dy

dx= x+ x x x+ x

1 2

sec tan i(sec tan sec )

= +

+

sec tan sec sec tan

x x x

x x

2

= +

+

sec (tan sec ) sec tan

x x x

x x

=secx

10. dy

dx= x+ x − x x− x

1 2

csc cot ( csc cot csc )

= − +

+

csc cot csc csc cot

x x x

x x

2

= − +

+

csc (cot csc ) csc cot

x x x

x x

= −cscx

Section 6.2 Exercises

1.

∫

(cosx−3x2)dx=sinx−x3+C

2.

∫

x dx−2 = −x−1+C

3. t

t dt t

t C

2 2

3 1 1

3

− ⎛

⎝⎜ ⎞⎠⎟ = + − +

∫

4. dt

t2 t

1

1 tan +

+ =

−

∫

C

5. (3 2 sec ) 3 tan

5

4 3 2 5 2

x − x− + x dx= x +x− + x+

∫

C

6. (2 sec tan 2 sec 2 /

3 3 2

ex+ x x− x dx= ex+ x− x +

∫

) C

7. ( cot− u+C)1= − −( csc2u)=csc2u

8. ( csc− u C+ )1= − −( cscucot )u =cscucotu

9. 1

2

1

2 2

2 1

2 2

e x+C e x e x ⎛

⎝⎜ ⎞⎠⎟ = ( )=

10. 1

55

1

5 5 5 5

1

ln ln (ln )

x x x

C + ⎛ ⎝⎜

⎞

⎠⎟ = =

11.(tan− + ) =

+

1 1

2 1 1

u C u

12.(sin ) 1

1

1 1

2 −

− u C

u + =

13.

∫

f u du( ) =

∫

u du=2u/ + =C x/ +C

3

2 3

3 2 3 2

f u dx( ) =

∫

u dx= x dx= x dx= x +C

∫

2

∫

1 2

2

14.

∫

f u du( ) =

∫

u du2 =1u3+ =C x15+C

3

1 3

f u dx( ) u dx x dx x C

∫

=

∫

2 =

∫

10 = 1 11+

11

15.

∫

f u du( ) =

∫

e duu =eu+ =C e7x+C f u du( ) =

∫

e dxu =

∫

e dxx = e x+C

∫

7 1 7

7

16.

∫

f u du( ) =

∫

sinu du= −cosu C+ = −cos4x+C f u dx( ) sinu dx sin x dx cos x C

∫

=

∫

=

∫

4 = −1 +

4 4

17.u=3x du=3dx

1 3du=dx

sin3 1 sin

3

x dx=

∫

u du

∫

= −1 +

3cosu C

= −1 +

3cos3x C

Check: d

dx − x+C x x

⎛

⎝⎜ ⎞⎠⎟= − =

1

3 3

1

3 3 3 3

(8)

18.u=2x2 du=4x dx

x dx=1du

4

xcos(2x )dx 1 cosu du

4

2 ₌

∫

=1 +

4sinu C

=1 +

4 2

2 sin( x ) C

Check: d

dx x C x x x

1

4 2

1

4 2 4 2

2 2

sin( )+ cos( )( ) cos(

⎛

⎝⎜ ⎞⎠⎟= = xx2)

19.u=2x du=2dx

1 2du=dx

sec2 tan2 1 sec tan 2

x x dx=

∫

u u du

∫

=1 +

2secu C

=1 +

2sec2x C

Check: d

dx x C x x x

1

2 2

1

2 2 2 2 2 2

sec + sec tan sec tan

⎛

⎝⎜ ⎞⎠⎟= i = xx

20.u=7x−2

du=7dx

1 7du=dx

28 7 2 1

7 28 7 2

3 3 4 4

( x− ) dx=

∫

u du=u + =C ( x− ) +C

∫

Check: d

dx (7x 2) C 4 7( x 2) ( )7 28 7( x 2)

4 3 3

− + ⎡

⎣ ⎤⎦ = − = −

21.u= x

3

du=1dx

3 3du=dx

dx x

du u

2 ₉ 2

3

9 9

+ =

∫

+

∫

= +

∫

3

9 2 1

du u

= +

∫

1

3 2 1

du u

=1 − +

3 1 tan u C

=1 − ⎛_⎝⎜ ⎞_⎠⎟+

3 3

1

tan x C

Check: d

dx

x C

x x

1

3 3

1 3

1

1 3

1 9 1

2 tan− +

⎛

⎝⎜ ⎞⎠⎟=

+ ⎛_⎝⎜ ⎞_⎠⎟ =

+

i ₂₂

22.u= −1 r3 du= −3r dr2

−1 =

3 2

du r dr

9 1

3 2

3

r dr

r

du

u − = −

⎛ ⎝⎜ ⎞⎠⎟

∫

= − −

∫

3 u1 2/du = −3 2( )u1 2/ +C = −6 1− +r3 C

Check: d

dx r C _r r

r

r − − +

(

)

= −

− ⎛ ⎝

⎜ ⎞_⎠⎟ −

= −

6 1 6 1

2 1

3

9 1

3

3 2

2

3

( )

23.u= −1 t 2 cos

du=1 tdt

2sin2

2

du=sintdt

1

2 2 2

2

− ⎛

⎝⎜ ⎞⎠⎟ =

∫

cost sintdt u du

= +

= ⎛ − ⎝⎜

⎞ ⎠⎟ +

2 3 2

3 1 2

3

u C

t C

cos

Check: d

dx

t C

2

3 1 2

3

− ⎛ ⎝⎜

⎞ ⎠⎟ + ⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

cos

=2 1⎛_⎝⎜ − ⎞_⎠⎟ ⎛_⎝⎜ ⎞_⎠⎟⎛_⎝⎜ ⎞_⎠⎟

2 2

1 2 2

cost sint

= −⎛ ⎝⎜

⎞ ⎠⎟

1

2 2

2 cost sint

24. u=y4+4y2+1

du=(4y3+8y dy)

du=4(y3+2y dy) 1

4 2

3

(9)

24. Continued

8 4 1 2 8 1

4

4 2 2 3 2

(y + y + ) (y + y dy) = ⎛ u du ⎝⎜ ⎞⎠⎟

∫

= +

= + + +

2 3 2

3 4 1

3

4 2 3

u C

y y C

( )

Check: d

dx y y C

2

3 4 1

4 2 3

( + + ) +

⎡ ⎣⎢

⎤ ⎦⎥ =2(y4+4y2+1) (24y3+8y)

=8(y4+4y2+1) (2 y3+2y)

25. Letu= −1 x du= −dx

dx x

du u

u C

x C

(1 )

1 1

2 2

1

− = − = + =

− +

∫

−

26.Letu= +x 2

du=dx

sec ( ) sec

tan

tan( )

2 ₂ 2

2

x dx u du

u C

x C

+ =

= +

= + +

∫

27.Letu=tanx du=sec2x dx

tan sec

(tan ) /

/

x x dx u du

u C

x

2 1 2

3 2

3 2 2 3 2 3

∫

=

∫

= +

= +CC

28.Letu= +θ π

2

du=dθ

sec⎛θ π+ tan θ π θ sec tan

⎝⎜ ⎞⎠⎟ ⎛⎝⎜ + ⎞⎠⎟ =

∫

2 2 d

∫

u u du

== + = ⎛_⎝⎜ + ⎞_⎠⎟+

sec sec

u C

C θ π

2

29. tan( )

tan

4 2

4 1 4 1 4

x dx

u x

du dx

u du + = +

= =

∫

== − + +

+ +

1

4 4 2

1

4 4 2

ln cos( )

ln sec ( )

x C

or

30.

∫

3(sin )x−2dx u=sinx du=cosx dx

du x dx

cos =

3 2

∫

_{u dx}−

= −3cotx+C

31.Letu=3z+4

du=3dz

1 3du=dz

cos(3 4) 1 cos 3

z+ dz=

∫

u du

∫

=1 +

3sinu C

=1 + +

3sin(3z 4) C

32.Letu=cotx du= −csc2x dx

cot cscx x dx u/ du

∫

2 = −

∫

1 2 = −2 +

3 3 2

u/ C

= −2 + 3

3 2 (cot )x / C

33.Letu=lnx du

xdx =1

ln6x 6

x dx u du

∫

=

∫

=1 +

7 7

u C

=1 +

7 7 (ln x) C

34.Letu=tanx 2

du=1 xdx

2 2

2 sec

tan sec

tan

7 2 7

8

2 2 2

2 1

8 1

4 2

∫

=

∫

= +

x x

dx u du

u C

x C

i

35.Letu=s4 3/ −8

du=4s ds

3 1 3/

3 4

1 3

(10)

35. Continued

s s ds u du

u C

1 3 4 3 ₈ 3

4 3 4 3 4

/ _cos( / ₎ _cos

sin

s

∫

− =

∫

= +

= iin(s4 3/ − +8) C

36. dx

x x dx

sin2 csc

2

3 =

∫

3

∫

Letu=3x du=3dx

1 3du=dx

csc23 1 csc2 3

∫

x dx=

∫

u du

= −1 + 3cotu C

= −1 + 3cot(3x) C

37.Letu=cos(2t+1)

du= −sin(2t+1 2)( )dt

−1 = +

2du sin(2t 1)dt

sin( )

cos ( )

2 1

1 2 1 2

1 2 2

2

1

t

t dt u du

u C

+

+ = −

= +

=

∫

−

ccos( )

sec( )

2 1

1

2 2 1

t C

+ +

= + +

38.Let u= +2 sint du=cost dt

6

2 6

6 6 2 2

2

1 cos

( sin )

sin

t

t dt u du

u C

t

+ =

= − + = −

+ +

∫

−

C C

39. dx

xlnx

∫

u=lnx

du dx x =

x du=dx du

u = u= x +C

∫

ln ln(ln )

40.

∫

tan2xsec2x dx

u x

du x dx

du x dx =

= =

tan sec

sec 2

2

u du u C

x C

2 3

3 1 3 1 3

∫

= +

+

tan

41. x dx

x2+1

∫

u x du x dx

du x dx = +

= =

2 ₁ 2

2 1

1 2 1

2 1

2

du

x u C

x C

+ = +

= + +

∫

ln

ln( )

42. 40

25 2

dx

x +

∫

u x a du dx

= = =

5

40 1

40

5 5 8 5

2 2

1

1 1

du

u a a

u a C x

C x

+ = +

+ =

−

− −

∫

tan

tan tan ++C

43. dx

x

x xdx

cot

sin cos 3

3 3

∫

=

∫

Letu=cos3x du= −3 sin 3x dx

− =

= −

= − +

= −

∫

1

3 3

3 1 3

1

1 3 1 3

du x dx

dx

x udu

u C

sin

cot

ln

lln cos3x+C

(An equivalent expression is1ln sec .)

3 3x+C

44. Letu=5x+8

du dx

dx

x u du

= =

+ =

−

∫

5 1 5

5 8

1 5

1 2/

= +

= + +

1

5 2

2

5 5 8

1 2

i u C

x C

(11)

45. sec sec sec tan sec tan

x dx x x x

x x dx

= +

+ ⎛

⎝⎜ ⎞⎠⎟

∫

i

=

sec sec tan

sec tan sec tan

2_x _x _x

x x dx

u x x

du +

+

= +

=

∫

Let

ssec tan sec

sec ln ln sec

x x x dx

x dx

udu u C x

+

=

_∫

= + =

∫

2

1

++tanx+C

46. csc csc csc cot

csc cot

x dx x x x

x x dx

= +

+ ⎛ ⎝⎜

⎞ ⎠⎟

∫

= +

+

∫

csccsc csccotcot

2_x _x _x

x x dx

Letu x x

du x x x dx

x dx

= +

= − −

= −

csc cot csc cot csc csc

2 1

u udu u C

x x C

∫

= − +

= − + +

ln

ln csc cot

47. sin

∫

32x dx=

∫

(sin22x) sin2x dx

=

∫

(1−cos22x) sin2x dx

u x

du x dx

= = −

cos sin

2 2

= −

= − +

∫

( )

cos cos

1

3

2 2

3 2

3

u du

u u C

x x C

48. sec

∫

4x dx=

∫

(sec2x)sec2x dx

= + =

= = +

∫

( tan )sec tan

sec

( )

1

2 2

x x dx

u x

du x dx

u du

∫∫

= + +

u u C

x x C

3

3 3

3 tan tan

49. 2

∫

sin2x dx=

∫

(1 2+ sin2x−1)dx

= + =

=

= +

∫

( cos )

(

1 2

2 2 1

2 1

1 2

x dx

u x

du dx

u du

u ssin ) sin

u C

x x C

+

= + 2 +

2

50. 4

∫

cos2x dx= −

∫

( 2 1 2( − cos2x)+2)dx

= − +

= =

= − +

=

∫

( cos )

2 2 2

2 2 1

2 2 2

1

x dx

u x

du dx

u du

2

2 2 2

2 2

( sin )

sin

− + +

= − +

u u C

x x C

51. tan

∫

4x=

∫

tan3x(sec2x−1)dx

= −

= =

= −

∫

(tan sec tan ) tan

sec (

3 2 2

2 2

x x x dx

u x

du x dx

u uu du

u u C

x

x x C

)

tan tan

∫

= − +

= − + +

3

3 3

3

52. (cos

∫

4x−sin4x dx)

= + −

=

∫

(cos sin )(cos sin )

( (cos ))

2 2 2 2

1 2

x x x x dx

x dx

∫∫

=1 +

2sin2x C

53. Letu y du dy

= + =

1

y dy u du

u

+ =

= ⎤_⎦

= −

∫

1

∫

2 3 2 3 4

2 3 1 0

3 _{1 2}

1 4

3 2 1 4

3 2 /

/

/ ( ) ( ))

( )

/ 3 2

2 8 8

2 3

14 3

= − =

54. Letu= −1 r2 du r dr

du r dr

r r dr u du

=

− =

− = −

∫

2 1 2

1 1

2 2 0

1 _{1 2}

1 0 _/

= − ⎤_⎦

= − + =

1 2

2 3 1 3 0

1 31

1 3 3 2

1 0 i u/

(12)

55. Letu=tanx

du x dx

x x dx u du =

= ₋

−

∫

sec tan sec

2

1 0 0

π/4

= ⎤_⎦

= − −

= −

− 1 2 1 2 0

1

2 1

1 2 2

1 0

2

u

( ) ( )

56. Letu= +4 r2 du r dr

du r dr

r

r dr u du

= =

+ = =

−

∫

2 1 2

5 4

5

2 0

2 2 1

1 ₂

5 5

( )

57. Letu= +1 ␪3 2/

du d

d =

=

+ =

∫

3 2 2 3

10 1

2 3 1 2

1 2

3 2 2 0

1

␪ ␪

␪

␪ ␪

/

( ) (110

2 1

2 )

∫

u− du

= − ⎤_⎦

= − ⎛ − ⎝⎜ ⎞⎠⎟

= − ⎛− ⎝⎜

⎞ ⎠⎟=

− 20

3 20

3 1

2 1

20 3

1 2

1 1

1 2

u

0 0 3

58. Letu= +4 3sinx

du x dx

x

xdx u

= =

+ =

−

∫

3 1 3

4 3

1 3 cos

cos cos

sin π

π ₁_{1 2}

4 4

0 / _du₌

∫

59. Letu= +t5 2t

du t dt

t t t dt u du

= +

+ + =

_∫

∫

( )

( ) /

5 2

2 5 2

4

5 4 1 2

0 3 0

1

= ⎤

⎦⎥ =

= =

2 3 2 3 3 2

3 27 2 3

3 2

0 3

3 2

u/

/ ( )

60.Letu=cos 2␪

du d

d = −

− =

= −

−

2 2

1

2 2

2 2 1

2 3

sin sin

cos sin

θ θ θ θ

θ θ θ

0 0

6 ₃

1 1 2

π/ /

∫

_u− _du

= − ⎛− ⎝⎜

⎞ ⎠⎟

⎤ ⎦ ⎥

= ⎛ ⎝⎜

⎞ ⎠⎟ − ⎛ ⎝ ⎜⎜

−

− 1 2

1 2 1 4

1

2 1

2

1 1 2

2

i u

/

⎞⎞ ⎠ ⎟⎟ =1 =

4 3 3 4 ( )

61. dx

x+

∫

0 2 7

u x du dx

du

u u x

= + =

= = + = ⎛

⎝⎜ ⎞⎠⎟

∫

2

2 9

2 0

7

0 7

ln ln ( ) ln ==1 504.

62. dx

x

2 3

2 5

−

∫

u x

du dx du

u u x

= − =

= = − =

∫

2 3

2 1 2

1 2

1

2 2 3

1 2

5

2 5

ln ln ( )

2

2ln ( )7 =0 973.

63. dt

t−

∫

1 3 2

u t du dt

du

u u t

= − =

= = − = ⎛

⎝⎜ ⎞ ⎠⎟

∫

3

3 1

2 1

2

1 2

ln ln( ) ln == −0 693.

64. cot cos

sin

x dx x dx

x =

_∫

∫

ππ ππ 4

/4

4

/4 3

3

u x

du x dx = =

sin cos

du

u u x

ππ _ππ _ππ

4 3 4

0 /

/ /

ln ln (sin )

∫

= = =

65. x dx

x2

1 3

1

+

−

∫

u x du x dx

= + =

2 ₁ 2 1 2

1 2

1

2 1

1

2 5 0

1 3

1

3 2

1 3

du

u u x

− − −

∫

= ln = ln ( + ) = ln ( )= .8805

66. e dx

e x

x

3 0

2

+

∫

u e

du e dx x x = +

=

3

du

u u e

x 0

2

0 2

3 0 954

(13)

67.Let u=x4+9,du=4x dx3 .

(a) x dx

x

u du u

3

4 0 1

9

10 _{1 2} _{1 2}

9 10

9 1 4

1 2

+ = =

⎤ ⎦⎥

∫

−/ /

= −

= − ≈

1 2 10

1 2 9 1

2 10 3 2 0 081.

(b) x

x dx u du

3

4

1 2 9

1 4

+ =

∫

−/

= +

= + +

1 2 1

2 9

1 2

4

u C

x C

/

x

x dx x

3

4 0

1 ₄

0 1

9 1

2 9

+ = + ⎤⎦⎥

∫

= −

= − ≈

1 2 10

1 2 9 1

2 10 3 2 0 081.

68. Letu= −1 cos3x du, =3sin3x dx.

(a) (1 cos3 )sin3 1 3

1 6 1

2 ₂

1 2

− = = ⎤

⎦⎥ =

∫

x x dx u du u

π π /6

/3

1 1 6 2

1 61

1 2

2 2

( ) − ( ) =

(b) (1 cos )sin 1

3

− =

_∫

∫

3x 3x dx u du

= +

= − +

1 6 1

6 1 3

2

u C

x C

( cos )

(1 cos3 )sin3 1( cos )

61 3

2

− = − ⎤

⎦⎥

∫

x x dx x

π π

/6 /3

/6 //3

=1

6 2 2

( ) −−1 = 61

1 2 2 ( )

69. We show thatf x′( )=tanxandf( )3 =5,where

f x( ) lncos cos .

= 3+

3 5

′ = +

= −

⎛

⎝⎜ ⎞⎠⎟ f x d

dx x

d dx ( ) lncos

cos

(ln cos ln co 3

5

3 ss )

ln cos

cos ( sin ) tan x d

dx x

x x x

+

= −

= − − =

5

1

f( ) cos

cos (ln ) 3 = 3+ =5 1+ =5 5

3

70. y=lnsinx + sin 2 6

u x v

du x dx

dy d dx

u v

dy d

= =

=

= ⎛ +

⎝⎜

⎞ ⎠⎟ =

sin sin

cos

ln

2

6

d

dx u v

dy du u

x

x x

f

(ln ln )

cos

sin cot

( ) cot(

− +

= = =

=

6

2 22)=6

71.False. The interval of integration should change from [ ,0 π/ ]4 to [0,1], resulting in a different numerical answer.

72. True. Using the substitutionu= f x du( ), = ′f x dx( ) , we have

∫

f x dx′ =

∫

=

f x

du

u u

a b

f a f b f a

f b

( )

( ) ( )

( ) ln

= − = ⎛

⎝⎜ ⎞ ⎠⎟

ln( ( )) ln( ( )) ln ( ) ( ) .

f b f a f b

f a

73. D.

74. E. e dxx e e x 2 0

2 2

0

2 ₄

2

1

∫

= = −

2

75.B.

∫

F x( −a dx) =F(5− −a) F(3− =a) 7 3

5

F x dx F a F a

a a

( ) = ( − −) ( − =) −

−

∫

5 3 7

3 5

76.A. d

dxsinx=cosx

cos

cos( )

cos

− ⎛ ⎝⎜

⎞ ⎠⎟= = ⎛ ⎝⎜ ⎞⎠⎟=

π

2 0

0 1

2 0

77.(a) Letu= +x 1 du=dx

x+ dx= u du

∫

1

∫

1 2/

= +

= + +

2 3 2

3 1

3 2

u C

x C

/

( )

Alternatively,d ( )/ .

dx x C x

2

3 1 1

3 2

+ +

⎛ ⎝⎜

⎞ ⎠⎟= +

(b) By Part 1 of the Fundamental Theorem of Calculus,

dy dx x

dy

dx x

1₌ ₊₁_and 2₌ ₊₁_{, so both are}

(14)

77. Continued

(c) Using NINT to find the values ofy₁andy₂, we have:

x 0 1 2 3 4

y₁ 0 1.219 2.797 4.667 6.787

y₂ –4.667 –3.448 –1.869 0 2.120

y₁−y₂ 4.667 4.667 4.667 4.667 4.667

C=42 3

(d)C= −y₁ y₂

= + − +

= + + +

= +

∫

x dx x dx

x

x x

x

1 1

1

0 3

0

3

0 3

d dx

78. (a) d

dx[ ( )F x +C]should equalf x( ).

(b) The slope field should help you visualize the solution curvey=F x( ).

(c) The graphs ofy₁ F x y₂ xf t dt

0

= ( )and =

_∫

( ) should differ only by a vertical shift C.

(d) A table of values fory₁−y₂should show thatt

y1−y2=Cfor any value ofxin the appropriate

domain.

(e) The graph of f should be the same as the graph of NDER of F(x).

(f) First, we need to find ( ). LetF x u=x2+1,du=22x dx.

x

dx u du

2

1 2 1

1 2

+ =

∫

−/

=

= + + u

x C

1 2

2 ₁ /

Therefore, we may letF x( )= x2+1.

a) d

dx( x C) _x ( x)

2

1 1

2 1

2

+ + = +

=

+ = x

x

f x

2 ₁ ( )

b)

c)

d)

x 0 1 2 3 4

y₁ 1.000 1.414 2.236 3.162 4.123

y₂ 0.000 0.414 1.236 2.162 3.123

y₁ – y₂ 1 1 1 1 1

e)

79. (a)

∫

2sin cosx x dx=

∫

2u du=u2+ =C sin2x+C

(b)

∫

2sin cosx x dx= −

∫

2u du= − + = −u2 C cos2x+C

(c) Sincesin2x− −( cos2x)=1, the two answers differ by a constant (accounted for in the constant of integration).

80. (a)

∫

2sec2xtanx dx=

∫

2u du=u2+ =C tan2x+C

(b)

∫

2sec2xtanx dx=

∫

2u du=u2+ =C sec2x+C

(c)Sincesec2x−tan2x=1,the two answers differ by a constant (accounted for in the constant of integration).

81. (a) dx

x

u du

u

u du

u du

1 1

1

2 2 2

− =

∫

− = =

∫

cos_sin cos_cos .

(Notecosu>0,so cos2u= cosu=cos .)u

(b) dx

x

du u C x C

1

1 2

1

− = = + = +

∫

_sin−

82. (a) dx

x

u du

u

u du

u du

1 2 1 1

2

+ = + = =

∫

sec_tan

∫

sec_sec

∫

(b) dx

x du u C x C

1 2 1

1

+ = = + = +

−

∫

tan

83. (a) xdx

x

y y y dy

y

1

2

1 2

0 1

1 1

− = − −

− _sin _{sin cos}

sin sin

sin /22 i 0

1 2

∫

/

=

_∫

2 =

_∫

2

0

4 ₂

0 4 sin cos

cos sin

/ y y dy /

y y dy

π π

(b) xdx

x y dy

1 2

2 0

4 0

1 2

− =

∫

/ π/ sin

=

_∫

/ (1−cos2 ) = −[ ( / )sin1 2 2 ]₀/4 0

4

y dy y yπ

π

(15)

84. (a) dx

x

u du

u

1 2 1

2

2 0

3 0

3

1 1

+ = − +

−

∫

tan sec_tan tan

=

_∫

sec =

_∫

sec sec

/ 2 /

0 3

u du

u u du

π π

(b) dx

x

u du u u

1 2 0

3

0 3 0

3

+ =

∫

=⎡⎣ + ⎤⎦

∫

π/ sec lnsec tan π/

=ln

(

2+ 3

)

−ln(1 0+ =) ln

(

2+ 3

)

Section 6.3

Antidifferentiation by Parts

(pp. 341–349)

Exploration 1 Choosing the Right

u

and

dv

1. u du

dv x x v x x dx

= =

_∫

1 0

cos cos

Using 1 for u is never a good idea because it places us back where we started.

2. u x x du x x x

dv dx

= = −

=

cos cos sin

v=

∫

dx=1

The selection ofu=xcos will place a more difficultx

integral into

∫

v du.

3. u x u x

dv x dx v x dx x

= = −

= =

_∫

=

cos d sin

2

The selection of dv=x dxwill place a more difficult integral into

∫

v du.

4. u=xand dv=cosx dxare good choices because the integral is simplified.

1. dy

dx=(x )(cos x)( ) (sin+ x)( x )

3 ₂ ₂ ₂ ₃ 2

=2x3cos2x+3x2sin2x

2. dy

dx e x x e

x x

=

+ ⎛ ⎝⎜

⎞

⎠⎟+ +

( 2 ) 3 ln ( )( 2 )

3 1 3 1 2

=

+ + +

3

3 1 2 3 1

2 2

e

x e x

x

x_{ln (} ₎

3. dy

dx= + x

1 1 (2 )2 i2

= +

2 1 4x2

4. dy

dx= _{− +}_x

1 1 ( 3)2

5. y=tan−13x

tan tan

y x

x y

= =

3 1 3

6. y x

y x

x y

= +

= −

−

cos ( )

cos cos

1 ₁

1 1

7. sinπ cos

π π x dx= − x⎤

⎦⎥

∫

1

0 1

= − +

= − − + =

1 1

0 1

1 1 2

π π π

cos cos

( )

8. dy

dx e x = 2

dy=e dx2x

Integrate both sides.

dy e dx

y e C

x x =

= +

∫

2

2 1 2

9. dy

dx= +x sinx dy= +(x sin )x dx

Integrate both sides.

dy x x dx

y x x C

y C

C

y

= +

= − +

= − + = =

∫

( sin )

cos ( )

1 2

0 1 2

3 2

==1 − +

2 3

2

x cosx

10. d

dx e x x

x

1

2 (sin −cos )

⎛ ⎝⎜

⎞ ⎠⎟

= + + −

= +

1 2

1 2 1

2

1

e x x x x e

e x

x x

x

(cos sin ) (sin cos )

cos 2 2

1 2

e x e x e x

e x

x x x

x

sin sin cos

sin

+ −

=

Section 6.3 Exercises

1.

∫

xsinx dx

dv x dx v x dx x

u x du dx

x x

= = = −

= =

− − −

∫

sin sin cos

cos

∫

cosxx dx= −xcosx+sinx+C

2.

∫

xe dxx

dv e dx v e dx e

u x du dx

xe e dx xe e C

x x x

x x x x

= = =

= =

− = − +