Calculus one and several variables 10E Salas solutions manual ch08

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU

JWDD027-08 JWDD027-Salas-v1 December 4, 2006 20:43

404 SECTION 8.1

CHAPTER 8

SECTION 8.1

1.∫

e2−xdx = −e2−x + C 2.∫

cos23x dx =

32

sin23x + C

3.∫ 1

0

sinπx dx =[− 1π

cos πx

]10

=2π

4.∫ t

0

secπx tanπx dx =1π

[secπx

]t0

=1π

(secπt− 1)

5.∫

sec2 (1 − x) dx = − tan (1 − x) + C

6.∫

dx

5x=∫

5−x dx =−1ln 5

5−x + C = − 15x ln 5

+ C

7.∫ π/3

π/6

cot x dx =[ln (sinx)

]π/3π/6

= ln√

32

− ln12

=12

ln 3

8.∫ 1

0

x3

1 + x4dx =

14

[ln(1 + x4)

]10

=14

ln 2

9.

{u = 1 − x2

du = −2x dx

};

∫x dx√1 − x2

= −12

∫u−1/2 du = −u1/2 + C = −

√1 − x2 + C

10.∫ π/4

−π/4

dx

cos2 x=∫ π/4

−π/4

sec2 x dx =[tanx

]π/4−π/4

= 2

11.∫ π/4

−π/4

sin x

cos2 xdx =

∫ π/4

−π/4

sec x tan x dx =[secx

]π/4−π/4

= 0

12.∫

e√x

√xdx = 2e

√x + C

13.

{u = 1/x

du = dx/x2

∣∣∣∣ x = 1 ⇒ u = 1

x = 2 ⇒ u = 1/2

};

∫ 2

1

e1/x

x2dx =

∫ 1/2

1

−eu du =[− eu

]1/21

= e−√e

14.∫

x3

√1 − x4

dx = −14

∫du√u

= −12√u + C = −1

2

√1 − x4 + C

15.∫ c

0

dx

x2 + c2=[1c

arctan(xc

)]c0

=π

4c

16.∫

axex dx =∫

(ae)x dx =(ae)x

ln(ae)+ C =

axex

1 + ln a+ C



SECTION 8.1 405

17.

{u = 3 tan θ + 1

du = 3 sec2 θ dθ

};

∫sec2 θ√

3 tan θ + 1d θ =

13

∫u−1/2 du =

23u1/2 + C =

23

√3 tan θ + 1 + C

18.∫

sinφ

3 − 2 cosφdφ =

12

∫du

u=

12

ln |u| + C =12

ln(3 − 2 cosφ) + C

19.∫

ex

aex − bdx =

1a

ln |aex − b| + C

20.∫

dx

x2 − 4x + 13=∫

dx

(x− 2)2 + 9=

13

arctan(x− 2

3

)+ C

21.

{u = x + 1

du = dx

};

∫x

(x + 1)2 + 4dx =

∫u− 1u2 + 4

du =∫

u

u2 + 4du−

∫du

u2 + 4

=12

ln |u2 + 4| − 12

arctanu

2+ C

=12

ln |(x + 1)2 + 4| − 12

arctan(x + 1

2

)+ C

22.∫

lnx

xdx =

12

(lnx)2 + C

23.

{u = x2

du = 2x dx

};

∫x√

1 − x4dx =

12

∫du√

1 − u2=

12

arcsinu + C = 12 arcsin (x2) + C

24.∫

ex

1 + e2xdx =

∫du

1 + u2= arctanu + C = arctan ex + C

25.

{u = x + 3

du = dx

};∫

dx

x2 + 6x + 10=∫

dx

(x + 3)2 + 1=∫

du

u2 + 1= arctanu + C = arctan (x + 3) + C

26.∫

ex tan ex dx =∫

tanu du = ln | secu| + C = ln | sec ex| + C

27.∫

x sin x2 dx = −12

cos x2 + C

28.∫

x + 1√1 − x2

dx =∫

x√1 − x2

dx +∫

dx√1 − x2

= −√

1 − x2 + arcsinx + C

29.∫

tan2 x dx =∫

(sec2 x− 1) dx = tan x− x + C



406 SECTION 8.1

30.∫

cosh 2x sinh3 2x dx =18

sinh4 2x + C

31.

{u = lnx

du = dx/x

∣∣∣∣ x = 1 ⇒ u = 0x = e ⇒ u = 1

};

∫ e

1

ln x3

xdx =

∫ e

1

3 lnx

xdx = 3

∫ 1

0

u du = 3[u2

2

]10

=32

32.∫ π/4

0

arctanx

1 + x2dx =

12

[(arctanx)2

]π/40

=12

33.

⎧⎨⎩

u = arcsinx

du =dx√

1 − x2

⎫⎬⎭ ;

∫arcsinx√

1 − x2dx =

∫u du =

12u2 + C =

12

(arcsinx)2 + C

34.∫

ex cosh(2 − ex) dx = −∫

coshu du = − sinhu + C = − sinh(2 − ex) + C

35.

{u = lnx

du = dx/x

};

∫1

x lnxdx =

∫1udu = ln |u| + C = ln | lnx| + C

36.∫ 1

−1

x2

x2 + 1dx =

∫ 1

−1

x2 + 1 − 1x2 + 1

dx =∫ 1

−1

(1 − 1

x2 + 1

)dx =

[x− arctanx

]1−1

= 2 − π

2

37.

{u = cos x

du = − sin x dx

∣∣∣∣ x = 0 ⇒ u = 1x = π/4 ⇒ u =

√2/2

};

∫ π/4

0

1 + sinx

cos2 xdx =

∫ π/4

0

sec2 x dx +∫ π/4

0

sinx

cos2 xdx

=[tanx

]π/40

−∫ √

2/2

1

du

u2= 1 +

[1u

]√2/2

1

=√

2

38.∫ 1/2

0

1 + x√1 − x2

dx =[arcsinx−

√1 − x2

]1/20

=π

6+ 1 −

√3

2

39. (formula 99)∫ √

x2 − 4 dx =x

2

√x2 − 4 − 2 ln |x +

√x2 − 4| + C

40. (formula 87)∫ √

4 − x2 dx =x

2

√4 − x2 + 2 arcsin

x

2+ C

41. (formula 18)∫

cos3 2t dt =12[sin 2t− 1

3sin3 2t] + C

42. (formula 38)∫

sec4 t dt =13

sec2 t sin t +23

∫sec2 t dt =

13

sec2 t sin t +23

tan t + C

43. (formula 108)∫

1x(2x + 3)

dx =13

ln∣∣∣∣ x

2x + 3

∣∣∣∣+ C



SECTION 8.1 407

44. (formula 106)∫

x

2 + 3xdx =

19(2 + 3x− 2 ln |2 + 3x|) + C

45. (formula 81)∫ √

x2 + 9x2

dx = −√x2 + 9x

+ ln |x +√x2 + 9| + C

46. (formula 103)∫

1x2

√x2 − 2

dx =√x2 − 22x

+ C

47. (formula 11)∫

x3 lnx dx = x4

(lnx

4− 1

16

)+ C

48. (formulas 23,24)∫

x3 sinx dx = −x3 cosx + 3∫

x2 cosx dx

= −x3 cosx + 3x2 sinx− 6∫x sinx dx = −x3 cosx + 3x2 sinx + 6x cosx− 6 sinx + C

49.

∫ π

0

√1 + cosx dx =

∫ π

0

√2 cos2

(x2

)dx

=√

2∫ π

0

cos(x

2

)dx

[cos(x

2

)≥ 0 on [0, π]

]

= 2√

2[sin(x

2

)]π0

= 2√

2

50. (a) Clear since du = sec2 x dx. (b) Clear since du = secx tanx dx

(c) Since sec2 x = 1 + tan2 x, 12 tan2 x + C1 = 1

2 sec2 x + C2 with C2 = C1 − 12

51. (a)∫ π

0

sin2 nx dx =∫ π

0

[12− cos 2nx

2

]dx =

[x

2− sin 2nx

4n

]π0

=π

2

(b)∫ π

0

sinnx cosnx dx =12

∫ π

0

sin 2nx dx = −[cos 2nx

4n

]π0

= 0

(c)∫ π/n

0

sinnx cosnx dx =12

∫ π/n

0

sin 2nx dx = −[cos 2nx

4n

]π/n0

= 0

52. (a)∫

sin3 x dx =∫

(1 − cos2 x) sinx dx = − cosx +13

cos3 x + C

(b)∫

sin5 x dx =∫

(1 − cos2 x)2 sinx dx =∫

(1 − 2 cos2 x + cos4 x) sinx dx

= − cosx +23

cos3 x− 15

cos5 x + C

(c) Write sin2k+1 x as sin2k x sinx = (1 − cos2 x)k sinx, expand (1 − cos2 x)k and then

integrate.



408 SECTION 8.1

53. (a)∫

tan3 x dx =∫

tan2 x tanx dx =∫ (

sec2 x− 1)tanx dx

=∫

sec2 x tanx dx−∫

tanx dx

=∫

u du−∫

tanx dx(u = tanx, du = sec2 x dx

)=

12u2 − ln | secx| + C =

12

tan2 x− ln | secx| + C

(b)∫

tan5 x dx =∫

tan3 x tan2 x dx =∫

tan3 x(sec2 x− 1

)dx

=∫

tan3 x sec2 x dx−∫

tan3 x dx

=∫

u3 du−∫

tan3 x dx(u = tanx du = sec2 x dx

)=

14u4 − 1

2tan2 x + ln | secx| + C

=14

tan4 x− 12

tan2 x + ln | secx| + C

(c)∫

tan7 x dx =∫

tan5 x sec2 x dx−∫

tan5 x dx

=16

tan6 x− 14

tan4 x +12

tan2 x− ln | secx| + C

(d)∫

tan2k+1 x dx =∫

tan2k−1 x tan2 x dx =∫

tan2k−1 x sec2 x dx−∫

tan2k−1 x dx

=12k

tan2k x−∫

tan2k−1 x dx

54. (a) (b) A =∫ π/2

π/6

(cscx− sinx) dx

= [ln | cscx− cotx| + cosx]π/2π/6

= − ln(2 −√

3) −√

32

(c) V =∫ π/2

π/6

π(csc2 x− sin2 x) dx

= π

[− cotx− x

2+

14

sin 2x]π/2π/6

=7π

√3

8− π2

6



SECTION 8.1 409

55. (a) (b) sinx + cosx =√

2 [sinx cos(π/4) + cosx sin(π/4)]

=√

2 sin(x +

π

4

);

A =√

2, B = π/4

(c) Area =∫ π/2

0

1sinx + cosx

dx =1√2

∫ π/2

0

1sin(x + π

4

) dx{

u = x + π/4

du = dx

∣∣∣∣ x = 0 ⇒ u = π/4

x = π/2 ⇒ u = 3π/4

};

1√2

∫ π/2

0

1sin(x + π

4

) dx =√

22

∫ 3π/4

π/4

1sinu

du

=√

22

∫ 3π/4

π/4

cscu du

=√

22

[ln | cscu− cotu|

]3π/4π/4

=√

22

ln

[√2 + 1√2 − 1

]

56. (a) (b) V =∫ a

0

2πxe−x2dx

=[−πe−x2

]a0

= π(1 − e−a2)

(c) π(1 − e−a2) = 2

a =√− ln(1 − 2/π) ∼= 1.0061

57. (a) (b) x1∼= −0.80, x2

∼= 5.80



410 SECTION 8.2

(c)Area ∼=

∫ 5.80

−0.80

[8 − 1

2x− x2 + 1

x + 1

]dx =

∫ 5.80

−0.80

[8 − 1

2x− x + 1 − 2

x + 1

]dx

=∫ 5.80

−0.80

[9 − 3

2x− 2

x + 1

]dx

=[9x− 3

4x2 − 2 ln |x + 1|

]5.80−0.80

∼= 27.6

58. (a) (b) Let u = 1 − x, du = −dx, u(0) = 1, u(1) = 0

A = 2∫ 1

0

x√

1 − x dx

A = −2∫ 0

1

(1 − u)√u du

= −2[23u

3/2 − 25u

5/2]01

= 815

SECTION 8.2

1.u = x, dv = e−x dx

du = dx, v = −e−x

∫xe−x dx = −xe−x −

∫−e−x dx = −xe−x − e−x + C

2.∫ 2

0

x2x dx =[x2x

ln 2

]20

−∫ 2

0

2x

ln 2dx

u = x, dv = 2x dx

du = dx, v =2x

ln 2

=8

ln 2−[

2x

(ln 2)2

]20

=8

ln 2− 3

(ln 2)2

3.

{t = −x3

dt = −3x2 dx

};∫

x2e−x3dx = −1

3

∫et dt = −1

3et + C = −1

3e−x3

+ C

4.∫

x lnx2 dx =12

∫lnu du =

12u lnu− 1

2u + C =

12x2 lnx2 − x2

2+ C = x2 lnx− x2

2+ C



SECTION 8.2 411

5.∫

x2e−x dx = −x2e−x −∫

−2xe−x dx = −x2e−x + 2∫

xe−x dx

u = x2,

du = 2x dx,

dv = e−x dx

v = −e−x= −x2e−x + 2

[−xe−x −

∫−e−x dx

]

u = x,

du = dx,

dv = e−x dx

v = −e−x= −x2e−x + 2 (−xe−x − e−x) + C

= −e−x(x2 + 2x + 2

)+ C∫ 1

0

x2e−x dx =[−e−x

(x2 + 2x + 2

)]10

= 2 − 5e−1

6.∫

x3e−x2dx = −x2

2e−x2 −

∫−1

2· 2xe−x2

dx

u = x2,

du = 2x dx,

dv = xe−x2dx

v = −12e−x2 = −x2e−x2

2+∫

xe−x2dx

= −x2e−x2

2− 1

2e−x2

+ C

7.∫

x2 (1 − x)−1/2 dx = −2x2 (1 − x)1/2 + 4∫

x (1 − x)1/2 dx

u = x2,

du = 2x dx,

dv = (1 − x)−1/2 dx

v = −2 (1 − x)1/2= −2x2 (1 − x)1/2 + 4

[−2x

3(1 − x)3/2 +

∫23

(1 − x)3/2 dx

]

u = x,

du = dx,

dv = (1 − x)1/2 dx

v = − 23 (1 − x)3/2

= −2x2 (1 − x)1/2 − 8x3

(1 − x)3/2 − 1615

(1 − x)5/2 + C

Or, use the substitution t = 1 − x (no integration by parts needed) to obtain:

−2 (1 − x)1/2 + 43 (1 − x)3/2 − 2

5 (1 − x)5/2 + C.

8.∫

dx

x(lnx)3=∫

du

u3= − 1

2u2+ C =

−12(lnx)2

+ C

9.∫

x ln√x dx =

12

∫x lnx dx =

12

[12x2 lnx− 1

2

∫x dx

]

u = lnx,

du =dx

x,

dv = x dx

v =12x2

= 14x

2 lnx− 18x

2 + C

∫ e2

1

x ln√x dx =

[14x

2 lnx− 18x

2]e21

= 38e

4 + 18



412 SECTION 8.2

10. {u = x + 1

du = dx

∣∣∣∣ u(0) = 1

u(3) = 4

} ∫ 3

0

x√x + 1 dx =

∫ 4

1

(u− 1)√u du =

[25u5/2 − 2

3u3/2

]41

=11615

11.∫

ln (x + 1)√x + 1

dx = 2√x + 1 ln (x + 1) −

∫2 dx√x + 1

u = ln (x + 1) ,

du =dx

x + 1,

dv =dx√x + 1

v = 2√x + 1

= 2√x + 1 ln (x + 1) − 4

√x + 1 + C

12.∫

x2(ex − 1) dx =∫

(x2ex − x2) dx =∫

x2ex dx− x3

3= x2ex −

∫2xex dx− x3

3

= x2ex − 2xex +∫

2ex dx− x3

3= ex(x2 − 2x + 2) − x3

3+ C

13.∫

(lnx)2 dx = x (lnx)2 − 2∫

lnx dx

u = (lnx)2 ,

du =2 lnx

xdx,

dv = dx

v = x

= x (lnx)2 − 2[x lnx−

∫dx

]u = lnx,

du =dx

x,

dv = dx

v = x

= x (lnx)2 − 2x lnx + 2x + C

14.{

u = x + 5

du = dx

} ∫x(x + 5)−14 dx =

∫(u− 5)u−14 du =

∫(u−13 − 5u−14) du

= − 112

u−12 +513

u−13 + C

= − 112

(x + 5)−12 +513

(x + 5)−13 + C



SECTION 8.2 413

15. ∫x3 3x dx =

x3 3x

ln 3− 3

ln 3

∫x2 3x dx

u = x3,

du = 3x2 dx,

dv = 3x dx

v =3x

ln 3

=x3 3x

ln 3− 3

ln 3

[x2 3x

ln 3− 2

ln 3

∫x 3x dx

]u = x2,

du = 2x dx

dv = 3x dx

v =3x

ln 3

=x3 3x

ln 3− 3x2 3x

(ln 3)2+

6(ln 3)2

∫x 3x dx

=x3 3x

ln 3− 3x2 3x

(ln 3)2+

6(ln 3)2

[x3x

ln 3− 1

ln 3

∫3x dx

]

u = x,

du = dx,

dv = 3x dx

v =3x

ln 3

= 3x[x3

ln 3− 3x2

(ln 3)2+

6x(ln 3)3

− 6(ln 3)4

]+ C

16.u = lnx,

du =1xdx,

dv =√x dx

v =23x3/2

∫ √x lnx dx =

23x3/2 lnx−

∫23x3/2 · 1

xdx =

23x3/2 lnx− 4

9x3/2 + C

17.∫

x (x + 5)14 dx =x

15(x + 5)15 − 1

15

∫(x + 5)15 dx

u = x,

du = dx,

dv = (x + 5)14 dx

v = 115 (x + 5)15

= 115x (x + 5)15 − 1

240 (x + 5)16 + C

Or, use the substitution t = x + 5 (integration by parts not needed) to obtain:

116 (x + 5)16 − 1

3 (x + 5)15 + C.

18.∫

(2x + x2)2 dx =∫

(22x + 2x22x + x4) dx =22x

2 ln 2+

x5

5+∫

2x22x dx

=4x

ln 4+

x5

5+

2x22x

ln 2−∫

4x2x

ln 2dx

=4x

ln 4+

x5

5+

2x22x

ln 2− 4x2x

(ln 2)2+∫

4 · 2x(ln 2)2

dx

=4x

ln 4+

x5

5+ 2x

[2x2

ln 2− 4x

(ln 2)2+

4(ln 2)3

]+ C



414 SECTION 8.2

19.∫

x cosπx dx =1πx sinx− 1

π

∫sinπx dx

u = x,

du = dx,

dv = cosπx dx

v = 1π sinπx

=1πx sinπx +

1π2

cosπx + C

∫ 1/2

0

x cosπx dx =[

1πx sinπx +

1π2

cosπx]1/20

=12π

− 1π2

20.∫ π/2

0

x2 sinx dx =[−x2 cosx

]π/20

+∫ π/2

0

2x cosx dx =[−x2 cosx

]π/20

+ [2x sinx]π/20 −∫ π/2

0

2 sinx dx

=[−x2 cosx + 2x sinx + 2 cosx

]π/20

= π − 2

21. ∫x2 (x + 1)9 dx =

x2

10(x + 1)10 − 1

5

∫x (x + 1)10 dx

u = x2,

du = 2x dx,

dv = (x + 1)9 dx

v = 110 (x + 1)10

=x2

10(x + 1)10 − 1

5

[x

11(x + 1)11 − 1

11

∫(x + 1)11 dx

]

u = x,

du = dx,

dv = (x + 1)10 dx

v = 111 (x + 1)11

=x2

10(x + 1)10 − x

55(x + 1)11 +

1660

(x + 1)12 + C

22. ∫x2(2x− 1)−7 dx = −x2

12(2x− 1)−6 +

∫x

6(2x− 1)−6 dx

u = x2 dv = (2x− 1)−7 dx

du = 2xdx v =(2x− 1)−6

−12

= −x2

12(2x− 1)−6 − x

60(2x− 1)−5 +

∫160

(2x− 1)−5 dx

= −x2

12(2x− 1)−6 − x

60(2x− 1)−5 − 1

480(2x− 1)−4 + C

23. ∫ex sinx dx = −ex cosx +

∫ex cosx dx

u = ex, dv = sinx dx

du = exdx, v = − cosx

= −ex cosx + ex sinx−∫

ex sinx dx

u = ex, dv = cosx dx

du = exdx, v = sinx



SECTION 8.2 415

Adding∫

ex sinx dx to both sides, we get

2∫

ex sinx dx = −ex cosx + ex sinx

so that ∫ex sinx dx =

12ex(sinx− cosx) + C.

24.∫

(ex + 2x)2 dx =∫

(e2x + 4xex + 4x2) dx =12e2x +

43x3 +

∫4xex dx

=12e2x +

43x3 + 4xex −

∫4ex dx =

12e2x +

43x3 + 4xex − 4ex + C

25.∫

ln(1 + x2

)dx = x ln

(1 + x2

)− 2

∫x2

1 + x2dx

u = ln(1 + x2), dv = dx

du =2x

1 + x2dx, v = x

= x ln(1 + x2

)− 2

∫x2 + 1 − 1

1 + x2dx

= x ln(1 + x2

)− 2

∫ (1 − 1

1 + x2

)dx

= x ln(1 + x2

)− 2x + 2 arctanx + C∫ 1

0

ln(1 + x2) dx =[x ln(1 + x2) − 2x + 2 arctanx

]10

= ln 2 − 2 +π

2

26.∫

x ln(x + 1) dx =∫

(x + 1) ln(x + 1) dx−∫

ln(x + 1) dx

(from Example 3) =12(x + 1)2 ln(x + 1) − 1

4(x + 1)2 − (x + 1) ln(x + 1) + (x + 1) + C1

=12(x2 − 1) ln(x + 1) − 1

4x2 +

12x + C

27.∫

xn lnx dx =xn+1 lnx

n + 1− 1

n + 1

∫xn dx

u = lnx, dv = xn dx

du =dx

x, v =

xn+1

n + 1

=xn+1 lnx

n + 1− xn+1

(n + 1)2+ C

28.∫

e3x cos 2x dx =12e3x sin 2x−

∫32e3x sin 2x dx

u = e3x dv = cos 2x dx

du = 3e3x dx v =12

sin 2x=

12e3x sin 2x +

34e3x cos 2x−

∫94e3x cos 2x dx

=⇒ 134

∫e3x cos 2x dx =

12e3x sin 2x +

34e3x cos 2x

=⇒∫

e3x cos 2x dx =213

e3x sin 2x +313

e3x cos 2x + C



416 SECTION 8.2

29. {t = x2, dt = 2x dx

};

∫x3 sinx2 dx =

12

∫t sin t dt

u = t, dv = sin t dt

du = dt, v = − cos t=

12

[−t cos t +

∫cos t dt

]

= 12 (−t cos t + sin t) + C

= − 12x

2 cosx2 + 12 sinx2 + C

30.∫

x3sinx dx = −x3 cosx +∫

3x2 cosx dx

u = x3 dv = sinx dx

du = 3x2 dx v = − cosx= −x3 cosx + 3x2 sinx−

∫6x sinx dx

= −x3 cosx + 3x2 sinx + 6x cosx−∫

6 cosx dx

= −x3 cosx + 3x2 sinx + 6x cosx− 6 sinx + C

31.{

u = 2x x = 0 =⇒ u = 0

du = 2 dx x = 1/4 =⇒ u = 1/2

};

∫ 1/4

0

arcsin 2x dx =12

∫ 1/2

0

arcsinu du

=12

[u arcsinu +

√1 − u2

]1/20

[by(8.2.5)]

=12

[π

12+

√3

2− 1

]=

π

24+

√3 − 24

32.⎧⎨⎩

u = arcsin 2x

du =2√

1 − 4x2dx

⎫⎬⎭

∫arcsin 2x√

1 − 4x2dx =

12

∫u du =

14u2 + C =

14

(arcsin 2x)2 + C

33.{

u = x2 x = 0 =⇒ u = 0

du = 2x dx x = 1 =⇒ u = 1

};∫ 1

0

x arctan (x)2 dx =12

∫ 1

0

arctanu du

=12[u arctanu− 1

2 (1 + u2)]10

[by 8.2.6]

=π

8− 1

4ln 2

34.∫

cos√x dx =

∫ √x

cos√x√

xdx

ux =√x dv =

cos√x√

xdx

du =1

2√xdx v = 2 sin

√x

= 2√x sin

√x−

∫sin

√x√

xdx

= 2√x sin

√x + 2 cos

√x + C



SECTION 8.2 417

35.∫

x2 cosh 2x dx = 12x

2 sinh 2x−∫x sinh 2x dx

u = x2, dv = cosh 2x dx

du = 2x dx, v = 12 sinh 2x

= 12x

2 sinh 2x− 12x cosh 2x + 1

2

∫cosh 2x dx

u = x, dv = sinh 2x dx

du = dx, v = 12 cosh 2x

= 12x

2 sinh 2x− 12x cosh 2x + 1

4 sinh 2x + C

36.∫ 1

−1

x sinh(2x2) dx =14[cosh(2x2)

]1−1

= 0

37. Let u = lnx, du = 1x dx. Then∫

1x

arcsin (lnx) dx =∫

arcsinu du = u arcsinu +√

(1 − u2) + C

= (lnx) arcsin (lnx) +√

1 − (lnx)2 + C

38.u = cos(lnx) dv = dx

du = − sin(lnx)x

dx v = x

∫cos(lnx) dx = x cos(lnx) +

∫sin(lnx) dx

u = sin(lnx) dv = dx

du =cos(lnx)

xdx v = x

= x cos(lnx) + x sin(lnx) −∫

cos(lnx) dx

=⇒∫

cos(lnx) dx =12x [cos(lnx) + sin(lnx)] + C

39.∫

sin(lnx) dx = x sin(lnx) −∫

cos(lnx) dx

u = sin(lnx), dv = dx

du = cos(lnx)1xdx, v = x

= x sin(lnx) − x cos(lnx) −∫

sin(lnx) dx

u = cos(lnx), dv = dx

du = − sin(lnx)1xdx, v = x

Adding∫

sin(lnx) dx to both sides, we get

2∫

sin(lnx) dx = x sin(lnx) − x cos(lnx)

so that ∫sin(lnx) dx = 1

2 [x sin(lnx) − x cos(lnx)] + C



418 SECTION 8.2

40.u = (lnx)2 dv = x2dx

du = 2lnx

xdx v =

x3

3

∫ 2e

1

x2(lnx)2 dx =[x3

3(lnx)2

]2e1

−∫ 2e

1

2x2

3lnx dx

u = lnx dv =2x2

3dx

du =1xdx v =

2x3

9

=[x3

3(lnx)2 − 2x3

9lnx

]2e1

+∫ 2e

1

2x2

9dx

=[x3

3(lnx)2 − 2x3

9lnx +

2x3

27

]2e1

=8e3

3

[(ln 2e)2 − 2

3ln 2e +

29

]− 2

27

41.u = lnx dv = dx

du =1xdx v = x

∫lnx dx = x lnx−

∫dx = x lnx− x + C

42.u = arctanx dv = dx

du =1

1 + x2dx v = x

∫arctanx dx = x arctanx−

∫x

1 + x2dx

= x arctanx− 12

ln (1 + x2) + C

43.u = lnx dv = xkdx

du =1xdx v =

1k + 1

xk+1

∫xk lnx dx =

1k + 1

xk+1 lnx−∫

1k + 1

xk dx

=1

k + 1xk+1 lnx− xk+1

(k + 1)2+ C

44.u = eax dv = cos bxdx

du = aeax dx v =1b

sin bx

∫eax cos bx dx =

1beax sin bx− a

b

∫eax sin bx dx

u = eax dv = sin bxdx

du = aeax dx v = −1b

cos bx=

1beax sin bx +

a

b2eax cos bx− a2

b2

∫eax cos bx dx.

=⇒∫

eax cos bx dx =eax

a2 + b2(a cos bx + b sin bx) + C

45.u = eax dv = sin bxdx

du = aeax dx v = −1b

cos bx

∫eax sin bx dx = −1

beax cos bx +

a

b

∫eax cos bx dx



SECTION 8.2 419

u = eax dv = cos bxdx

du = aeax dx v =1b

sin bx= −1

beax cos bx +

a

b2eax sin bx− a2

b2

∫eax sin bx dx.

=⇒∫

eax sin bx dx =eax

a2 + b2(a sin bx− b cos bx) + C

46. The integrals cancel each other if you integrate by parts.∫eax cosh ax dx =

e2ax

4a+

x

2+ C

47.∫ π

0

x sinx dx =[− x cosx + sinx

]π0

= π

48.∫ π

0

x cos(x/2) dx =[2x sin(x/2) + 4 cos(x/2)

]π0

= 2π − 4

49. A =∫ 1/2

0

arcsinx dx =[x arcsinx +

√1 − x2

]1/20

=π

12+

√3 − 22

50.A =

∫ 2

0

xe−2x dx =[−xe−2x

2

]20

+∫ 2

0

e−2x

2dx

=[−xe−2x

2− e−2x

4

]20

=14− 5

4e−4

51. (a) A =∫ e

1

lnx dx = [x lnx− x]e1 = 1

(b) xA =∫ e

1

x lnx dx =[12x2 lnx− 1

4x2

]e1

=14(e2 + 1

), x =

14(e2 + 1

)yA =

∫ e

1

12

(lnx)2 dx =12

[x (lnx)2 − 2x lnx + 2x

]e1

=12e− 1, y =

12e− 1

(c) Vx = 2πyA = π (e− 2) , Vy = 2πxA = 12π(e2 + 1

)52. (a) A =

∫ 2e

1

lnx

xdx =

[12(lnx)2

]2e1

=(ln 2e)2

2=

(1 + ln 2)2

2

(b)

V =∫ 2e

1

π

(lnx

x

)2

dx

u = (lnx)2, dv =1x2

dx

du =2 lnx

x, dx v = − 1

x

= π

[− (lnx)2

x

]2e1

− π

∫ 2e

1

−2 lnx

x2dx

= π

[− (ln 2e)2

2e

]+ 2π

[− lnx

x

]2e1

− 2π∫ 2e

1

− 1x2

dx

= π

[− (ln 2e)2

2e− ln 2e

e− 1

e+ 2]



420 SECTION 8.2

53. x =1

e− 1, y =

14

(e + 1) 54. x =e− 2e− 1

, y =14

(e + 1e

)

55. x = 12π, y = 1

8π 56. x =π

2− 1, y =

π

8

57. (a) M =∫ 1

0

ekx dx =1k

(ek − 1

)

(b) xMM =∫ 1

0

xekx dx =(k − 1) ek + 1

k2, xM =

(k − 1) ek + 1k (ek − 1)

58. (a) M =∫ 3

2

lnx dx = [x lnx− x]32 = 3 ln 3 − 2 ln 2 − 1

(b) xM M =∫ 3

2

x lnx dx =[x2

2lnx− x2

4

]32

=92

ln 3 − 2 ln 2 − 54

=⇒ xM =18 ln 3 − 8 ln 2 − 5

4(3 ln 3 − 2 ln 2 − 1)

59. Vy =∫ 1

0

2πx cos12πx dx =

[4x sin

12πx +

8π

cos12πx

]10

= 4 − 8π

60. Vy =∫ π

0

2πx2 sinx dx = 2π[−x2 cosx + 2x sinx + 2 cosx

]π0

= 2π(π2 − 4)

61. Vy =∫ 1

0

2πx2ex dx = 2π (e− 2) (see Example 6)

62. Vy =∫ π/2

0

2πx2 cosx dx = 2π[x2 sinx + 2x cosx− 2 sinx

]π/20

=π

2(π2 − 8)

63. Vx =∫ 1

0

πe2x dx = π

[12e2x

]10

=12π(e2 − 1

)xVx =

∫ 1

0

πxe2x dx = π

[12xe2x − 1

4e2x

]10

=14π(e2 + 1

);

x =e2 + 1

2 (e2 − 1)

64. xVx =∫ π/2

0

πx sin2 x dx =18π[2x2 − 2x sin 2x− cos 2x

]π/20

=116

π(π2 + 4)

Vx =∫ π/2

0

π sin2 x dx = π

[12x− 1

4sin 2x

]π/20

=14π2;

x =π2 + 4

4π



SECTION 8.2 421

65. A =∫ 1

0

coshx dx = [sinhx]10 = sinh 1 =e− e−1

2=

e2 − 12e

xA =∫ 1

0

x coshx dx = [x sinhx− coshx]10 = sinh 1 − cosh 1 + 1 =2 (e− 1)

2e

yA =∫ 1

0

12

cosh2 x dx =14

[sinhx coshx + x]10 =14

(sinh 1 cosh 1 + 1) =e4 + 4e2 − 1

16e2

Therefore x =2

e + 1and y =

e4 + 4e2 − 18e (e2 − 1)

.

66. (a) xVx =∫ 1

0

πx cosh2 x dx =π

8[2x2 + 2x sinh 2x− cosh 2x

]10

=π

8(3 + 2 sinh 2 − cosh 2)

Vx =∫ 1

0

π cosh2 x dx = π

[x

2+

14

sinh 2x]10

=π

4(2 + sinh 2)

x =3 + 2 sinh 2 − cosh 2

2(2 + sinh 2)

(b) yVy =∫ 1

0

2πx cosh2 x dx =π

8(1 + 2 sinh 2 − cosh 2)

Vy =∫ 1

0

2πx coshx dx = 2π [x sinhx− coshx]10 = 2π(sinh 1 − cosh 1 + 1)

y =1 + 2 sinh 2 − cosh 2

16(sinh 1 − cosh 1 + 1)

67.u = x2,

du = 2x dx,

dv = e−x dx

v = −e−x

∫xneax dx =

xneax

a− n

a

∫xn−1eax dx

68. ∫(lnx)n dx = x(lnx)n − n

∫x(lnx)n−1

xdx

u = (lnx)n

du =n(lnx)n−1

xdx

dv = dx

v = x= x(lnx)n − n

∫(lnx)n−1 dx

69.∫

x3e2x dx =12x3e2x − 3

2

∫x2e2x dx =

12x3e2x − 3

2

[12x2e2x −

∫xe2x dx

]

=12x3e2x − 3

4x2e2x − 3

4xe2x − 3

4

∫e2x dx

=12x3e2x − 3

4x2e2x +

34xe2x − 3

8e2x + C

70.∫

x2e−x dx = −x2e−x + 2∫

xe−x dx

= −x2e−x + 2[−xe−x +

∫e−x dx

]= −x2e−x − 2xe−x − 2e−x + C



422 SECTION 8.2

71. ∫(lnx)3 dx = x(lnx)3 − 3

∫(lnx)2 dx = x(lnx)3 − 3x(lnx)2 + 6

∫lnx dx

= x(lnx)3 − 3x(lnx)2 + 6x lnx− 6∫

dx

= x(lnx)3 − 3x(lnx)2 + 6x lnx− 6x + C

72. ∫(lnx)4 dx = x(lnx)4 − 4

∫(lnx)3 dx

= x(lnx)4 − 4[x(lnx)3 − 3

∫(lnx)2 dx

]

= x(lnx)4 − 4x(lnx)3 + 12∫

(lnx)2 dx

= x(lnx)4 − 4x(lnx)3 + 12[x(lnx)2 − 2

∫lnx dx

]= x(lnx)4 − 4x(lnx)3 + 12x(lnx)2 − 24x lnx + 24x + C

73. (a) Differentiating, x3ex = Ax3ex + 3Ax2ex + 2Bxex + Bx2ex + Cex + Cxex + Dex

=⇒ A = 1, B = −3, C = 6, and D = −6

(b) ∫x3ex dx = x3ex − 3

∫x2ex dx

= x3ex − 3x2ex + 6∫

xex dx

= x3ex − 3x2ex + 6xex − 6∫

ex dx

= x3ex − 3x2ex + 6xex − 6ex + C

74. Use induction on k. For k = 0, P (x) is a constant and the result follows immediately.Now suppose the statement is true for k = n and let P (x) have degree n + 1.Then P (x) = axn+1 + Q(x) for Q(x) of degree n.∫

P (x)ex dx = a

∫xn+1ex dx +

∫Q(x)ex dx

= axn+1ex − a(n + 1)∫

xnex dx +∫

Q(x)ex dx

=[axn+1 − a(n + 1)(xn − n(n− 1)xn−1 + · · · ± a

]ex +

[Q(x) −Q′(x) + · · · ±Q(n)(x)

]ex + C

=[axn+1 + Q(x) − [a(n + 1)xn + Q′(x)] + · · · ±Q(n)(x)

]ex + C

=[P (x) − P ′(x) + · · · ± P (n)(x)

]ex + C

75. (a) Set∫ (

x2 − 3x + 1)ex dx = Ax2ex + Bxex + Cex and differentiate both sides of the equation.



SECTION 8.2 423

x2ex − 3xex + ex = Ax2ex + 2Axex + Bxex + Bex + Cex

= Ax2ex + (2A + B)xex + (B + C)ex

Equating the coefficients, we find that A = 1, B = −5, C = 6

Thus∫ (

x2 − 3x + 1)ex dx = x2ex − 5xex + 6ex + K

(b) Set∫ (

x3 − 2x)ex dx=Ax3ex + Bx2ex + Cxex + Dex and differentiate both sides of the equation.

x3ex − 2xex = Ax3ex + 3Ax2ex + Bx2ex + 2Bxex + Cxex + cex + Dex

= Ax3ex + (3A + B)x2ex + (2B + C)xex + (C + D)ex

Equating the coefficients, we find that A = 1, B = −3, C = 4, D = −4

Thus∫ (

x3 − 2x)ex dx = x3ex − 3x2ex + 4xex − 4ex + K

76. Set u = f (−1)(x) and dv = dx.

77. Let u = f(x), dv = g′′(x) dx. Then du = f ′(x) dx, v = g′(x), and∫ b

a

f(x)g′′(x) dx = [f(x)g′(x)]ba −∫ b

a

f ′(x)g′(x) dx = −∫ b

a

f ′(x)g′(x) dx since f(a) = f(b) = 0

Now let u = f ′(x), dv = g′(x) dx. Then du = f ′′(x) dx, v = g(x), and

−∫ b

a

f ′(x)g′(x) dx = [−f ′(x)g(x)]ba +∫

f ′′(x)g(x) dx =∫

g(x)f ′′(x) dx since g(a) = g(b) = 0

Therefore, if f and g have continuous second derivatives, and if f(a) = g(a) = f(b) = g(b) = 0, then

∫ b

a

f(x)g′′(x) dx =∫ b

a

g(x)f ′′(x) dx

78. (a)

f(b) − f(a) =∫ b

a

f ′(x) dx = [f ′(x)(x− b)]ba −∫ b

a

f ′′(x)(x− b) dx (a)

u = f ′(x)

du = f ′′(x) dx

dv = dx

v = (x− b)= f ′(a)(b− a) −

∫ b

a

f ′′(x)(x− b) dx

(b)f(b) − f(a) = f ′(a)(b− a) −

[f ′′(x)

(x− b)2

2

]ba

+∫ b

a

f ′′′(x)2

(x− b)2 dx

u = f ′′(x)

du = f ′′′(x) dx

dv = (x− b) dx

v =(x− b)2

2

= f ′(a)(b− a) +f ′′(a)

2(b− a)2 +

∫ b

a

f ′′′(x)2

(x− b)2 dx



424 SECTION 8.2

79. (a) From Exercise 47, A =∫ π

0

x sinx dx = π

(b) f(x) = x sinx < 0 on (π, 2π). Therefore

A = −∫ 2π

π


]2ππ

= 3π

(c) A =∫ 3π

2π


]3π2π

= 5π

(d) A = (2n + 1)π, n = 1, 2, 3, . . .

80. (a)∫ 3π/2

π/2

x cosx dx =[x sinx + cosx

]3π/2π/2

= −2π; A = 2π

(b) A = 4π (c) A = 6π (d) A = 2nπ

81. (a) area of R:∫ π

0

(1 − sinx) dx =[x + cosx

]π0

= π − 2

(b) V =∫ π

0

2πx (1 − sinx) dx = 2π[12 x

2 + x cosx− sinx]π0

= π3 − 2π2

(c) The region is symmetric about the line x = 12π. Therefore x = 1

2π.

yA = 12

∫ π

0

(1 − sinx)2 dx

= 12

∫ π

0

(1 − 2 sinx + sin2 x

)dx

= 12

∫ π

0

(32 − 2 sinx− 1

2 cos 2x)dx

= 12

[32x + 2 cosx− 1

4 sin 2x]π0

= 34π − 2

Therefore, y =34π − 2π − 2

∼= 0.31202

82. (a) A =∫ 10

0

xe−x dx ∼= 0.9995

x =

∫ 10

0

x2e−x dx

A∼= 1.9955 y =

12

∫ 10

0

[xe−x

]2dx

A∼= 0.1251

(b) around x-axis: V ∼= 0.7854; around y-axis: V ∼= 12.5316

PROJECT 8.2

1.∫ 2π

0

sin2 nx dx =∫ 2π

0

(12− 1

2cos 2nx

)dx =

[12x− 1

4nsin 2nx

]2π0

= π.

∫ 2π

0

cos2 nx dx =∫ 2π

0

(12

+12

cos 2nx)

dx =[12x +

14n

sin 2nx]2π0

= π



SECTION 8.3 425

2.∫ 2π

0

cos [(m + n)x] dx =[

1m + n

sin [(m + n)x]]2π0

= 0.

0 =∫ 2π

0

cos [(m + n)x] dx =∫ 2π

0

cos (mx + nx) dx =∫ 2π

0

cos mx cos nx dx−∫ 2π

0

sin mx sin nx dx

Thus,∫ 2π

0

cos mx cos nx dx =∫ 2π

0

sin mx sin nx dx

SECTION 8.3

1.∫

sin3 x dx =∫

(1 − cos2 x) sinx dx =13

cos3 x− cosx + C

2.∫ π/8

0

cos2 4x dx =[x

2+

116

sin 8x]π/80

=π

16

3.∫ π/6

0

sin2 3x dx =∫ π/6

0

1 − cos 6x2

dx =[12x− 1

12sin 6x

]π/60

=π

12

4.∫

cos3 x dx =∫

(1 − sin2 x) cosx dx = sinx− 13

sin3 x + C

5. ∫cos4 x sin3 x dx =

∫cos4 x (1 − cos2 x) sinx dx

=∫

(cos4 x− cos6 x) sinx dx

= − 15 cos5 x + 1

7 cos7 x + C

6.∫

sin3 x cos2 x dx =∫

cos2(1 − cos2 x) sinx dx = −13

cos3 x +15

cos5 x + C

7. ∫sin3 x cos3 x dx =

∫sin3 x (1 − sin2 x) cosx dx =

∫(sin3 x− sin5 x) cosx dx

= 14 sin4 x− 1

6 sin6 x + C

8. ∫sin2 x cos4 x dx =

∫(sinx cosx)2 cos2 x dx =

∫14

sin2 2x(12

+12

cos 2x) dx

=18

∫sin2 2x dx +

18

∫sin2 2x cos 2x dx

=18

(12x− 1

8sin 4x

)+

148

sin3 2x + C

9.∫

sec2 πx dx =1π

tanπx + C

10.∫

csc2 2x dx = −12

cot 2x + C



426 SECTION 8.3

11.∫

tan3 x dx =∫

(sec2 x− 1) tanx dx

=∫

tanx sec2 x dx−∫

tanx dx

= 12 tan2 x + ln | cosx| + C

12.∫

cot3 x dx =∫

cotx(csc2 x− 1) dx = −12

cot2 x− ln | sinx| + C

13.∫

sin4 x dx =∫ (

1 − cos 2x2

)2

dx

=14

∫ (1 − 2 cos 2x + cos2 2x

)dx

=14

∫ (1 − 2 cos 2x +

1 + cos 4x2

)dx

=∫ (

38− 1

2cos 2x +

18

cos 4x)

dx

= 38x− 1

4 sin 2x + 132 sin 4x + C∫ π

0

sin4 x dx =[38x− 1

4 sin 2x + 132 sin 4x

]π0

= 38π

14.∫

cos3 x cos 2x dx =∫

(1 − sin2 x)(1 − 2 sin2 x) cosx dx =∫

(1 − 3 sin2 x + 2 sin4 x) cosx dx

= sinx− sin3 x +25

sin5 x + C

15.∫

sin 2x cos 3x dx =∫

12

[sin (−x) + sin 5x] dx

=∫

12

(− sinx + sin 5x) dx

= 12 cosx− 1

10 cos 5x + C

16.

∫ π/2

0

cos 2x sin 3x dx =∫ π/2

0

12

[sin(3x− 2x) + sin(3x + 2x)] dx =12

∫ π/2

0

(sinx + sin 5x) dx

=[−1

2cosx− 1

10cos 5x

]π/20

=35

17.∫

tan2 x sec2 x dx =13

tan3 x + C

18.∫

cot2 x csc2 x dx = −13

cot3 x + C

19.∫

sin2 x sin 2x dx =∫

sin2 x (2 sinx cosx) dx

= 2∫

sin3 x cosx dx

= 12 sin4 x + C



SECTION 8.3 427

20. ∫ π/2

0

cos4 x dx =[14

cos3 x sinx

]π/20

+34

∫ π/2

0

cos2 x dx

=[14

cos3 x sinx +38x +

316

sin 2x]π/20

=3π16

21. ∫sin6 x dx =

∫ (1 − cos 2x

2

)3

dx

=18

∫(1 − 3 cos 2x + 3 cos2 2x− cos3 2x) dx

=18

∫ [1 − 3 cos 2x + 3

(1 + cos 4x

2

)− cos 2x (1 − sin2 2x)

]dx

=18

∫ (52− 4 cos 2x +

32

cos 4x + sin2 2x cos 2x)

dx

= 516x− 1

4 sin 2x + 364 sin 4x + 1

48 sin3 2x + C

22.∫

cos5 x sin5 x dx =∫

cos5 x sin4 x sinx dx =∫

cos5 x(1 − cos2 x)2 sinx dx

=∫

cos5 x(1 − 2 cos2 x + cos4 x) sinx dx

= −cos6 x6

+14

cos8 x− 110

cos10 x + C

Equivalently,∫

cos5 x sin4 x sinx dx gives − 16 cosx + 1

4 cos8 x− 110 cos10 x + C.

23.∫ π/2

π/6

cot2 x dx =∫ π/2

π/6

(csc2 x− 1) dx = [− cotx− x]π/2π/6 =√

3 − π

3

24.

∫tan4 x dx =

∫tan2 x(sec2 x− 1) dx =

∫tan2 x sec2 x dx−

∫tan2 x dx

=13

tan3 x−∫

(sec2 x− 1) dx =13

tan3 x− tanx + x + C

25.

∫cot3 x csc3 x dx =

∫(csc2 x− 1) csc3 x cotx dx

=∫

(csc4 x− csc2 x) cscx cotx dx

= −15

csc5 x +13

csc3 x + C

26.

∫tan3 x sec3 x dx =

∫(sec2 x− 1) sec2 x secx tanx dx

=15

sec5 x− 13

sec3 x + C



428 SECTION 8.3

27.∫

sin 5x sin 2x dx =∫

12(cos 3x− cos 7x) dx

= 16 sin 3x− 1

14 sin 7x + C

28.∫

sec4 3x dx =∫

(1 + tan2 3x) sec2 3x dx =19

tan3 3x +13

tan 3x + C

29. ∫sin5/2 x cos3 x dx =

∫sin5/2 x

(1 − sin2 x

)cosx dx

=∫

sin5/2 x cosx dx−∫

sin9/2 x cosx dx

= 27 sin7/2 x− 2

11 sin11/2 x + C

30.∫

sin3 x

cosxdx =

∫ (1 − cos2 x

)sinx

cosxdx =

∫ (tanx− sin 2x

2

)dx = ln | secx| + 1

4 cos 2x + C

31.∫

tan5 3x dx =∫

tan3 3x (sec2 3x− 1) dx

=∫

tan3 3x sec2 3x dx−∫

tan3 3x dx

=∫

tan3 3x sec2 3x dx−∫

(tan 3x sec2 3x− tan 3x) dx

= 112 tan4 3x− 1

6 tan2 3x + 13 ln | sec 3x| + C

32.∫

cot5 2x dx =∫

(cot3 2x csc2 2x− cot3 2x) dx

=∫

(cot3 2x csc2 2x− cot 2x csc2 2x + cot 2x) dx

= −18

cot4 2x +14

cot2 2x +12

ln | sin 2x| + C

33.∫ 1/3

−1/6

sin4 3πx cos3 3πx dx =∫ 1/3

−1/6

sin4 3πx cos2 3πx cos 3πx dx

=∫ 1/3

−1/6

sin4 3πx (1 − sin2 3πx cos 3πx) dx

=∫ 0

−1

u4(1 − u2)13π

du [u = sin 3πx, du = 3π cos 3πx dx]

=13π[15u

5 − 17u

7]0−1

=2

105π



SECTION 8.3 429

34. ∫ 1/2

0

cosπx cosπ

2x dx =

12

∫ 1/2

0

[cos(πx− π

2x)

+ cos(πx +

π

2x)]

dx

=12

∫ 1/2

0

(cos

π

2x + cos

3π2x

)dx =

12

[2π

sinπ

2x +

23π

sin3π2x

]1/20

=2√

23π

35.∫ π/4

0

cos 4x sin 2x dx =∫ π/4

0

12(sin 6x− sin 2x) dx

=[− 1

12 cos 6x + 14 cos 2x

]π/40

= − 16

36.∫

(sin 3x− sinx)2 dx =∫

(sin2 3x− 2 sin 3x sinx + sin2 x) dx

=∫

(12− 1

2cos 6x− cos 2x + cos 4x +

12− 1

2cos 2x) dx

= x− 112

sin 6x +14

sin 4x− 34

sin 2x + C

37.∫

tan4 x sec4 x dx =∫

tan4 x (tan2 x + 1) sec2 x dx

=∫

(tan6 x + tan4 x) sec2 x dx

=17

tan7 x +15

tan5 x + C

38.∫

cot4 x csc4 x dx =∫

cot4 x(cot2 x + 1) csc2 x dx = −17

cot7 x− 15

cot5 x + C

39.∫

sin(x/2) cos 2x dx =∫

12 (sin

(52x)− sin

(32x)dx = 1

3 cos(

32x)− 1

5 cos(

52x)

+ C

40.∫ 2π

0

sin2 ax dx =∫ 2π

0

(12− 1

2cos 2ax

)dx =

[x

2− 1

4asin 2ax

]2π0

= π − sin 4πa4a

41.

{u = tan x

du = sec2 x dx

∣∣∣∣ x = 0 ⇒ u = 0

x = π/4 ⇒ u = 1

};∫ π/4

0

tan3 x sec2 x dx =∫ 1

0

u3 du =[

14u

4]10

= 14

42.∫ π/2

π/4

csc3 x cotx dx =∫ π/2

π/4

csc2 x cscx cotx dx =[−1

3csc3 x

]π/2π/4

=2√

2 − 13



430 SECTION 8.3

43.∫ π/6

0

tan2 2x dx =∫ π/6

0

(sec2 2x− 1

)dx =

[12 tan 2x− x

]π/60

=√

32

− π

6

44.∫ π/3

0

tanx sec3/2 x dx =∫ π/3

0

sec1/2 x secx tanx dx =[23

sec3/2 x

]π/30

=4√

2 − 23

45. A =∫ π

0

sin2 x dx =∫ π

0

12 (1 − cos 2x) dx = 1

2

[x− 1

2 sin 2x]π0

=π

2

46. V =∫ π/2

−π/2

π cos2 x dx = π

[x

2+

sin 2x4

]π/2−π/2

=π2

2

47.V =

∫ π

0

π(sin2 x

)2dx = π

∫ π

0

sin4 x dx = π

∫ π

0

(12 (1 − cos 2x

)2dx

=π

4

∫ π

0

(1 − 2 cos 2x + cos2 2x

)dx

=π

4[x− sin 2x]π0 +

π

8

∫ π

0

(1 + cos 4x) dx

=π2

4+

π

8[x + 1

4 sin 4x]π0

=3π2

8

48. V =∫ π/4

0

π(cos2 x− sin2 x) dx = π

∫ π/4

0

cos 2x dx =π

2[sin 2x]π/40 =

π

2

49. V =∫ π/4

0

π[12 − tan2 x

]dx = π

∫ π/4

0

[2 − sec2 x

]dx = π [2x− tanx]π/40 =

π2

2− π

50. V =∫ π/4

0

π tan4 x dx = π

[13

tan3 x− tanx + x

]π/40

= π

(π

4− 2

3

)

51.V =

∫ π/4

0

π[(tanx + 1)2 − 12

]dx = π

∫ π/4

0

[tan2 x + 2 tanx

]dx

= π

∫ π/4

0

(sec2 x + 2 tanx− 1

)dx

= π [tanx + 2 ln | secx| − x]π/40 = π[ln 2 + 1 − π

4

]

52. V =∫ π/4

0

π sec4 x dx = π

[13

tan3 x + tanx

]π/40

=4π3

53. (a)∫

sinn x dx =∫

sinn−1 x sin x dx;

{u = sinn−1 x dv = sin x dx

du = (n− 1) sinn−2 x dx v = − cos x

}∫

sinn x dx = − sinn−1 x cos x + (n− 1)∫

sinn−2 x cos2 x dx

= − sinn−1 x cos x + (n− 1)∫

sinn−2 x (1 − sin2 x) dx



SECTION 8.3 431

Therefore,

n

∫sinn x dx = − sinn−1 x cos x + (n− 1)

∫sinn−2 x dx

and ∫sinn x dx = − 1

nsinn−1 x cos x +

n− 1n

∫sinn−2 x dx.

(b)∫ π/2

0

sinn x dx =[− 1n

sinn−1 x cos x

]π/20

+n− 1n

∫ π/2

0

sinn−2 x dx =n− 1n

∫ π/2

0

sinn−2 x dx

(c) n even: ∫ π/2

0

sinn x dx =n− 1n

∫ π/2

0

sinn−2 x dx =n− 1n

n− 3n− 2

∫ π/2

0

sinn−4 x dx

and so on,∫ π/2

0

sinn x dx =n− 1n

n− 3n− 2

· · · 34

12

∫ π/2

0

1 dx =n− 1n

n− 3n− 2

· · · 34

12π

2.

n odd:∫ π/2

0

sinn x dx =n− 1n

n− 3n− 2

· · · 45

23

∫ π/2

0

sin x dx

=n− 1n

n− 3n− 2

· · · 45

23

[− cos x

]π/20

=n− 1n

n− 3n− 2

· · · 45

23

.

54.∫ π/2

0

cosn x dx =∫ π/2

0

sinn(π

2− x)dx = −

∫ 0

π/2

sinn u du =∫ π/2

0

sinn u du

55. (a)∫ π/2

0

sin7 x dx =6 · 4 · 27 · 5 · 3 =

1635

(b)∫ π/2

0

cos6 x dx =(

5 · 3 · 16 · 4 · 2

)π

2=

5π32

56. (a)∫ π

0

π (x + sin 2x)2 dx =π4

3− π2

2∼= 27.5349

(b) ∫ π

0

π (x + sin 2x)2 dx = π

∫ π

0

(x2 + 2x sin 2x + sin2 2x

)dx

= π[

13 x

3 − x cos 2x + 12 sin 2x + 1

2x− 18 sin 4x

]π0

=π4

3− π2

2

57. (a) V ∼= 4.9348

(b) V =∫ √

π

0

2πx sin2(x2)dx = π

∫ π

0

sin2 u du = 12π

∫ π

0

(1 − cos 2u) du

u = x2∧

= 12π[u− 1

2 sin 2u]π0

= 12π

2 ∼= 4.9348



432 SECTION 8.4

58. (a) The x-coordinates of the points of intersection are: x1 = 1.7918, x2 = 4.4914;

A =∫ x2

x1

[sin(x/2) − 1 − cosx] dx ∼= 1.751

(b) V =∫ x2

x1

π[sin2(x/2) − (1 + cosx)2

]dx ∼= 6.173

SECTION 8.4

1.

{x = a sinu

dx = a cosu du

};

∫dx√

a2 − x2=∫

a cosu dua cosu

=∫

du = u + C = arcsin(xa

)+ C

2.

{u = x2 − 4

du = 2x dx

};

∫ 4

5/2

x√x2 − 4

dx =12

∫ 12

9/4

u−1/2 du =[u1/2

]129/4

=√

12 − 32

=4√

3 − 32

3.

{x = secu

dx = secu tanu du

};

∫ √x2 − 1 dx =

∫tan2 u secu du

=∫

(sec3 u− secu) du

=12

secu tanu− 12

ln | secu + tanu| + C

Example 8, Section 8.3

=12x√x2 − 1 − 1

2ln |x +

√x2 − 1| + C

4.∫

x√4 − x2

dx = −∫ −2x

2√

4 − x2dx = −

∫du

2√u

= −√u + C = −

√4 − x2 + C

5.

{x = 2 sinu

dx = 2 cosu du

};

∫x2

√4 − x2

dx =∫

4 sin2 u

2 cosu2 cosu du

= 2∫

(1 − cos 2u) du

= 2u− sin 2u + C

= 2u− 2 sinu cosu + C

= 2 arcsin(x

2

)− 1

2x√

4 − x2 + C

6.

{x = 2 secu

dx = 2 secu tanu du

};∫

x2

√x2 − 4

dx =∫

8 sec3 u tanu√4(sec2 u− 1)

du

= 4∫

sec3 u du

= 2 secu tanu + 2 ln | secu + tanu| + C1

=12x√x2 − 4 + 2 ln

∣∣∣∣x +√x2 − 42

∣∣∣∣+ C1

=12x√x2 − 4 + 2 ln |x +

√x2 − 4| + C (C = C1 − 2 ln 2)



SECTION 8.4 433

7.

{u = 1 − x2

du = −2x dx

};

∫x

(1 − x2)3/2dx = −1

2

∫du

u3/2= u−1/2 + C =

1√1 − x2

+ C

8.{

x = 2 tanu

dx = 2 sec2 u du

};

∫x2

√4 + x2

dx =∫

8 tan2 u sec2 u√4(1 + tan2 u)

du = 4∫

tan2 u secu du

= 4∫

(sec3 u− secu) du

= 4(

12

secu tanu− 12

ln | secu + tanu|)

+ C

=12x√x2 + 4 − 2 ln

(x +

√x2 + 4

)+ C

(absorbing 2 ln 2 into C)

9.{

x = sinu

dx = cosu du

};

∫x2

(1 − x2)3/2dx =

∫sin2 u

cos3 ucosu du =

∫tan2 u du

=∫

(sec2 u− 1) du = tanu− u + C

=x√

1 − x2− arcsinx + C

∫ 1/2

0

x2

(1 − x2)3/2dx =

[x√

1 − x2− arcsinx

]1/20

=2√

3 − π

6

10.{

u = a2 + x2

dx = 2x dx

};

∫x

a2 + x2dx =

12

∫du

u=

12

ln |u| + C

=12

ln(a2 + x2) + C

11.{

u = 4 − x2

du = −2x dx

};

∫x√

4 − x2 dx = −12

∫u1/2 du = −1

3u3/2 + C

= −13(4 − x2)3/2 + C

12.{

x = 4 sinu

dx = 4 cosu du

};

∫x3

√16 − x2

dx =256 sin3 u cosu√

16 − 16 sin2 udu = 64

∫sin3 u du

= 64∫

(1 − cos2 u) sinu du = 64(− cosu +

cos3 u3

)+ C

=13

(√16 − x2

)3

− 16√

16 − x2 + C∫ 2

0

x3

√16 − x2

dx =[13(16 − x2)3/2 + 64

√16 − x2

]20

=1283

− 12√

12



434 SECTION 8.4

13.

{x = 5 sinu| x = 0 =⇒ u = 0

dx = 5 cosu du| x = 5 =⇒ u = π/2

};

∫ 5

0

x2√

25 − x2 dx =∫ π/2

0

(5 sinu)2(5 cosu)2 du

= 625∫ π/2

0

(sin2 u− sin4 u

)du

= 625[12· π

2− 3 · 1

4 · 2 · π2

]=

625π16

[see Exercise 53, Section 8.3]

14.

{x = sinu

dx = cosu du

};

∫ √1 − x2

x4dx =

∫cos2 usin4 u

du =∫

cot2 u csc2 u du

= −cot3 u3

+ C = − (1 − x2)3/2

3x3+ C

15.

{x =

√8 tanu

dx =√

8 sec2 u du

};∫

x2

(x2 + 8)3/2dx =

∫8 tan2 u

(8 sec2 u)3/2√

8 sec2 u du

=∫

tan2 u

secudu =

∫sec2 u− 1

secudu

=∫

(secu− cosu) du

= ln | secu + tanu| − sinu + C

= ln

(√x2 + 8 + x√

8

)− x√

x2 + 8+ C

(absorb − ln√

8 in C)

= ln(√

x2 + 8 + x)− x√

x2 + 8+ C

16.∫ a

0

√a2 − x2 dx =

14(Area of circle of radius a) =

πa2

4

17.

{x = a sinu

dx = a cosu du

};

∫dx

x√a2 − x2

=∫

a cosu dua sinu (a cosu)

=1a

∫cscu du

=1a

ln | cscu− cotu| + C

=1a

ln∣∣∣∣a−

√a2 − x2

x

∣∣∣∣+ C

18.

{x = secu

dx = secu tanu du

};∫ √

x2 − 1x

dx =∫

tanu

secusecu tanu du

=∫

tan2 u du =∫

(sec2 u− 1) du

= tanu− u + C =√x2 − 1 − arctan

√x2 − 1 + C



SECTION 8.4 435

19.

{x = 3 tanu

dx = 3 sec2 u du

};

∫x3

√9 + x2

dx =∫

27 tan3 u

3 secu· 3 sec2 u du

= 27∫

tan3 u secu du

= 27∫ (

sec2 u− 1)secu tanu du

= 27[13 sec3 u− secu

]+ C

= 13

(9 + x2

)3/2 − 9(9 + x2

)1/2 + C∫ 3

0

x3

√9 + x2

dx =[

13

(9 + x2

)3/2 − 9(9 + x2

)1/2]30

= 18 − 9√

2

20.

{x = a sinu

dx = a cosu du

};

∫dx

x2√a2 − x2

=∫

a cosua2 sin2 u a cosu

du

=1a2

∫csc2 u du = − 1

a2cotu + C

= −√a2 − x2

a2x+ C

21.

{x = a tanu

dx = a sec2 u du

};

∫dx

x2√a2 + x2

=∫

a sec2 u du

a2 tan2 u (a secu)

=1a2

∫secutan2 u

du

=1a2

∫cotu cscu du

= − 1a2

cosu + C = − 1a2x

√a2 + x2 + C

22.

{x =

√2 tanu

dx =√

2 sec2 u du

};

∫dx

(x2 + 2)3/2=∫ √

2 sec2 u du

(2 tan2 u + 2)3/2

=∫ √

2 sec2 u

2√

2 sec3 udu

=12

∫cosu du =

12

sinu + C

=12

x√x2 + 2

+ C

23.

{x =

√5 sinu

dx =√

5 cosu du

};

∫dx

(5 − x2)3/2=∫ √

5 cosu du(5 cos2 u )3/2

=15

∫sec2 u du

=15

tanu + C =x

5√

5 − x2+ C

∫ 1

0

dx

(5 − x2)3/2=[

x

5√

5 − x2

]10

=110



436 SECTION 8.4

24.

{ex = 2 tanu

ex dx = 2 sec2 u du

};∫

dx

ex√

4 + e2x=∫

ex

e2x√

4 + e2xdx =

∫2 sec2 u

4 tan2 u · 2 secudu

=14

∫cosusin2 u

du = −14· 1sinu

+ C

= −14

√4 + e2x

ex+ C

25.

{x = a secu

dx = a secu tanu du

};

∫dx

x2√x2 − a2

=∫

a secu tanu du

a2 sec2 u (a tanu)

=1a2

∫cosu du

=1a2

sinu + C

=1

a2x

√x2 − a2 + C

26.

{u = ex

du = ex dx

};

∫ex√

9 − e2xdx =

∫du√

9 − u2

= arcsin(u

3

)+ C = arcsin

(ex

3

)+ C

27.

{ex = 3 secu

exdx = 3 secu tanu du

};∫

dx

ex√e2x − 9

=∫

tanu du

3 secu (3 tanu)

=19

∫cosu du

= 19 sinu + C

= 19e

−x√e2x − 9 + C

28.

{x− 1 = 2 secu

dx = 2 secu tanu du

};∫

dx√x2 − 2x− 3

=∫

dx√(x− 1)2 − 4

=∫

2 secu tanu

2 tanudu

=∫

secu du = ln | secu + tanu| + C

= ln∣∣∣x− 1 +

√(x− 1)2 − 4

∣∣∣+ C

29. (x2 − 4x + 4)32 =

⎧⎨⎩(x− 2)3, x > 2

(2 − x)3, x < 2

∫dx

(x2 − 4x + 4)3/2=

⎧⎪⎨⎪⎩

− 12(x− 2)2

+ C, x > 2

12(2 − x)2

+ C, x < 2



SECTION 8.4 437

30.

{x− 3 = 3 sinu

dx = 3 cosu du

};

∫x√

6x− x2dx =

∫x√

9 − (x− 3)2dx =

∫3 + 3 sinu

3 cosu· 3 cosu du

= 3∫

(1 + sinu) du = 3u− 3 cosu + C

= 3 arcsin(x− 3

3

)−√

6x− x2 + C

31.

{x− 3 = sinu

dx = cosu du

};

∫x√

6x− x2 − 8 dx =∫

x√

1 − (x− 3)2 dx

=∫

(3 + sinu)(cosu) cosu du

=∫

(3 cos2 u + cos2 u sinu) du

=∫ [

3(

1 + cos 2u2

)+ cos2 u sinu

]du

=3u2

+34

sin 2u− 13

cos3 u + C

=32

arcsin (x− 3) +32(x− 3)

√6x− x2 − 8 − 1

3(6x− x2 − 8)3/2 + C

32.

{x + 2 = 3 tanu

dx = 3 sec2 u du

};

∫x + 2√

x2 + 4x + 13dx =

∫x + 2√

(x + 2)2 + 9dx

=∫

3 tanu

3 secu· 3 sec2 u du = 3

∫secu tanu du

= 3 secu + C =√x2 + 4x + 13 + C

33.

{x + 1 = 2 tanu

dx = 2 sec2 u du

};

∫x

(x2 + 2x + 5)2dx =

∫x

[(x + 1)2 + 4]2dx

=∫

2 tanu− 1(4 sec2 u)2

2 sec2 u du

=18

∫2 tanu− 1

sec2 udu

=18

∫(2 sinu cosu− cos2 u) du

=18

∫ (2 sinu cosu− 1 + cos 2u

2

)du

=18

(sin2 u− u

2− sin 2u

4

)+ C



438 SECTION 8.4

=18

⎡⎢⎢⎣(

x + 1√x2 + 2x + 5

)2

− 12

arctan(x + 1

2

)− 1

4

sin 2u=2sinu cosu︷︸︸︷(2)(

x + 1√x2 + 2x + 5

)(2√

x2 + 2x + 5

)⎤⎥⎥⎦+ C

=x2 + x

8(x2 + 2x + 5)− 1

16arctan

(x + 1

2

)+ C

34.

{x− 1 = 2 secu

dx = 2 secu tanu du

};

∫x√

x2 − 2x− 3dx =

∫x√

(x− 1)2 − 4dx

=∫

1 + 2 secu2 tanu

· 2 secu tanu du

=∫

(secu + 2 sec2 u) du

= ln | secu + tanu| + 2 tanu + C

= ln∣∣∣x− 1 +

√x2 − 2x− 3

∣∣∣+√x2 − 2x− 3 + C

35.

u = sec−1 x

du =1

x√x2 − 1

dx

dv = dx

v = x

∫sec−1 x dx = x sec−1 x−

∫1√

x2 − 1dx

x = secu dx = secu tanu du = x sec−1 x−∫

secu du

= x sec−1 x− ln [secu + tanu] + C

= x sec−1 x− ln[x +

√x2 − 1

]+ C

36. (a) Set u =√a2 − x2. Then u2 = a2 − x2 and 2u du = −2x dx.

∫ √a2 − x2

xdx =

∫ √a2 − x2

x2x dx = −

∫u

a2 − u2du

=∫

u2

u2 − a2du

=∫ (

1 − a2

a2 − u2

)du

= u +a

2ln∣∣∣∣a− u

a + u

∣∣∣∣+ C Formula 96

=√a2 − x2 +

a

2ln

∣∣∣∣∣a−√a2 − x2

a +√a2 − x2

∣∣∣∣∣+ C



SECTION 8.4 439

(b) Set x = a sinu. Then dx = a cosu du and√a2 − x2 = a cosu.

∫ √a2 − x2

xdx =

∫a cosua sinu

a cosu du

= a

∫cos2 usinu

du = a

∫1 − sin2 usinu du

= a

∫(cscu− sinu) du

= a ln |cscu− cotu| + a cosu + C

=√a2 − x2 + a ln

∣∣∣∣∣ax −√a2 − x2

x

∣∣∣∣∣+ C

=√a2 − x2 + a ln

∣∣∣∣∣a−√a2 − x2

x

∣∣∣∣∣+ C

(c)a−

√a2 − x2

a +√a2 − x2

=

(a−

√a2 − x2

x

)2

37. Let x = a tanu. Then dx = a sec2 u du,√x2 + a2 = a secu, and

∫dx

(x2 + a2)n=∫

a sec2 u

a2n sec2n udu =

1a2n−1

∫cos2n−2 u du

38.

∫1

(x2 + 1)2dx =

∫cos2 u du =

12

∫(1 + cos 2u) du

=12

[u +

12

sin 2u]

+ C

=12

[u + sinu cosu] + C

=12

[arctanx +

x

x2 + 1

]+ C

39.

∫1

(x2 + 1)3dx =

∫cos4 u du

=14

cos3 u sinu +38

cosu sinu +38u + C [by the reduction formula (8.3.2)]

=38

arctanx +3x

8 (x2 + 1)+

x

4 (x2 + 1)2+ C



440 SECTION 8.4

40.u = arctanx

du =1

1 + x2dx

dv = xdx

v =x2

2

∫x arctanx dx =

x2

2arctanx− 1

2

∫x2

1 + x2dx

x = tanu dx = sec2 u du =x2

2arctanx− 1

2

∫tan2 u

1 + tan2 usec2 u du

=x2

2arctanx− 1

2

∫tan2 u du

=x2

2arctanx− 1

2tanu +

12u + C

=x2

2arctanx− 1

2x +

12

arctanx + C

= arctanx

(x2

2+

12

)− 1

2x + C

41.u = arcsinx

du =1√

1 − x2dx

dv = x dx

v =x2

2

∫x arcsinx dx =

x2

2arcsinx− 1

2

∫x2

√1 − x2

dx

x = sinu dx = cosu du =x2

2arcsinx− 1

2

∫sin2 u du

=x2

2arcsinx− 1

2

[12u− 1

4sin 2u

]+ C

=x2

2arcsinx− 1

4arcsinx +

18(2x√

1 − x2) + C

= arcsinx

(x2

2− 1

4

)+

14x√

1 − x2 + C

42.x = 3 secu dx = 3 secu tanu du

∫ 5

3

√x2 − 9x

dx = 3∫ b

0

tan2 u du (where b = sec−1 53)

= 3 [tanu− u]b0

= 4 − 3 sec−1 53



SECTION 8.4 441

43.V =

∫ 1

0

π

(1

1 + x2

)2

dx = π

∫ 1

0

1(1 + x2)2

dx

= π

∫ π/4

0

cos2 u du [x = tanu, see Ex.45]

=π

2

∫ π/4

0

(1 + cos 2u) du

=π

2[u + 1

2 sin 2u]π/40

=π2

8+

π

4

44. A = 2∫ √

r2−h2

0

(√r2 − x2 − h

)dx = 2

[12x√r2 − x2 +

12r2 arcsin

(xr

)− hx

]√r2−h2

0

= r2 arcsin

(√r2 − h2

r

)− h√r2 − h2

45. We need only consider angles θ between 0 and π. Assume first that 0 ≤ θ ≤ π2 .

The area of the triangle is: 12r

2 sin θ cos θ.

x

y

The area of the other region is given by:

∫ r

r cos θ

√r2 − x2 dx =

[x

2

√r2 − x2 +

r2

2arcsin

x

r

]rr cos θ

=πr2

4− r2

2sin θ cos θ − r2

2arcsin (cos θ)

=r2θ

2− r2

2sin θ cos θ

Thus, the area of the sector is A = 12r

2θ. Ifπ

2< θ ≤ π, then

A = 12πr

2 − 12r

2(π − θ) = 12r

2θ.

46. A = 4∫ b

0

a

b

√b2 + y2dy =

4ab

[y

2

√b2 + y2 +

b2

2ln(y +

√b2 + y2

)]b0

= 2ab[√

2 + ln(1 +

√2)]

47. A = 2∫ 5

3

4

√x2

9− 1dx = 24

∫ 53

1

√u2 − 1du (where u = x

3 )

= 24[u

2

√u2 − 1 − 1

2ln |u +

√u2 − 1|

] 53

1

=803

− 12 ln 3.

48. V = 2∫ b+a

b−a

2πx√a2 − (x− b)2dx = 4π

∫ a

−a

(u + b)√a2 − u2 du = 2π2a2b.



442 SECTION 8.4

49. M =∫ a

0

dx√x2 + a2

=[ln (x +

√x2 + a2)

]a0

= ln (1 +√

2)

xMM =∫ a

0

x√x2 + a2

dx =[√

x2 + a2]a0

= (√

2 − 1)a xM =(√

2 − 1)aln (1 +

√2)

50. M =∫ a

0

(x2 + a2)−3/2 dx =[

1a2

sinu

]π/40

=√

22a2

xM M =∫ a

0

x

(x2 + a2)3/2dx =

[−(x2 + a2)−1/2

]a0

=2 −

√2

2a=⇒ xM = (

√2 − 1)a

51.

A =∫ √

2a

a

√x2 − a2 dx =

[12x√x2 − a2 − 1

2a2 ln

∣∣x +√x2 − a2

∣∣]√2a

a

= 12a

2[√

2 − ln (√

2 + 1)]

xA =∫ √

2a

a

x√x2 − a2 dx =

13a3, yA =

∫ √2a

a

[12(x2 − a2)

]dx =

16a3(2 −

√2)

x =2a

3[√

2 − ln (√

2 + 1)], y =

(2 −√

2)a3[√

2 − ln (√

2 + 1)]

52. Using y and A found in Exercise 51,

Vx = 2πyA = 2π(2 −

√2)a

3[√

2 − ln(√

2 + 1)] · 1

2a2[√

2 − ln(√

2 + 1)]

=13πa3(2 −

√2)

xVx =∫ √

2a

a

πx(x2 − a2) dx =14πa4 =⇒ x =

3a4(2 −

√2)

=38a(2 +

√2)

53. Vy = 2πRA =23πa3 yVy =

∫ √2a

a

πx(x2 − a2) dx =14πa4, y =

38a

54. Let x = a tanu. Then dx = a sec2 u du and√x2 − a2 = a secu.∫

1√a2 + x2

dx =∫


= ln

∣∣∣∣∣√a2 + x2

a+

x

a

∣∣∣∣∣+ C

= ln |x +√a2 + x2| + K, K = C − ln a

55. Let x = a secu. Then dx = a secu tanu du and√x2 − a2 = a tanu.∫

1√x2 − a2

dx =∫


= ln

∣∣∣∣∣xa +√x2 − a2

a

∣∣∣∣∣+ C

= ln |x +√x2 − a2| + K, K = C − ln a



SECTION 8.5 443

56. (a)(b) A =

∫ 1/2

0

x2

√1 − x2

dx

=∫ π/6

0

sin2 u du =[u

2− sin 2u

4

]π/60

=2π − 3

√3

24

(c) V =∫ 1/2

0

πx4

1 − x2dx =

∫ π/6

0

π sin4 u

cos2 ucosu du

= π

∫ π/6

0

(secu− cosu− sin2 u cosu) du

= π

[ln | secu + tanu| − sinu− 1

3sin3 u

]π/60

= π

(ln√

3 − 1324

)

57. (a)

(b) A =∫ 6

3

√x2 − 9x2

dx

=∫ π/3

0

3 tanu

9 sec2 u· 3 secu tanu du [x = 3 secu]

=∫ π/3

0

(secu− cosu) du

= [ ln | secu + tanu| − sinu]π/30

= ln(2 +√

3) −√

32

(c) xA =∫ 6

3

x ·√x2 − 9x2

dx =∫ π/3

0

(3 sec2 u− 3

)du = [3 tanu− 3u]π/30 = 3

√3 − π

Thus, x =2(3√

3 − π)

2 ln(2 +

√3)−√

3.

yA =∫

12

(√x2 − 9x2

)2

dx =12

∫ 6

3

(1x2

− 9x4

)dx =

12

[− 1

x+

3x3

]63

=5

144

Thus, y =5

72[2 ln

(2 +

√3)−√

3] .

SECTION 8.5

1.1

x2 + 7x + 6=

1(x + 1)(x + 6)

=A

x + 1+

B

x + 6

1 = A(x + 6) + B(x + 1)

x = −6: 1 = −5B =⇒ B = −1/5

x = −1: 1 = 5A =⇒ A = 1/5

1x2 + 7x + 6

=1/5x + 1

− 1/5x + 6



444 SECTION 8.5

2.x2

(x− 1)(x2 + 4x + 5)=

A

x− 1+

Bx + C

x2 + 4x + 5

x2 = A(x2 + 4x + 5) + (Bx + C)(x− 1)

x = 1 : 1 = 10A =⇒ A = 1/10

x = 0 : 0 = 5A− C =⇒ C = 5A = 1/2

Coefficient of x2 is A + B = 1 =⇒ B = 1 −A = 9/10

R(x) =110

x− 1+

910x + 1

2

x2 + 4x + 5

3.x

x4 − 1=

x

(x2 + 1)(x + 1)(x− 1)=

Ax + B

x2 + 1+

C

x + 1+

D

x− 1

x = (Ax + B)(x2 − 1) + C(x− 1)(x2 + 1) + D(x + 1)(x2 + 1)

x = 1: 1 = 4D =⇒ D = 1/4

x = −1: −1 = −4C =⇒ C = 1/4

x = 0: −B − C + D = 0 =⇒ B = 0

x = 2: 6A + 5C + 15D = 2 =⇒ A = −1/2x

x4 − 1=

1/4x− 1

+1/4x + 1

− x/2x2 + 1

4.x4

(x− 1)3=

[(x− 1) + 1]4

(x− 1)3=

(x− 1)4 + 4(x− 1)3 + 6(x− 1)2 + 4(x− 1) + 1(x− 1)3

= x + 3 +6

x− 1+

4(x− 1)2

+1

(x− 1)3

5.x2 − 3x− 1x3 + x2 − 2x

=x2 − 3x− 1

x(x + 2)(x + 1)=

A

x+

B

x + 2+

C

x− 1

x2 − 3x− 1 = A(x + 2)(x− 1) + Bx(x− 1) + Cx(x + 2)

x = 0: −1 = −2A =⇒ A = 1/2

x = −2: 9 = 6B =⇒ B = 3/2

x = 1: −3 = 3C =⇒ C = −1

x2 − 3x− 1x3 + x2 − 2x

=1/2x

+3/2x + 2

− 1x− 1

6.x3 + x2 + x + 2x4 + 3x2 + 2

=x3 + x2 + x + 2(x2 + 1)(x2 + 2)

=Ax + B

x2 + 1+

Cx + D

x2 + 2

x3 + x2 + x + 2 = (Ax + B)(x2 + 2) + (Cx + D)(x2 + 1)

= (A + C)x3 + (B + D)x2 + (2A + C)x + 2B + D

A + C = 1 = 2A + 1 =⇒ A = 0, C = 1; B + D = 1, 2B + D = 2 =⇒ B = 1, D = 0

R(x) =1

x2 + 1+

x

x2 + 2

7.2x2 + 1

x3 − 6x2 + 11x− 6=

2x2 + 1(x− 1)(x− 2)(x− 3)

=A

x− 1+

B

x− 2+

C

x− 3

2x2 + 1 = A(x− 2)(x− 3) + B(x− 1)(x− 3) + C(x− 1)(x− 2)



SECTION 8.5 445

x = 1: 3 = 2A =⇒ A = 3/2

x = 2: 9 = −B =⇒ B = −9

x = 3: 19 = 2C =⇒ C = 19/2

2x2 + 1x3 − 6x2 + 11x− 6

=3/2x− 1

− 9x− 2

+19/2x− 3

8.1

x(x2 + 1)2=

A

x+

Bx + C

x2 + 1+

Dx + E

(x2 + 1)2

1 = A(x2 + 1)2 + (Bx + C)x(x2 + 1) + (Dx + E)x

Constant term: A=1, x4 term: 0 = A + B =⇒ B = −1

x3 term: 0 = C, x2 term: 0 = 2A + B + D =⇒ D = −1

x term: 0 = E

R(x) =1x− x

x2 + 1− x

(x2 + 1)2

9.7

(x− 2)(x + 5)=

A

x− 2+

B

x + 57 = A(x + 5) + B(x− 2)

x = −5: 7 = −7B =⇒ B = −1

x = 2: 7 = 7A =⇒ A = 1∫7

(x− 2)(x + 5)dx =

∫ (1

x− 2− 1

x + 5

)= ln |x− 2| − ln |x + 5| + C = ln

∣∣∣∣x− 2x + 5

∣∣∣∣+ C

10.∫

x

(x + 1)(x + 2)(x + 3)dx =

∫ (−1/2x + 1

+2

x + 2− 3/2

x + 3

)dx

= −12

ln |x + 1| + 2 ln |x + 2| − 32

ln |x + 3| + C

11. We carry out the division until the numerator has degree smaller than the denominator:

2x4 − 4x3 + 4x2 + 3x3 − x2

= 2x− 2 +2x2 + 3x2(x− 1)

2x2 + 3x2(x− 1)

=A

x+

B

x2+

C

x− 1=⇒ 2x2 + 3 = Ax(x− 1) + B(x− 1) + Cx2

x = 0: 3 = −B =⇒ B = −3

x = 1: 5 = C =⇒ C = 5

x = −1: 5 = 2A− 2B + C =⇒ A = −3

∫ (2x− 2 +

2x2 + 3x2(x− 1)

)dx = x2 − 2x +

∫ (− 3x− 3

x2+

5x− 1

)dx

= x2 − 2x− 3 ln |x| + 3x

+ 5 ln |x− 1| + C

12.∫

x2 + 1x(x2 − 1)

dx =∫ (−1

x+

1x− 1

+1

x + 1

)dx = − ln |x| + ln |x− 1| + ln |x + 1| + C = ln

∣∣∣∣x2 − 1x

∣∣∣∣+ C



446 SECTION 8.5

13. We carry out the division until the numerator has degree smaller than the denominator:

x5

(x− 2)2=

x5

x2 − 4x + 4= x3 + 4x2 + 12x + 32 +

80x− 128(x− 2)2

.

Then,

80x− 128(x− 2)2

=80x− 160 + 32

(x− 2)2=

80x− 2

+32

(x− 2)2

∫x5

(x− 2)2dx =

∫ (x3 + 4x2 + 12x + 32 +

80x− 2

+32

(x− 2)2

)dx

=14x4 +

43x3 + 6x2 + 32x + 80 ln |x− 2| − 32

x− 2+ C.

14.∫

x5

x− 2dx =

∫ (x4 + 2x3 + 4x2 + 8x + 16 +

32x− 2

)dx

=15x5 +

12x4 +

43x3 + 4x2 + 16x + 32 ln |x− 2| + C

15.x + 3

x2 − 3x + 2=

A

x− 1+

B

x− 2

x + 3 = A(x− 2) + B(x− 1)

x = 1: 4 = −A =⇒ A = −4

x = 2: 5 = B =⇒ B = 5∫x + 3

x2 − 3x + 2dx =

∫ ( −4x− 1

+5

x− 2

)dx = −4 ln |x− 1| + 5 ln |x− 2| + C

16.∫

x2 + 3x2 − 3x + 2

dx =∫ (

1 +7

x− 2− 4

x− 1

)dx = x− 7 ln |x− 2| − 4 ln |x− 1| + C

17.∫

dx

(x− 1)3=∫

(x− 1)−3 dx = −12(x− 1)−2 + C = − 1

2(x− 1)2+ C

18.∫

dx

x2 + 2x + 2=∫

dx

(x + 1)2 + 1= arctan (x + 1) + C

19.x2

(x− 1)2(x + 1)=

A

x− 1+

B

(x− 1)2+

C

x + 1

x2 = A(x− 1)(x + 1) + B(x + 1) + C(x− 1)2

x = 1: 1 = 2B =⇒ B = 1/2

x = −1: 1 = 4C =⇒ C = 1/4



SECTION 8.5 447

x = 0: 0 = −A + B + C =⇒ A = 3/4

∫x2

(x− 1)2(x + 1)dx =

∫ (3/4x− 1

+1/2

(x− 1)2+

1/4x + 1

)dx

=34

ln |x− 1| − 12(x− 1)

+14

ln |x + 1| + C

20.∫

2x− 1(x + 1)2(x− 2)2

dx =∫ [ −1/3

(x + 1)2+

1/3(x− 2)2

]dx =

13

(1

x + 1− 1

x− 2

)+ C

21. x4 − 16 = (x2 − 4)(x2 + 4) = (x− 2)(x + 2)(x2 + 4)

1x4 − 16

=A

x− 2+

B

x + 2+

Cx + D

x2 + 4

1 = A(x + 2)(x2 + 4) + B(x− 2)(x2 + 4) + (Cx + D)(x2 − 4)

x = 2: 1 = 32A =⇒ A = 1/32

x = −2: 1 = −32B =⇒ B = −1/32

x = 0: 1 = 8A− 8B − 4D =⇒ D = −1/8

x = 1: 1 = 15A− 5B − 3C − 3D =⇒ C = 0

∫dx

x4 − 16=∫ (

1/32x− 2

− 1/32x + 2

− 1/8x2 + 4

)dx

=132

ln |x− 2| − 132

ln |x + 2| − 18

(12

arctanx

2

)+ C

=132

ln∣∣∣∣x− 2x + 2

∣∣∣∣− 116

arctanx

2+ C

22. Divide the denominator into the numerator:3x5 − 3x2 + x

x3 − 1= 3x2 +

x

x3 − 1

∫ (3x2 +

x

x3 − 1

)dx = x3 +

13

∫ (1

x− 1+

1 − x

x2 + x + 1

)dx

= x3 +13

∫1

x− 1dx− 1

6

∫2x + 1

x2 + x + 1dx +

12

∫1

x2 + x + 1dx

= x3 +13

ln |x− 1| − 16

ln |x2 + x + 1) +12

∫1

(x + 12 )2 + 3

4

dx

= x3 +16

ln(x2 − 2x + 1x2 + x + 1

)+

1√3

arctan[

2√3

(x +

12

)]+ C



448 SECTION 8.5

23.x3 + 4x2 − 4x− 1

(x2 + 1)2=

Ax + B

x2 + 1+

Cx + D

(x2 + 1)2

x3 + 4x2 − 4x− 1 = (Ax + B)(x2 + 1) + (Cx + D)

x = 0: −1 = B + D =⇒ D = −B − 1

=⇒ B = 4, D = −5

x = 1: 0 = 2A + 2B + C + D =⇒ 6 = 4B + 2D

x = −1: 6 = −2A + 2B − C + D 6 = −2A + 8 − C − 5=⇒ A = 1,

C = −5x = 2: 15 = 10A + 5B + 2C + D 15 = 10A + 20 + 2C − 5

∫x3 + 4x2 − 4x− 1

(x2 + 1)2dx =

∫ (x

x2 + 1+

4x2 + 1

− 5x(x2 + 1)2

− 5(x2 + 1)2

)dx

(∗) =12

ln (x2 + 1) + 4 arctanx +5

2(x2 + 1)− 5

∫dx

(x2 + 1)2

For this last integral we set

{x = tanu

dx = sec2 u du

};∫

dx

(x2 + 1)2=∫

sec2 u du

(1 + tan2 u)2=∫

cos2 u du

=12

∫(1 + cos 2u) du

= 12

(u + 1

2 sin 2u)

+ C = 12 (u + sinu cosu) + C

=12

(arctanx +

x

1 + x2

)+ C.1

1 + x

2

x

u

Substituting this result in (∗) and rearranging the terms, we get

∫x3 + 4x2 − 4x + 1

(x2 + 1)2dx =

12

ln (x2 + 1) +32

arctanx +5(1 − x)2(1 + x2)

+ C.

24.∫

dx

(x2 + 16)2=

164

∫cos2 u du =

1128

[u +

sin 2u2

]+ C

=1

128arctan

(x4

)+

132

· x

(x2 + 16)+ C



SECTION 8.5 449

25.1

x4 + 4=

Ax + B

x2 + 2x + 2+

Cx + D

x2 − 2x + 2(using the hint)

1 = (Ax + B)(x2 − 2x + 2) + (Cx + D)(x2 + 2x + 2)

x = 0: 1 = 2B + 2D

x = 1: 1 = A + B + 5C + 5D

x = −1: 1 = −5A + 5B − C + D

x = 2: 1 = 4A + 2B + 20C + 10D

⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭

=⇒

A = 1/8

B = 1/4

C = −1/8

D = 1/4

∫dx

x4 + 4=

18

∫x + 2

x2 + 2x + 2dx− 1

8

∫x− 2

x2 − 2x + 2dx

=18

∫x + 1

x2 + 2x + 2dx +

18

∫dx

(x + 1)2 + 1− 1

8

∫x− 1

x2 − 2x + 2dx +

18

∫dx

(x− 1)2 + 1

=116

ln (x2 + 2x + 2) +18

arctan (x + 1) − 116

ln (x2 − 2x + 2) +18

arctan (x− 1) + C

=116

ln(x2 + 2x + 2x2 − 2x + 2

)+

18

arctan (x + 1) +18

arctan (x− 1) + C

26.∫

dx

x4 + 16=

√2

64

∫ (2x + 4

√2

x2 + 2√

2x + 4− 2x− 4

√2

x2 − 2√

2x + 4

)dx

=√

264

∫ (2x + 2

√2

x2 + 2√

2x + 4+

2√

2x2 + 2

√2x + 4

)dx

−√

264

∫ (2x− 2

√2

x2 − 2√

2x + 4− 2

√2

x2 − 2√

2x + 4

)dx

=√

264

ln |x2 + 2√

2x + 4| + 116

· 1√2

arctan

(x +

√2√

2

)

−√

264

ln |x2 − 2√

2x + 4| + 116

· 1√2

arctan

(x−

√2√

2

)+ C

=√

264

ln

(x2 + 2

√2x + 4

x2 − 2√

2x + 4

)+

√2

32arctan

(x +

√2√

2

)+

√2

32arctan

(x−

√2√

2

)+ C

27.x− 3x3 + x2

=x− 3

x2(x + 1)=

A

x+

B

x2+

C

x + 1

x− 3 = Ax(x + 1) + B(x + 1) + Cx2

x = 0: −3 = B

x = −1: −4 = C



450 SECTION 8.5

x = 1: −2 = 2A + 2B + C =⇒ A = 4∫x− 3x3 + x2

dx = 4∫

1xdx− 3

∫1x2

dx− 4∫

1x + 1

dx

= 4 ln |x| + 3x− 4 ln |x + 1| + C =

3x

+ 4 ln∣∣∣∣ x

x + 1

∣∣∣∣+ C

28.∫

1(x− 1)(x2 + 1)2

dx

=14

∫ [1

x− 1− x + 1

x2 + 1− 2x + 2

(x2 + 1)2

]dx

=14

∫ [1

x− 1− x

x2 + 1− 1

x2 + 1− 2x

(x2 + 1)2− 2

(x2 + 1)2

]dx

=14

ln |x− 1| − 18

ln(x2 + 1) − 14

arctanx +14· 1(x2 + 1)

− 14

(arctanx +

x

x2 + 1

)+ C

=18

ln(x2 − 2x + 1

x2 + 1

)− 1

2arctanx− x− 1

4(x2 + 1)+ C

29.x + 1

x3 + x2 − 6x=

x + 1x(x− 2)(x + 3)

=A

x+

B

x− 2+

C

x + 3

x + 1 = A(x− 2)(x + 3) + Bx(x + 3) + Cx(x− 2)

x = 0: 1 = −6A =⇒ A = −1/6

x = 2: 3 = 10B =⇒ B = 3/10

x = −3: −2 = 15C =⇒ C = −2/15

∫x + 1

x3 + x2 − 6xdx = − 1

6

∫1x dx + 3

10

∫1

x−2 dx− 215

∫1

x+3 dx

= − 16 ln |x| + 3

10 ln |x− 2| − 215 ln |x + 3| + C

30.∫

x3 + x2 + x + 3(x2 + 1)(x2 + 3)

dx =∫ (

1x2 + 1

+x

x2 + 3

)dx = arctanx +

12

ln(x2 + 3) + C

31.∫ 2

0

x

x2 + 5x + 6dx =

∫ 2

0

x

(x + 2)(x + 3)dx =

∫ 2

0

(3

x + 3− 2

x + 2

)dx

= [3 ln |x + 3| − 2 ln |x + 2|]20 = ln(

125108

)

32.∫ 3

1

1x3 + x

dx =∫ 3

1

(1x− x

x2 + 1

)dx =

[ln |x| − 1

2ln(x2 + 1)

]31

= ln(

3√5

)

33. ∫ 3

1

x2 − 4x + 3x3 + 2x2 + x

dx =∫ 3

1

x2 − 4x + 3x(x + 1)2

dx =∫ 3

1

(3x− 2

x + 1− 8

(x + 1)2

)dx

=[3 ln |x| − 2 ln |x + 1| + 8

x + 1

]31

= ln(

274

)− 2



SECTION 8.5 451

34.∫ 2

0

x3

(x2 + 2)2dx =

∫ 2

0

(x

x2 + 2− 2x

(x2 + 2)2

)dx =

[12

ln(x2 + 2) +1

x2 + 2

]20

= ln√

3 − 13

35. ∫cos θ

sin2 θ − 2 sin θ − 8dθ =

16

∫cos θ

sin θ − 4dθ − 1

6

∫cos θ

sin θ + 2dθ

=16

ln | sin θ − 4| − 16

ln | sin θ + 2| + C

=16

ln∣∣∣∣ sin θ − 4sin θ + 2

∣∣∣∣+ C

36. ∫et

e2t + 5et + 6dt =

∫et

et + 2dt−

∫et

et + 3dt

= ln |et + 2| − ln |et + 3| + C

= ln∣∣∣∣et + 2et + 3

∣∣∣∣+ C

37. ∫1

t[(ln t)2 − 4]dt =

14

∫1

t(ln t− 2)dt− 1

4

∫1

t(ln t + 2)dt

=14

ln | ln t− 2| − 14

ln | ln t + 2| + C

=14

ln∣∣∣∣ ln t− 2ln t + 2

∣∣∣∣+ C

38. ∫sec2 θ

tan3 θ − tan2 θdθ =

∫sec2 θ

tan θ − 1dθ −

∫sec2 θ

tan θdθ −

∫sec2 θ

tan2 θdθ

= ln | tan θ − 1| − ln | tan θ| + cot θ + C

= ln∣∣∣∣ tan θ − 1

tan θ

∣∣∣∣+ cot θ + C

39. First we carry out the division:u

a + bu=

1b− a/b

a + bu.

∫u

a + budu =

∫ (1b− a/b

a + bu

)du

=1bu− a

b2ln |a + bu| + K

=1b2

(a + bu− a ln |a + bu|) + C, C = K − a

b2



452 SECTION 8.5

40. ∫1

u(a + bu)du =

1a

∫1udu− 1

a

∫b

a + budu

=1a(ln |u| − ln |a + bu|) + C

=1a

ln∣∣∣∣ u

a + bu

∣∣∣∣+ C

41. ∫1

u2(a + bu)du = − b

a2

∫1udu +

1a

∫1u2

du +b

a2

∫b

a + budu

=b

a2(ln |a + bu| − ln |u|) − 1

au+ C

=b

a2ln∣∣∣∣ u

a + bu

∣∣∣∣− 1au

+ C

42. ∫1

u(a + bu)2du =

1a2

∫1udu− 1

a2

∫b

a + budu− b

ab

∫b

(a + bu)2du

=1a2

(ln |u| − ln |a + bu|) +b

ab(a + bu)+ C

=1

a(a + bu)− 1

a2ln∣∣∣∣a + bu

u

∣∣∣∣+ C

43. ∫1

a2 − u2du =

12a

∫1

a + udu +

12a

∫1

a− udu

=12a

(ln∣∣∣∣a + u

a− u

∣∣∣∣)

+ C

44. ∫u

a2 − u2du = −1

2

∫1vdv, where v = a2 − u2

= −12

ln v + C

= −12

ln |a2 − u2| + C

45. ∫u2

a2 − u2du =

∫a2

a2 − u2du−

∫du

= a2

(12a

)(ln∣∣∣∣a + u

a− u

∣∣∣∣)− u + C by Exercise 44

= −u +a

2

(ln∣∣∣∣a + u

a− u

∣∣∣∣)

+ C



SECTION 8.5 453

46. If ad = bc, thena

b=

c

dand

1(a + bu)(c + du)

=1

bd(a/b + u)2.

Therefore

∫1

(a + bu)(c + du)du =

1bd

∫1

(a/b + u)2du

= − 1bd(a/b + u)

+ C = − 1d(a + bu)

+ C.

47.∫

1(a + bu)(c + du)

du = − 1ad− bc

∫b

a + budu +

1ad− bc

∫d

c + dudu

=1

ad− bc

(ln∣∣∣∣c + du

a + bu

∣∣∣∣)

+ C

48. Note that y =1

x2 − 1=

12

[1

x− 1− 1

x + 1

]

and thusd0y

dx0=(

12

)(−1)0 0!

[1

(x− 1)0+1− 1

(x + 1)0+1

].

The rest is a routine induction.

49. (a) V =∫ 3/2

0

π

(1√

4 − x2

)2

dx = π

∫ 3/2

0

14 − x2

dx =π

4ln∣∣∣∣2 + x

2 − x

∣∣∣∣3/20

= 14π ln 7

(b) V =∫ 3/2

0

2πx1√

4 − x2dx = π

∫ 3/2

0

2x√4 − x2

dx = −2π[√

4 − x2]3/20

= π[4 −

√7]

50.∫

x3 arctanx dx =x4

4arctanx− 1

4

∫x4

x2 + 1dx

u = arctanx

du =1

x2 + 1dx

dv = x3 dx

v =x4

4

=x4

4arctanx− 1

4

∫ (x2 − 1 +

1x2 + 1

)dx

51.A =

∫ 1

0

dx

x2 + 1= [arctanx]10 =

14π

xA =∫ 1

0

x

x2 + 1dx =

12[ln (x2 + 1)

]10

=12

ln 2

yA =∫ 1

0

dx

2(x2 + 1)2=

12

[arctanx +

x

x2 + 1

]10

=18(π + 2)

x =2 ln 2π

; y =π + 22π



454 SECTION 8.5

52. xVx = yVy =∫ 1

0

πx

(x2 + 1)2dx =

12

[− 1x2 + 1

]10

=14π

(a) Vx =∫ 1

0

π

(x2 + 1)2dx = π

[arctanx +

x

x2 + 1

]10

=14π(π + 2); x =

1π + 2

(b) Vy =∫ 1

0

2πxx2 + 1

dx =[ln(x2 + 1)

]10

= π ln 2; y =1

4 ln 2

53. (a)6x4 + 11x3 − 2x2 − 5x− 2

x2(x + 1)3=

1x− 2

x2+

5x + 1

− 4(x + 1)3

(b) −x3 + 20x2 + 4x + 93(x2 + 4)(x2 − 9)

=1

x2 + 4+

3x + 3

− 4x− 3

(c)x2 + 7x + 12x(x2 + 2x + 4)

=2x− 1

x2 + 2x + 4− 3

x

54.2x6 − 13x5 + 23x4 − 15x3 + 40x2 − 24x + 9

x5 − 6x4 + 9x3= 2x− 1 +

3x− 2

x2+

1x3

+2

(x− 3)2− 4

x− 3

55.x8 + 2x7 + 7x6 + 23x5 + 10x4 + 95x3 − 19x2 + 133x− 52

x6 + 2x5 + 5x4 + 16x3 − 8x2 + 32x− 48

= x2 + 2 +2

x− 1+

1x + 3

+x

(x2 + 4)2+

3x2 + 4

56. n = 0 :∫

1x2 + 2x

dx =12

ln∣∣∣∣ x

x + 2

∣∣∣∣+ C; n = 1 :∫

1x2 + 2x + 1

dx =−1

x + 1+ C

n = 2 :∫

1x2 + 2x + 2

dx = arctan (x + 1) + C

57. (a)

-4 -3 -2 -1x

-20

20

y (b) A =∫ 4

0

x

x2 + 5x + 6dx

=∫ 4

0

x

(x + 2)(x + 3)dx

=∫ 4

0

(3

x + 3− 2

x + 2

)dx

=[3 ln |x + 3| − 2 ln |x + 2|

]40

= 3 ln 7 − 5 ln 3

58. (a) Vy =∫ 4

0

2πx2

x2 + 5x + 6dx = 2π

∫ 4

0

(1 +

4x + 2

− 9x + 3

)dx

= 2π [x + 4 ln |x + 2| − 9 ln |x + 3|]40 = 2π(4 + 13 ln 3 − 9 ln 7)



SECTION 8.6 455

(b)yVy =

∫ 4

0

πx3

(x2 + 5x + 6)2dx = π

∫ 4

0

[28

x + 2− 8

(x + 2)2− 27

x + 3− 27

(x + 3)2

]dx

= π

[28 ln |x + 2| + 8

x + 2− 27 ln |x + 3| + 27

x + 3

]40

= π

[43

+277

− 13 + 55 ln 3 − 27 ln 7]

y =− 164

21 + 55 ln 3 − 27 ln 72(4 + 13 ln 3 − 9 ln 7)

59. (a)

-4 -3 -2 -1x

50

y(b) A =

∫ 9

−2

9 − x

(x + 3)2dx

=∫ 9

−2

(12

(x + 3)2− 1

x + 3

)dx

=[− ln |x + 3| − 12

x + 3

]9−2

= 11 − ln 12

60. (a)Vx =

∫ 9

−2

π(9 − x)2

(x + 3)4dx = π

∫ 9

−2

[1

(x + 3)2− 24

(x + 3)3+

144(x + 3)4

]dx

= π

[ −1x + 3

+12

(x + 3)2− 48

(x + 3)3

]9−2

=133136

π

(b)

xVx =∫ 9

−2

πx(9 − x)2

(x + 3)4dx = π

∫ 9

−2

[1

x + 3− 27

(x + 3)2+

216(x + 3)2

− 432(x + 3)4

]dx

= π

[ln |x + 3| + 27

x + 3− 108

(x + 3)2+

144(x + 3)3

]9−2

= π

(ln 12 − 737

12

)

x =36

1331

(ln 12 − 737

12

)

SECTION 8.6

1. {x = u2

dx = 2u du

};

∫dx

1 −√x

=∫

2u du1 − u

= 2∫ (

11 − u

− 1)

du

= −2(ln |1 − u| + u) + C = −2(√x + ln |1 −

√x| ) + C



456 SECTION 8.6

2. {u =

√x

du = 12√xdx

};

∫ √x

1 + xdx =

∫2u2

1 + u2du =

∫ (2 − 2

1 + u2

)du

= 2u− 2 arctanu + C

= 2√x− 2 arctan

√x + C

3.

{u2 = 1 + ex

2u du = ex dx

};

∫ √1 + ex dx =

∫u · 2u du

u2 − 1= 2

∫ (1 +

1u2 − 1

)du

= 2∫ (

1 +12

[1

u− 1− 1

u + 1

])du

= 2u + ln |u− 1| − ln |u + 1| + C

= 2√

1 + ex + ln[√

1 + ex − 1√1 + ex + 1

]+ C

= 2√

1 + ex + ln

[(√1 + ex − 1

)2ex

]+ C

= 2√

1 + ex + 2 ln (√

1 + ex − 1) − x + C

4.

{u3 = x

3u2 du = dx

};

∫dx

x(x1/3 − 1)=∫

3 duu(u− 1)

= 3∫ (

1u− 1

− 1u

)du

= 3 ln |u− 1| − 3 ln |u| + C = 3 ln |u− 1| − ln |u3| + C

= 3 ln |x1/3 − 1| − ln |x| + C

5. (a)

{u2 = 1 + x

2u du = dx

};

∫x√

1 + x dx =∫

(u2 − 1)(u)2u du

=∫

(2u4 − 2u2) du

= 25u

5 − 23u

3 + C

= 25 (x + 1)5/2 − 2

3 (x + 1)3/2 + C

(b)

{u = 1 + x

du = dx

};∫

x√

1 + x dx =∫

(u− 1)√u du =

∫ (u3/2 − u1/2

)du

= 25u

5/2 − 23u

3/2 + C

= 25 (1 + x)5/2 − 2

3 (1 + x)3/2 + C

6. (a)

{u2 = 1 + x

2u du = dx

};∫

x2√

1 + x dx =∫

(u2 − 1)2u · 2u du

=∫

(2u6 − 4u4 + 2u2) du

=27u7 − 4

5u5 +

23u3 + C

=27(1 + x)7/2 − 4

5(1 + x)5/2 +

23(1 + x)3/2 + C



SECTION 8.6 457

(b) {u = 1 + x

du = dx

};∫

x2√

1 + x dx =∫

(u− 1)2√u du =

∫(u2 − 2u + 1)

√u du

=27u7/2 − 4

5u5/2 +

23u3/2 + C

=27(1 + x)7/2 − 4

5(1 + x)5/2 +

23(1 + x)3/2 + C

7.{

u2 = x− 1

2u du = dx

};

∫(x + 2)

√x− 1 dx =

∫(u + 3)(u)2u du

=∫

(2u4 + 6u2) du

= 25u

5 + 2u3 + C

= 25 (x− 1)5/2 + 2(x− 1)3/2 + C

8.{

u = x + 2

du = dx

};

∫(x− 1)

√x + 2 dx =

∫(u− 3)

√u du

=25u5/2 − 2u3/2 + C

=25(x + 2)5/2 − 2(x + 2)3/2 + C

9.{

u2 = 1 + x2

2u du = 2x dx

};

∫x3

(1 + x2)3dx =

∫x2

(1 + x2)3x dx =

∫u2 − 1u6

u du

=∫

(u−3 − u−5) du =12u−2 +

14u−4 + C

=1

4(1 + x2)2− 1

2(1 + x2)+ C

= − 1 + 2x2

4(1 + x2)2+ C

10.{

u = 1 + x

du = dx

};

∫x(1 + x)1/3 dx =

∫(u− 1)u1/3 du

=37u7/3 − 3

4u4/3 + C

=37(1 + x)7/3 − 3

4(1 + x)4/3 + C

11.{

u2 = x

2u du = dx

};

∫ √x√

x− 1dx =

∫ (u

u− 1

)2u du = 2

∫ (u + 1 +

1u− 1

)du

= u2 + 2u + 2 ln |u− 1| + C

= x + 2√x + 2 ln |

√x− 1| + C



458 SECTION 8.6

12. {u = x + 1

du = dx

};

∫x√

1 + xdx =

∫u− 1√

udu

=23u3/2 − 2u1/2 + C

=23(x + 1)3/2 − 2(x + 1)1/2 + C

13.{

u2 = x− 1

2u du = dx

};

∫ √x− 1 + 1√x− 1 − 1

dx =∫

u + 1u− 1

2u du =∫ (

2u + 4 +4

u− 1

)du

= u2 + 4u + 4 ln |u− 1| + C

= x− 1 + 4√x− 1 + 4 ln |

√x− 1 − 1| + C

(absorb −1 in C)

= x + 4√x− 1 + 4 ln |

√x− 1 − 1| + C

14.{

u = ex

du = ex dx

};

∫1 − ex

1 + exdx =

∫1 − u

u(1 + u)du

=∫ (

1u− 2

1 + u

)du

= ln |u| − 2 ln |1 + u| + C

= x− 2 ln(1 + ex) + C

15.{

u2 = 1 + ex

2u du = ex dx

};

∫dx√

1 + ex=∫ (

1u

)2u duu2 − 1

=∫ [

1u− 1

− 1u + 1

]du

= ln |u− 1| − ln |u + 1| + C

= ln[√

1 + ex − 1√1 + ex + 1

]+ C

= ln[(√

1 + ex − 1)2

ex

]+ C

= 2 ln (√

1 + ex − 1) − x + C

16.{

u = e−x

du = −e−x dx

};

∫dx

1 + e−x= −

∫du

u(1 + u)

= −∫ (

1u− 1

1 + u

)du

= − ln |u| + ln |1 + u| + C

= ln∣∣∣∣1 + u

u

∣∣∣∣+ C

= ln∣∣∣∣1 + ex

e−x

∣∣∣∣+ C = ln(1 + ex) + C



SECTION 8.6 459

17. {u2 = x + 4

2u du = dx

};

∫x√x + 4

dx =∫

u2 − 4u

2u du =∫

(2u2 − 8) du

= 23u

3 − 8u + C

= 23 (x + 4)3/2 − 8(x + 4)1/2 + C

= 23 (x− 8)

√x + 4 + C

18.

{u2 = x− 2

2u du = dx

};

∫x + 1

x√x− 2

dx = 2∫

u2 + 3u2 + 2

du = 2∫ (

1 +1

u2 + 2

)du

= 2u +√

2 arctan(

u√2

)+ C

= 2√x− 2 +

√2 arctan

(√x− 2

2

)+ C

19.

{u2 = 4x + 1

2u du = 4 dx

};

∫2x2(4x + 1)−5/2 dx =

∫2(u2 − 1

4

)2

(u−5)u

2du

=116

∫(1 − 2u−2 + u−4) du =

116

u +18u−1 − 1

48u−3 + C

= 116 (4x + 1)1/2 + 1

8 (4x + 1)−1/2 − 148 (4x + 1)−3/2 + C

20.

{u = x− 1

du = dx

};

∫x2

√x− 1 dx =

∫(u + 1)2

√u du

=27u7/2 +

45u5/2 +

23u3/2 + C

=27(x− 1)7/2 +

45(x− 1)5/2 +

23(x− 1)3/2 + C

21.

{u2 = ax + b

2u du = a dx

};

∫x

(ax + b)3/2dx =

∫ u2 − b

au3

2ua

du

=2a2

∫(1 − bu−2) du

=2a2

(u + bu−1) + C =2u2 + 2b

a2u+ C

=4b + 2axa2√ax + b

+ C

22.

{u = ax + b

du = a dx

} ∫x√

ax + bdx =

1a2

∫u− b√

udu

=1a2

[23u3/2 − 2bu1/2

]+ C

=2

3a2(ax− 2b)

√ax + b + C



460 SECTION 8.6

23. ⎧⎪⎨⎪⎩u = tan(x/2), dx =

21 + u2

du

sinx =2u

1 + u2, cosx =

1 − u2

1 + u2

⎫⎪⎬⎪⎭ ;

∫1

1 + cosx− sinxdx =

∫1

1 +1 − u2

1 + u2− 2u

1 + u2

· 21 + u2

du

=∫

11 − u

du

= − ln |1 − u| + C = − ln∣∣∣∣1 − tan

(x2

) ∣∣∣∣+ C

24.

⎧⎪⎨⎪⎩

u = tan(x/2), dx =2

1 + u2du

cosx =1 − u2

1 + u2

⎫⎪⎬⎪⎭ ;

∫1

2 + cosxdx =

∫1

2 + 1−u2

1+u2

· 21 + u2

du =∫

22 + 2u2 + 1 − u2

du =∫

23 + u2

du

=2√3

arctan(

u√3

)+ C =

2√3

arctan(

tan(x/2)√3

)+ C

25.

⎧⎪⎨⎪⎩

u = tan(x/2), dx =2

1 + u2du

sinx =2u

1 + u2

⎫⎪⎬⎪⎭ ;

∫1

2 + sinxdx =

∫1

2 +2u

1 + u2

· 21 + u2

du

=∫

1u2 + u + 1

du =∫

1(u + 1

2

)2 +(√

32

)2 du

=2√3

arctan

(u + 1

2√3

2

)+ C

=2√3

arctan[

1√3

(2 tan(x/2) + 1)]

+ C

26.⎧⎪⎨⎪⎩

u = tan(x/2), dx =2

1 + u2du

sinx =2u

1 + u2

⎫⎪⎬⎪⎭ ;



SECTION 8.6 461

∫sinx

1 + sin2 xdx =

∫ 2u1+u2

1 + 4u2

(1+u2)2

· 21 + u2

du

{v = u2

dv = 2u du

}=∫

4u1 + 6u2 + u4

du

=∫

21 + 6v + v2

dv =∫

2(v + 3)2 − 8

dv

=2√8

ln∣∣∣∣ v + 3 −

√8√

(v + 3)2 − 8

∣∣∣∣+ C

=2√8

ln∣∣∣∣ tan2(x/2) + 3 −

√8√

[tan2(x/2) + 3]2 − 8

∣∣∣∣+ C

27. ⎧⎪⎨⎪⎩

u = tan(x/2), dx =2

1 + u2du

sinx =2u

1 + u2, tanx =

2u1 − u2

⎫⎪⎬⎪⎭ ;

∫1

sinx + tanxdx =

∫1

2u1 + u2

+2u

1 − u2

· 21 + u2

du

=∫

1 − u2

2udu =

12

∫ (1u− u

)du

= 12

(ln |u| − 1

2u2)

+ C = 12 ln | tan(x/2)| − 1

4 [tan(x/2)]2 + C

28. ∫1

1 + sinx + cosxdx =

∫2

1 + 2u1+u2 + 1−u2

1+u2

· 21 + u2

du

=∫

22 + 2u

du =∫

du

1 + u

=∫

ln |1 + u| + C = ln |1 + tan(x

2

)| + C

29.⎧⎪⎨⎪⎩

u = tan(x/2), dx =2

1 + u2du

sinx =2u

1 + u2, cosx =

1 − u2

1 + u2

⎫⎪⎬⎪⎭ ;



462 SECTION 8.6

∫1 − cosx1 + sinx

dx =∫ 1 − 1 − u2

1 + u2

1 +2u

1 + u2

· 21 + u2

du

=∫

4u2

(1 + u2)(u + 1)2du

=∫ [

2u1 + u2

− 2u + 1

+2

(u + 1)2

]du

= ln(u2 + 1) − 2 ln |u + 1| − 2u + 1

+ C

= ln[

u2 + 1(u + 1)2

− 2u + 1

]+ C

= ln[

tan2(x/2) + 1(tan(x/2) + 1)2

]− 2

tan(x/2) + 1+ C = ln

∣∣∣∣ 11 + sinx

∣∣∣∣− 2tan(x/2) + 1

+ C

30.∫

15 + 3 sinx

dx =∫

15 + 6u

1+u2

· 21 + u2

du

=∫

25 + 5u2 + 6u

du =25

∫du

(u + 35 )2 + 16

25

=25· 54

arctan(u + 3/5

4/5

)+ C

=12

arctan[54

(tan

(x2

)+

35

)]+ C

31.{u2 = x

2u du = dx

};

∫x3/2

x + 1dx =

∫u3

u2 + 12u du

=∫

2u4

u2 + 1du =

∫ [2u2 − 2 +

2u2 + 1

]du

=23u3 − 2u + 2 arctanu + C =

23x3/2 − 2x1/2 + 2 arctanx1/2 + C∫ 4

0

x3/2

x + 1dx =

[23x3/2 − 2x1/2 + 2 arctanx1/2

]40

=43

+ 2 arctan 2

32.{u2 = x

2u du = dx

} ∫ 4

0

11 +

√xdx =

∫ 2

0

2u1 + u

du

=∫ 2

0

(2 − 2

1 + u

)du = [2u− 2 ln |1 + u|]20

= 4 − 2 ln 3



SECTION 8.6 463

33. ∫ π/2

0

sin 2x2 + cosx

dx =∫ π/2

0

2 sinx cosx2 + cosx

dx

=∫ 1

0

2u2 + u

du [u = cosx, du = − sinx dx]

=∫ 1

0

(2 − 4

2 + u

)du

= [2u− 4 ln |2 + u| ]10 = 2 + 4 ln(

23

)

34.

⎧⎪⎨⎪⎩

u = tan(x/2), dx =2

1 + u2du

sinx =2u

1 + u2

⎫⎪⎬⎪⎭ ;

∫ π/2

0

11 + sinx

dx =∫ 1

0

11 + 2u

1+u2

· 21 + u2

du

=∫ 1

0

21 + 2u + u2

du =∫ 1

0

2(u + 1)2

du

= −[

2u + 1

]10

= 1

35.⎧⎪⎨⎪⎩

u = tan(x/2), dx =2

1 + u2du

sinx =2u

1 + u2, cosx =

1 − u2

1 + u2

⎫⎪⎬⎪⎭ ;

∫1

sinx− cosx− 1dx =

∫1

2u1 + u2

− 1 − u2

1 + u2− 1

· 21 + u2

du

=∫

1u− 1

du

= ln |u− 1| + C = ln | tan(x/2) − 1| + C∫ π/3

0

1sinx− cosx− 1

dx = [ ln | tan(x/2) − 1| ]π/30 = ln

(√3 − 1√

3

)

36.

{u2 = x

2u du = dx

} ∫ 1

0

√x

1 +√xdx =

∫ 1

0

2u2

1 + udu

=∫ 1

0

(2u− 2 +

21 + u

)du

=[u2 − 2u + 2 ln |1 + u|

]10

= 2 ln 2 − 1

37.⎧⎪⎨⎪⎩

u = tan(x/2), dx =2

1 + u2du

sinx =2u

1 + u2, cosx =

1 − u2

1 + u2

⎫⎪⎬⎪⎭ ;



464 SECTION 8.6∫

secx dx =∫

1cosx

dx =∫

11 − u2

1 + u2

· 21 + u2

du

= 2∫

11 − u2

du = 2∫ [

1/21 − u

+1/2

1 + u

]du

=∫ [

11 − u

+1

1 + u

]du = − ln |1 − u| + ln |1 + u| + C

= ln∣∣∣∣1 + tan(x/2)1 − tan(x/2)

∣∣∣∣+ C

38. (a) {u = sinx

du = cosx dx

} ∫secx dx =

∫du

1 − u2=

12

∫ (1

1 − u+

11 + u

)du

= −12

ln |1 − u| + 12

ln |1 + u| + C

= ln

√1 + sinx

1 − sinx+ C

(b)

√1 + sinx

1 − sinx=

√(1 + sinx)2

1 − sin2 x=∣∣∣∣1 + sinx

cosx

∣∣∣∣ = | secx + tanx|

39. ∫cscx dx =

∫sinx

sin2 xdx =

∫sinx

1 − cos2 xdx

= −∫

11 − u2

du [u = cosx, du = − sinx dx]

=12

∫ [1

u− 1− 1

u + 1

]du

=12

[ ln |u− 1| − ln |u + 1|] + C = ln

√1 − cosx1 + cosx

+ C

40. cosh(x

2

)=

1sech

(x2

) =1√

1 − tanh2(x2 )=

1√1 − u2

sinh(x

2

)=√

cosh2(x

2

)− 1 =

u√1 − u2

sinhx = 2 sinh(x

2

)cosh

(x2

)=

2u1 − u2

, coshx = cosh2(x

2

)+ sinh2

(x2

)=

1 + u2

1 − u2

du =12

sech 2(x

2

)dx =⇒ dx = 2 cosh2

(x2

)du =

21 − u2

du

41.⎧⎪⎪⎨⎪⎪⎩

u = tanh(x/2), dx =2

1 − u2du

coshx =1 + u2

1 − u2, sechx =

1 − u2

1 + u2

⎫⎪⎪⎬⎪⎪⎭ ;



SECTION 8.7 465

∫sechx dx =

∫1 − u2

1 + u2· 21 − u2

du

=∫

21 + u2

du = 2 arctanu + C = 2 arctan (tanh(x/2)) + C

42.∫

11 + coshx

dx =∫

11 + 1+u2

1−u2

· 21 − u2

du =∫

du = u + C = tanh(x

2

)+ C

43.⎧⎪⎪⎨⎪⎪⎩

u = tanh(x/2), dx =2

1 − u2du

sinhx =2u

1 − u2, coshx =

1 + u2

1 − u2

⎫⎪⎪⎬⎪⎪⎭ ;

∫1

sinhx + coshxdx =

∫1

2u1 − u2

+1 + u2

1 − u2

· 21 − u2

du

=∫

2(1 + u)2

du =−2

u + 1+ C =

−2tanh(x/2) + 1

+ C

44. ∫1 − ex

1 + exdx =

∫1 − coshx− sinhx

1 + coshx + sinhxdx

=∫ 1 − 1 + u2

1 − u2− 2u

1 − u2

1 +1 + u2

1 − u2+

2u1 − u2

· 21 − u2

du

=∫

− 2u1 − u2

du = ln |1 − u2| + C = ln |1 − tanh2(x

2

)| + C

SECTION 8.7

1. (a) L12 = 1212 [0 + 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 + 121] = 506

(b) R12 = 1212 [1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 + 121 + 144] = 650

(c) M6 = 126 [1 + 9 + 25 + 49 + 81 + 121] = 572

(d) T12 = 1224 [0 + 2(1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 + 121) + 144] = 578

(e) S6 = 1236 [0 + 144 + 2(4 + 16 + 36 + 64 + 100) + 4(1 + 9 + 25 + 49 + 81 + 121)] = 576∫ 12

0

x2 dx =[13x3

]120

= 576

2. (a) 0.500000 (b) 0.500000 (c) 0.500000

∫ 1

0

sin2 πx dx =1π

[πx

2− sin 2πx

4

]10

=12



466 SECTION 8.7

3. (a) L6 =36

[1

1 + 0+

11 + 1/8

+1

1 + 1+

11 + 27/8

+1

1 + 8+

11 + 125/8

]= 1

2

[1 + 8

9 + 12 + 8

35 + 19 + 8

133

] ∼= 1.394

(b) R6 =36

[1

1 + 1/8+

11 + 1

+1

1 + 27/8+

11 + 8

+1

1 + 125/8+

11 + 27

]= 1

2

[89 + 1

2 + 835 + 1

9 + 8133 + 1

28

] ∼= 0.9122

(c) M3 =33

[1

1 + 1/8+

11 + 27/8

+1

1 + 125/8

]=

89

+835

+8

133∼= 1.1776

(d) T6 = 312

[1 + 2

(89 + 1

2 + 835 + 1

9 + 8133

)+ 1

28

] ∼= 1.1533

(e) S3 = 318

{1 + 1

28 + 2[12 + 1

9

]+ 4

[89 + 8

35 + 8133

]} ∼= 1.1614

4. (a) 0.4229 (b) 0.4339

5. (a)14π ∼= T4 =

18

[1 + 2

(1

1 + 1/16+

11 + 1/4

+1

1 + 9/16

)+

11 + 1

]= 1

8

[1 + 2

(1617 + 4

5 + 1625

)+ 1

2

] ∼= 0.7828

π ∼= 4(0.7828) = 3.1312

(b) 14π

∼= S4 = 124

[1 + 1

2 + 2(

1617 + 4

5 + 1625

)+ 4

(6465 + 64

73 + 6489 + 64

113

)] ∼= 0.7854

π ∼= 4(0.7854) = 3.1416

6. (a) 0.8511 (b) 0.8542

7. (a) M4 =24

[cos(−3

4

)2 + cos(−1

4

)2 + cos(

14

)2 + cos(

34

)2] ∼= 1.8440

(b) T8 =216

[cos(−1)2 + 2 cos

(−34

)2 + 2 cos(−1

2

)2 + 2 cos(−1

4

)2 +

2 cos(0)2 + 2 cos(

14

)2 + 2 cos(

12

)2 + 2 cos(

34

)2 + cos(1)2]∼= 1.7915

(c) S4 =224

{cos(−1)2 + cos(1)2 + 2

[cos(−1

2

)2 + cos(0)2 + cos(

12

)2]+

4[cos(−3

4

)2 + cos(−1

4

)2 + cos(

14

)2 + cos(

34

)2]} ∼= 1.8090

8. (a) 3.0543 (b) 3.0615 (c) 3.0591

9. (a) T10 =220

[e−02

+ 2e−(1/5)2 + 2e−(2/5)2 + 2e−(3/5)2 + 2e−(4/5)2 + 2e−12+ 2e−(6/5)2

+ 2e−(7/5)2 + 2e−(8/5)2 + 2e−(9/5)2 + e−22]∼= 0.8818



SECTION 8.7 467

(b) S5 =230

{e−02

+ e−22+ 2

[e−(2/5)2 + e−(4/5)2 + e−(6/5)2 + e−(8/5)2

]

+ 4[e−(1/5)2 + e−(3/5)2 + e−12

+ e−(7/5)2 + e−(9/5)2]}

∼= 0.8821

10. (a) 1.9133 (b) 1.9271 (c) 1.9225

11. Such a curve passes through the three points

(a1, b1), (a2, b2), (a3, b3)

iff

b1 = a12A + a1B + C, b2 = a2

2A + a2B + C, b3 = a32A + a3B + C,

which happens iff

A =b1(a2 − a3) − b2(a1 − a3) + b3(a1 − a2)

(a1 − a3)(a1 − a2)(a2 − a3),

B = −b1(a22 − a3

2) − b2(a12 − a3

2) + b3(a12 − a2

2)(a1 − a3)(a1 − a2)(a2 − a3)

,

C =a1

2(a2b3 − a3b2) − a22(a1b3 − a3b1) + a3

2(a1b2 − a2b1)(a1 − a3)(a1 − a2)(a2 − a3)

.

12. b− a

6

[g(a) + 4g

(a + b

2

)+ g(b)

]

=b− a

6

{(Aa2 + Ba + C) + 4

[A

(a + b

2

)2

+ B

(a + b

2

)+ C

]+ (Ab2 + Bb + C)

}

=b− a

6{A(b2 + a2) + B(b + a) + 2C + A(a2 + 2ab + b2) + 2B(a + b) + 4C

}=

b− a

6{2A(b2 + ab + a2) + 3B(b + a) + 6C

}=

13A(b3 − a3) +

12B(b2 − a2) + C(b− a)∫ b

a

Ax2 dx +∫ b

a

Bxdx +∫ b

a

C dx =∫ b

a

g(x) dx

13. (a)∣∣∣∣ (b− a)3

12n2f ′′(c)

∣∣∣∣ = 2712n2

14c3/2

≤ 916n2

< 0.01 =⇒ n2 >

(152

)2

=⇒ n ≥ 8

(b)∣∣∣∣ (b− a)5

2880n4f (4)(c)

∣∣∣∣ = 2432880n4

1516c7/2

≤ 811024n4

< 0.01 =⇒ n ≥ 2

14. (a)∣∣∣∣ (b− a)3

12n2f ′′(c)

∣∣∣∣ = 812n2

20c3 ≤ 8 · 20 · 2712n2

< 0.01 =⇒ n ≥ 190



468 SECTION 8.7

(b)∣∣∣∣ (b− a)5

2880n4f (4)(c)

∣∣∣∣ = 322880n4

· 120c ≤ 32 · 120 · 32880n4

< 0.01 =⇒ n ≥ 5

15. (a)∣∣∣∣ (b− a)3

12n2f ′′(c)

∣∣∣∣ = 2712n2

14c3/2

≤ 916n2

< 0.00001 =⇒ n > 75√

10 =⇒ n ≥ 238

(b)∣∣∣∣ (b− a)5

2880n4f (4)(c)

∣∣∣∣ = 2432880n4

1516c7/2

≤ 811024n4

< 0.00001 =⇒ n ≥ 10

16. (a)8 · 20 · 27

12n2< 0.00001 =⇒ n ≥ 6000

(b)32 · 120 · 3

2880n4< 0.00001 =⇒ n ≥ 26

17. (a)∣∣∣∣ (b− a)3

12n2f ′′(c)

∣∣∣∣ = π3

12n2sin c ≤ π3

12n2< 0.001 =⇒ n > 5π

√10π3

=⇒ n ≥ 51

(b)∣∣∣∣ (b− a)5

2880n4f (4)(c)

∣∣∣∣ = π5

2880n4sin c ≤ π5

2880n4< 0.001 =⇒ n ≥ 4

18. (a)∣∣∣∣ (b− a)3

12n2f ′′(c)

∣∣∣∣ = π3

12n2cos c ≤ π3

12n2< 0.001 =⇒ n ≥ 51

(b)∣∣∣∣ (b− a)5

2880n4f (4)(c)

∣∣∣∣ = π5

2880n4cos c ≤ π5

2880n4< 0.001 =⇒ n ≥ 4

19. (a)∣∣∣∣ (b− a)3

12n2f ′′(c)

∣∣∣∣ = 812n2

ec ≤ 812n2

e3 < 0.01 =⇒ n > 10e

√2e3

=⇒ n ≥ 37

(b)∣∣∣∣ (b− a)5

2880n4f (4)(c)

∣∣∣∣ = 322880n4

ec ≤ 190n4

e3 < 0.01 =⇒ n ≥ 3

20. (a)∣∣∣∣ (b− a)3

12n2f ′′(c)

∣∣∣∣ = (e− 1)3

12n2· 1c2

≤ (e− 1)3

12n2< 0.01 =⇒ n ≥ 7

(b)∣∣∣∣ (b− a)5

2880n4f (4)(c)

∣∣∣∣ = (e− 1)5

2880n4· 6c4

≤ 6(e− 1)5

2880n4< 0.01 =⇒ n ≥ 2

21. (a)∣∣∣∣ (b− a)3

12n2f ′′(c)

∣∣∣∣ =∣∣∣∣ 812n2

2e−c2(2c2 − 1)∣∣∣∣ ≤ 8

3n2e−3/2 < 0.0001

=⇒ n > 100√

83 e

−3/2 =⇒ n ≥ 78



SECTION 8.7 469

(b)∣∣∣∣ (b− a)5

2880n4f (4)(c)

∣∣∣∣ =∣∣∣∣ 322880n4

4e−c2(4c4 − 12c2 + 3

)∣∣∣∣ ≤ 322880n4

12 < 0.0001

=⇒ n > 10[32 · 122880

]1/4=⇒ n ≥ 7

22. (a)∣∣∣∣ (b− a)3

12n2f ′′(c)

∣∣∣∣ = 812n2

· ec ≤ 812n2

· e2 < 0.00001 =⇒ n ≥ 702

(b)∣∣∣∣ (b− a)5

2880n4f (4)(c)

∣∣∣∣ = 322880n4

· ec ≤ 322880n4

· e2 < 0.00001 =⇒ n ≥ 10

23. f (4)(x) = 0 for all x; therefore by (8.7.3) the theoretical error is zero

24. If f is linear, f ′′(x) = 0 for all x, so the theoretical error is zero

25. (a)∣∣∣∣T2 −

∫ 1

0

x2 dx

∣∣∣∣ = 38− 1

3=

124

= ET2

(b)∣∣∣∣S1 −

∫ 1

0

x4 dx

∣∣∣∣ = 524

− 15

=1

120= ES

1

26. Since mi ≤ f(xi−1) ≤ Mi, mi ≤ f(xi) ≤ Mi, and mi ≤ f(xi−1+xi

2

)≤ Mi, we get

mi ≤12

[f(xi−1) + f(xi)] ≤ Mi and mi ≤16

[f(xi−1) + 4f

(xi−1 + xi

2

)+ f(xi)

]≤ Mi

So by the intermediate value theorem, we can find ai, bi ∈ [xi−1, xi] such that

f(ai) =12

[f(xi−1) + f(xi)] , f(bi) =16

[f(xi−1) + 4f

(xi−1 + xi

2

)+ f(xi)

]

So using ai or bi as x∗i , we can write Tn and Sn as Riemann sums.

27. (a) Let f be twice differentiable on [a, b] with f(x) > 0 and f ′′(x) > 0, and let P = {x0, x1, x2, . . . , xn}be a regular partition of [a, b]. Figure A shows a typical subinterval with the approximating trape-

zoid ABCD. Since the area under the curve is less than the area of the trapezoid, we can conclude

that

∫ b

a

f(x) dx ≤ Tn.



470 SECTION 8.7

Figure A Figure B

Now consider Figure B. Since the triangles EBP and PFC are congruent, the area of the rectangle

ABCD equals the area of the trapezoid AEFD, and since the area under the curve is greater than

the area of AEFD it follows that

Mn ≤∫ b

a

f(x) dx.

(b) A similar argument in the case f(x) > 0, f ′′(x) < 0 gives

Tn ≤∫ b

a

f(x) dx ≤ Mn.

28. 13Tn +

23Mn

=13b− a

2n[f(x0) + 2f(x1) + · · · + 2f(xn−1) + f(xn)] +

23· b− a

n

[f

(x0 + x1

2

)+ · · · + f

(xn−1 + xn

2

)]

=b− a

6n

{f(x0) + 2f(x1) + · · · + 2f(xn−1) + f(xn) + 4

[f

(x0 + x1

2

)+ · · · + f

(xn−1 + xn

2

)]}= Sn

29. (a)∫ 10

0

(x + cosx) dx ∼= 49.4578

(b)∫ 7

−4

(x5 − 5x4 + x3 − 3x2 − x + 4

)dx ∼= 1280.56

30. (a)∫ 3

−4

x2

x2 + 4dx ∼= 2.8201 (b)

∫ π/6

0

(x + tanx) dx ∼= 0.2809

31.∣∣ES

20

∣∣ ≤ 45

2880(20)4max

∣∣∣f (4)(t)∣∣∣ ≤ 1024

2880(20)4(0.1806) ≤ 4.01 × 10−7

32.∣∣ET

30

∣∣ ≤ 0.00036



REVIEW EXERCISES 471

REVIEW EXERCISES

1. Substitution: let u = sinx, du = cosx dx∫cosx

4 + sin2 xdx =

∫1

4 + u2du =

12

arctan(u

2

)+ C =

12

arctan(

sinx

2

)+ C

2.x2

1 + x2= 1 − 1

x2∫ π/4

0

x2

1 + x2dx =

∫ π/4

0

(1 − 1

1 + x2

)dx =

[x− arctanx

]π/40

=π

4− 1

3. Integration by parts:

{u = x dv = sinhx dx

du = dx v = coshx

}

∫2x sinhx dx = 2

∫x sinhx dx = 2

[x coshx−

∫coshx dx

]= 2x coshx− 2 sinhx + C

4.∫

(tanx + cotx)2dx =∫ (

tan2 x + cot2 x + 2)dx =

∫ (sec2 x + csc2 x

)dx = tanx− cotx + C

5. Partial fractions:x− 3

x2(x + 1)=

4x− 3

x2− 4

x + 1∫x− 3

x2(x + 1)dx =

∫ (4x− 3

x2− 4

x + 1

)dx = 4 ln |x| + 3

x− 4 ln |x + 1| + C


⎧⎨⎩u = arctanx dv = x dx

du =1

1 + x2dx v = 1

2x2

⎫⎬⎭

∫x arctan x dx =

12

[x2 arctan −

∫x2

1 + x2dx

]

=12

[x2 arctan x−

∫ (1 − 1

1 + x2

)dx

]=

12(x2 + 1) arctan x− 1

2x + C

7.∫

sin 2x cosx dx =∫

(2 sinx cosx) cosx dx = 2∫

cos2 x sinx dx

Substitution: u = cosx, du = − sinx dx

2∫

cos2 x sinx dx = −2∫

u2 du = − 23 u

3 + C = − 23 cos3 x + C


{u = x dv = e−3x dx

du = dx v = − 13e

−3x

}

∫3xe−3x dx = 3

∫xe−3x dx− = 3

[−1

3xe−3x +

13

∫e−3x dx

]

= 3[−1

3xe−3x − 1

9e−3x

]+ C

= −xe−3x − 13e

−3x + C



472 REVIEW EXERCISES

9.∫

ln√x + 1 dx =

∫ln(x + 1)1/2 dx = 1

2

∫ln(x + 1) dx

Integration by parts:

⎧⎨⎩u = ln(x + 1) dv = dx

du =1

x + 1dx v = x + 1

⎫⎬⎭

12

∫ln(x + 1) dx =

12

[(x + 1) ln(x + 1) −

∫dx

]=

12

[(x + 1) ln(x + 1) − x] + C

∫ 3

0

ln√x + 1dx =

12

[(x + 1) ln(x + 1) − x] |30 = 4 ln 2 − 32

Note: This answer can also be written: 12 [(x + 1) ln(x + 1) − (x + 1)] + C;

set w = x + 1, dw = dx, and integrate∫

lnw dw by parts.

10. Partial fractions:2

x(1 + x2)=

2x− 2x

x2 + 1∫2

x(1 + x2)dx =

∫2xdx−

∫2x

x2 + 1dx = 2 ln |x| − ln |x2 + 1| + C = ln

x2

x2 + 1+ C

11.∫

sin3 x

cosxdx =

∫ (1 − cos2 x

)sinx

cosxdx =

∫tanx dx−

∫cosx sinx dx = ln | secx| − 1

2sin2 x + C

12. Substitution: u = sinx, du = cosx dx∫cosxsin3 x

dx =∫

u−3 du = − 12u

−2 + C = − 12 csc2 x + C

13.∫ 1

0

e−x coshx dx =∫ 1

0

e−x

[ex + e−x

2

]dx =

12

∫ 1

0

[1 + e−2x

]dx

=12

[x− e−2x

2

]10

=12

[1 − e−2

2+

12

]=

34− e−2

2

14. Trigonometric substitution: set x = 3 tanu, dx = 3 sec2 u du.∫x2 + 3√x2 + 9

dx =∫

3 tan2 u + 33 secu

3 sec2 u du

= 3∫

sec3 u du = 3[ 12 secu tanu + 12 ln | secu + tanu|] + C

15.∫

dx

ex − 4e−x=∫

ex

e2x − 4dx; substitution: u = ex, du = ex dx

∫dx

ex − 4e−x=∫

1u2 − 4

du =∫ (

1/4u− 2

− 1/4u + 2

)du =

14

[ln |u− 2| − ln |u + 2|

]+ C

=14

[ln |ex − 2| − ln |ex + 2|

]+ C =

14

ln∣∣∣∣ex − 2ex + 2

∣∣∣∣+ C




16. Partial fractions:1

x3 − 1=

1(x− 1)(x2 + x + 1)

=13

x− 1+

− 13x− 2

3

x2 + x + 1

∫1

x3 − 1dx =

∫ 13

x− 1dx +

∫ − 13x− 2

3

x2 + x + 1dx

=13

ln |x− 1| − 13

∫x + 2

x2 + x + 1dx

=13

ln |x− 1| − 16

∫2x + 1 − 1 + 4x2 + x + 1

dx

=13

ln |x− 1| − 16

ln |x2 + x + 1| − 12

∫1

(x + 12 )2 + (

√3

2 )2dx

=13

ln |x− 1| − 16

ln |x2 + x + 1| −√

33

arctan2x + 1√

3+ C

17.Integration by parts:

{u = x dv = 2x dx

du = dx v = 2x/ ln 2

}

∫x2x dx =

1ln 2

x2x − 1ln 2

∫2x dx =

1ln 2

(x2x − 1

ln 22x)

+ C

18.∫

ln(x√x)dx =

∫lnx3/2 dx =

32

∫lnx dx =

32

[x lnx− x] + C

19. Trigonometric substitution: set x = a sinu, dx = a cosu du

∫ √a2 − x2

x2dx =

∫a cosua2 sin2 u

a cosu du =∫

cos2 usin2 u

du =∫

1 − sin2 u

sin2 udu

=∫

csc2 u du−∫

du = − cotu− u + c

= −√a2 − x2

x− arcsin

(xa

)+ C

20. substitution: u = x3, du = 3x2 dx; u(0) = 0, u(2) = 8

∫ 2

0

x2ex3dx =

13

∫ 8

0

eu du =13

[eu]80

=13(e8 − 1)

21.∫

x3ex2dx =

∫x2 xex

2dx; integration by parts:

{u = x2 dv = xex

2dx

du = 2x dx v = 12 e

x2

}

∫x3ex

2dx =

12

[x2ex

2 −∫

2xex2dx2

]=

12

[x2ex

2 − ex2]

+ C

22.∫

sin 2x sin 3xdx =12

∫[cosx− cos 5x]dx =

12

sinx− 110

sin 5x + C

23.∫

sin5 x

cos7 xdx =

∫tan5 sec2 dx =

16

tan6 x + C (substitution: u = tan x)




24.∫ (√

4 + x2

x− x√

4 + x2

)dx =

∫4

x√

4 + x2dx

Trigonometric substitution: set x = 2 tanu, dx = 2 sec2 u du∫4

x√

4 + x2dx = 2

∫secutanu

du = 2∫

cscu du

= 2 ln | cscu− cotu| + C = 2 ln

∣∣∣∣∣√

4 + x2 − 2x

∣∣∣∣∣+ C

25.∫ π/3

π/6

sinx

sin 2xdx =

∫ π/3

π/6

sinx

2 sinx cosxdx =

12

∫ π/3

π/6

secx dx

=12

[ln | secx + tanx|

]π/3π/6

=12

ln(2 +√

3) − 14

ln 3

26.

∫x + 3√

x2 + 2x− 8dx =

12

∫2x + 6√

x2 + 2x− 8dx =

12

∫2x + 2 + 4√x2 + 2x− 8

dx

=12

∫2x + 2√

x2 + 2x− 8dx + 2

∫1√

(x + 1)2 − 9dx

=√x2 + 2x− 8 + 2

∫1√

(x + 1)2 − 9dx

Use the trigonometric substitution: x + 1 = 3 secu, dx = 3 secu tanu du, on the remaining integral.

∫1√

(x + 1)2 − 9dx =

∫secu du

= ln | secu + tanu| = ln

∣∣∣∣∣x + 1 +√x2 + 2x− 83

∣∣∣∣∣Thus

∫x + 3√

x2 + 2x− 8dx =

√x2 + 2x− 8 + ln |x + 1 +

√x2 + 2x− 83

| + C

27.∫

x2 + x√1 − x2

dx =∫

x2

√1 − x2

dx +∫

x√1 − x2

dx

∫x√

1 − x2dx = −

√1 − x2 + C1, substitution: set u = 1 − x2, du = −2xdx

For the first integral, use the trigonometric substitution: x = sinu, dx = cosu du

∫x2

√1 − x2

dx =∫

sin2 u

cosucosu du =

∫sin2 u du =

∫1 − cos 2u

2du

= 12u− 1

4 sin 2u + C2 = 12u− 1

2 sinu cosu + C2

=12

arcsinx− 12x√

1 − x2 + C2




Therefore, ∫x2 + x√1 − x2

dx = 12 arcsinx− 1

2x√

1 − x2 −√

1 − x2 + C

28.∫

x tan2 2x dx =∫

x(sec2 2x− 1

)dx = −

∫x dx +

∫x sec2 2x dx

= − 12x

2 + 12x tan 2x− 1

4 ln | sec 2x| + C (integration by parts)

29. Sincecos4 xsin2 x

=(1 − sin2 x)2

sin2 x=

1 − 2 sin2 x + sin4 x

sin2 x= cscx −2 + sinx

∫cos4 xsin2 x

dx =∫

(cscx −2 + sinx)dx = − cotx− 32x− 1

4sin 2x + C

30.∫

x ln√x2 + 1dx =

12

∫ln√x2 + 1d(x2) =

12

[x2 ln

√x2 + 1 −

∫x3

x2 + 1dx

]

=12

[x2 ln

√x2 + 1 −

∫xdx +

∫x

x2 + 1dx

]

=12

[x2 ln

√x2 + 1 − 1

2x2 +

12

ln(x2 + 1)]

+ C

Hence ∫ 3

0

x ln√x2 + 1dx =

12{x2 ln

√x2 + 1 − 1

2x2 +

12

ln(x2 + 1)}|30 = 5 ln√

10 − 94

31.∫

(sin 2x + cos 2x)2dx =∫

(1 + 2 sin 2x cos 2x) dx = x +∫

sin 4x dx = x− 14

cos 4x + C

32.∫ √

cosx sin3 xdx = −∫ √

cosx(1 − cos2 x)d(cosx) = −∫

(√

cosx− cos5/2 x)d(cosx)

= −23

cos3/2 x +27

cos7/2 x + C

33.5x + 1

(x + 2)(x2 − 2x + 1)=

−1x + 2

+1

x− 1+

2(x− 1)2

∫5x + 1

(x + 2)(x2 − 2x + 1)dx =

∫ { −1x + 2

+1

x− 1+

2(x− 1)2

}dx

= − ln |x + 2| + ln |x− 1| − 2x− 1

+ C

34.∫

tan3/2 x sec4 xdx =∫

tan3/2(1 + tan2 x)d(tanx) =∫{tan3/2 x + tan7/2 x}d(tanx)

=25

tan5/2 x +29

tan9/2 x + C

35.∫

1√x + 1 −√

xdx =

∫ √x + 1 +

√x

(√x + 1 −√

x)(√x + 1 +

√x)

dx

=∫

(√x + 1 +

√x)dx =

23(x + 1)3/2 +

23x3/2 + C




36.∫

cosπx cos12πxdx =

12

∫ {cos

32πx + cos

12πx

}dx =

13π

sin3π2x +

1π

sinπ

2x + C

Hence ∫ 1/2

0

cosπx cos12πxdx =

[13π

sin3π2x +

1π

sinπ

2x

] ∣∣∣1/20

=√

26π

+√

22π

37.integration by parts:

{u = x2 dv = cos 2x dx

du = 2x dx v = 12 sin 2x

}

∫x2 cos 2x dx =

12x2 sin 2x−

∫x sin 2x dx

integration by parts again:

{u = x dv = sin 2x dx

du = dx v = − 12 cos 2x

}

∫x2 cos 2x dx =

12x2 sin 2x−

[−1

2x cos 2x +

12

∫cos 2x dx

]=

12x2 sin 2x +

12x cos 2x− 1

4sin 2x + C

38. See Exercise 45, Section 8.2.∫e2x sin 4x dx =

e2x(2 sin 4x− 4 cos 4x)22 + 42

=e2x(2 sin 4x− 4 cos 4x)

20+ C

39. 1 − sin 2x1 + sin 2x

=1 − sin 2x1 + sin 2x

1 − sin 2x1 − sin 2x

=1 − 2 sin 2x + sin2 2x

1 − sin2 2x=

1 − 2 sin 2x + sin2 2xcos2 2x

= sec2 2x− 2 sin 2xcos2 2x

+ tan2 2x = 2 sec2 2x− 1 − 2 sin 2xcos2 2x

Therefore ∫1 − sin 2x1 + sin 2x

dx = tan 2x− x− 1cos 2x

+ C = tan 2x− x− sec 2x + C

40. partial fraction decomposition:5x + 3

(x− 1)(x2 + 2x + 5)=

1x− 1

+−x + 2

x2 + 2x + 5,

∫5x + 3

(x− 1)(x2 + 2x + 5)dx =

∫ {1

x− 1+

−x + 2x2 + 2x + 5

}dx

= ln |x− 1| +∫ −x + 2

x2 + 2x + 5dx

= ln |x− 1| −∫ −x− 1 + 3

x2 + 2x + 5dx

= ln |x− 1| − 12

∫2x + 2

x2 + 2x + 5+ 3

∫1

(x + 1)2 + 5dx

= ln |x− 1| − 12

ln(x2 + 2x + 5) +32

arctanx + 1

2+ C




41. (a) integration by parts:

{u = xn dv = cos ax dx

du = nxn−1 dx v = 1a sin ax

}∫

xn cos ax dx =xn sin ax

a− n

a

∫xn−1 sin ax dx

(b) integration by parts:

{u = xn dv = sin ax dx

du = nxn−1 dx v = − 1a cos ax

}∫

xn sin ax dx = −xn cos axa

+n

a

∫xn−1 cos ax dx

42. (a) Using 41, ∫x2 cos 3xdx =

x2 sin 3x3

− 23

∫x sin 3xdx

=x2 sin 3x

3− 2

3

{−x cosx

3+

13

∫cos 3xdx

}

=x2 sin 3x

3+

29x cos 3x− 2

27sin 3x + C

(b) Similarly,∫x3 sin 4xdx = −x3 cos 4x

4+

34

∫x2 cos 4xdx

= −x3 cos 4x4

+34

{x2 sin 4x

4− 1

2

∫x sin 4xdx

}

= −x3 cos 4x4

+316

x2 sin 4x− 38

∫x sin 4xdx

= −x3 cos 4x4

+316

x2 sin 4x− 38

{−x cos 4x

4+

14

∫cos 4xdx

}

= −x3 cos 4x4

+316

x2 sin 4x +332

x cos 4x− 3128

sin 4x + C

43. (a) integration by parts:

{u = (lnx)n dv = xm dx

du = n(lnx)n−1 1x dx v = 1

m+1xm+1

}∫

xm(lnx)n dx =xm+1(lnx)n

m + 1− n

m + 1

∫xm(lnx)n−1 dx

(b) From (a),

∫x4(lnx)3dx =

x5(lnx)3

5− 3

5

∫x4(lnx)2dx

=x5(lnx)3

5− 3

5

{x5(lnx)2

5− 2

5

∫x4 lnxdx

}

=15x5(lnx)3 − 3

25x5(lnx)2 +

625

{x5 lnx

5− 1

5

∫x4dx

}

=15x5(lnx)3 − 3

25x5(lnx)2 +

6125

x5 lnx− 6125

x5dx + C




44. Using integration by parts,

∫x2 arctanx dx =

13x3 arctanx− 1

3

∫x3

1 + x2dx

=13x3 arctanx− 1

3

∫(x− x

1 + x2)dx

=13x3 arctanx− 1

3

(12x2 − 1

2ln(1 + x2)

)

=13x3 arctanx− 1

6x2 +

16

ln(1 + x2) + C

Therefore, A =∫ 1

0

x2 arctanxdx =π

12− 1

6− ln 2

6

45.A =

∫ 12

0

(1 − x2)−1/2 dx =[arcsinx

]1/20

=π

6

xA =∫ 1/2

0

x(1 − x2)−1/2 dx = −[√

1 − x2]1/20

= 1 −√

32

=⇒ x =6π

(1 −

√3

2

)

yA =∫ 1/2

0

12(1 − x2)−1 dx =

14

∫ 1/2

0

[1

1 − x+

11 + x

]dx =

14

[ln(

1 + x

1 − x

)]1/20

=14

ln 3

Therefore, y =3 ln 32π

46. (a)

2

x

y

ππ

π

(b) A =∫ π

0

(x + sinx− x) dx =∫ π

0

sinx dx = −[cosx

]π0

= 2

(c) xA =∫ π

0

x(x + sinx− x) dx =∫ π

0


]π0

= π

integration by parts∧

Therefore x = π2

yA =∫ π

0

12[(x + sinx)2 − x2

]dx =

12

∫ π

0

[2x sinx + sin2 x

]dx

=∫ π

0

x sinx dx +14

∫ π

0

(1 + cos 2x) dx

= π +14

[x +

12

sin 2x]π0

=5π4

Therefore, y =5π8




47. Use Pappus’s theorem and Problem 45.

(a) about the x-axis: V = 2π(

3 ln 32π

)π6 = π ln 3

2

(b) about the y-axis: V = 2π[

6π (1 −

√3

2 )]

π6 = 2π(1 −

√3

2 )

48. V =∫ e

1

2πx ln 2x dx =[πx2 ln(2x) − π

2x2]e1

=π

2[e2(1 + 2 ln 2) − 2 ln 2 + 1

].

∧integration by parts

49. Let f =√x3 + x.

(a)∫ 2

0

√x3 + xdx =

24

[f

(0 + 1/2

2

)+ f

(1/2 + 1

2

)+ f

(1 + 3/2

2

)+ f

(3/2 + 2

2

)]

=12[f(1/4) + f(3/4) + f(5/4) + f(7/4)] ≈ 3.0270

(b)∫ 2

0

√x3 + x dx

=216

[f(0) + 2f(1/4) + 2f(1/2) + 2f(3/4) + 2f(1) + 2f(5/4) + 2f(3/2) + 2f(7/4) + f(2)]

=18[2f(1/4) + 2f(1/2) + 2f(3/4) + 2f(1) + 2f(5/4) + 2f(3/2) + 2f(7/4) + f(2)] ≈ 3.0120

(c)∫ 2

0

√x3 + x dx

=224

{f(0) + 2[f(1/2) + 2f(1) + f(3/2)] + 4[f(1/4) + f(3/4) + f(5/4) + f(7/4)] + f(2)]}

=112

{2[f(1/2) + 2f(1) + f(3/2)] + 4[f(1/4) + f(3/4) + f(5/4) + f(7/4)] + f(2)]} ≈ 3.0170

50. Let f =√

1 + 3x.

(a)∫ 2

0

√1 + 3x dx

=216

[f(0) + 2f(1/4) + 2f(1/2) + 2f(3/4) + 2f(1) + 2f(5/4) + 2f(3/2) + 2f(7/4) + f(2)]

=18[1 + 2f(1/4) + 2f(1/2) + 2f(3/4) + 2f(1) + 2f(5/4) + 2f(3/2) + 2f(7/4) + f(2)] ≈ 3.8886

(b)∫ 2

0

√1 + 3x dx

=224

{f(0) + 2[f(1/2) + 2f(1) + f(3/2)] + 4[f(1/4) + f(3/4) + f(5/4) + f(7/4)] + f(2)]}

=112

{1 + 2[f(1/2) + 2f(1) + f(3/2)] + 4[f(1/4) + f(3/4) + f(5/4) + f(7/4)] + f(2)]} ≈ 3.8932

(c)∫ 2

0

√1 + 3x dx =

29(1 + 3x)3/2|20 =

29[73/2 − 1] ≈ 3.8934




51. (a) |ETn | ≤

(b− a)3

12n2M .

With a = 0, b = 2 and

|f ′′(x)| =94(1 + 3x)−3/2 ≤ 9

4= M on [0, 2],

we have

|ETn | ≤

23

12n2

94

=3

2n2.

Solving3

2n2≤ 0.0001 for n gives n ≥ 123.

(b) |ESn | ≤

(b− a)5

2880n4M .

With a = 0, b = 2 and

|f (4)(x)| =121516

(1 + 3x)−7/2 ≤ 121516

= M,

we have

|ESn | ≤=

25

2880n4

121516

=27

32n4.

Solving27

32n4≤ 0.0001 for n gives n ≥ 11.

52. Let f(x) =e−x

x.

(a)∫ 3

1

e−x

xdx

=216

[f(1) + 2f(1.25) + 2f(1.5) + 2f(1.75) + 2f(2) + 2f(2.25) + 2f(2.5) + 2f(2.75) + f(3)]

≈ 0.2100

(b)∫ 3

1

e−x

xdx

=224

{f(0) + 2[f(1/2) + 2f(1) + f(3/2)] + 4[f(1/4) + f(3/4) + f(5/4) + f(7/4)] + f(2)]}

≈ 0.2064

Calculus one and several variables 10E Salas solutions manual ch08

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