Ism et chapter_6

Chapter 6

IntegrationTechniques

6.1 Review of Formulasand Techniques

1.

∫eaxdx =

1

aeax + c, for a 6= 0.

2.

∫cos(ax)dx =

1

asin(ax) + c, for a 6= 0.

3.

∫1√

a2 − x2dx =

∫1√

1−(xa

)2(

1

a

)dx

Let u =x

a, du =

1

adx.

=

∫1√

1− u2du = sin−1 (u) + c

= sin−1(xa

)+ c, a > 0.

4.

∫b

|x|√x2 − a2

dx

=

∫b

|x|√(

xa

)2 − 1

(1

a

)dx

Let u =x

a, du =

1

adx and |au| = |x| .

=

∫b

|au|√u2 − 1

du

=b

|a|

∫1

|u|√u2 − 1

du

=b

|a|sec−1 (u) + c

=b

|a|sec−1

(xa

)+ c, a > 0.

5.

∫sin(6t)dt = −1

6cos(6t) + c

6.

∫sec 2t tan 2t dt =

1

2sec 2t+ c

7.

∫(x2 + 4)2dx =

∫(x4 + 8x2 + 16)dx

=x5

5+

8

3x3 + 16x+ c

8.

∫x(x2 + 4)2dx =

∫(x5 + 8x3 + 16x)dx

=x6

6+ 2x4 + 8x2 + c

9.3

16 + x2dx =

3

4tan−1 x

4+ c

10.2

4 + 4x2dx =

1

2tan−1 x+ c

11.

∫1√

3− 2x− x2dx

=

∫1√

4− (x+ 1)2dx = arcsin

(x+ 1

2

)+ c

12.

∫x+ 1√

3− 2x− x2dx

= −1

2

∫−2(x+ 1)√4− (x+ 1)2

dx

= −1

2· 2[4− (x+ 1)2]1/2 + C

= −√

4− (x+ 1)2 + c

13.

∫4

5 + 2x+ x2dx

= 4

∫1

4 + (x+ 1)2dx = 2 tan−1

(x+ 1

2

)+ c

14.

∫4x+ 4

5 + 2x+ x2dx

= 2

∫2(x+ 1)

4 + (x+ 1)2dx = 2 ln | 4 + (x+ 1)2|+ c

15.

∫4t

5 + 2t+ t2dt

=

∫4t+ 4

5 + 2t+ t2dt−

∫4

5 + 2t+ t2dt

= 2 ln∣∣∣4 + (t+ 1)

2∣∣∣− 2tan−1

(t+ 1

2

)+ c

16.

∫t+ 1

t2 + 2t+ 4dt =

∫2 (t+ 1)

(t+ 1)2

+ 3dt

=1

2ln∣∣∣(t+ 1)

2+ 3∣∣∣+ c

17.

∫e3−2xdx = −1

2e3−2x + c

18.

∫3e−6xdx = −3

6e−6x + c

360

6.1. REVIEW OF FORMULAS AND TECHNIQUES 361

19. Let u = 1 + x2/3, du =2

3x−1/3dx∫

4

x1/3(1 + x2/3)dx = 4

(3

2

)∫u−1du

= 6 ln |u|+ C = 6 ln |1 + x2/3|+ c

20. Let u = 1 + x3/4, du =3

4x−1/4dx∫

2

x1/4 + xdx =

∫2

x1/4(1 + x3/4)dx

= 2

(4

3

)∫u−1du =

8

3ln |u|+ C

=8

3ln |1 + x3/4|+ c

21. Let u =√x, du =

1

2√xdx∫

sin√x√

xdx = 2

∫sinudu

= −2 cosu+ C = −2 cos√x+ c

22. Let u =1

x, du = − 1

x2dx∫

cos(1/x)

x2dx = −

∫cosudu

= − sinu+ C = − sin1

x+ c

23. Let u = sinx, du = cosxdx∫ π

0

cosxesin xdx =

∫ 0

0

eudu = 0

24. Let u = tanx, du = sec2 xdx∫ π/2

0

sec2 xetan xdx =

∫ 1

0

eudu

= eu∣∣∣10

= e− 1

25.

∫ 0

−π/4secx tanxdx

= secx∣∣∣0−π/4

= 1−√

2

26.

∫ π/2

π/4

csc2 xdx = − cotx∣∣∣π/2π/4

= 1

27. Let u = x3, du = 3x2dxx2

1 + x6dx =

1

3

∫1

1 + u2du

=1

3tan−1 u+ C =

1

3tan−1 x3 + c

28.

∫x5

1 + x6dx =

1

6ln(1 + x6) + c

29.1√

4− x2dx = sin−1 x

2+ c

30. Let u = ex, du = exdxex√

1− e2xdx =

∫1√

1− u2du

= sin−1 u+ C = sin−1 ex + c

31. Let u = x2, du = 2xdx∫x√

1− x4dx =

1

2

∫1√

1− u2du

=1

2sin−1 u+ C =

1

2sin−1 x2 + c

32. Let u = 1− x4, du = −4x3dx∫2x3

√1− x4

dx = −1

2

∫u−1/2du

= −u1/2 + C = −(1− x4)1/2 + c

33.

∫1 + x

1 + x2dx

=

∫1

1 + x2dx+

1

2

∫2x

1 + x2dx

= tan−1 x+1

2ln |1 + x2|+ c

34.

∫1√x+ x

dx

=

∫x−1/2 · 1

1 + x1/2dx

= 2 ln | 1 + x1/2|+ c

35.

∫lnx2

xdx = 2

∫lnx

(1

x

)dx

Let u = lnx, du =1

xdx.

= 2

∫u du = u2 + c = (lnx)

2+ c

36.

∫ 3

1

e2 ln xdx =

∫ 3

1

x2dx =x3

3

∣∣∣∣31

=26

3

37.

∫ 4

3

x√x− 3dx

=

∫ 4

3

(x− 3 + 3)√x− 3dx

=

∫ 4

3

(x− 3)3/2dx+ 3

∫ 4

3

(x− 3)1/2dx

=2

5(x− 3)5/2

∣∣∣∣43

+ 3 · 2

3(x− 3)3/2

∣∣∣∣43

=12

5

38.

∫ 1

0

x(x− 3)2dx

=

∫ 1

0

(x3 − 6x2 + 9x)dx

=

(x4

4− 2x3 +

9

2x2

)∣∣∣∣10

=11

4

362 CHAPTER 6. INTEGRATION TECHNIQUES

39.

∫ 4

1

x2 + 1√x

dx

=

∫ 4

1

x3/2dx+

∫ 4

1

x−1/2dx

=2

5x5/2

∣∣∣∣41

+ 2x1/2∣∣∣41

=72

5

40.

∫ 0

−2

xe−x2

dx = − 1

2e−x

2

∣∣∣∣0−2

=e−4 − 1

2

41.

∫5

3 + x2dx =

5√3

arctanx√3

+ c∫5

3 + x3dx: N/A

42.

∫sin(3x)dx =

1

3

∫sin(3x)3dx

Let u = 3x, du = 3dx.

=1

3

∫(sinu)du = −1

3cosu+ c

= −1

3cos(3x) + c.∫

sin3xdx =

∫(sin2x) sinxdx

=

∫(1− cos2x)sinxdx

Let u = cosx, du = − sinxdx.

=

∫ (1− u2

)(−du) =

∫u2du−

∫du

=u3

3− u =

cos3x

3− cosx.

43.

∫lnxdx: N/A

Substituting u = lnx,∫lnx

2xdx =

1

4ln2 x+ c

44. Substituting u = x4∫x3

1 + x8dx =

1

4arctanx4 + c∫

x4

1 + x8dx: N/A

45.

∫e−x

2

dx: N/A

Substituting u = −x2∫xe−x

2

dx = −1

2e−x

2

+ c

46.

∫secxdx: N/A∫sec2 xdx = tanx+ c

47.

∫ 2

0

f(x)dx

=

∫ 1

0

x

x2 + 1dx+

∫ 2

1

x2

x2 + 1dx

=1

2ln |x2 + 1|

∣∣∣10

+

∫ 2

1

(1− 1

x2 + 1

)dx

=1

2ln 2 + (x− arctanx)

∣∣∣21

=ln 2

2+ 1 +

π

4− arctan 2

48.

∫4x+ 1

2x2 + 4x+ 10dx

=

∫4x+ 4

2x2 + 4x+ 10dx−

∫3

2x2 + 4x+ 10dx

= ln |2x2 + 4x+ 10| − 3

2

∫1

(x+ 1)2 + 4dx

= ln |2x2 + 4x+ 10| − 3

4tan−1

(x+ 1

2

)+ c

49.

∫1

(1 + x2)dx = tan−1 (x) + c.∫

x

(1 + x2)dx =

1

2

∫2x

(1 + x2)dx

=1

2ln(1 + x2

)+ c.∫

x2

(1 + x2)dx =

∫x2 + 1− 1

(1 + x2)dx

=

∫ (x2 + 1

)(x2 + 1)

dx−∫

1

(1 + x2)dx

=

∫dx−

∫1

(1 + x2)dx

= x− tan−1 (x) + c.∫x3

(1 + x2)dx =

1

2

∫x2

(1 + x2)2xdx

Let u = x2, du = 2xdx.

=1

2

∫u

1 + udu =

1

2

∫u+ 1− 1

1 + udu

=1

2

∫u+ 1

1 + udu−

∫1

1 + udu

=

1

2

∫du −

∫1

1 + udu

=

1

2(u− ln (1 + u)) + c

=1

2x2 − 1

2ln(1 + x2

)+ c.

Hence we can generalize this as follows,∫ (xn

1 + x2

)dx

=1

n− 1xn−1 −

∫ (xn−2

1 + x2

)dx

50.

∫x

1 + x4dx =

1

2

∫1

1 + x42xdx

6.2. INTEGRATION BY PARTS 363


=1

2

∫1

1 + u2du =

1

2tan−1 (u) + c

=1

2tan−1

(x2)

+ c.∫x3

1 + x4dx =

1

4

∫1

1 + x44x3dx

Let u = 1 + x4, du = 4x3.

=1

4

∫1

udu =

1

4ln (u) + c

=1

4ln(1 + x4

)+ c.∫

x5

1 + x4dx

=1

2

∫x4

1 + x42xdx


=1

2

∫u2

1 + u2du =

1

2

∫u2 + 1− 1

1 + u2du

=1

2

∫u2 + 1

1 + u2du−

∫1

1 + u2du

=

1

2

∫du−

∫1

1 + u2du

=

1

2

u− tan−1 (u)

+ c

=1

2

x2 − tan−1

(x2)

+ c.

Hence we can generalize this as follows,∫x4n+1

1 + x4dx =

1

2

x2n−2

n− 1

−∫x4(n−1)+1

1 + x4dx

and∫x4n+3

1 + x4dx =

1

4

x2n

n

−∫x4(n−1)+3

1 + x4dx

6.2 Integration by Parts

1. Let u = x, dv = cosxdxdu = dx, v = sinx.∫x cosxdx = x sinx−

∫sinxdx

= x sinx+ cosx+ c

2. Let u = x, dv = sin 4xdx

du = dx, v = −1

4cos 4x∫

x sin 4x dx

= −1

4x cos 4x−

∫−1

4cos 4x dx

= −1

4x cos 4x+

1

16sin 4x+ c.

3. Let u = x, dv = e2xdx

du = dx, v =1

2e2x.

∫xe2xdx =

1

2xe2x −

∫1

2e2xdx

=1

2xe2x − 1

4e2x + c.

4. Let u = lnx, dv = x dx

du =1

xdx and v =

x2

2.∫

x lnx dx =1

2x2 lnx−

∫1

2x dx

=1

2x2 lnx− 1

4x2 + c.

5. Let u = lnx, dv = x2dx

du =1

xdx, v =

1

3x3.∫

x2 lnxdx =1

3x3 lnx−

∫1

3x3 · 1

xdx

=1

3x3 lnx− 1

3

∫x2dx

=1

3x3 lnx− 1

9x3 + c.

6. Let u = lnx, du =1

xdx.∫

lnx

xdx =

∫udu =

u2

2+ c =

1

2(lnx)2 + c.

7. Let u = x2, dv = e−3xdx

du = 2xdx, v = −1

3e−3x

I =

∫x2e−3xdx

= −1

3x2e−3x −

∫ (−1

3e−3x

)· 2xdx

= −1

3x2e−3x +

2

3

∫xe−3xdx

Let u = x, dv = e−3xdx

du = dx, v = −1

3e−3x

I = −1

3x2e−3x

+2

3

[−1

3xe−3x −

∫ (−1

3e−3x

)dx

]= −1

3x2e−3x − 2

9xe−3x +

2

9

∫e−3xdx

= −1

3x2e−3x − 2

9xe−3x − 2

27e−3x + c

8. Let u = x3, du = 3x2 dx.∫x2ex

3

dx =1

3

∫eu dx =

1

3eu + c

=1

3ex

3

+ c.

9. Let I =

∫ex sin 4xdx

u = ex, dv = sin 4xdx

du = exdx, v = −1

4cos 4x


I = −1

4ex cos 4x−

∫ (−1

4cos 4x

)exdx

= −1

4ex cos 4x+

1

4

∫ex cos 4xdx

Use integration by parts again, this time letu = ex, dv = cos 4xdx

du = exdx, v =1

4sin 4x

I = −1

4ex cos 4x

+1

4

(1

4ex sin 4x−

∫1

4(sin 4x)exdx

)I = −1

4ex cos 4x+

1

16ex sin 4x− 1

16I

So,17

16I = −1

4ex cos 4x+

1

16ex sin 4x+ c1

I = − 4

17ex cos 4x+

1

17ex sin 4x+ c

10. Let, u = e2x, dv = cosx dx so that,du = 2e2x dx and v = sinx.∫e2x cosx dx

= e2x sinx− 2

∫e2x sinx dx

Let, u = e2x, dv = sinx dx so that,du = 2e2x dx and v = − cosx.∫e2x sinx dx

= −e2x cosx+ 2

∫e2x cosx dx∫

e2x cosx dx

= e2x sinx+ 2e2x cosx− 4

∫e2x cosx dx

Now we notice that the integral on both ofthese is the same, so we bring them to one sideof the equation.

5

∫e2x cosx dx

= e2x sinx+ 2e2x cosx+ c1∫e2x cosx dx

=1

5e2x sinx+

2

5e2x cosx+ c

11. Let I =

∫cosx cos 2xdx

and u = cosx, dv = cos 2xdx

du = sinxdx, v =1

2sin 2x

I =1

2cosx sin 2x−

∫1

2sin 2x(− sinx)dx

=1

2cosx sin 2x+

1

2

∫sinx sin 2xdx

Let,u = sinx, dv = sin 2xdx

du = cosxdx v = −1

2cos 2x

I =1

2cosx sin 2x+

1

2

[−1

2cos 2x sinx

−∫ (−1

2cos 2x

)cosxdx

]=

1

2cosx sin 2x − 1

4cos 2x sinx+

1

4Idx

So,3

4I =

1

2cosx sin 2x− 1

4cos 2x sinx+ c1

I =2

3cosx sin 2x− 1

3cos 2x sinx+ c

12. Here we use the trigonometric identity:sin 2x = 2 sinx cosx.

We then make the substitutionu = sinx, du = cosx dx.∫

sinx sin 2x dx =

∫2 sin2 x cosx dx

=

∫2u2 du =

2

3u3 + c =

2

3sin3 x+ c

This integral can also be done by parts, twice.If this is done, an equivalent answer is ob-tained:1

3cosx sin 2x− 2

3cos 2x sinx+ c

13. Let u = x, dv = sec2 xdxdu = dx, v = tanx∫x sec2 xdx = x tanx−

∫tanxdx

= x tanx−∫

sinx

cosxdx

Let u = cosx, du = − sinxdx∫x sec2 xdx = x tanx+

∫1

udu

= x tanx+ ln |u|+ c= x tanx+ ln |cosx|+ c

14. Let u = (lnx)2, dv = dx

du = 2lnx

xdx, v = x

I =

∫(lnx)2dx

= x(lnx)2 −∫x · 2 lnx

xdx

= x(lnx)2 − 2

∫lnxdx

Integration by parts again,

u = lnx, dv = dxdu =1

xdx, v = x

I = x(lnx)2 − 2

[x lnx−

∫x · 1

xdx

]= x(lnx)2 − 2x lnx+ 2

∫dx


= x(lnx)2 − 2x lnx+ 2x+ c

15. Let u = x2, dv = xex2

dx so that, du = 2x dx

and v =1

2ex

2

(v is obtained using substitu-

tion).∫x3ex

2

dx =1

2x2ex

2

−∫xex

2

dx

=1

2x2ex

2

− 1

2ex

2

+ c

16. Let u = x2, dv =

(x

(4 + x2)3/2

)dx

du = 2xdx, v = − 1√4 + x2∫

x3

(4 + x2)3/2

dx =

∫x2

(x

(4 + x2)3/2

)dx

= − x2

√4 + x2

+

∫1√

4 + x22xdx

= − x2√(4 + x2)

+ 2√

(4 + x2) + c.

17. Let u = ln(sinx), dv = cosxdx

du =1

sinx· cosxdx, v = sinx

I =

∫cosx ln(sinx)dx

= sinx ln(sinx)

−∫

sinx · 1

sinx· cosxdx

= sinx ln(sinx)−∫

cosxdx

= sinx ln(sinx)− sinx+ c

18. This is a substitution u = x2.∫x sinx2dx =

1

2

∫sinudu

= −1

2cosu+ c = −1

2cosx2 + c.

19. Let u = x, dv = sin 2xdx

du = dx, v = −1

2cos 2x∫ 1

0

x sin 2xdx

= −1

2x cos 2x

∣∣∣∣10

−∫ 1

0

(−1

2cos 2x

)dx

= −1

2(1 cos 2− 0 cos 0) +

1

2

∫ 1

0

cos 2xdx

= −1

2cos 2 +

1

2

[1

2sin 2x

]1

0

= −1

2cos 2 +

1

4(sin 2− sin 0)

= −1

2cos 2 +

1

4sin 2

20. Let u = 2x, dv = cosx dxdu = 2 dxandv = sinx.∫ π

0

2x cosxdx = 2x sinx|π0 − 2

∫ π

0

sinxdx

= (2x sinx+ 2 cosx)|π0 = −4.

21.

∫ 1

0

x2 cosπxdx

Let u = x2, dv = cosπxdx,

du = 2xdx, v =sinπx

π.∫ 1

0

x2cosπxdx = x2 sinπx

π

∣∣∣∣10

−∫ 1

0

sinπx

π2xdx

= (0− 0)− 2

π

∫ 1

0

x sin (πx) dx

= − 2

π

∫ 1

0

x sin (πx) dx

Let u = x, dv = sin(πx)dx,

du = dx, v = −cos(πx)

π.

− 2

π

∫ 1

0

xsin(πx)dx

= − 2

π

−x cos(πx)

π

∣∣∣∣10

−∫ 1

0

−cos(πx)

πdx

= − 2

π

(−cosπ

π− 0) +

1

π

[sin(πx)

π

]1

0

= − 2

π

1

π+

1

π(0− 0)

= − 2

π2

22.

∫ 1

0

x2e3xdx

Let u = x2, dv = e3xdx,

du = 2xdx, v =e3x

3.∫ 1

0

x2e3xdx =x2e3x

3

∣∣∣∣10

−∫ 1

0

e3x

32xdx

=1

3

(e3 − 0

)− 2

3

∫ 1

0

xe3xdx.

Let u = x, dv = e3xdx,

dv = dx, v =e3x

3.

e3

3− 2

3

∫ 1

0

xe3xdx

=e3

3− 2

3

xe3x

3

∣∣∣∣10

−∫ 1

0

e3x

3dx

=e3

3− 2

3

(e3

3

)−∫ 1

0

e3x

3dx

=e3

3− 2

3

(e3

3

)−[e3x

9

]1

0

=e3

3− 2

3

(e3

3

)− 1

9

(e3 − 1

)


=e3

3− 2e3

9+

2

27

(e3 − 1

)=e3

3− 2e3

9+

2e3

27− 2

27=

5e3

27− 2

27

23.

∫ 10

1

ln 2xdx

Let u = ln 2x, dv = dx

du =1

xdx, v = x.∫ 10

1

ln (2x)dx = x ln (2x)|101 −

∫ 10

1

x1

xdx

= (10 ln(20)− ln 2)−∫ 10

1

dx

= (10 ln(20)− ln 2)− [x]101

= (10 ln(20)− ln 2)− (10− 1)= (10 ln(20)− ln 2)− 9.

24. Let, u = lnx, dv = x dx

du =1

xdx, v =

x2

2.∫ 2

1

x lnxdx =1

2x2 lnx

∣∣∣∣21

−∫ 2

1

1

2xdx

=

(1

2x2 lnx− 1

4x2

)∣∣∣∣21

= 2 ln 2− 3

4.

25.

∫x2eaxdx

Let u = x2, dv = eaxdx,

du = 2xdx, v =eax

a.∫

x2eaxdx = x2 eax

a−∫eax

a2xdx

=x2eax

a− 2

a

∫xeaxdx.

Let u = x, dv = eaxdx,

dv = dx, v =eax

a.

x2eax

a− 2

a

∫xeaxdx

=x2eax

a− 2

a

xeax

a−∫eax

adx

=x2eax

a− 2

a

xeax

a− eax

a2

+ c

=x2eax

a− 2xeax

a2+

2eax

a3+ c, a 6= 0.

26.

∫x sin (ax) dx

Let u = x, dv = sin axdx,

du = dx, v = −cos ax

a.∫

xsin (ax) dx

= x− cos (ax)

a−∫−cos (ax)

adx

= −x cos (ax)

a+

sin (ax)

a2+ c, a 6= 0.

27.

∫(xn) (lnx) dx =

∫(lnx) (xn) dx

Let u = lnx, dv = xndx,

du =1

xdx, v =

xn+1

(n+ 1).∫

(lnx)(xn) dx

= (lnx)xn+1

(n+ 1)−∫

xn+1

(n+ 1)

dx

x

=xn+1 (lnx)

(n+ 1)−∫

xn

(n+ 1)dx

=xn+1 (lnx)

(n+ 1)− xn+1

(n+ 1)2 + c, n 6= −1.

28.

∫(sin ax) (cos bx) dx

Let u = sin ax, dv = (cos bx) dx

du = a (cos ax) dx, v =sin bx

b.∫

sin ax cos bx dx

= (sin ax)sin bx

b−∫a

(sin bx

b

)(cos ax) dx

=(sin ax) (sin bx)

b− a

b

∫(cos ax) (sin bx) dx

Let u = cos ax, dv = sin bxdx,

du = −a (sin ax) dx, v = −cos bx

b.

sin ax sin bx

b− a

b

∫cos ax sin bx dx

=sin ax sin bx

b− a

b

cos ax

− cos bx

b

−∫− cos bx

b(− sin ax) adx

=

sin ax sin bx

b− a

b

− cos ax cos bx

b

−ab

∫cos bx sin ax dx

=

sin ax sin bx

b+a cos ax cos bx

b2

+(ab

)2∫

sin ax cos bx dx∫sin ax cos bx dx

=sin ax sin bx

b+a cos ax cos bx

b2

+(ab

)2∫

sin ax cos bx dx∫sin ax cos bx dx−

(ab

)2∫

sin ax cos bx dx

=sin ax sin bx

b+a cos ax cos bx

b2


(1− a2

b2

)∫sin ax cos bx dx

=sin ax sin bx

b+a cos ax cos bx

b2∫sin ax cos bx dx

=

(b2

b2 − a2

)(sin ax sin bx

b+a cos ax cos bx

b2

)∫

sin ax cos bx dx

=

(1

b2 − a2

)(b sin ax sin bx+ a cos ax cos bx) ,

a 6= 0 b 6= 0.

29. Letu = cosn−1 x, dv = cosxdxdu = (n− 1)(cosn−2 x)(− sinx)dx, v = sinx∫

cosn xdx

= sinx cosn−1 x

−∫

(sinx)(n− 1)(cosn−2 x)(− sinx)dx

= sinx cosn−1 x

+

∫(n− 1)(cosn−2 x)(sin2 x)dx

= sinx cosn−1 x

+

∫(n− 1)(cosn−2 x)(1− cos2 x)dx

= sinx cosn−1 x

+

∫(n− 1)(cosn−2 x− cosn x)dx

Thus,

∫cosn xdx

= sinx cosn−1 x+

∫(n− 1) cosn−2 xdx

− (n− 1)

∫cosn xdx.

n

∫cosn xdx = sinx cosn−1 x

+ (n− 1)

∫cosn−2 xdx∫

cosn xdx

=1

nsinx cosn−1 x+

n− 1

n

∫cosn−2 xdx

30. Let u = sinn−1 x, dv = sinx dxdu = (n− 1) sinn−2 x cosx, v = − cosx.∫

sinn xdx

= − sinn−1 x cosx

+ (n− 1)

∫cos2 x sinn−2 xdx


+ (n− 1)

∫(1− sin2 x) sinn−2 xdx


− (n− 1)

∫sinn−2 xdx

+ (n− 1)

∫sinn xdx

n

∫sinn xdx


− (n− 1)

∫sinn−2 xdx∫

sinn xdx = − 1

nsinn−1 x cosx

− n− 1

n

∫sinn−2 xdx

31.

∫x3exdx = ex(x3 − 3x2 + 6x− 6) + c

32.

∫cos5 xdx

=1

5cos4 sinx+

4

5

∫cos3 xdx

=1

5cos4 sinx

+4

5

(1

3cos2 x sinx+

2

3

∫cosxdx

)=

1

5cos4 sinx+

4

15cos2 x sinx

+8

15sinx+ c

33.

∫cos3 xdx

=1

3cos2 x sinx+

2

3

∫cosxdx

=1

3cos2 x sinx+

2

3sinx+ c

34.

∫sin4 xdx

= −1

4sin3 x cosx+

3

4

∫sin2 xdx

= −1

4sin3 x cosx+

3

4

(1

2x− 1

4sin 2x

)

35.

∫ 1

0

x4exdx

= ex(x4 − 4x3 + 12x2 − 24x+ 24)∣∣10

= 9e− 24

36. Using the work done in Exercise 34,∫ π/2

0

sin4 xdx

=

(−1

4sin3 x cosx+

3

8x− 3

16sin 2x

)∣∣∣∣π/20

=3π

16

37.

∫ π/2

0

sin5 xdx


= −1

5sin4 x cosx

∣∣∣∣π/20

+4

5

∫ π/2

0

sin3 xdx

= −1

5sin4 x cosx

∣∣∣∣π/20

+4

5

(−1

3sin2 x cosx− 2

3cosx

)∣∣∣∣π/20

(Using Exercise 30)

= −1

5

(sin4

(π2

)cos

π

2− sin4 0 cos 0

)+

4

5

(−1

3sin2

(π2

)cos

π

2− 2

3cos

π

2

)=

8

15

38. Here we will again use the work we did in Ex-ercise 34.∫

sin6 xdx

= −1

6sin5 x cosx+

5

6

∫sin4 xdx

= −1

6sin5 x cosx

+5

6

(−1

4sin3 x cosx+

3

8x− 3

16sin 2x

)+ c

= −1

6sin5 x cosx− 5

24sin3 x cosx

+15

48x− 15

96sin 2x+ c

We now just have to plug in the endpoints:∫ π/2

0

sin6 xdx

=

(−1

6sin5 x cosx− 5

24sin3 x cosx

+15

48x− 15

96sin 2x

)∣∣∣∣π/20

=15π

96

39. m even :∫ π/2

0

sinm xdx

=(m− 1)(m− 3) . . . 1

m(m− 2) . . . 2· π

2m odd:∫ π/2

0

sinm xdx

=(m− 1)(m− 3) . . . 2

m(m− 2) . . . 3

40. m even:∫ π/2

0

cosm xdx

=π(n− 1)(n− 3)(n− 5) · · · 1

2n(n− 2)(n− 4) · · · 2m odd:

∫ π/2

0

cosm xdx

=(n− 1)(n− 3)(n− 5) · · · 2n(n− 2)(n− 4) · · · 3

.

41. Let u = cos−1 x, dv = dx

du = − 1√1− x2

dx, v = x

I =

∫cos−1 xdx

= x cos−1 x−∫x

(− 1√

1− x2

)dx

= x cos−1 x+

∫x√

1− x2dx

Substituting u = 1− x2, du = −2xdx

I = x cos−1 x+

∫1√u

(−1

2du

)= x cos1 x− 1

2

∫u−1/2du

= x cos−1 x− 1

2· 2u1/2 + c

= x cos−1 x−√

1− x2 + c

42. Let u = tan−1 x, dv = dx

du =1

1 + x2dx, v = x

I =

∫tan−1 xdx = x tan−1 x−

∫x

1 + x2dx

Substituting u = 1 + x2,

I = x tan−1 x− 1

2ln(1 + x2) + c.

43. Substituting u =√x, du =

1

2√xdx

I =

∫sin√xdx = 2

∫u sinudu

= 2(−u cosu+ sinu) + c= 2(−

√x cos

√x+ sin

√x) + c

44. Substituting w =√x

dw =1

2√xdx =

1

2wdx

I =

∫e√xdx =

∫2wewdx

Next, using integration by partsu = 2w, dv = ewdwdu = 2dw, v = ew

I = 2wew − 2

∫ewdw

= 2wew − 2ew + c = 2√xe√x − 2e

√x + c

45. Let u = sin(lnx), dv = dx

du = cos(lnx)dx

x, v = x

I =

∫sin(lnx)dx

= x sin(lnx)−∫

cos(lnx)dx


Integration by parts again,u = cos(lnx), dv = dx

du = − sin(lnx)dx

x, v = x∫

cos(lnx)dx

= x cos(lnx) +

∫sin(lnx)dx

I = x sin(lnx)− x cos(lnx)− I2I = x sin(lnx)− x cos(lnx) + c1

I =1

2x sin(lnx)− 1

2x cos(lnx) + c

46. Let u = 4 + x2, du = 2xdx

I =

∫x ln(4 + x2)dx

=1

2

∫lnudu =

1

2(u lnu− u) + C

=1

2[(4 + x2) ln(4 + x2)− 4− x2] + c

47. Let u = e2x, du = 2e2xdx

I =

∫e6x sin(e2x)dx =

1

2

∫u2 sinudu

Let v = u2, dw = sinududv = 2udu, w = − cosu

I =1

2

(−u2 cosu+ 2

∫u cosudu

)= −1

2u2 cosu+

∫u cosudu

= −1

2u2 cosu+ (u sinu+ cosu) + c

= −1

2e4x cos(e2x) + e2x sin(e2x)

+ cos(e2x) + c

48. Let u = 3√x = x1/3, du =

1

3x−2/3dx,

3u2du = dx

I =

∫cosx1/3dx = 3

∫u2 cosudu

Let v = u2, dw = cosududv = 2udu,w = sinu

I = 3

(u2 sinu− 2

∫u sinudu

)= 3u2 sinu− 6

∫u sinudu

= 3u2 sinu− 6

(−u cosu+

∫cosudu

)= 3u3 sinu+ 6u cosu− 6 sinu+ c= 3x sin 3

√x+ 6 3

√x cos 3

√x− 6 sin 3

√x+ c

49. Let u = 3√x = x1/3, du =

1

3x−2/3dx,

3u2du = dx

I =

∫e

3√xdx = 3

∫u2eudu

= 3

(u2eu − 2

∫ueudu

)= 3u2eu − 6

(ueu −

∫eudu

)= 3u2eu − 6ueu + 6eu + c

Hence

∫ 8

0

e3√xdx =

∫ 2

0

3u2eudu

=(3u2eu − 6ueu + 6eu

)∣∣20

= 6e2 − 6

50. Let u = tan−1 x, dv = xdx

du =dx

1 + x2, v =

x2

2

I =

∫x tan−1 xdx

= tan−1 xx2

2− 1

2

∫x2

1 + x2dx

= tan−1 xx2

2

− 1

2

[∫1dx−

∫1

1 + x2dx

]= tan−1 x

x2

2− 1

2

(x− tan−1 x

)+ C

= tan−1 xx2

2− x

2+

1

2tan−1 x+ c

Hence

∫ 1

0

x tan−1 xdx

=

(tan−1 x

x2

2− x

2+

1

2tan−1 x

)∣∣∣∣10

=π

4− 1

2

51. n times. Each integration reduces the power ofx by 1.

52. 1 time. The first integration by parts gets ridof the lnx and turns the integrand into a sim-ple integral. See, for example, Problem 4.

53. (a) As the given problem,∫x sinx2dx can

be simplified by substituting x2 = u, wecan solve the example using substitutionmethod.

(b) As the given integral,∫x2 sinx dx can not

be simplified by substitution method andcan be solved using method of integrationby parts.

(c) As the integral,∫x lnx dx can not be sim-

plified by substitution and can be solvedusing the method of integration by parts.

(d) As the given problem,

∫lnx

xdx can be

simplified by substituting , lnx = u wecan solve the example by substitutionmethod.

54. (a) As this integral,∫x3e4xdx can not be

simplified by substitution method and can


be solved by using the method of integra-tion by parts.

(b) As the given problem,∫x3ex

4

dx can besimplified by substituting x4 = u, we cansolve the example using the substitutionmethod.

(c) As the given problem,

∫x−2e

4x dx can be

simplified by substituting1

x= u, we can

solve the example using the substitutionmethod.

(d) As this integral,∫x2e−4xdx can not be

simplified by substitution and can besolved by using the method of integrationby parts.

55. First column: each row is the derivative of theprevious row; Second column: each row is theantiderivative of the previous row.

56.sinx

x4 − cosx +4x3 − sinx −

12x2 cosx +24x sinx −

24 − cosx +∫x4 sinxdx

= −x4 cosx+ 4x3 sinx+ 12x2 cosx− 24x sinx− 24 cosx+ c

57.cosx

x4 sinx +4x3 − cosx −

12x2 − sinx +24x cosx −

24 sinx +∫x4 cosxdx

= x4 sinx+ 4x3 cosx− 12x2 sinx− 24x cosx+ 24 sinx+ c

58.ex

x4 ex +4x3 ex −

12x2 ex +24x ex −

24 ex +∫x4exdx

= (x4 − 4x3 + 12x2 − 24x+ 24)ex + c

59.e2x

x4 e2x/2 +4x3 e2x/4 −

12x2 e2x/8 +24x e2x/16 −

24 e2x/32 +∫x4e2xdx

=

(x4

2− x3 +

3x2

2− 3x

2+

3

4

)e2x + c

60.cos 2x

x5 sin 2x/2 +5x4 − cos 2x/4 −

20x3 − sin 2x/8 +60x2 cos 2x/16 −120x sin 2x/32 +

120 − cos 2x/64 −∫x5 cos 2xdx

=1

2x5 sin 2x +

5

4x4 cos 2x

− 20

8x3 sin 2x− 60

16x2 cos 2x

+120

32x sin 2x+

120

64cos 2x+ c

61.e−3x

x3 −e−3x/3 +3x2 e−3x/9 −6x −e−3x/27 +

6 e−3x/81 −∫x3e−3xdx

=

(−x

3

3− x2

3− 2x

9− 2

27

)e−3x + c

62.x2

lnx x3/3 +x−1 x4/12 +−x−2 x5/60 +

The table will never terminate.

63. (a) Use the identitycosA cosB

=1

2[cos(A−B) + cos(A+B)]

This identity gives∫ π

−πcos(mx) cos(nx)dx


=

∫ π

−π

1

2[cos((m− n)x)

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

=1

2

[sin((m− n)x)

m− n

+sin((m+ n)x)

m+ n

]∣∣∣∣π−π

= 0

It is important that m 6= n because oth-erwise cos((m− n)x) = cos 0 = 1

(b) Use the identitysinA sinB

=1

2[cos(A−B)− cos(A+B)]


−πsin(mx) sin(nx)dx

=

∫ π

−π

1

2[cos((m− n)x)

− cos((m+ n)x)]dx

=1

2

[sin((m− n)x)

m− n

− sin((m+ n)x)

m+ n

]∣∣∣∣π−π

= 0

It is important that m 6= n because oth-erwise cos((m− n)x) = cos 0 = 1

64. (a) Use the identitycosA sinB

=1

2[sin(B +A)− sin(B −A)]


−πcos(mx) sin(nx) dx

=

∫ π

−π

1

2[sin((n+m)x)

− sin((n−m)x)] dx

=1

2

[−cos((n+m)x)

n+m

+cos((n−m)x)

n−m

]∣∣∣∣π−π

= 0

(b) We have seen that∫cos2 xdx =

1

2x+

1

4cos(2x) + c

Hence by letting u = nx:∫ π

−πcos2(nx)dx

=1

n

∫ nπ

−nπcos2 udu

=1

n

(1

2u+

1

4cos(2u)

)∣∣∣∣nπ−nπ

= π

And then

∫ π

−πsin2(nx)dx

=

∫ π

−π(1− cos2(nx))dx

=

∫ π

−πdx−

∫ π

−πcos2(nx)dx

= 2π − π = π

65. The only mistake is the misunderstanding of

antiderivatives. In this problem,

∫exe−xdx

is understood as a group of antiderivatives ofexe−x, not a fixed function. So the subtraction

by

∫exe−xdx on both sides of∫

exe−xdx = −1 +

∫exe−xdx

does not make sense.

66. V = π

∫ π

0

(x√

sinx)2dx = π

∫ π

0

x2 sinxdx

Using integration by parts twice we get∫x2 sinxdx

= −x2 cosx+ 2

∫x cosxdx

= −x2 cosx+ 2(x sinx−∫

sinxdx)

= −x2 cosx+ 2x sinx+ 2 cosx+ c

Hence,V = (−x2 cosx+ 2x sinx+ 2 cosx)

∣∣π0

= π2 − 4 ≈ 5.87

67. Let u = lnx, dv = exdx

du =dx

x, v = ex∫

ex lnxdx = ex lnx−∫ex

xdx∫

ex lnxdx+

∫ex

xdx = ex lnx+ C

Hence,∫ex(

lnx+1

x

)dx = ex lnx+ c

68. We can guess the formula:∫ex(f(x) + f ′(x))dx = exf(x) + c

and prove it by taking the derivative:d

dx(exf(x)) = exf(x) + exf ′(x)


= ex(f(x) + f ′(x))

69. Consider,

∫ 1

0

f ′′(x)g (x) dx

Choose u = g (x) and dv = f ′′(x)dx,so that du = g′ (x) dx and , v = f ′ (x) .

Hence, we have∫ 1

0

g (x)f ′′(x)dx

= g (x) f ′ (x)|10 −∫ 1

0

f ′ (x)g′ (x) dx

= (g (1) f ′ (1)− g (0) f ′ (0))

−∫ 1

0

g′ (x)f ′ (x) dx

From the given data.

= (0− 0)−∫ 1

0

g′ (x)f ′ (x) dx.

Choose, u = g′ (x) and dv = f ′(x)dx,so that,du = g′′ (x) dx and v = f (x) .

Hence, we have

−∫ 1

0

g′ (x)f ′ (x) dx

= −g′ (x) f (x)|10 −

∫ 1

0

f (x) g′′ (x) dx

= −(g′ (1) f (1)− g′ (0) f (0) )

−∫ 1

0

f (x) g′′ (x) dx

From the given data.

= −

(0− 0 )−∫ 1

0

f (x) g′′ (x) dx

=

∫ 1

0

f (x) g′′ (x) dx.

70. Consider,∫ b

a

f ′′ (x) (b− x) dx =

∫ b

a

(b− x) f ′′ (x) dx

Choose u = (b− x) and dv = f ′′ (x) dx,so that du = −dx and v = f ′ (x) .

Hence, we have:∫ b

a

(b− x)f ′′ (x) dx

= (b− x) f ′ (x)|ba +

∫ b

a

f ′ (x) dx

= (0− [(b− a) f ′ (a)]) +

∫ b

a

f ′ (x) dx

= − [(b− a) f ′ (a)] + f (x)|ba= − [(b− a) f ′ (a)] + f (b)− f (a)∫ b

a

f ′′ (x) (b− x) dx

= − [(b− a) f ′ (a)] + f (b)− f (a)

f (b) = f (a) + (b− a) f ′ (a)

+

∫ b

a

f ′′ (x) (b− x) dx

Consider

∫ b

0

x sin (b− x) dx

=

∫ b

0

(b− x) sinxdx =

∫ b

0

(sinx) (b− x) dx

Now, considerf (x) = x− sinx⇒ f ′ (x) = 1− cosx

and f ′′ (x) = sinx.Therefore, usingf (b) = f (a) + f ′ (a) (b− a)

+

∫ b

a

f ′′ (x) (b− x) dx,

we getb− sin b = 0− sin 0 + f ′ (0) (b− 0)

+

∫ b

0

(sinx) (b− x) dx

⇒ |sin b− b| =

∣∣∣∣∣∫ b

0

x sin (b− x) dx

∣∣∣∣∣.Further,

|sin b− b| =

∣∣∣∣∣∫ b

0

x sin (b− x) dx

∣∣∣∣∣ ≤∣∣∣∣∣∫ b

0

xdx

∣∣∣∣∣,as sin (b− x) ≤ 1.

Thus, |sin b− b| ≤ b2

2.

Therefore the error in the approximation

sinx ≈ x is at most1

2x2.

6.3 TrigonometricTechniques ofIntegration

1. Let u = sinx, du = cosxdx∫cosx sin4 xdx =

∫u4du

=1

5u5 + c =

1

5sin5 x+ c

2. Let u = sinx, du = cosxdx∫cos3 x sin4 xdx =

∫(1− u2)u4du

=u5

5− u7

7+ c

=sin5 x

5− sin7 x

7+ c

3. Let u = sin 2x, du = 2 cos 2xdx.∫ π/4

0

cos 2xsin32xdx

=1

2

∫ 1

0

u3du =1

2

[u4

4

]1

0

=1

8

6.3. TRIGONOMETRIC TECHNIQUES OF INTEGRATION 373

4. Let u = cos 3x, du = −3 sinxdx.∫ π/3

π/4

(cos33x

) (sin33x

)dx

= −1

3

∫ −1

−1/√

2

u3(1− u2

)du

= −1

3

[u4

4− u6

6

]−1

−1√2

= −1

3

(3

16− 7

48

)= − 1

72

5. Let u = cosx, du = − sinxdx∫ π/2

0

cos2 x sinxdx =

∫ 0

1

u2(−du)

=

(−1

3u3

)∣∣∣∣01

=1

3

6. Let u = cosx, du = − sinxdx∫ 0

−π/2cos3 x sinxdx = −

∫ 1

0

u3du = −1

7.

∫cos2 (x+ 1) dx

=1

2

∫(1 + cos 2 (x+ 1))dx

=1

2x+

1

4(sin 2 (x+ 1)) + c.

8. Let u = x− 3, du = dx∫sin4(x− 3)dx =

∫sin4udu

=

∫ (sin2u

)2du

=

∫(1− cos 2u)

2× (1− cos 2u)

2du

=

∫1

4

(1− 2 cos 2u+ cos22u

)=

1

4

∫ [1− 2 cos 2u+

1

2(1 + cos 4u)

]du

=3

8u− 1

4sin 2u+

1

32cos 4u+ c

=3

8(x− 3)− 1

4sin 2 (x− 3)

+1

32cos 4 (x− 3) + c.

9. Let u = secx, du = secx tanxdx∫tanx sec3 xdx

=

∫tanx secx sec2 xdx

=

∫u2du =

1

3u3 + c =

1

3sec3 x+ c

10. Let u = cotx, du = − csc2 xdx∫cotx csc4 xdx

= −∫

cotx(1 + cot2 x) · csc2 xdx

= −∫

(u+ u3)du = −u2

2− u4

4+ C

= −cot2 x

2− cot4

4+ c

11. Let u = x2 + 1, so that du = 2xdx.∫xtan3

(x2 + 1

) (sec(x2 + 1

))dx

=1

2

∫tan3u (secu) du

=1

2

∫ [(sec2u− 1

)tanu (secu)

]du

Let secu = t, dt = tanu secudu

=1

2

∫ (t2 − 1

)dt =

1

2

[t3

3− t]

+ c

=1

2

[sec3u

3− secu

]+ c

=1

6sec3

(x2 + 1

)− 1

2sec(x2 + 1

)+ c.

12. Let u = 2x+ 1, so that du = 2dx.∫tan (2x+ 1) .sec3 (2x+ 1) dx

=1

2

∫tanu. secu.sec2udu

=1

2

∫sec2utanu secudu

Let t = secu, so that dt = tanu secudu.

=1

2

∫t2dt =

1

2

[t3

3

]+ c

=1

2

[sec3u

3

]+ c =

1

6sec3 (2x+ 1) + c.

13. Let u = cotx, du =(−csc2x

)dx∫

cot2x csc4xdx =

∫cot2x

(1 + cot2x

)csc2xdx

= −∫u2(1 + u2

)du

= −u3

3− u5

5+ c

= − (cotx)3

3− (cotx)

5

5+ c.

14. Let u = cotx, du =(−csc2x

)dx.∫

cot2x csc2xdx = −∫u2du

= −u3

3+ c =

cot3x

3+ c.

15. Let u = tanx, du = sec2 xdx∫ π/4

0

tan4 x sec4 xdx

=

∫ π/4

0

tan4 x sec2 x sec2 xdx


=

∫ π/4

0

tan4 x(1 + tan2 x) sec2 xdx

=

∫ 1

0

u4(1 + u2) du

=

∫ 1

0

(u4 + u6)du =u5

5+u7

7

∣∣∣∣10

=12

35

16. Let u = tanx, du = sec2 xdx.∫ π/4

π/4

tan4 x sec2 xdx

=

∫ 1

−1

u4du =u5

5

∣∣∣∣1−1

=2

5

17.

∫cos2 x sin2 xdx

=

∫1

2(1 + cos 2x) · 1

2(1− cos 2x)dx

=1

4

∫(1− cos2 2x)dx

=1

4

∫ [1− 1

2(1 + cos 4x)

]dx

=1

4

(1

2x− 1

8sin 4x

)+ c

=1

8x− 1

32sin 4x+ c

18.

∫(cos2 x+ sin2 x)dx =

∫1dx = x+ c

19. Let u = cosx, du = − sinxdx∫ 0

−π/3

√cosx sin3 xdx

=

∫ 0

−π/3

√cosx(1− cos2 x) sinxdx

=

∫ 1

1/2

√u(1− u2)(−du)

=

∫ 1

1/2

(u5/2 − u1/2)du

=

[2

7u7/2 − 2

3u3/2

]∣∣∣∣11/2

=25

168

√2− 8

21

20. Let u = cotx, du = − csc2 xdx∫ π/2

π/4

cot2 x csc4 xdx

=

∫ π/2

π/4

cot2 x csc2 x csc2 xdx

=

∫ π/2

π/4

cot2 x(1 + cot2 x) csc2 xdx

= −∫ 0

1

u2(1 + u2)du

= −[u3

3+u5

5

]∣∣∣∣01

=1

3+

1

5=

8

15

21. Let x = 3 sin θ,−π2< θ <

π

2dx = 3 cos θ dθ∫

1

x2√

9− x2dx =

∫3 cos θ

9 sin2 θ · 3 cos θdθ

=1

9

∫csc2 θdθ = −1

9cot θ + C

By drawing a diagram, we see that if

x = sin θ, then cot θ =

√9− x2

x.

Thus the integral = −√

9− x2

9x+ c

22. Let x = 4 sin θ,−π2< θ <

π

2,

dx = 4 cos θdθ∫1

x2√

16− x2dx =

∫cos θ

16 sin2 θ cos θdθ

=1

16

∫csc2 θdθ = − 1

16cot θ + c

= −√

16− x2

16x+ c

23. Let x = 4sinθ, so that dx = 4 cos θdθ.∫x2

√16− x2

dx =

∫ (16sin2θ

)4 cos θ√

16− (4 sin θ)2dθ

= 64

∫ (sin2θ

)cos θ√

16− 16sin2θdθ

= 64

∫ (sin2θ

)cos θ

4√(

1− sin2θ)dθ

= 16

∫sin2θ cos θ

cos θdθ = 16

∫sin2θdθ

= 16

∫ (1− cos 2θ

2

)dθ

= 8

[∫dθ −

∫(cos 2θ) dθ

]= 8

[θ − sin 2θ

2

]+ c

= 8sin−1(x

4

)− 4 sin

[2sin−1

(x4

)]+ c.

= 8sin−1(x

4

)− x√

16− x2

2+ c

24. Let x = 3 sin θ, so that dx = 3 cos θdθ.∫x3

√9− x2

dx

=

∫27(sin3θ

)√9− (3 sin θ)

2(3 cos θ) dθ

= 81

∫sin3θ√

9− 9sin2θ(cos θ) dθ

= 81

∫ (sin3θ

3 cos θ

)cos θdθ = 27

∫sin3θdθ

= 27

∫ (3 sin θ − sin 3θ

4

)dθ


=27

4

[3

∫sin θdθ −

∫sin 3θdθ

]=

27

4

[−3 cos θ +

cos 3θ

3

]+ c

=27

4

−3 cos

[sin−1

(x3

)]+

cos[sin−1

(x3

)]3

+ c.

25. This is the area of a quarter of a circle of radius2,∫ 2

0

√4− x2dx = π

26. Let u = 4− x2, du = −2xdx∫ 1

0

x√4− x2

dx = −∫ 3

4

du

2√u

= −u1/2∣∣∣34

= 2−√

3

27. Let x = 3 sec θ, dx = 3 sec θ tan θdθ.

I =

∫x2

√x2 − 9

dx

=

∫27 sec2 θ sec θ tan θ√

9 sec2 θ − 9dθ

=

∫9 sec3 θdθ

Use integration by parts.Let u = sec θ and dv = sec2 θdθ. This gives∫

sec3 θdθ

= sec θ tan θ −∫

sec θ tan2 θdθ

= sec θ tan θ −∫

sec θ(sec2 θ − 1)dθ

= sec θ tan θ +

∫sec θdθ −

∫sec3 θdθ

2

∫sec3 θdθ

= sec θ tan θ +

∫sec θdθ∫

sec3 θdθ

=1

2sec θ tan θ +

1

2

∫sec θ dθ

This leaves us to compute

∫sec θdθ.

For this notice if u = sec θ + tan θ thendu = sec θ tan θ + sec2 θ.∫

sec θdθ

=

∫sec θ(sec θ + tan θ)

sec θ + tan θdθ

=

∫1

udu = ln |u|+ c

= ln | sec θ + tan θ|+ c

Putting all these together and using

sec θ =x

3, tan θ =

√x2 − 9

3:∫

x2

√x2 − 9

dx =

∫9 sec3 θ dθ

=9

2sec θ tan θ +

9

2

∫sec θ dθ

=9

2sec θ tan θ +

9

2ln | sec θ + tan θ|+ c

=9

2

(x3

)(√x2 − 9

3

)

+9

2ln

∣∣∣∣∣x3 +

√x2 − 9

3

∣∣∣∣∣+ c

=x√x2 − 9

2+

9

2ln

∣∣∣∣∣x+√x2 − 9

3

∣∣∣∣∣+ c

28. Let u = x2 − 1, du = 2xdx∫x3√x2 − 1dx

=1

2

∫x2√x2 − 1(2x)dx

=1

2

∫(u+ 1)

√udu

=1

2

∫u3/2 + u1/2 du

=1

2

(2u5/2

5+

2u3/2

3

)+ c

=1

5(x2 − 1)5/2 +

1

3(x2 − 1)3/2 + c

29. Let x = 2 sec θ, dx = 2 sec θ tan θdθ∫2√

x2 − 4dx =

∫4 sec θ tan θ

2 tan θdθ

= 2

∫sec θdθ

= 2 ln |2 sec θ + 2 tan θ|+ c

= 2 ln∣∣∣x+

√x2 − 4

∣∣∣+ c

30. Let x = 2 sec θ, dx = 2 sec θ tan θdθ∫x√

x2 − 4dx =

∫4 sec2 θ tan θ

2 tan θdθ

= 2

∫sec2 θdθ = 2 tan θ + C =

√x2 − 4 + c

31.

∫ √4x2 − 9

xdx =

∫ √4x2 − 9

4x24xdx

Let u =√

4x2 − 9,

du =1

2√

4x2 − 98xdx =

1

2u8xdx

or udu = 4xdx.

Hence, we have∫ √4x2 − 9

xdx


=

∫u

u2 + 9udu =

∫u2

u2 + 9du

=

∫u2 + 9− 9

u2 + 9du =

∫du−

∫9

u2 + 9du

= u− 9tan−1(u

3

)+ c

=√

4x2 − 9− 9tan−1

(√4x2 − 9

3

)+ c.

32. Let x = 2 sec θ, dx = 2 tan θ sec θdθ.∫ √x2 − 4

x2dx

=

∫ √4sec2θ − 4

4sec2θ(2 tan θ sec θ) dθ

=

∫2 tan θ

4sec2θ(2 tan θ sec θ) dθ

=

∫tan2θ

sec θdθ =

∫sec2θ − 1

sec θdθ

=

∫sec θdθ −

∫1

sec θdθ

=

∫sec θdθ −

∫cos θdθ

= ln |sec θ + tan θ| − sin θ + c

= ln∣∣∣sec

[sec−1

(x2

)]+ tan

[sec−1

(x2

)]∣∣∣− sin

[sec−1

(x2

)]+ c

= ln∣∣∣(x

2

)+ tan

[sec−1

(x2

)]∣∣∣− sin

[sec−1

(x2

)]+ c.

33. Let x = 3 tan θ, dx = 3 sec2 θdθ∫x2

√9 + x2

dx

=

∫27 tan2 θ sec2 θ√

9 + 9 tan2 θdθ

=

∫9 tan2 θ sec θdθ

= 9

∫(sec2 θ − 1) sec θdθ

= 9

∫sec3 θdθ − 9

∫sec θdθ

=9

2sec θ tan θ − 9


=9

2

(√9 + x2

3

)(x3

)− 9

2ln

∣∣∣∣∣√

9 + x2

3+x

3

∣∣∣∣∣+ c

=x√

9 + x2

2− 9

2ln

∣∣∣∣∣x+√

9 + x2

3

∣∣∣∣∣+ c

34. Let x = 2√

2 tan θ, dx = 2√

2 sec2 θdθ∫x3√

8 + x2dx

=

∫(16√

2 tan3 θ)(2√

2 sec θ)dθ

= 64

∫tan3 θ sec θ dθ

= 64

∫(sec2 θ − 1)(sec θ tan θ dθ)

= 64

∫(u2 − 1)du =

64

3u3 − 64u+ c

=64

3sec3 θ − 64 sec θ + c

=64

3

(√8 + x2

2√

2

)3

− 64

(√8 + x2

2√

2

)+ c

=2√

2

3(8 + x2)3/2 − 16

√2(8 + x2)1/2 + c

35. Let x = 4 tan θ, dx = 4 sec2 θdθ∫ √16 + x2dx

=

∫ √16 + 16 tan2 θ · 4 sec2 θdθ

= 16

∫sec3 θdθ

= 16

(1

2sec θ tan θ +

1

2

∫sec θdθ

)= 8 sec θ tan θ + 8

∫sec θdθ

= 8 sec θ tan θ + 8 ln |sec θ + tan θ|+ c

=1

2x√

16 + x2

+ 8 ln

∣∣∣∣14√16 + x2 +x

4

∣∣∣∣+ c

36. Let x = 2 tan θ, dx = 2 sec2 θdθ∫1√

4 + x2dx =

∫2 sec2 θ

2 sec θdθ

=

∫sec θdθ = ln | sec θ + tan θ|+ c

= ln

∣∣∣∣∣x+√

4 + x2

2

∣∣∣∣∣+ c

37. Let u = x2 + 8, du = 2xdx∫ 1

0

x√x2 + 8dx =

1

2

∫ 9

8

u1/2du

=1

3u3/2

∣∣∣∣98

=27− 16

√2

3

38. Let x = 3 tan θ, dx = 3 sec2 θ dθ

I =

∫x2√x2 + 9dx

=

∫27 tan2 θ sec2 θ

√9 tan2 θ + 9dx

= 81

∫tan2 θ sec3 θdx

= 81

∫(sec2 θ − 1) sec3 θdx


= 81

∫(sec5 θ − sec3 θ)dx

To compute∫

sec5 θ dθ, we use integration byparts with u = sec3 θ and dv = sec2 θdθ.∫

sec5 θ dθ

= sec3 θ tan θ −∫

3 sec3 θ tan2 θdθ

= sec3 θ tan θ − 3

∫sec3 θ(sec2 θ − 1)dθ

= sec3 θ tan θ − 3

∫(sec5 θ − sec3 θ)dθ

4

∫sec5 θdθ

= sec3 θ tan θ + 3

∫sec3 θdθ

∫sec5 θdθ

=1

4sec3 θ tan θ +

3

4

∫sec3 θdθ

To compute∫

sec3 θdθ and∫

sec θ dθ, see Ex-ercise 27.Putting all this together gives:

I = 81

∫(sec5 θ − sec3 θ)dx

=81

4sec3 θ tan θ +

243

4

∫sec3 θdθ

− 81

∫sec3 θdθ

=81

4sec3 θ tan θ − 81

4

∫sec3 θdθ

=81


8sec θ tan θ

− 81


We don’t worry about the result being in termsof x since this is a definite integral. Our lim-its of integration are x = 0 and x = 2. Interms of θ this means the limits of integration

correspond to θ = 0 and tan θ =2

3.

∫ 2

0

x2√x2 + 9dx

=

(81


8sec θ tan θ

−81

8ln | sec θ + tan θ|

)∣∣∣∣x=2

x=0

=

81

4

(√13

3

)3(2

3

)− 81

8

(√13

3

)(2

3

)

−81

8ln

∣∣∣∣∣√

13

3+

2

3

∣∣∣∣∣)

−(

81

4(1)(0)− 81

8(1)(0)− 81

8ln |1 + 0|

)

=17√

13

4− 81

8ln

∣∣∣∣∣2 +√

13

3

∣∣∣∣∣39. Let x = tan θ, dx = sec2θdθ.∫

x3

√1 + x2

dx =

∫ (tan3θ

sec θ

)sec2θdθ

=

∫ (tan2θ

)(tan θ sec θ) dθ

Let t = sec θ, dt = tan θ sec θdθ.

=

∫ (sec2θ − 1

)tan θ sec θdθ

=

∫ (t2 − 1

)dt =

[t3

3− t]

+ c

=

[sec3θ

3− sec θ

]+ c

=

[sec3

(tan−1x

)3

− sec(tan−1x

)]+ c.

40. Let x = 2 tan θ, dθ =(2sec2θ

)dθ.∫

x+ 1√4 + x2

dx

=

∫ (2 tan θ + 1√4 + 4tan2θ

)2sec2θdθ

=

∫ (2 tan θ + 1

2 sec θ

)(2sec2θ

)dθ

=

∫(2 tan θ + 1) (sec θ) dθ

= 2

∫sec θ tan θdθ +

∫sec θdθ

= 2 sec θ + ln |sec θ + tan θ|+ c

= 2 sec[tan−1

(x2

)]+ ln

∣∣∣sec[tan−1

(x2

)]+ tan

[tan−1

(x2

)]∣∣∣+ c

= 2 sec[tan−1

(x2

)]+ ln

∣∣∣sec[tan−1

(x2

)]+(x

2

)∣∣∣+ c.

41.

∫x√

x2 + 4xdx

=1

2

∫2x+ 4− 4√x2 + 4x

dx

=1

2

∫2x+ 4√x2 + 4x

dx− 1

2

∫4√

x2 + 4xdx

Let u = x2 + 4x, du = (2x+ 4) dx.

=1

2

∫du√u− 1

2

∫4√

x2 + 4x− 4 + 4dx

= u1/2 − 1

2

∫4√

(x+ 2)2 − 4

dx

=√

(x2 + 4x)

− 2 log

[(x2 + 4x

)+

√(x+ 2)

2 − 4

]+ c.

42.

∫2√

x2 − 6xdx =

∫2√

x2 − 6x+ 9− 9dx


=

∫2√

(x− 3)2 − 9

dx

Let u = x− 3, du = dx.

=

∫2√

u2 − 9du

Let u = 3 sec θ, du = 3 sec θ tan θdθ.

=

∫2√

(3 sec θ)2 − 9

3 sec θ tan θdθ

= 2

∫1√

sec2θ − 1sec θ tan θdθ

= 2

∫1

tan θsec θ tan θdθ

= 2

∫sec θdθ = 2 ln |sec θ + tan θ|+ c

= 2 ln∣∣∣sec

(sec−1

(u3

))+ tan

(sec−1

(u3

))∣∣∣+ c

= 2 ln∣∣∣(u

3

)+ tan

(sec−1

(u3

))∣∣∣+ c

= 2 ln

∣∣∣∣(x− 3

3

)+ tan

(sec−1

(x− 3

3

))∣∣∣∣+ c.

43.

∫x√

10 + 2x+ x2dx

=

∫x√

9 + 1 + 2x+ x2dx

=

∫x√

(x+ 1)2

+ 9dx

=

∫x+ 1− 1√(x+ 1)

2+ 9

dx

=

∫x+ 1√

(x+ 1)2

+ 9dx−

∫1√

(x+ 1)2

+ 9dx

Let u = x+ 1, du = dx.

=

∫u√

u2 + 9du−

∫1√

u2 + 9du

=1

2

∫2u√u2 + 9

du−∫

1√u2 + 32

du

Let t = u2 + 9, dt = 2udu.

=1

2

∫dt√tdt− log

[u+

√u2 + 32

]+ c

=√t− log

[u+

√u2 + 32

]+ c

=√u2 + 9− log

[u+

√u2 + 9

]+ c

=

√(x+ 1)

2+ 9

− log

[(x+ 1) +

√(x+ 1)

2+ 9

]+ c.

44.

∫2√

4x− x2dx =

∫2√

4− 4 + 4x− x2dx

=

∫2√

4− (x− 2)2dx

Let u = x− 2, du = dx.

=

∫2√

4− u2du

Let u = 2 sin θ, du = 2 cos dθ.

=

∫2√

4− (2 sin θ)2

2 cos θdθ

= 2

∫1√

1− sin2θcos θdθ

= 2

∫1

cos θcos θdθ = 2

∫dθ = 2θ + c

= 2sin−1(u

2

)+ c = 2sin−1

(x− 2

2

)+ c.

45. Using u = tanx, gives∫tanx sec4 xdx

=

∫tanx(1 + tan2 x) sec2 xdx

=

∫u(1 + u2)du =

∫(u+ u3)du

=1

2u2 +

1

4u4 + c

=1

2tan2 x+

1

4tan4 x+ c

Using u = secx, gives∫tanx sec4 xdx

=

∫tanx secx sec3 xdx

=

∫u3du =

1

4u4 + c =

1

4sec4 x+ c

46. Using u = tanx gives∫tan3 x sec4 xdx =

∫u3(u2 + 1)du

=u6

6+u4

4+ c1

=tan6 x

6+

tan4 x

4+ c2

Using u = secx gives∫tan3 x sec4 xdx =

∫(u2 − 1)u3 du

=u6

6− u4

4=

sec6 x

6− sec4 x

4

=(tan2 x+ 1)3

6− (tan2 x+ 1)2

4

=tan6 x

6+

tan4 x

4− 1

12+ c1

=tan6 x

6+

tan4 x

4+ c2

47. (a) This is using integration by parts followedby substitution


u = secn−2 x, dv = sec2 xdxdu = (n − 2) secn−2 x tanxdx, v = tanx

I =

∫secn xdx = secn−2 x tanx

− (n− 2)

∫secn−2(sec2 x− 1)dx

= secn−2 x tanx

− (n− 2)

∫(secn x− secn−2 x)dx

= secn−2 x tanx− (n− 2)I

+ (n− 2)

∫secn−2 xdx (n− 1)I

= secn−2 x tanx + (n − 2)

∫secn−2 xdx

I =secn−2 x tanx

n− 1+n− 2

n− 1

∫secn−2 xdx

(b)

∫sec3 xdx

=1

2secx tanx+

1

2

∫secxdx

=1

2secx tanx+

1

2ln | secx+ tanx|+ c

(c)

∫sec4 xdx

=1

3sec3 x tanx+

2

3

∫sec2 xdx

=1

3sec3 x tanx+

2

3tanx+ c

(d)

∫sec5 xdx

=1

4sec3 x tanx+

3

4

∫sec3 xdx

=1

4sec3 x tanx+

3

8secx tanx

+3

8ln | secx+ tanx|+ c

48. Make the substitution x = a sin θ.

4b

a

∫ a

0

√a2 − x2dx =

4b

a

∫ a

0

√a2 − x2dx

=4b

a

∫ π/2

0

a cos θ√a2 − a2 sin2 θdθ

= 4b

∫ π/2

0

a cos2 θdθ

= 4ab

(1

2x+

1

4sin 2x

)∣∣∣∣π/20

= abπ

49.

∫cscxdx =

∫cscx

cscx+ cotx

cscx+ cotxdx

=

∫(cscx) cotx+ csc2x

cscx+ cotxdx

Letu = cscx+ cotx,du = − (cscx) cotx− csc2x.

= −∫

1

udu = − ln |u|+ c

= − ln |cscx+ cotx|+ c.= ln |cscx− cotx|+ c.∫

csc3xdx =

∫cscx.csc2xdx

u = cscx, dv = csc2xdxdu = − cscx. cotx, v = − cotx∫

csc3xdx

= − cscx. cotx

−∫

(− cotx) (− cscx. cotx)dx

= − cscx cotx−∫ (

cscx.cot2x)dx

= − cscx cotx−∫

cscx.(csc2x− 1

)dx

= − cscx cotx−∫ (

csc3x)dx+

∫cscxdx

2

∫csc3xdx = − cscx cotx+

∫cscxdx

= − cscx cotx+ ln |cscx− cotx|+ c∫csc3xdx

=1

2(− cscx cotx+ ln |cscx− cotx|) + c

50.

∫1

cosx− 1dx

=

∫cosx+ 1

(cosx− 1) (cosx+ 1)dx

= −∫

cosx+ 1

sin2xdx

= −∫ (

1

sinx

)(cosx

sinx+

1

sinx

)dx

= −∫

cscx (cotx+ cscx) dx

=

∫− cscx cotx− csc2xdx

=

∫(− cscx cotx) dx+

∫ (−csc2x

)dx

= cscx+ cotx+ c and,∫1

cosx+ 1dx

=

∫cosx− 1

(cosx− 1) (cosx+ 1)dx

= −∫

cosx− 1

sin2xdx

= −∫ (

1

sinx

)(cosx

sinx− 1

sinx

)dx

= −∫

cscx (cotx− cscx) dx

=

∫− cscx cotx+ csc2xdx

=

∫(− cscx cotx) dx−

∫ (−csc2x

)dx

= cscx− cotx+ c

51. Using a CAS we get


(Ex 3.2)

∫cos4 x sin3 xdx

= −1

7sinx2 cosx5 − 2

35cosx5 + c

(Ex 3.3)

∫ √sinx cos5 xdx

=2

11sinx11/2 − 4

7sinx7/2

+2

3sinx3/2 + c

(Ex 3.5)

∫cos4 xdx

=1

4cosx3 sinx+

3

8cosx sinx+

3

8x+ c

(Ex 3.6)

∫tan3 x sec3 xdx

= 1/5sinx4

cosx5+ 1/15

sinx4

cosx3

− 1/15sinx4

cosx− 1/15 sinx2 cosx

− 2/15 cosx+ c

Obviously my CAS used different tech-niques. The answers given by the bookare simpler.

52. (a) −1

7sin2 x cos5 x− 2

35cos5 x

= −1

7(1− cos2 x) cos5 x− 2

35cos5 x

=1

7cos7 x− 1

5cos5 x

The conclusion is c = 0

(b) − 2

15tanx− 1

15sec2 x tanx

+1

5sec4 x tanx

= − 2

15tanx− 1

15(1 + tan2 x) tanx

+1

5(1 + tan2 x)2 tanx

=1

3tan3 x+

1

5tan5 x

The conclusion is c = 0

53. The average power

=12πω

∫ 2π/ω

0

RI2 cos2(ωt) dt

=ωRI2

2π

∫ 2π/ω

0

1

2[1 + cos(2ωt)] dt

=ωRI2

4π

[t+

1

2ωsin(2ωt)

]∣∣∣∣2π/ω0

=ωRI2

4π

[2π

ω+

1

2ωsin

(4ωπ

ω

)− 0

]=

1

2RI2

6.4 Integration ofRational FunctionsUsing PartialFractions

1.x− 5

x2 − 1=

x− 5

(x+ 1)(x− 1)

=A

x+ 1+

B

x− 1

x− 5 = A(x− 1) +B(x+ 1)x = −1 : −6 = −2A;A = 3x = 1 : −4 = 2B;B = −2

x− 5

x2 − 1=

3

x+ 1− 2

x− 1∫x− 5

x2 − 1dx =

∫ (3

x+ 1− 2

x− 1

)dx

= 3 ln |x+ 1| − 2 ln |x− 1|+ c

2.5x− 2

x2 − 4=

5x− 2

(x+ 2)(x− 2)

=A

x+ 2+

B

x− 2

5x− 2 = A(x− 2) +B(x+ 2)x = −2 : −12 = −4A;A = 3x = 2 : 8 = 4B;B = 2

5x− 2

x2 − 4=

3

x+ 2+

2

x− 2∫5x− 2

x2 − 4dx =

∫ (3

x+ 2+

2

x− 2

)dx

= 3 ln |x+ 2|+ 2 ln |x− 2|+ c

3.6x

x2 − x− 2=

6x

(x− 2)(x+ 1)

=A

x− 2+

B

x+ 1

6x = A(x+ 1) +B(x− 2)x = 2 : 12 = 3A;A = 4x = −1 : −6 = −3B;B = 2

6x

x2 − x− 2=

4

x− 2+

2

x+ 1∫6x

x2 − x− 2dx

=

∫ (4

x− 2+

2

x+ 1

)dx

= 4 ln |x− 2|+ 2 ln |x+ 1|+ c

4.3x

x2 − 3x− 4=

3x

(x+ 1)(x− 4)

=A

x+ 1+

B

x− 4

6.4. INTEGRATION OF RATIONAL FUNCTIONS USING PARTIAL FRACTIONS 381

3x = A(x− 4) +B(x+ 1)

x = −1 : −3 = −5A;A =3

5

x = 3 : 12 = 5B;B =12

5

3x

x2 − 3x− 4=

3/5

x+ 1+

12/5

x− 4∫3x

x2 − 3x− 4dx

=

∫ (3/5

x+ 1+

12/5

x− 4

)dx

=3

5ln |x+ 1|+ 12

5ln |x− 4|+ c

5.−x+ 5

x3 − x2 − 2x=

−x+ 5

x(x− 2)(x+ 1)

=A

x+

B

x− 2+

C

x+ 1

− x+ 5 = A(x− 2)(x+ 1) +Bx(x+ 1)+ cx(x− 2)

x = 0 : 5 = −2A : A = −5

2

x = 2 : 3 = 6B : B =1

2x = −1 : 6 = 3C : C = 2

−x+ 5

x3 − x2 − 2x= −5/2

x+

1/2

x− 2+

2

x+ 1∫−x+ 5

x3 − x2 − 2xdx

=

∫ (−5/2

x+

1/2

x− 2+

2

x+ 1

)dx

= −5

2ln |x|+ 1

2ln |x− 2|

+ 2 ln |x+ 1|+ c

6.3x+ 8

x3 + 5x2 + 6x=

3x+ 8

x(x+ 2)(x+ 3))

=A

x+

B

x+ 2+

C

x+ 3

3x+ 8 = A(x+ 2)(x+ 3) +Bx(x+ 3)+ cx(x+ 2)

x = 0 : 8 = 6A;A =4

3x = −2 : 2 = −2B;B = −1

x = −3 : −1 = 3C;C = −1

3

3x+ 8

x3 + 5x2 + 6x=

4/3

x− 1

x+ 2− 1/3

x+ 3∫3x+ 8

x3 + 5x2 + 6xdx

=

∫ (4/3

x− 1

x+ 2− 1/3

x+ 3

)dx

=4

3ln |x| − ln |x+ 2| − 1

3ln |x+ 3|+ c

7.5x− 23

6x2 − 11x− 7=

5x− 23

(2x+ 1)(3x− 7)

=A

2x+ 1+

B

3x− 7

5x− 23 = A(3x− 7) +B(2x+ 1)

x = −1

2: −51

2= −17

2A;A = 3

x =7

3: −34

3=

17

3B;B = −2

5x− 23

6x2 − 11x− 7=

3

2x+ 1− 2

3x− 7∫5x− 23

6x2 − 11x− 7dx

=

∫ (3

2x+ 1− 2

3x− 7

)dx

=3

2ln | 2x+ 1| − 2

3ln | 3x− 7|+ c

8.3x+ 5

5x2 − 4x− 1=

3x+ 5

(5x+ 1)(x− 1)

=A

5x+ 1+

B

x− 1

3x+ 5 = A(x− 1) +B(5x+ 1)

x = −1

5:

22

5= −6

5A;A = −11

3

x = 1 : 8 = 6B;B =4

3

3x+ 5

5x2 − 4x− 1= − 11/3

5x+ 1+

4/3

x− 1∫3x+ 5

5x2 − 4x− 1dx

=

∫ (− 11/3

5x+ 1+

4/3

x− 1

)dx

= −11

15ln |5x+ 1|+ 4

3ln |x− 1|+ c

9.x− 1

x3 + 4x2 + 4x=

x− 1

x(x+ 2)2

=A

x+

B

x+ 2+

C

(x+ 2)2

x− 1 = A(x+ 2)2 +Bx(x+ 2) + Cx

x = 0 : −1 = 4A;A = −1

4

x = −2 : −3 = −2C;C =3

2

x = 1 : 0 = 9A+ 3B + C;B =1

4

x− 1

x3 + 4x2 + 4x

= −1/4

x+

1/4

x+ 2+

3/2

(x+ 2)2

382 CHAPTER 6. INTEGRATION TECHNIQUES∫x− 1

x3 + 4x2 + 4xdx

=

∫ (−1/4

x+

1/4

x+ 2+

3/2

(x+ 2)2

)dx

= −1

4ln |x|+ 1

4ln |x+ 2| − 3

2(x+ 2)+ c

10.4x− 5

x3 − 3x2=

4x− 5

x2(x− 3)

=A

x+B

x2+

C

x− 3

4x− 5 = Ax(x− 3) +B(x− 3) + Cx2

= (A+ C)x2 + (−3A+B)x+ (−3B)

B =5

3;A = −7

9;C =

7

9

4x− 5

x3 − 3x2= −7/9

x+

5/3

x2+

7/9

x− 3∫4x− 5

x3 − 3x2dx

=

∫ (−7/9

x+

5/3

x2+

7/9

x− 3

)dx

= −7/9

ln|x| − 5

3

1

x+

7

9ln |x− 3|+ c

11.x+ 2

x3 + x=

x+ 2

x(x2 + 1)

=A

x+Bx+ C

x2 + 1

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

= Ax2 +A+Bx2 + Cx= (A+B)x2 + Cx+A

A = 2;C = 1;B = −2

x+ 2

x3 + x=

2

x+−2x+ 1

x2 + 1∫x+ 2

x3 + xdx =

∫ (2

x+−2x+ 1

x2 + 1

)dx

=

∫ (2

x− 2x

x2 + 1+

1

x2 + 1

)dx

= 2 ln |x| − ln(x2 + 1) + tan−1 x+ c

12.1

x3 + 4x=

1

x(x2 + 4)

=A

x+Bx+ C

x2 + 4

1 = A(x2 + 1) + (Bx+ C)x1 = (A+B)x2 + Cx+A

A = 1;B = −1;C = 0

1

x3 + 4x=

1

x+−x

x2 + 4

∫1

x3 + 4xdx

=

∫ (1

x+−x

x2 + 4

)dx

= ln |x| − 1

2ln(x2 + 4) + c

13.4x2 − 7x− 17

6x2 − 11x− 10

=2

3+

1

3

x− 31

(2x− 5)(3x+ 2)

=2

3+

1

3

[A

2x− 5+

B

3x+ 2

]x− 31 = A(3x+ 2) +B(2x− 5)

x =5

2: −57

2=

19

2A,A = −3;

x = −2

3: −95

3= −19

3B,B = 5;

4x2 − 7x− 17

6x2 − 11x− 10

=2

3+

1

3

[−3

2x− 5+

5

3x+ 2

]∫

4x2 − 7x− 17

6x2 − 11x− 10dx

=

∫ (2

3− 1

2x− 5+

5/3

3x+ 2

)dx

=2

3x− 1

2ln | 2x− 5|+ 5

9ln | 3x+ 2|+ c

14.x3 + x

x2 − 1= x+

2x

(x+ 1)(x− 1)

= x+A

x+ 1+

B

x− 1

2x = A(x− 1) +B(x+ 1)A = B = 1

x3 + x

x2 − 1= x+

1

x+ 1+

1

x− 1∫x3 + x

x2 − 1dx

=

∫ (x+

1

x+ 1+

1

x− 1

)dx

=x2

2+ ln |x+ 1|+ ln |x− 1|+ c

15.2x+ 3

x2 + 2x+ 1=

2x+ 3

(x+ 1)2

=A

x+ 1+

B

(x+ 1)2

2x+ 3 = A(x+ 1) +B

x = −1 : B = 1;A = 2


2x+ 3

x2 + 2x+ 1=

2

x+ 1+

1

(x+ 1)2∫2x+ 3

x2 + 2x+ 1dx

=

∫ (2

x+ 1+

1

(x+ 1)2

)dx

= 2 ln |x+ 1| − 1

x+ 1+ c

16.2x

x2 − 6x+ 9=

2x

(x− 3)2

=A

x− 3+

B

(x− 3)2

2x = A(x− 3) +BA = 2;B = 6

2x

x2 − 6x+ 9=

2

x− 3+

6

(x− 3)2∫2x

x2 − 6x+ 9dx

=

∫ (2

x− 3+

6

(x− 3)2

)dx

= 2 ln |x− 3| − 6

x− 3+ c

17.x3 − 4

x3 + 2x2 + 2x= 1 +

−2x2 − 2x− 4

x(x2 + 2x+ 2)

= 1 +A

x+

Bx+ c

x2 + 2x+ 2

− 2x2 − 2x− 4 = A(x2 + 2x+ 2) + (Bx+ c)x= (A+B)x2 + (2A+ c)x+ 2AA = −2;B = 0;C = 2

x3 − 4

x3 + 2x2 + 2x

= 1 +−2

x+

2

x2 + 2x+ 2∫x3 − 4

x3 + 2x2 + 2xdx

=

∫ (1 +−2

x+

2

(x+ 1)2 + 1

)dx

= x− 2 ln |x|+ 2 tan−1(x+ 1) + c

18.4

x3 − 2x2 + 4x=

4

x(x2 − 2x+ 4)

=A

x+

Bx+ C

x2 − 2x+ 4

4 = A(x2 − 2x+ 4) + (Bx+ C)x= (A+B)x2 + (−2A+ C)x+ 4AA = 1;B = −1;C = 2

4

x3 − 2x2 + 4x=

1

x+

−x+ 2

x2 − 2x+ 4∫4

x3 − 2x2 + 4xdx

=

∫ (1

x+

−x+ 2

x2 − 2x+ 4

)dx

=

∫ (1

x− 1

2

2x− 2

x2 − 2x+ 4+

1

(x− 1)2 + 3

)dx

= ln |x| − 1

2ln(x2 − 2x+ 4)

+1√3

tan−1

(x− 1√

3

)+ c

19.3x3 + 1

x3 − x2 + x− 1

= 3 +3x2 − 3x+ 4

x3 − x2 + x− 1

= 3 +3x2 − 3x+ 4

(x2 + 1)(x− 1)

= 3 +Ax+B

x2 + 1+

C

x− 1

3x2 − 3x+ 4 = (Ax+B)(x− 1) + C(x2 + 1)= Ax2 −Ax+Bx−B + Cx2 + Cx = 1 : 4 = 2C;C = 2A+ c = 3 : A = 1−A+B = −3 : B = −2

3x3 + 1

x3 − x2 + x− 1= 3 +

x− 2

x2 + 1+

2

x− 1∫3x3 + 1

x3 − x2 + x− 1dx

=

∫ (3 +

x− 2

x2 + 1+

2

x− 1

)dx

=

∫ (3 +

x

x2 + 1− 2

x2 + 1+

2

x− 1

)dx

= 3x+1

2ln(x2 + 1)− 2 tan−1 x

+ 2 ln |x− 1|+ c

20.2x4 + 9x2 + x− 4

x3 + 4x= 2x+

x2 + x− 4

x(x2 + 4)

= 2x+A

x+Bx+ C

x2 + 4

x2 + x− 4 = A(x2 + 4) + (Bx+ C)x= (A+B)x2 + Cx+ 4AA = −1;B = 2;C = 1

2x4 + 9x2 + x− 4

x3 + 4x= 2x− 1

x+

2x+ 1

x2 + 4

= 2x− 1

x+

2x

x2 + 4+

1

x2 + 4∫2x4 + 9x2 + x− 4

x3 + 4xdx

=

∫ (2x− 1

x+

2x

x2 + 4+

1

x2 + 4

)dx

= x2 − ln |x|+ ln(x2 + 4) +1

2tan−1 x

2+ c


21.x3 + x+ 2

x2 + 2x− 8= x− 2 +

11

x+ 4+

2

x− 2∫x3 + x+ 2

x2 + 2x− 8dx

=

∫ (x− 2 +

11

x+ 4+

2

x− 2

)dx

=x2

2− 2x+ 11 ln |x+ 4|

+ 2 ln |x− 2|+ c

22.x2 + 1

x2 − 5x− 6= − 2/7

x+ 1+

37/7

x− 6∫x2 + 1

x2 − 5x− 6dx

=

∫ (− 2/7

x+ 1+

37/7

x− 6

)dx

= −2

7ln |x+ 1|+ 37

7ln |x− 6|+ c

23.x+ 4

x3 + 3x2 + 2x=

2

x+

1

x+ 2− 3

x+ 1∫x+ 4

x3 + 3x2 + 2xdx

=

∫ (2

x+

1

x+ 2− 3

x+ 1

)dx

= 2 ln |x|+ ln |x+ 2| − 3 ln |x+ 1|+ c

24.1

x3 − 1=

1/3

(x− 1)− (x+ 2)/3

(x2 + x+ 1)∫1

x3 − 1dx

=1

3

∫1

(x− 1)− x+ 2

(x2 + x+ 1)dx

=1

3

∫1

(x− 1)− 1

2

2x+ 4

(x2 + x+ 1)dx

=1

3

∫1

(x− 1)− 1

2

2x+ 1

(x2 + x+ 1)

− 1

2

3

(x2 + x+ 1)dx

=1

3

∫1

(x− 1)− 1

2

2x+ 1

(x2 + x+ 1)

− 1

2

3

(x+ 1/2)2

+ 3/4dx

=1

3

[ln |x− 1| − 1

2ln∣∣x2 + x+ 1

∣∣−√

3tan−1

(2x+ 1√

3

)]+ c

25. Let u = x4 − x, du =(4x3 − 1

)dx.∫ (

4x3 − 1)

x4 − xdx =

∫du

u= ln |u|+ c = ln

∣∣x4 − x∣∣+ c.

26. Let u = x2, du = (2x) dx.∫x

x4 + 1dx =

1

2

∫2x

x4 + 1dx

=1

2

∫du

u2 + 1=

1

2tan (u) + c

=1

2tan

(x2)

+ c.

27.4x− 2

16x4 − 1=−4x+ 1

4x2 + 1+

1

2x+ 1∫4x− 2

16x4 − 1dx

=

∫ (−4x+ 1

4x2 + 1+

1

2x+ 1

)dx

=

∫ (−1

2

8x

4x2 + 1+

1

4x2 + 1+

1

2x+ 1

)dx

= −1

2ln |4x2 + 1|+ 1

2tan−1(2x)

+1

2ln |2x+ 1|+ c

28.3x+ 7

x4 − 16=

13/32

x− 2− 1/32

x+ 2− 3x/8 + 7/8

x2 + 4∫3x+ 7

x4 − 16dx

=

∫ (13/32

x− 2− 1/32

x+ 2− 3x/8 + 7/8

x2 + 4

)dx

=

∫ (13/32

x− 2− 1/32

x+ 2− 3

16

2x

x2 + 4

−7

8

1

x2 + 4

)dx

=13

32ln |x− 2| − 1

32ln |x+ 2|

− 3

16ln(x2 + 4)− 7

16tan−1 x

2+ c

29.x3 + x

3x2 + 2x+ 1

=x

3− 2

9+

1

9

10x+ 2

3x2 + 2x+ 1∫x3 + x

3x2 + 2x+ 1dx

=

∫ (x

3− 2

9+

1

9

10x+ 2

3x2 + 2x+ 1

)dx

=

∫ (x

3− 2

9+

1

9

5

3

6x+ 2

3x2 + 2x+ 1

−1

9

4

3

1

3(x+ 1/3)2 + 2/3

)dx

=x2

6− 2

9x+

5

27ln(3x2 + 2x+ 1)

− 2√

2

27tan−1

(3x+ 1√

2

)+ c


30.x3 − 2x

2x2 − 3x+ 2

=x

2+

4

3+

1

4

21x− 6

2x2 − 3x+ 2∫x3 − 2x

2x2 − 3x+ 2dx

=

∫ (x

2+

4

3+

1

4

21x− 6

2x2 − 3x+ 2

)dx

=

∫ (x

2+

4

3+

21

16

4x− 3

2x2 − 3x+ 2

+39

32

1

(x− 3/4)2 + 7/16

)dx

=x2

4+

4

3x+

21

16ln(2x2 − 3x+ 2)

− 39√

7

56tan−1

(4x− 3√

7

)+ c

31.4x2 + 3

x3 + x2 + x=

3

x+

x− 3

x2 + x+ 1∫4x2 + 3

x3 + x2 + xdx

=

∫ (3

x+

x− 3

x2 + x+ 1

)dx

=

∫ (3

x+

x+ 1/2

x2 + x+ 1− 7/2

x2 + x+ 1

)dx

= 3 ln |x|+ 1

2ln |x2 + x+ 1|

− 7√3

tan−1

(2x+ 1√

3

)+ c

32.4x+ 4

x4 + x3 + 2x2=

1

x+

2

x2+−x− 3

x2 + x+ 2∫4x+ 4

x4 + x3 + 2x2dx

=

∫ (1

x+

2

x2+−x− 3

x2 + x+ 2

)dx

= ln |x| − 2

x− 1

2ln(x2 + x+ 2)

− 5√7

tan−1

(2x+ 1√

7

)+ c

33. Let u = x2, dv = (sinx) dx

So that du = (2x) dx and v = − cosx.∫x2 sinxdx

= x2 (− cosx)−∫

(− cosx) (2x) dx

= −x2 cosx+ 2

∫x (cosx) dx

Let u = x, dv = cosxdx,so that du = dx and v = sinx.∫x2 sinxdx

= −x2 cosx+ 2

∫x cosxdx

= −x2 cosx+ 2 x sinx+ cosx+ c.

34. Let u = x, dv = e2xdx .

so that du = dx and v =e2x

2.∫

xe2xdx = xe2x

2−∫e2x

2dx

= xe2x

2− e2x

4+ c

35. Let u =(sin2x− 4

),

so that du = 2 sinx cosx dx.∫sinx cosx

sin2x− 4dx =

1

2

∫du

u

=1

2ln |u|+ c =

1

2ln∣∣sin2x− 4

∣∣+ c

36. Let t = ex, dt = exdx and e3x = t3∫2ex

e3x + exdx =

∫2

t3 + tdt.

=

∫2

t− 2t

t2 + 1dt = 2 ln |t| − ln

∣∣t2 + 1∣∣+ c

= 2 ln |ex| − ln∣∣e2x + 1

∣∣+ c

37.4x2 + 2

(x2 + 1)2=Ax+B

x2 + 1+

Cx+D

(x2 + 1)2

4x2 + 2 = (Ax+B)(x2 + 1) + (Cx+D)= Ax3 +Bx2 + (A+ C)x+ (B +D)A = 0;B = 4;C = 0;D = −2

4x2 + 2

(x2 + 1)2=

4

x2 + 1+

−2

(x2 + 1)2

38.x3 + 2

(x2 + 1)2=Ax+B

x2 + 1+

Cx+D

(x2 + 1)2

x3 + 2 = (Ax+B)(x2 + 1) + cx+D= Ax3 +Bx2 + (A+ c)x+ (B +D)A = 1;B = 0;C = −1;D = 2

x3 + 2

(x2 + 1)2=

x

x2 + 1+−x+ 2

(x2 + 1)2

39.4x2 + 3

(x2 + x+ 1)2

=Ax+B

x2 + x+ 1+

Cx+D

(x2 + x+ 1)2

4x2 + 3 = (Ax+B)(x2 + x+ 1) + cx+D= Ax3 +Ax2 +Ax+Bx2 +Bx+B + cx+DA = 0A+B = 4 : B = 4A+B + c = 0 : C = −4B +D = 3 : D = −1


4x2 + 3

(x2 + x+ 1)2

=4

x2 + x+ 1− 4x+ 1

(x2 + x+ 1)2

40.x4 + x3

(x2 + 4)2= 1 +

x3 − 8x2 − 8

(x2 + 4)2

= 1 +Ax+B

x2 + 4+

Cx+D

(x2 + 4)2

x3 − 8x2 − 8 = (Ax+B)(x2 + 4) + cx+D= Ax3 +Bx2 + (4A+ c)x+ (4B +D)A = 1;B = −8;C = −4;D = 24

x4 + x3

(x2 + 4)2= 1 +

x− 8

x2 + 4+−4x+ 24

(x2 + 4)2

41. Let u = x3 + 1, du = 3x2dx∫3

x4 + xdx =

∫3x2

x3(x3 + 1)dx

=

∫1

(u− 1)udu

=

∫ (1

u− 1− 1

u

)du

= ln |u− 1| − ln |u|+ c

= ln

∣∣∣∣u− 1

u

∣∣∣∣+ c

= ln

∣∣∣∣ x3

x3 + 1

∣∣∣∣+ c

On the other hand, we can let

u =1

x, du = − 1

x2dx∫

3

x4 + xdx = −

∫3u2

1 + u3du

= − ln |1 + u3|+ c = − ln |1 + 1/x3|+ c

To see that the two answers are equivalent,note that

ln

∣∣∣∣ x3

x3 + 1

∣∣∣∣ = − ln

∣∣∣∣x3 + 1

x3

∣∣∣∣ = − ln |1 + 1/x3|

42. Let u = x2 + 1, du = 2xdx∫2

x3 + xdx =

∫2x

x2(x2 + 1)dx

=

∫du

u(u− 1)= ln

∣∣∣∣u− 1

u

∣∣∣∣+ c

= ln

∣∣∣∣ x2

x2 + 1

∣∣∣∣+ c

Let u =1

x, du = − 1

x2dx∫

2

x3 + xdx = −

∫2u

1 + u2du

= − ln |1 + u2|+ c = − ln

∣∣∣∣1 +1

x2

∣∣∣∣+ c

To see that the two answers are equivalent,note that

ln

∣∣∣∣ x2

x2 + 1

∣∣∣∣ = − ln

∣∣∣∣x2 + 1

x2

∣∣∣∣ = − ln

∣∣∣∣1 +1

x2

∣∣∣∣43. (a) Partial fractions

(b) Substitution method

(c) Substitution and Partial fractions.

(d) Substitution

44. (a) Partial fractions

(b) Substitution and Partial fractions.

(c) Partial fractions

(d) Partial fractions

45.

∫sec3xdx =

∫cosx(

1− sin2x)2 dx

Letu = sinx, so that du = cosxdx.∫cosx dx(

1− sin2x)2 =

∫du

(1− u2)2

=

∫1

(1− u)2(1 + u)

2 du

By partial fractions,

1

(1− u)2(1 + u)

2 =1

4

(1

(1− u)+

1

(1− u)2

+1

(1 + u)+

1

(1 + u)2

)Hence,

∫sec3xdx

=1

4

[− ln |1− u|+ 1

(1− u)+ ln |1 + u|

− 1

(1 + u)

]+ c

=1

4

[− ln |1− sinx|+ 1

(1− sinx)

+ ln |1 + sinx| − 1

(1 + sinx)

]+ c

6.5 Integration Tableand ComputerAlgebra Systems

1.

∫x

(2 + 4x)2dx

=2

16(2 + 4x)+

1

16ln | 2 + 4x|+ c

=1

8(2 + 4x)+

1

16ln | 2 + 4x|+ c

6.5. INTEGRATION TABLE AND COMPUTER ALGEBRA SYSTEMS 387

2.

∫x2

(2 + 4x)2dx

=1

64

(2 + 4x− 4

2 + 4x− 4 ln |2 + 4x|

)+ c

3. Substitute u = 1 + ex∫e2x√

1 + exdx =

∫(u− 1)

√u du

=

∫(u3/2 − u1/2) du

=2

5u5/2 − 2

3u3/2 + c

=2

5(1 + ex)5/2 − 2

3(1 + ex)3/2 + c

4. Substitute u = ex∫e3x√

1 + e2x dx =

∫u2√

1 + u2 du

=1

8u(1 + 2u2)

√1 + u2

− 1

8ln |u+

√1 + u2|+ c

=1

8ex(1 + 2e2x)

√1 + e2x

− 1

8ln |ex +

√1 + e2x|+ c

5. Substitute u = 2x∫x2

√1 + 4x2

dx

=1

8

∫u2

√1 + u2

du

=1

8

[u2−√

1 + u2

−1

2ln(u+

√1 + u2)

]+ c

=1

8x√

1 + 4x2

− 1

16ln(2x+

√1 + 4x2) + c

6. Substitute u = sinx∫cosx

sin2 x(3 + 2 sinx)dx

=

∫1

u2(3 + 2u)du

=2

9ln

∣∣∣∣3 + 2u

u

∣∣∣∣− 1

2u+ c

=2

9

∣∣∣∣3 + 2 sinx

sinx

∣∣∣∣− 1

3 sinx+ c

7. Substitute u = t3∫t8√

4− t6dt

=1

3

∫u2(√

4− u2)du

=1

3

[u

8

(2u2 − 4

)√4− u2 +

16

8sin−1u

2

]+ c

=1

24t3(2t6 − 4

)√4− t6 +

2

3sin−1 t

3

2+ c∫ 1

0

t8√

4− t6dt =π

9−√

3

12

8. Substitute u = et∫ √16− e2tdt =

∫ √16− u2

udu

=√

16− u2 − 4 ln

∣∣∣∣∣4 +√

16− u2

u

∣∣∣∣∣+ c

=√

16− e2t − 4 ln

∣∣∣∣∣4 +√

16− e2t

et

∣∣∣∣∣+ c∫ ln 4

0

√16− e2tdt = −

√15 + 4 ln

(√15 + 4

)9. Substitute u = ex∫

ex√e2x + 4

dx =

∫1√

u2 + 4du

= ln(u+√

4 + u2) + c

= ln(ex +√

4 + e2x) + c∫ ln 2

0

ex√e2x + 4

dx = ln

(2√

2 + 2

1 +√

5

)

10. Substitute u = x2∫ 2

√3

x√x4 − 9

x2dx =

1

2

∫ 4

3

√u2 − 9

udu

=1

2

(√u2 − 9− 3 sec−1 |u|

3

)∣∣∣∣43

=

√7

2− 3

2sec−1

(4

3

)11. Substitute u = x− 3∫ √

6x− x2

(x− 3)2dx

=

∫ √(u+ 3)(6− (u+ 3))

u2du

=

∫ √9− u2

u2du

= − 1

u

√9− u2 − sin−1 u

3+ c

= − 1

x− 3

√9− (x− 3)2

− sin−1

(x− 3

3

)+ c

12. Substitute u = tanx∫sec2 x

tanx√

8 tanx− tan2 xdx

=

∫1

u√

8u− u2du

= −√

8u− u2

4u+ c


= −√

8 tanx− tan2 x

4 tanx+ c

13.

∫tan6udu

=1

5tan5u−

∫tan4udu

=1

5tan5u−

[1

3tan3u−

∫tan2udu

]=

1

5tan5u− 1

3tan3u+ tanu− u+ c.

14.

∫csc4udu

= −1

3csc2u cotu+

2

3

∫csc2udu

= −1

3csc2u cotu− 2

3cotu+ c.


sinx√

4 + sinxdx =

∫1

u√

4 + udu

=1√4

ln

∣∣∣∣ √4 + u− 2√4 + u+ 2

∣∣∣∣+ c

=1

2ln

∣∣∣∣ √4 + sinx− 2√4 + sinx+ 2

∣∣∣∣+ c

16. Substitute u = x2∫x5

√4 + x2

dx =1

2

∫u2

√4 + u2

du

=

(1

2

)2

15(3u2 − 16u+ 128)

√4 + u+ c

=1

15(3x4 − 16x2 + 128)

√4 + x2 + c

17. Substitute u = x2∫x3 cosx2dx =

1

2

∫u cosu du

=1

2(cosu+ u sinu) + c

=1

2cosx2 +

1

2x2 sinx2 + c

18. Substitute u = x2∫x sin(3x2) cos(4x2) dx

=1

2

∫sin(3u) cos(4u) du

=1

2

(cosu

2− cos 7u

14

)+ c

=cosx2

4− cos 7x2

28+ c

19. Substitute u = cosx∫sin 2x√1 + cosx

dx =

∫2 sinx cosx√

1 + cosxdx

= −2

∫u√

1 + udu

= −2

[2

3(u− 2)

√1 + u

]+ c

= −4

3(cosx− 2)

√1 + cosx+ c

20. Substitute u = x2∫x√

1 + 4x2

x4dx =

1

2

∫ √1 + 4u

u2du

= −√

1 + 4u

2u+ ln

[√1 + 4u− 1√1 + 4u+ 1

]+ c

= −√

1 + 4x2

2x2 + ln

[√1 + 4x2 − 1√1 + 4x2 + 1

]+ c

21. Substitute u = sin t∫sin2 t cos t√

sin2 t+ 4dt

=

∫u2

√u2 + 4

du

=u

2

√4 + u2 − 4

2ln(u+

√4 + u2) + c

=1

2sin t

√4 + sin2 t

− 2 ln(

sin t+√

4 + sin2 t)

+ c

22. Substitute u =√t∫

ln√t√tdt = 2

∫lnu du

= 2u lnu− 2u+ c = 2√t ln√t− 2

√t+ c

23. Substitute u = − 2

x2∫e−2/x2

x3dx =

1

4

∫eu du

=1

4eu + c =

1

4e−2/x2

+ c

24. Substitute u = 2x2∫x3e2x2

dx =1

8

∫ueu du

=1

8(u− 1)eu + c =

1

8(2x2 − 1)e2x2

+ c

25.

∫x√

4x− x2dx

= −√

4x− x2 + 2 cos−1

(2− x

2

)+ c

26.

∫e5x cos 3x dx

=1

34(5 cos 3x+ 3 sin 3x)e5x + c

27. Substitute u = ex∫ex tan−1(ex)dx =

∫tan−1 u du

6.6. IMPROPER INTEGRALS 389

= u tan−1 u− 1

2ln(1 + u2) + c

= ex tan−1 ex − 1

2ln(1 + e2x) + c

28. Substitute u = 4x∫(ln 4x)3 dx =

1

4

∫(lnu)3 dx

=1

4

(u(lnu)3 − 3

∫(lnu)2 dx

)=

1

4u(lnu)3

− 3

4

(u(lnu)2 − 2u lnu+ 2u

)+ c

= x(ln 4x)3 − 3x(lnu)2 + 6x ln 4x− 6x+ c

29. Answer depends on CAS used.


31. Any answer is wrong because the integrand isundefined for all x 6= 1.





36. Maple gives the result:bπ√

1

a2

37. If the CAS is unable to compute an antideriva-tive,

∫f(x) dx is generally printed showing this

inability.

6.6 Improper Integrals

1. (a) improper, function not defined at x = 0

(b) not improper, function continuous onentire interval

(c) not improper, function continuous onon entire interval

2. (a) improper, interval is infinite

(b) improper, function not defined at x = 0

(c) improper, interval is infinite

3. (a)

∫ 1

0

x−1/3dx = limR→0+

∫ 1

R

x−1/3dx

= limR→0+

3

2x2/3

∣∣∣∣1R

= limR→0+

3

2

(1−R2/3

)=

3

2

(b)

∫ 1

0

x−4/3dx = limR→0+

∫ 1

R

x−4/3dx

= limR→0+

(−3x−1/3)∣∣∣1R

= limR→0+

(−3)(1−R−1/3) =∞So the original integral diverges.

4. (a)

∫ ∞1

x−4/5dx = limR→∞

∫ R

1

x−4/5dx

= limR→∞

5x1/5∣∣∣R1

= limR→∞

5R1/5 − 5 =∞So the original integral diverges.

(b)

∫ ∞1


∫ R

1

dx

= limR→∞

−5x−1/5∣∣∣R1

= limR→∞

−5R−1/5 + 5 = 5

5. (a)

∫ 1

0

1√1− x

dx = limR→1−

∫ R

0

1√1− x

dx

= limR→1−

− 2√

1− x∣∣R0

= limR→1−

−2(√

1−R− 1) = 2

(b)

∫ 5

1

2√5− x

dx = limR→5−

∫ R

1

2√5− x

dx

= limR→5−

− 4√

5− x∣∣R1

= limR→5−

−4(√

5−R− 2) = −8

6. (a)

∫ 1

0

2√1− x2

dx = limR→1−

∫ R

0

2√1− x2

dx

= limR→1−

2 sin−1 x

∣∣∣∣R0

= limR→1−

2(sin−1R− sin−1 0)

= 2(π

2− 0)

= π

(b)

∫ 1/2

0

2

x√

1− x2dx

= limR→0+

∫ 1/2

R

2

x√

1− x2dx

= limR→0+

−2 ln

(1 +√

1− x2

x

)∣∣∣∣∣1/2

R

=∞

Therefore the original integral diverges.

7. (a)

∫ ∞0

xexdx = limR→∞

∫ R

0

xexdx

= limR→∞

(xex − ex)∣∣∣R0


= limR→∞

eR(R− 1) + 1 =∞So the original integral diverges.

(b) Substitute u = −2x

I =

∫ ∞1

x2e−2xdx = −1

8

∫ −∞−2

u2eudu

=1

8

∫ −2

−∞u2eudu

=1

8lim

R→−∞

∫ −2

R

u2eudu

=1

8lim

R→−∞(u2eu − 2ueu + 2eu)

∣∣∣∣−2

R

=10

8e−2 +

1

8lim

R→−∞eR(−R2 + 2R− 2)

But, limR→−∞

eR(−R2 + 2R− 2)

= limR→∞

e−R(−R2 − 2R− 2)

= limR→∞

−R2 − 2R− 2

eR

= limR→∞

−2R− 2

eR= limR→∞

−2

eR= 0

Hence, I =5

4e−2

8. (a) Substitute u = 3x

I = lim−∞→1

∫ 1

−∞x2e3xdx

=1

27

∫ 3

−∞u2eudu

=1

27lim

R→−∞(u2eu − 2ueu + 2eu)

∣∣∣∣3R

=5

27e3 − 1

27lim

R→−∞eR(R2 − 2R+ 2)

But, limR→∞

eR(R2 − 2R+ 2)

= limR→∞

e−R(R2 + 2R+ 2)

= limR→∞

R2 + 2R+ 2

eR= 0

Hence, I =5

27e3

(b) Substitute u = −4x

I =

∫ 0

−∞xe−4xdx

=1

16

∫ 0

−∞ueudu

=1

16lim

R→−∞

∫ 0

R

ueudu

=1

16lim

R→−∞(ueu − eu)

∣∣∣∣0R

= − 1

16+

1

16lim

R→−∞eR(R− 1)

But, limR→−∞

eR(R− 1)

= limR→∞

e−R(−R− 1) = 0

Hence, I = − 1

16

9. (a)

∫ −1

−∞

1

x2dx = lim

R→−∞

∫ −1

R

1

x2dx

= limR→−∞

− 1

x

∣∣∣∣−1

R

= 1 + limR→−∞

1

R= 1∫ 0

−1

1

x2dx = lim

R→0+

∫ R

−1

1

x2dx

= limR→0+

− 1

x

∣∣∣∣R−1

= −1− limR→0+

1

R=∞

So the original integral diverges.

(b)

∫ −1

−∞

13√xdx = lim

R→−∞

∫ −1

R

13√xdx

= limR→−∞

3

2x2/3

∣∣∣∣−1

R

=3

2+

3

2lim

R→−∞R2/3 =∞


10. (a)

∫ ∞0

cosxdx = limR→∞

∫ R

0

cosxdx

= limR→∞

sinx∣∣∣R0

= limR→∞

(sinR− sin 0)


(b)

∫ ∞0

cosxe− sin xdx

= limR→∞

∫ R

0

cosxe− sin xdx

= limR→∞

−e− sin x∣∣∣R0

= limR→∞

−e− sinR + 1


11. (a)

∫ 1

0

lnxdx

= limR→0+

∫ 1

R

lnxdx

= limR→0+

(x lnx− x)

∣∣∣∣1R

= limR→0+

(−1−R lnR+R)

= −1− limR→0+

lnR

1/R

= −1− limR→0+

1/R

−1/R2

= −1 + limR→0+

R = −1


(b)

∫ π/2

0

sec2 xdx

= limR→π/2−

∫ R

0

sec2 xdx

= limR→π/2−

tanx

∣∣∣∣R0

= limR→π/2−

tanR− tan 0 =∞


12. (a)

∫ π/2

0

cotxdx

= limR→0+

∫ π/2

R

cosx

sinxdx

= limR→0+

ln | sinx|∣∣∣∣π/2R

= ln | sin(π/2)| − limR→0+

ln | sinR| =∞So the original integral diverges.

(b)

∫ π/2

0

tanxdx

= limR→π/2

∫ R

0

sinx

cosxdx

= limR→π/2

− ln | cosx|∣∣∣∣R0

= limR→π/2

(− ln | cosR|+ ln 1) =∞


13. (a)

∫ 3

0

2

x2 − 1dx

=

∫ 3

0

(− 1

x+ 1+

1

x− 1

)dx

= limR→1−

∫ R

0

(− 1

x+ 1+

1

x− 1

)dx

+ limR→1+

∫ 3

R

(− 1

x+ 1+

1

x− 1

)dx

Both of these integrals behave like

limR→0+

∫ 1

R

1

xdx

= limR→0+

(ln 1− lnR)

= limR→0+

ln

(1

R

)=∞


(b)

∫ 4

1

2x

x2 − 1dx

= limR→1+

∫ 4

R

2x

x2 − 1dx

= limR→1+

ln(x2 − 1)∣∣4R

= limR→1+

ln 15− ln(R2 − 1) =∞∫ 1

−4

2x

x2 − 1dx

= limR→1−

∫ R

−4

2x

x2 − 1dx

= limR→1−

ln(x2 − 1)∣∣R−4

= limR→1−

ln(R2 − 1)− ln 15 =∞So the original integral diverges.

14. (a)

∫ π

0

xsec2xdx

=

∫ π/2

0

xsec2xdx+

∫ π

π/2

xsec2xdx

= limR→π/2−

(x tanx+ ln |cosx|)|R0+ limR→π/2+

(x tanx+ ln |cosx|)|πR=∞So the original integral diverges.

(b)

∫ 2

0

2

x3 − 1dx

=

∫ 1

0

2

x3 − 1dx+

∫ 2

1

2

x3 − 1dx

= limR→1−

∫ R

0

2

x3 − 1dx

+ limR→1+

∫ 2

R

2

x3 − 1dx

= limR→1−

2

(−

ln(x2 + x+ 1

)6

−tan−1

(2x+1√

3

)√

3+

ln (x− 1)

3

R

0

+ limR→1+

2

(−

ln(x2 + x+ 1

)6

−tan−1

(2x+1√

3

)√

3+

ln (x− 1)

3

R

0=∞

15. (a)

∫ ∞−∞

1

1 + x2dx

=

∫ 0

−∞

1

1 + x2dx+

∫ ∞0

ds1

1 + x2dx

= limR→−∞

∫ 0

R

1

1 + x2dx

+ limR→∞

∫ R

0

1

1 + x2dx

= limR→∞

tan−1 x|0R + limR→∞

tan−1 x|R0= limR→−∞

(tan−1 0− tan−1R)

+ limR→∞

(tan−1R− tan−1 0)

= 0−(−π

2

)+π

2− 0 = π


(b)

∫ 2

1

1

x2 − 1dx

= limR→1+

∫ 1

R

1

x2 − 1dx

= limR→1+

1

2ln

(x− 1

x+ 1

)∣∣∣∣2R

= limR→1+

1

2ln

(1

3

)− 1

2ln

(R− 1

R+ 1

)=∞Therefore the original integral diverges.

16. (a)

∫ 2

0

x

x2 − 1dx

=

∫ 1

0

x

x2 − 1dx+

∫ 2

1

x

x2 − 1dx

= limR→1−

∫ R

0

x

x2 − 1dx

+ limR→1+

∫ 2

R

x

x2 − 1dx

= limR→1−

1

2ln |x2 − 1|

∣∣∣∣R0

+ limR→1+

1

2ln |x2 − 1|

∣∣∣∣2R

= limR→1−

(1

2ln |R2 − 1| − 1

2ln |−1|

)= −∞So the original integral diverges.

(b)

∫ 2

0

1

(x− 2)2dx

= limR→2−

∫ R

0

1

(x− 2)2dx

= limR→2−

1

2− x

∣∣∣∣R0

= limR→2−

1

2−R− 1

2=∞


17. (a) Substitute u =√x∫

1√xe√xdx =

∫2e−udu

Hence

∫ ∞0

1√xe√xdx

= limR→0+

∫ 1

R

1√xe√xdx

+ limR→∞

∫ R

1

1√xe√xdx

= limR→0+

(−2

e√x

)∣∣∣∣1R

+ limR→∞

(−2

e√x

)∣∣∣∣R1

= limR→0+

(−2

e+

2

eR

)

+ limR→∞

(−2

e+

2

eR

)= 1 + 1 = 2

(b)

∫ π/2

0

tanxdx

= limR→π/2−

∫ R

0

tanxdx

= limR→π/2−

− ln cosx

∣∣∣∣R0

= limR→π/2−

(− ln cosR) =∞


18. (a) Substitute u = ex

I =

∫ ∞0

ex

e2x + 1dx

=

∫ ∞1

1

u2 + 1dx

= limR→∞

∫ R

1

1

u2 + 1dx

= limR→∞

tan−1u∣∣∣R1

= limR→∞

(tan−1R− tan−11

)=π

2− π

4=π

4

(b) Substitute u = tan−1x

I =

∫ ∞0

x√x2 + 1

dx

=

∫ π/2

0

tanu(√

tan2u+ 1)du

= limR→π/2

∫ R

0

tanu (secu) du

= limR→π/2

secu

∣∣∣∣R0

= limR→π/2−

secR− sec 0 =∞


19. (a) Ip =

∫ 1

0

x−pdx = limR→0+

∫ 1

R

x−pdx

= limR→0+

(x−p+1

−p+ 1

)∣∣∣∣1R

= limR→0+

1−R−p+1

−p+ 1We need p < 1 for the above limit to con-verge. If this is the case,

Ip =1

−p+ 1.

(b) Ip =

∫ ∞1

x−pdx = limR→∞

∫ R

1

x−pdx

= limR→∞

x−p+1

−p+ 1

∣∣∣∣R1

= limR→∞

R−p+1 − 1

−p+ 1

We need p > 1 for the above limit to


converge.

(c) There are three cases.Case 1: p > −1∫ ∞

0

xpdx = limR→∞

∫ R

0

xpdx

= limR→∞

xp+1

p+ 1

∣∣∣∣R0

= limR→∞

Rp+1

p+ 1=∞

So

∫ ∞−∞

xpdx diverges.

Case 2: p = −1

We have already seen that

∫ ∞−∞

1

xdx

diverges.

Case 3: p < −1∫ 1

0

xpdx = limR→0+

∫ 1

R

xpdx

= limR→0+

xp+1

p+ 1

∣∣∣∣1R

= limR→0+

1−Rp+1

p+ 1=∞

So

∫ ∞−∞

xpdx diverges.

20. (a) Case1: If r ≥ 0Substitute u = rx.

I =

∫ ∞0

xerxdx

=1

r2limR→∞

∫ R

0

ueudu

=1

r2limR→∞

(ueu − eu)|R0

=1

r2limR→∞

eR (R− 1)− 1

r2=∞

So

∫ ∞0

xerxdx diverges for r ≥ 0.

Case2: For r < 0,Substitute u = −rxI =

∫ ∞0

xerxdx

=1

r2limR→∞

∫ −R0

ueudu

= − 1

r2limR→∞

∫ 0

−Rueudu

= − 1

r2limR→∞

(ueu − eu)|0−R

= − 1

r2limR→∞

e−R (−R− 1)− 1

r2

= 0− 1

r2= − 1

r2

Therefore, for all r < 0 the integral∫ ∞0

xerxdx converges.

(b) Case1: If r > 0Substitute u = rx

I =

∫ 0

−∞xerxdx

=1

r2lim

R→−∞

∫ 0

R

ueudu

=1

r2lim

R→−∞(ueu − eu)|0R

=1

r2− 1

r2lim

R→−∞eR (R− 1)

=1

r2− 0 =

1

r2

So

∫ 0

−∞xerxdx converges for r > 0.

Case2: For r ≤ 0,

the integral

∫ 0

−∞xerxdx diverges.

Therefore, for all r < 0the integral

∫∞0xerxdx converges.

21. 0 <x

1 + x3<

x

x3=

1

x2∫ ∞1

1

x2dx = lim

R→∞

∫ R

1

1

x2dx

= limR→∞

(− 1

x

)∣∣∣∣R1

= limR→∞

(− 1

R+ 1

)= 1

So

∫ ∞1

x

1 + x3dx converges.

22.x2 − 2

x4 + 3≤ 3x2

x4= 3x−2

∫ ∞1

3x−2dx = limR→∞

∫ R

1

3x−2dx

= limR→∞

−3

x

∣∣∣∣R1

= limR→∞

−3

R+ 3 = 3

So

∫ ∞1

x2 − 2

x4 + 3dx converges.

23.x

x3/2 − 1>

x

x3/2=

1

x1/2> 0∫ ∞

2


∫ R

2

x−1/2dx

= limR→∞

2√x∣∣∣R2

= limR→∞

(2√R− 2

√2) =∞

So

∫ ∞2

x

x3/2 − 1dx diverges.


24.2 + sec2 x

x≥ 1

x∫ ∞1

1

xdx = lim

R→∞

∫ R

1

1

xdx

= limR→∞

ln |x||R1 = limR→∞

ln |R| =∞

So

∫ ∞1

2 + sec2 x

xdx diverges.

25. 0 <3

x+ ex<

3

ex∫ ∞0

3

exdx = lim

R→∞

∫ R

0

3

exdx

= limR→∞

(− 3

ex

)∣∣∣∣R0

= limR→∞

(− 3

eR+ 3

)= 3

So

∫ ∞0

3

x+ exdx converges.

26. e−x3

< e−x∫ ∞1

e−xdx = limR→∞

∫ R

1

e−xdx

= limR→∞

−e−x∣∣∣R1

= limR→∞

−e−R + e−1

= e−1.

So

∫ ∞1

e−x3

dx converges.

27.sin2 x

1 + ex≤ 1

1 + ex<

1

ex∫ ∞0

1

exdx = lim

R→∞

∫ R

0

1

exdx

= limR→∞

(−e−x)∣∣∣R0

= limR→∞

(−e−R + 1) = 1

So

∫ ∞0

sin2 x

1 + exdx converges.

28.lnx

ex + 1<

x

ex∫ ∞2

x

exdx = lim

R→∞

∫ R

2

xe−xdx

= limR→∞

(−xe−x − e−x)∣∣∣R2

= limR→∞

e−R(−R− 1) + 3e−2

and limR→∞

e−R(−R− 1)

= limR→∞

−R− 1

eR= limR→∞

−1

eR= 0

So

∫ ∞2

lnx

ex + 1dx converges.

29.x2ex

lnx> ex

∫ ∞2

exdx = limR→∞

∫ R

2

exdx

= limR→∞

ex∣∣∣R2

= limR→∞

(eR − e2) =∞

So

∫ ∞2

x2ex

lnxdx diverges.

30. ex2+x+1 > ex∫ ∞

1

exdx = limR→∞

∫ R

1

exdx

= limR→∞

ex∣∣∣R1

= limR→∞

(eR − e) =∞

So

∫ ∞1

ex2+x+1dx diverges.

31. Let u = ln 4x, dv = xdx

du =dx

x, v =

x2

2∫x ln 4xdx =

1

2x2 ln 4x− 1

2

∫xdx

=1

2x2 ln 4x− x2

4+ c

I =

∫ 1

0

x ln 4xdx = limR→ 0+

∫ 1

R

x ln 4xdx

= limR→ 0+

(1

2x2 ln 4x− x2

4

)∣∣∣∣1R

= −1

4− limR→ 0+

(1

2R2 ln 4R− R2

4

)= −1

4− 1

2lim

R→ 0+R2 ln 4R

limR→ 0+

R2 ln 4R = limR→ 0+

ln 4R

R−2

= limR→ 0+

R−1

−2R−3= limR→ 0+

R2

−2= 0

Hence I = −1

4

32. Let u = x, dv = e−2xdx

du = dx, v = −1

2e−2x∫

xe−2xdx = −x2e−2x +

1

2

∫e−2xdx

= −x2e−2x − e−2x

∫ ∞0

xe−2xdx = limR→∞

∫ R

0

xe−2xdx

= limR→∞

(−x

2e−2x − e−2x

)∣∣∣R0

= limR→∞

e−2R(−R/2− 1) + 1

= limR→∞

−R/2− 1

e2R+ 1


= limR→∞

−2

2e2R+ 1 = 1

33. The volume is finite:

V = π

∫ ∞1

1

x2dx = π lim

R→∞

∫ R

1

1

x2dx

= π limR→∞

− 1

x

∣∣∣∣R1

= π limR→∞

(− 1

R+ 1

)= π

The surface area is infinite:

S = 2π

∫ ∞1

1

x

√1 +

1

x4dx

1

x

√1 +

1

x4>

1

x

and

∫ ∞1

1

xdx = lim

R→∞

∫ R

1

1

xdx

= limR→∞

ln |x|∣∣∣R1

= limR→∞

lnR =∞

34. The integral

∫ ∞−∞

x3dx diverges:∫ ∞0

x3dx = limR→∞

∫ R

0

x3dx

= limR→∞

x4

4

∣∣∣∣R0

= limR→∞

R4

4=∞

The limit limR→∞

∫ R

−Rx3dx = 0:

limR→∞

∫ R

−Rx3dx = lim

R→∞

x4

4

∣∣∣∣R−R

= limR→∞

(R4

4− R4

4

)= limR→∞

0 = 0

35. True, this statement can be proved using theintegration by parts:∫f(x)dx = xf(x)−

∫g(x)dx,

where g(x) is some function related to f(x).

36. False, consider f(x) =1

x

37. False, consider f(x) = lnx

38. True, this statement is best understood graph-ically.

39. (a) Substitute u =√kx∫ ∞

−∞e−kx

2

dx =1√k

∫ ∞−∞

e−u2

du =

√π√k

(b) We use integration by parts

(u = e−x2

, v = x):∫e−x

2

dx = xe−x2

+ 2

∫x2e−x

2

dx

Since the graph of the function xe−x2

isanti-symmetric across the y-axis,

limR→∞

(xe−x

2∣∣∣0−R

+ xe−x2∣∣∣R0

)= 0

Then we have∫ ∞−∞

e−x2

dx = 2

∫ ∞−∞

x2e−x2

dx

And the conclusion is∫ ∞−∞

x2e−x2

dx =

√π

2

40. (a) Since k > 0, we have∫ ∞0

sin kx

xdx =

∫ ∞0

sin kx

kx(k) dx

Let u = kx, du = kdx.

=

∫ ∞0

sinu

udu =

π

2.

(b) Since k < 0 , assume k = −m , wherem > 0.∫ ∞

0

sin kx

xdx =

∫ ∞0

sin (−m)x

xdx

= −∫ ∞

0

sinmx

xdx

= −∫ ∞

0

sinmx

mx(m) dx

Let u = mx, so that du = mdx.

= −∫ ∞

0

sinu

udu = −π

2.

(c) Since k > 0 , we have∫ ∞0

sin2kx

x2dx =

∫ ∞0

sin2kx

(kx)2

(k2)dx

Let u = kx, du = kdx.

=

∫ ∞0

sinu

ukdu = k

π

2.

(d) Since k < 0, assume k = −m, wherem > 0.∫ ∞

0

sin2kx

x2dx =

∫ ∞0

sin2 [(−m)x]

x2dx

=

∫ ∞0

sin2mx

x2dx

=

∫ ∞0

sin2mx

(mx)2

(m2)dx

Let u = mx, du = mdx.

=

∫ ∞0

m

(sin2u

u2

)du = m

π

2= −kπ

2.

41. Sincex

x5 + 1≈ 1

x4,∫ ∞

1

x

x5 + 1dx ≈

∫ ∞1

1

x4dx.

we have

∫ ∞1

1

x4dx converges to -

1

3.


Hence

∫ ∞1

x

x5 + 1dx also converges.

Let f(x) =x

x5 + 1and g(x) =

1

x4.

So that, we have, 0 < f(x) < g(x) .By Comparison test,∫ ∞

1

x

x5 + 1dx <

∫ ∞1

1

x4dx = −1

3.

42. (a) Let f(x) =1√x2

and g(x) =x√

x3 − 1.

So that, we have, 0 < f(x) < g(x).By Comparison test,∫ ∞

2

1√x2dx <

∫ ∞2

x√x3 − 1

dx, and

f(x) =1√x2

diverges.

Hence g(x) =x√

x3 − 1also diverges.

(b) Let f(x) =x√

x5 − 1and g(x) =

1

x5/4.


2

x√x5 − 1

dx <

∫ ∞2

1

x5/4dx, and

g(x) =1

x5/4converges.

Hence f(x) =x√

x5 − 1also converges.

(c) Let f(x) =x√

x5 + x− 1and

g(x) =x√

x5 − 1.


2

x√x5 + x− 1

dx <

∫ ∞2

x√x5 − 1

dx ,

and g(x) =x√

x5 − 1converges.

Hence f(x) =x√

x5 + x− 1also con-

verges.

43. Substitute u =π

2− x∫ π/2

0

ln(sinx)dx = −∫ 0

π/2

ln(sin(π/2− u))du

=

∫ π/2

0

ln(cosu)du =

∫ π/2

0

ln(cosx)dx

Moreover,

2

∫ π/2

0

ln(sinx)dx

=

∫ π/2

0

ln(cosx)dx+

∫ π/2

0

ln(sinx)dx

=

∫ π/2

0

[ln(cosx) + ln(sinx)] dx

=

∫ π/2

0

ln(sinx cosx)dx

=

∫ π/2

0

[ln(sin(2x))− ln 2]dx

=

∫ π/2

0

ln(sin(2x))dx− π

2ln 2

=1

2

∫ π

0

ln(sinx)dx− π

2ln 2

Hence,

2

∫ π/2

0

ln(sinx)dx

=1

2

∫ π

0

ln(sinx)dx− π

2ln 2

On the other hand, we notice that the graph ofsinx is symmetric over the interval [0, π] acrossthe line x = π/2, hence∫ π

0

ln(sinx)dx = 2

∫ π/2

0

ln(sinx)dx

and then1

2

∫ π

0

ln(sinx)dx =

∫ π/2

0

ln(sinx)dx

So we get

∫ π/2

0

ln(sinx)dx = −π2

ln 2

44.

∫ 1

0

(lnx)ndx = lim

t→0+

∫ 1

t

(lnx)ndx

= limt→0+

[x(lnx)

n|1t − n∫ 1

t

(lnx)n−1

dx

]Using#112 from the table.

= limt→0+

[(0− t(ln t)2

)]− limt→0+

[n

∫ 1

t

(lnx)n−1

dx

]= 0− lim

t→0+

[n

∫ 1

t

(lnx)n−1

dx

].

Continuing in the same manner,∫ 1

0

(lnx)ndx

= (−1)n−1

n!

[limt→0+

∫ 1

t

(lnx) dx

]= (−1)

n−1n!

[limt→0+

(x lnx− x)

]1

t

= (−1)n−1

n!

[limt→0+

((0− t ln t)− (1− t))]

= (−1)nn!.

45. Improper because tan(π/2) is not defined. Thetwo integrals∫ π/2

0

1

1 + tanxdx =

∫ π/2

0

f(x)dx

because the two integrand only differ at onevalue of x, and that except for this value, ev-


erything is proper.

g(x) =

tanx

1 + tanxif 0 ≤ x < π

20 if x =

π

2

Substitute u = x− π

2followed by w = −u∫ π/2

0

1

1 + tanxdx =

∫ 0

−π/2

1

1− cotudu

= −∫ 0

π/2

1

1 + cotwdw

=

∫ π/2

0

1

1 +cosw

sinw

dw

=

∫ π/2

0

sinw

sinw + coswdw

=

∫ π/2

0

tanw

tanw + 1dw

=

∫ π/2

0

tanx

tanx+ 1dx

Moreover,∫ π/2

0

tanx

tanx+ 1dx+

∫ π/2

0

1

1 + tanxdx

=

∫ π/2

0

(tanx

1 + tanx+

1

1 + tanx

)dx

=

∫ π/2

0

1 dx = x∣∣∣π/20

=π

2

Hence

∫ π/2

0

1

1 + tanxdx =

1

2

(π2

)=π

4

46. As in exercise.45, We have∫ π/2

0

1

1 + tankxdx =

∫ π/2

0

tankx

1 + tankxdx

hence,

2

∫ π/2

0

1

1 + tankxdx

=

∫ π/2

0

1

1 + tankxdx+

∫ π/2

0

tankx

1 + tankxdx

=

∫ π/2

0

1dx =π

2therefore,∫ π/2

0

1

1 + tankxdx =

π

4

47. Use integration by parts twice, first time

let u = −1

2x3, dv = ds− 2xe−x

2

dx

second time

let u = −1

2x, dv = −2xe−x

2

dx∫x4e−x

2

dx

= −1

2x3e−x

2

+

∫3

2x2e−x

2

dx

= −1

2x3e−x

2

+3

2

(−1

2xe−x

2

+1

2

∫e−x

2

dx

)= −1

2x3e−x

2

− 3

4xe−x

2

+3

4

∫e−x

2

dx

Putting integration limits to all the above, andrealizing that when taking limits to ±∞, allmultiples of e−x

2

as shown in above will goto 0 (we have seen this a lot of times before).Then we get∫ ∞−∞

x4e−x2

dx =3

4

∫ ∞−∞

e−x2

dx =3

4

√π

This means when n = 2, the statement∫ ∞−∞

x2ne−x2

dx =(2n− 1) · · · 3 · 1

2n√π

is true. (We can also check that the case forn = 1 is correct.) For general n, supposingthat the statement is true for all m < n, thenintegration by parts gives∫x2ne−x

2

dx

= −1

2x2n−1e−x

2

+2n− 1

2

∫x2n−2e−x

2

dx

and hence∫ ∞−∞

x2ne−x2

dx

=2n− 1

2

∫ ∞−∞

x2n−2e−x2

dx

=2n− 1

2· (2n− 3) · · · 3 · 1

2n−1

√π

=(2n− 1) · · · 3 · 1

2n√π

48. Substitute u =√ax∫

e−ax2

dx =1√a

∫e−u

2

du

and then∫ ∞−∞

e−ax2

dx =1√a

∫ ∞−∞

e−u2

du =

√π

aIgnoring issues of convergence, the derivativescan be taken from the integrand, we get thefollowing:1st derivatived

da

∫ ∞−∞

e−ax2

dx =d

da

√π

a

−∫ ∞−∞

x2e−ax2

dx = −1

2

√π

a3

2nd derivatived2

da2

∫ ∞−∞

e−ax2

dx =d2

da2

√π

a∫ ∞−∞

x4e−ax2

dx =3

4

√π

a5

. . . nth derivativedn

dan

∫ ∞−∞

e−ax2

dx =dn

dan

√π

a


(−1)n∫ ∞−∞

x2ne−ax2

dx

= (−1)n(2n− 1) · · · 3 · 1

2n

√π

a2n+1

Setting a = 1, we get the result of Exercise 47.

49. (a)

∫ ∞0

ke−2xdx = limR→∞

∫ R

0

ke−2xdx

= −k2

limR→∞

e−2x

∣∣∣∣R0

= −k2

limR→∞

(e−2R − 1) =k

2= 1

So k = 2

(b)

∫ ∞0

ke−4xdx = limR→∞

∫ R

0

ke−4xdx

= −k4

limR→∞

e−4x

∣∣∣∣R0

= −k4

limR→∞

(e−4R − 1) =k

4= 1

So k = 4

(c) If r > 0:∫ ∞0

ke−rxdx = limR→∞

∫ R

0

ke−rxdx

= −kr

limR→∞

e−rx∣∣∣∣R0

= − k

r2limR→∞

(e−rR − 1) =k

r= 1

So k = r

50. (a) Substitute u = 2x∫ ∞0

kxe−2xdx = limR→∞

∫ R

0

kxe−2xdx

=k

4limR→∞

∫ R

0

ue−udx

=k

4limR→∞

(ue−u + e−u)

∣∣∣∣R0

=k

4limR→∞

(R+ 1

eR− 1

)= −k

4= 1

So k = −4

(b) Substitute u = 4x∫ ∞0

kxe−4xdx = limR→∞

∫ R

0

kxe−4xdx

=k

16limR→∞

∫ R

0

ue−udx

=k

16limR→∞

(ue−u + e−u)

∣∣∣∣R0

=k

16limR→∞

(R+ 1

eR− 1

)= − k

16= 1

So k = −16

(c) If r > 0:Substitute u = rx∫ ∞

0

kxe−rxdx = limR→∞

∫ R

0

kxe−rxdx

=k

r2limR→∞

∫ R

0

ue−udx

=k

r2limR→∞

(ue−u + e−u)

∣∣∣∣R0

=k

r2limR→∞

(R+ 1

eR− 1

)= − k

r2= 1

So k = −r2

If r ≤ 0:

The integral

∫ ∞0

kxe−rxdx diverges for

any value of k, so there is no value of k tomake the function f(x) = k a pdf.

51. From Exercise 49 (c) we know that this r hasto be positive.Substitute u = rx

µ =

∫ ∞0

xf(x)dx =

∫ ∞0

rxe−rxdx

= limR→∞

∫ R

0

rxe−rxdx

=1

rlimR→∞

∫ R

0

ue−udu

=1

rlimR→∞

e−u(−u− 1)

∣∣∣∣R0

= limR→∞

−R− 1

eR+

1

r= 0 +

1

r=

1

r

52. From Exercise 50 (c) we know that this r hasto be positive.Substitute u = rx

µ =

∫ ∞0

xf(x)dx =

∫ ∞0

r2x2e−rxdx

= limR→∞

∫ R

0

r2x2e−rxdx

=1

rlimR→∞

∫ R

0

u2e−udu

=1

rlimR→∞

e−u(−u2 − 2u− 2)∣∣∣R0

= limR→∞

−R2 − 2R− 2

eR+

2

r= 0 +

2

r=

2

r

53.

∫ 35

0

1

40e−x/40dx = −e−x/40

∣∣∣35

0= 1− e−35/40

P (x > 35) = 1− above = e−35/40∫ 40

0

1

40e−x/40dx = −e−x/40

∣∣∣40

0= 1− e−40/40

P (x > 40) = 1− above = e−40/40∫ 45

0

1

40e−x/40dx = −e−x/40

∣∣∣45

0= 1− e−45/40

P (x > 45) = 1− above = e−45/40

Hence,


P (x > 40|x > 35) =P (x > 40)

P (x > 35)

=e−40/40

e−35/40= e−5/40 ≈ 0.8825

P (x > 45|x > 40) =P (x > 45)

P (x > 40)

=e−45/40

e−40/40= e−5/40 ≈ 0.8825

54. (a) Following Exercise 53, we get

P (x > m+ n|x > n) =P (x > m+ n)

P (x > n)

=1−

∫m+n

0140e−x/40dx

1−∫m

0140e−x/40dx

=e−(m+n)/40

e−n/40= e−m/40

(b)

∫ A

0

ce−cxdx = −e−cx∣∣A0

= 1 − e−cA

P (x > m+ n|x > n) =P (x > m+ n)

P (x > n)

=1−

∫m+n

0ce−cxdx

1−∫m

0ce−cxdxdx

=e−c(m+n)

e−cn

= e−cm

55. (a) For x ≥ 0,

F1(x) =

∫ x

−∞f1(t)dt =

∫ x

0

f1(t)dt

=

∫ x

0

2e−2tdt = −e−2t

∣∣∣∣x0

= 1− e−2x

Ω1(r) =

∫∞r

[1− F1(x)]dx∫ r−∞ F1(x)dx

=

∫∞re−2xdx∫ r

0(1− e−2x)dx

=12e−2r

r + 12e−2r − 1

2

=e−2r

2r + e−2r − 1

(b) For 0 ≤ x ≤ 1, F2(x) =

∫ x

−∞f2(t)dt

=

∫ x

0

f2(t)dt =

∫ x

0

1 dt = t

∣∣∣∣x0

= x

Ω2(r) =

∫∞r

[1− F2(x)]dx∫ r−∞ F2(x)dx

=

∫ 1

r(1− x)dx∫ r

0xdx

=12 − r + r2

2r2

2

=1− 2r + r2

r2

(c) µ1 =

∫ ∞−∞

xf1(x)dx

=

∫ ∞0

2xe−2xdx

= limR→∞

∫ R

0

2xe−2xdx

= limR→∞

e−2x(−x− 1/2)∣∣∣R0

= limR→∞

e−2R(R+ 1/2) +1

2=

1

2

µ2 =

∫ ∞−∞

xf2(x)dx

=

∫ 1

0

xdx =x2

2

∣∣∣∣10

=1

2µ1 = µ2 and when r = 1/2

Ω1(1/2) =e−2·1/2

2 · 1/2 + e−2·1/2 − 1= 1

Ω2(1/2) =1− 2 · 1/2 + (1/2)2

(1/2)2= 1

(d) The graph of f2(x) is more stable thanthat of f1(x).f1(x) > f2(x) for 0 < x < 0.34

and f1(x) < f2(x) for x > 1.

(e) Ω1(r) = 1− 2r − 1

e−2r + 2r − 1

Ω2r = 1− 2r − 1

r2

andr2− (e−2r + 2r− 1) = e−2r + (r− 1)2 > 0This meanswhen r < 1/2, Ω1(r) < Ω2(r)

when r > 1/2, Ω1(r) > Ω2(r)

In terms of this example, we see that theriskier investment is only disadvantageouswhen r small, and will be better when rlarge.

56. Following Exercise 54(b),R(t) = P (x ≥ t) = P (x > t)

= 1−∫ t

0

f(x)dx = 1−∫ t

0

ce−cxdx

= 1− (1− e−ct) = e−ct

57. Graph of p1(x):∫ 1

0

p1 (x) dx =

∫ 1

0

1dx = 1,

Graph of p2(x):

Similarly,∫ 1

0

p2 (x) dx =

∫ 1/2

0

4xdx+

∫ 1

1/2

(4− 4x) dx

= 2x2∣∣1/20

+(4x− 2x2

)11/2

=

(1

2− 0

)+

((4− 2)−

(2− 1

2

))= 1.


The Boltzmann integral

I(p1) =

∫ 1

0

p1 (x) ln p1 (x) dx

=

∫ 1

0

1 ln 1dx = 0.

Also, I(p2) =

∫ 1

0

p2 (x) ln p2 (x) dx

=

∫ 1/2

0

4x ln (4x) dx

+

∫ 1

1/2

(4− 4x) ln (4− 4x) dx

Let u = 4x, du = 4dxand t = 4− 4x, dt = −4dx

=1

4

∫ 2

0

u lnudu− 1

4

∫ 0

2

t ln tdt

=1

2

(1

2u2 lnu− 1

4u2

)2

0

=1

2(2 ln 2− 1) = 0.193147.

For the pdf p2(x) , the probability at x =1

2is

maximum which is equal to1

2.The probability

decreases as x tends to 0 or 1.

p3(x) =

2x 0 ≤ x < 1

4

10x− 21

4≤ x < 1

2

8− 10x1

2≤ x < 3

4

2− 2x1

4< x ≤ 1

Graph of p3(x):∫ 1

0

p3 (x) dx =

∫ 1/4

0

2xdx+

∫ 1/2

1/4

(10x− 2)dx

+

∫ 3/4

1/2

(8− 10x)dx+

∫ 1

3/4

(2− 2x) dx

= 1.

Also, The Boltzmann integral

I(p3) =

∫ 1

0

p3 (x) ln p3 (x) dx

=

∫ 1/4

0

2x ln (2x) dx

+

∫ 1/2

1/4

(10x− 1) ln(10x− 1)dx

+

∫ 3/4

1/2

(8− 10x) ln(8− 10x)dx

+

∫ 1

3/4

(2− 2x) ln (2− 2x) dx

= 0.42.

Ch. 6 Review Exercises

1. Substitute u =√x∫

e√x

√xdx = 2

∫eu du = 2eu + c = 2e

√x + c

2. Substitute u =1

x∫sin(1/x)

x2dx = −

∫sinu du

= cosu+ c = cos(1/x) + c

3. Use the table of integrals,∫x2

√1− x2

dx = −1

2x√

1− x2 +1

2sin−1 x+ c

4. Use the table of integrals,∫2√

9− x2dx = 2 sin−1 x

3+ c

5. Use integration by parts, twice:∫x2e−3x dx

= −1

3x2e−3x +

2

3

∫xe−3x dx

= −1

3x2e−3x

+2

3

(−1

3xe−3x +

1

3

∫e−3x dx

)= −1

3x2e−3x − 2

9xe−3x − 2

27e−3x + c

6. Substitute u = x3∫x2e−x

3

dx =1

3

∫e−u du =

1

3e−x

3

+ c

7. Substitute u = x2∫x

1 + x4dx =

1

2

∫du

1 + u2

=1

2tan−1 u+ c =

1

2tan−1 x2 + c

8.x3

1 + x4dx =

1

4ln(1 + x4) + c

9.x3

4 + x4dx =

1

4ln(4 + x4) + c

10. Substitute u = x2∫x

4 + x4dx =

1

2

∫du

4 + u2

=1

4tan−1 u

2+ c =

1

4tan−1 x

2

2+ c

11.

∫e2 ln x dx =

∫x2 dx =

x3

3+ c

12.

∫cos 4x dx =

1

4sin 4x+ c

CHAPTER 6 REVIEW EXERCISES 401

13. Integration by parts,∫ 1

0

x sin 3x dx

= −1

3x cos 3x

∣∣∣10

+1

3

∫ 1

0

cos 3x dx

= −1

3cos 3 +

1

9sin 3x

∣∣∣10

= −1

3cos 3 +

1

9sin 3

14. Substitute u = x2∫ 1

0

x sin 4x2 dx =

∫ 1

0

1

2sin 4u du

= −1

8cos 4u

∣∣∣10

=1

8(1− cos 4)

15. Use the table of integrals∫ π/2

0

sin4 x dx

= −1

4sin3 x cosx

∣∣∣π/20

+3

4

(x

2− 1

2sinx cosx

) ∣∣∣π/20

=3π

16

16. Use the table of integrals∫ π/2

0

cos3 x dx

=

(2

3sinx+

1

3sinx cos2 x

) ∣∣∣π/20

=2

3

17. Use integration by parts,∫ 1

−1

x sinπx dx

= − 1

πx cosπx

∣∣∣1−1

+1

π

∫ 1

−1

cosπx dx

=2

π+

1

π2sinπx

∣∣∣1−1

=2

π

18. Use integration by parts, twice∫ 1

0

x2 cosπx dx

=1

πx2 sinπx

∣∣∣10− 2

π

∫ 1

0

x sinπx dx

= − 2

π

(− 1

πx cosπx

∣∣∣10

+1

π

∫ 1

0

cosπx dx

)= − 2

π

(1

π+

1

π2sinπx

∣∣∣10

)= − 2

π2

19. Use integration by parts∫ 2

1

x3 lnx dx =x4

4lnx

∣∣∣21− 1

4

∫ 2

1

x3 dx

= 4 ln 2− x4

16

∣∣∣21

= 4 ln 2− 15

16

20.

∫ π/4

0

sinx cosx dx =

∫ π/4

0

1

2sin 2x dx

= −1

4cos 2x

∣∣∣π/40

=1

4

21. Substitute u = sinx∫cosx sin2 x dx =

∫u2 du

=u3

3+ c =

sin3 x

3+ c

22. Substitute u = sinx∫cosx sin3 x dx =

∫u3 du

=u4

4+ c =

sin4 x

4+ c

23. Substitute u = sinx∫cos3 x sin3 x dx =

∫(1− u2)u3 du

=u4

4− u6

6+ c =

3 sin4 x− 2 sin6 x

12+ c

24. Substitute u = cosx∫cos4 x sin3 x dx = −

∫u4(1− u2) du

= −u5

5+u7

7+ c =

−7u5 + 5u7

35+ c

25. Substitute u = tanx∫tan2 x sec4 x dx =

∫u2(1 + u2) du

=u3

3+u5

5+ c =

5 tan3 x+ 3 tan5 x

15+ c

26. Substitute u = tanx∫tan3 x sec2 x dx =

∫u3 du

=u4

4+ c =

tan4 x

4+ c

27. Substitute u = sinx∫ √sinx cos3 x dx

=

∫u1/2(1− u2) du =

2

3u3/2 − 2

7u7/2 + c

=2

3sin3/2 x− 2

7sin7/2 x+ c

28. Substitute u = secx∫tan3 x sec3 x dx =

∫(u2 − 1)u2 du

=u5

5− u3

3+ c =

3 sec5 x− 5 sec3 x

15+ c

29. Complete the square,∫2

8 + 4x+ x2dx

=

∫2

(x+ 2)2 + 22dx = tan−1

(x+ 2

2

)+ c


30. Complete the square,∫3√

−2x− x2dx

=

∫3√

1− (x− 1)2dx = 3 sin−1(x− 1) + c

31. Use the table of integrals,∫2

x2√

4− x2dx = −

√4− x2

2x+ c

32. Substitute u = 9− x2∫x√

9− x2dx = −1

2

∫du

u1/2

= −1

3u3/2 + c = −1

3(9− x2)3/2 + c

33. Substitute u = 9− x2∫x3

√9− x2

dx = −1

2

∫(9− u)

u1/2du

= −9

2

∫u−1/2 du+

1

2

∫u1/2 du

= −9u1/2 +1

3u3/2 + c

= −9(9− x2)1/2 +1

3(9− x2)3/2 + c

34. Substitute u = x2 − 9∫x3

√x2 − 9

dx =1

2

∫(u+ 9)u−1/2 du

=1

3u3/2 + 9u1/2 + c

=1

3(9− x2)3/2 + 9(9− x2)1/2 + c

35. Substitute u = x2 + 9∫x3

√x2 + 9

dx =1

2

∫(u− 9)u−1/2 du

=1

3u3/2 − 9u1/2 + c

=1

3(x2 + 9)3/2 − 9(x2 + 9)1/2 + c

36. Substitute u = x+ 9∫4√x+ 9

dx = 4

∫du√u

= 8u1/2 + c = 8√x+ 9 + c

37. Use the method of PFD∫x+ 4

x2 + 3x+ 2dx

=

∫ (3

x+ 1+−2

x+ 2

)dx

= 3 ln |x+ 1| − 2 ln |x+ 2|+ c

38. Use the method of PFD∫5x+ 6

x2 + x− 12dx

=

∫ (3

x− 3+

3

x+ 4

)dx

= 3 ln |x− 3|+ 3 ln |x+ 4|+ c

39. Use the method of PFD∫4x2 + 6x− 12

x3 − 4xdx

=

∫ (3

x+−1

x+ 2+

2

x− 2

)dx

= 3 ln |x| − ln |x+ 2|+ 2 ln |x− 2|+ c

40. Use the method of PFD∫5x2 + 2

x3 + xdx =

∫ (2

x+

3x

x2 + 1

)dx

= 2 ln |x|+ 3

2ln(x2 + 1) + c

41. Use the table of integrals,∫ex cos 2x dx

=(cos 2x+ 2 sin 2x)ex

5+ c

42. Substitute u = x2 followed by integration byparts∫x3 sinx2 dx =

1

2

∫u sinu du

= −1

2u cosu+

1

2

∫cosu du

= −1

2u cosu+

1

2sinu+ c

= −1

2x2 cosx2 +

1

2sinx2 + c

43. Substitute u = x2 + 1∫x√x2 + 1 dx =

1

2

∫u1/2 du

=1

3u3/2 + c =

1

3(x2 + 1)3/2 + c

44. Use the table of integrals∫ √1− x2 dx

=1

2

√1− x2 +

1

2sin−1 x+ c

45.4

x2 − 3x− 4=

A

x+ 1+

B

x− 44 = A(x− 4) +B(x+ 1)

= (A+B)x+ (−4A+B)

A = −4

5;B =

4

54

x2 − 3x− 4=−4/5

x+ 1+

4/5

x− 4

46.2x

x2 + x− 6=

A

x− 2+

B

x+ 3

2x = A(x+ 3) +B(x− 2)= (A+B)x+ (3A− 2B)


A =4

5;B =

6

52x

x2 + x− 6=

4/5

x− 2+

6/5

x+ 3

47.−6

x3 + x2 − 2x=A

x+

B

x− 1+

C

x+ 2− 6 = A(x− 1)(x+ 2) +Bx(x+ 2) + cx(x− 1)A = −3;B = −2;C = −1

−6

x3 + x2 − 2x=−3

x+−2

x− 1+−1

x+ 2

48.x2 − 2x− 2

x3 + x=A

x+Bx+ c

x2 + 1

x2 − 2x− 2 = A(x2 + 1) + (Bx+ c)x= (A+B)x2 + cx+AA = −2;B = 3;C = −2x2 − 2x− 2

x3 + x=−2

x+

3x− 2

x2 + 1

49.x− 2

x2 + 4x+ 4=

A

x+ 2+

B

(x+ 2)2

x− 2 = A(x+ 2) +BA = 1;B = −4

x− 2

x2 + 4x+ 4=

1

x+ 2+

−4

(x+ 2)2

50.x2 − 2

(x2 + 1)2=Ax+B

x2 + 1+

Cx+D

(x2 + 1)2

x2 − 2 = (Ax+B)(x2 + 1) + cx+DA = 0;B = 1;C = 0;D = −3

x2 − 2

(x2 + 1)2=

1

x2 + 1+

−3

(x2 + 1)2

51. Substitute u = e2x∫e3x√

4 + e2x dx

=

∫e2x√

4e2x + e4xdx =1

2

∫ √4u+ u2 du

=1

2

∫ √(u+ 2)2 − 4 du

=1

4(u+ 2)

√(u+ 2)2 − 4

− ln |(u+ 2) +√

(u+ 2)2 − 4|+ c

=(e2x + 2)

√4e2x + e4x

4

− ln∣∣∣(e2x + 2) +

√4e2x + e4x

∣∣∣+ c

52. Substitute u = x2∫x√x4 − 4 dx =

1

2

∫ √u2 − 4 du

=u√u2 − 4

4− ln |u+

√u2 − 4|+ c

=x2√x4 − 4

4− ln |x2 +

√x4 − 4|+ c

53.

∫sec4 x dx

=1

3sec2 x tanx+

2

3

∫sec2 x dx

=1

3sec2 x tanx+

2

3tanx+ c

54.

∫tan5 x dx

=1

4tan4 x−

∫tan3 x dx

=1

4tan4 x− 1

2tan2 x+

∫tanx dx

=1

4tan4 x− 1

2tan2 x− ln | cosx|+ c

55. Substitute u = 3− x∫4

x(3− x)2dx = −4

∫1

(3− u)u2du

=4

9ln

∣∣∣∣3− uu∣∣∣∣+

4

3u+ c

=4

9ln

∣∣∣∣ x

3− x

∣∣∣∣+4

3(3− x)+ c


sin2 x(3 + 4 sinx)dx =

∫du

u2(3 + 4u)

=4

9ln

∣∣∣∣3 + 4u

u

∣∣∣∣− 1

3u+ c

=4

9ln

∣∣∣∣3 + 4 sinx

sinx

∣∣∣∣− 1

3 sinx+ c

57.

∫ √9 + 4x2

x2dx =

∫ 2

√9

4+ x2

x2dx

= 2

−√

9

4+ x2

x+ ln

∣∣∣x+√

9/4 + x2∣∣∣+ c

= −√

9 + 4x2

x+ 2 ln

∣∣∣∣∣x+

√9

4+ x2

∣∣∣∣∣+ c

58.

∫x2

√4− 9x2

dx =1

3

∫x2√

4/9− x2dx

= −1

6x√

4/9− x2 +2

27sin−1 3x

2+ c

59.

∫ √4− x2

xdx

=√

4− x2 − 2 ln

∣∣∣∣∣2 +√

4− x2

x

∣∣∣∣∣+ c

60.

∫x2

(x6 − 4)3/2dx =

1

3

∫1

(u2 − 4)3/2dx


=1

3

∫2 sec θ tan θ

(4 sec2 θ − 4)3/2dx

=1

3

∫sec θ tan θ

tan3 θdx

=1

3

∫csc θ cot θ dx = − x3

3√x6 − 4

+ c

61. Substitute u = x2 − 1∫ 1

0

x

x2 − 1dx =

∫ 0

−1

du

2u

= limR→0−

∫ R

−1

du

2u= limR→0−

ln |u|∣∣∣0−1

This limit does not exist, so the integral di-verges.

62. Substitute u = x− 4∫ 10

4

2 dx√x− 4

=

∫ 6

0

2 du√u

= limR→0+

∫ 6

R

2u−1/2 du = limR→0+

4u1/2∣∣∣6R

= limR→0+

(4√

6− 4√R) = 4

√6

63.

∫ ∞1

3

x2dx = lim

R→∞

∫ R

1

3

x2dx

= limR→∞

− 3

x

∣∣∣∣R1

= limR→∞

− 3

R+ 3 = 3

64. Use integration by parts,∫ ∞1

xe−3x dx = limR→∞

∫ R

1

xe−3x dx

= limR→∞

[e−3x

(−x

3− 1

9

)]∣∣∣∣R1

= limR→∞

e−3R

(−R

3− 1

9

)+

4e−3

9

=4e−3

9

65.

∫ ∞0

4

4 + x2dx = lim

R→∞

∫ R

0

4

4 + x2dx

= limR→∞

2 tan−1 x

2

∣∣∣R0

= limR→∞

2 tan−1R = π

66.

∫ ∞0

xe−x2

dx = limR→∞

−e−x2

2

∣∣∣∣∣R

0

=1

2∫ 0

−∞xe−x

2

dx = limR→∞

−e−x2

2

∣∣∣∣∣0

−R

= −1

2

So

∫ ∞−∞

xe−x2

dx =1

2− 1

2= 0

67.

∫ 2

0

3

x2dx = lim

R→0+

∫ 2

R

3

x2dx

= limR→0+

− 3

x

∣∣∣∣2R

=∞


68.

∫ 2

1

x dx

1− x2= limR→1+

∫ 2

R

x dx

1− x2

= limR→1+

−1

2ln |1− x2|

∣∣∣∣2R

=∞


69. If c(t) = R, then the total amount of dye is∫ T

0

c(t) dt =

∫ T

0

Rdt = RT

If c(t) = 3te2Tt, then we can use integration byparts to get∫ T

0

3te2Tt dt

=3t

2Te2Tt

∣∣∣T0−∫ T

0

3

2Te2Tt dt

=3

2e2T 2

− 3

4T 2e2Tt

∣∣∣∣T0

=3

2e2T 2

− 3

4T 2e2T 2

+3

4T 2

Since R = c(T ) = 3Te2T 2

The cardiac output is

RT∫ T0c(t) dt

=3T 2e2T 2

32e

2T 2 − 34T 2 e2T 2 + 3

4T 2

=RT 3

3T 2e2T 2/2− 3e2T 2/4 + 3/4

70. With u = ln(x+ 1) and v = x∫lnx+ 1 dx

= x ln(x+ 1)−∫

x

x+ 1dx

= x ln(x+ 1)−∫ (

1− 1

x+ 1

)dx

= x ln(x+ 1)− x+ ln(x+ 1) + c

With u = ln(x+ 1) and v = x+ 1∫lnx+ 1 dx

= (x+ 1) ln(x+ 1)−∫x+ 1

x+ 1dx

= (x+ 1) ln(x+ 1)− x+ c

The two answers are the same.

71. fn,ave =1

en

∫ en

0

lnx dx

=1

enlimR→0

∫ en

R

lnx dx

=1

enlimR→0

(x lnx− x)∣∣∣enR

=1

enlimR→0

(nen − en −R lnR+R) = n− 1


72. First we notice that

lim∆t→0

P (t < x < t+ ∆t)

∆t

= lim∆t→0

1

∆t

∫ t+∆t

t

f(x) dx = f(t)

And then the failure rate function

lim∆t→0

P (x < t+ ∆t|x > t)

∆t

= lim∆t→0

1

∆t

P (t < x < t+ ∆t)

P (x > t)

= lim∆t→0

P (t < x < t+ ∆t)

∆t· 1

R(t)=f(t)

R(t)

73. R(t) = P (x > t) = 1−∫ t

0

ce−cx dx

= 1− (−e−cx)∣∣∣t0

= 1− (1− e−ct) = e−ct

Hencef(t)

R(t)=ce−ct

e−ct= c

74. (a) P (x > s) = 1−∫ s

0

xe−x dx

= 1− (−xe−x − e−x)∣∣∣s0

= 1− (1− se−s − e−s)= (s+ 1)e−s

P (x > s+ t|x > s)

=P (x > s+ t)

P (x > s)

=(s+ t+ 1)e−s−t

(s+ 1)e−s= e−t +

t

1 + se−t

(b) Take the derivative w.r.t s:d

ds

(e−t +

t

1 + se−t)

= −e−t t

(1 + s)2

When t > 0, since e−t > 0 and(1 + s)2 > 0, the above derivative is neg-ative, so the function P (x > s + t|x > s)is decreasing w.r.t. s.

75. We use a CAS to see that∫ 100

90

1√450π

e−(x−100)2/450 dx ≈ 24.75%

We can use substitution to get1√

450π

∫ ∞a

e−(x−100)2/450 dx

=1√π

∫ ∞a− 90√

450

e−u2

du

Since

∫ ∞−∞

e−x2

dx =√π,∫ ∞

0

e−x2

dx =

√π

2

So we want to find the value of a so that∫ a− 90√450

0

e−u2

du = 0.49√π

Using a CAS we finda− 90√

450≈ 1.645, a ≈ 125

Some body being called a genius need to havea IQ score of at least 125.

76. I (1) =

∫ ∞0

1

(1 + x2)dx = tan−1x

∣∣∞0

=π

2

Now, I (n+ 1)

=

[1

(1 + x2)n

∫1

(1 + x2)dx

]∞0

−∫ ∞

0

d((

1 + x2)−n)

dx

∫1

(1 + x2)dx

dx

=

[tan−1x

(1 + x2)n

]∞0

+ 2n

∫ ∞0

xtan−1x

(1 + x2)n dx

⇒ I (n+ 1) = 2n

∫ ∞0

xtan−1x

(1 + x2)n dx ... (1)

Also, I (n+ 1) =

∫ ∞0

1 + x2 − x2

(1 + x2)n+1 dx

= I (n)−∫ ∞

0

x2

(1 + x2)n+1 dx

= I (n)−[

1

(1 + x2)n

∫x2

(1 + x2)dx

]∞0

−∫ ∞

0

d((

1 + x2)−n)

dx

∫x2

(1 + x2)dx

dx

= I (n)−

[x− tan−1x

(1 + x2)n+1

]∞0

+2n

∫ ∞0

x(x− tan−1x

)(1 + x2)

n+1 dx

ds = I (n)− 2n

∫ ∞0

(x2 − xtan−1x

)(1 + x2)

n+1 dx

= I (n)− 2n

∫ ∞0

x2

(1 + x2)n+1 dx

+ 2n

∫ ∞0

xtan−1x

(1 + x2)n+1 dx

Therefore,I(n+ 1) = I (n)

−2n

∫ ∞0

x2

(1 + x2)n+1 dx+I (n+ 1)

(using (1))I (n+ 1)= I (n)− 2n (I (n)− I (n+ 1)) + I (n+ 1)

Hence proved.


As,I (n+ 1) =2n− 1

2nI (n)

I (n) =2n− 3

2n− 2I (n− 1)

I (n− 1) =2n− 5

2n− 4I (n− 2)

I (n− 2) =2n− 7

2n− 6I (n− 3)

and so on therefore,

I (2) =3

4I(1),

I (1) =3

4· π

2,

Thus, I (n) =2n− 3

2n− 2· 2n− 5

2n− 4· 2n− 7

2n− 6· · ·

I (1) I (n) =1

2· 3

4· · · 2n− 3

2n− 2· π

2

Ism et chapter_6

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