Formulas From Integration by Parts

1. Formulas From Integration by Parts
Integration by parts:
Let u = u(x), v = v(x), then
where dv = v'dx and du = u'dx.
∫ udv = uv – ∫ vdu

2. Formulas From Integration by Parts
Integration by parts:
Let u = u(x), v = v(x), then
where dv = v'dx and du = u'dx.
∫ udv = uv – ∫ vdu
The integrals of some functions require using
integration by parts many times.

3. Formulas From Integration by Parts
Integration by parts:
Let u = u(x), v = v(x), then
where dv = v'dx and du = u'dx.
∫ udv = uv – ∫ vdu
The integrals of some functions require using
integration by parts many times.
This leads to special methods and formulas
that summarize the results.

4. Formulas From Integration by Parts
Integration by parts:
Let u = u(x), v = v(x), then
where dv = v'dx and du = u'dx.
∫ udv = uv – ∫ vdu
The integrals of some functions require using
integration by parts many times.
This leads to special methods and formulas
that summarize the results.
These formulas usually reduce the complexity
of the integrands in stages until the final
answer is obtained.

5. Formulas From Integration by Parts
The Tableau Method

6. Formulas From Integration by Parts
The Tableau Method
Given
+u
∫ udv, set up the table as shown.
dv

7. Formulas From Integration by Parts
The Tableau Method
Given
+u
∫ udv, set up the table as shown.
differentiate
with alternate signs
–u' +u'' •–u'''
dv

8. Formulas From Integration by Parts
The Tableau Method
Given
+u
∫ udv, set up the table as shown.
differentiate
with alternate signs
integrate
v
–u' +u''
∫ v dx ∫[∫ v dx]dx
•
•
–u'''
dv

9. Formulas From Integration by Parts
The Tableau Method
Given
+u
∫ udv, set up the table as shown.
differentiate
with alternate signs
integrate
v
–u' +u''
∫ v dx ∫[∫ v dx]dx
•
•
–u'''
The integral ∫ u dv is the sum of the
diagonal products.
dv

10. Formulas From Integration by Parts
The Tableau Method
Given
+u
∫ udv, set up the table as shown.
differentiate
with alternate signs
integrate
v
–u' +u''
∫ v dx ∫[∫ v dx]dx
•
•
–u'''
The integral ∫ u dv is the sum of the
diagonal products.
dv
That is, ∫ u dv = uv

11. Formulas From Integration by Parts
The Tableau Method
Given
+u
∫ udv, set up the table as shown.
differentiate
with alternate signs
integrate
v
–u' +u''
∫ v dx ∫[∫ v dx]dx
•
•
–u'''
The integral ∫ u dv is the sum of the
diagonal products.
dv
That is, ∫ u dv = uv – u' ∫vdx

12. Formulas From Integration by Parts
The Tableau Method
Given
+u
∫ udv, set up the table as shown.
differentiate
with alternate signs
integrate
v
–u' +u''
∫ v dx ∫[∫ v dx]dx
•
•
–u'''
The integral ∫ u dv is the sum of the
diagonal products.
dv
That is, ∫ u dv = uv – u' ∫vdx + u'' ∫[∫ v dx]dx ..

13. Formulas From Integration by Parts
Example: Find ∫(x3 + 2x) ex/2dx

14. Formulas From Integration by Parts
Example: Find ∫(x3 + 2x) ex/2dx
x3+2x
ex/2 dx
Set u = x3 + 2x, dv = ex/2dx

15. Formulas From Integration by Parts
Example: Find ∫(x3 + 2x) ex/2dx
x3+2x
differentiate
with alternate signs
Set u = x3 + 2x, dv = ex/2dx
ex/2 dx

16. Formulas From Integration by Parts
Example: Find ∫(x3 + 2x) ex/2dx
x3+2x
differentiate
with alternate signs
Set u = x3 + 2x, dv = ex/2dx
–(3x2+2)
ex/2 dx

17. Formulas From Integration by Parts
Example: Find ∫(x3 + 2x) ex/2dx
x3+2x
differentiate
with alternate signs
6x
Set u = x3 + 2x, dv = ex/2dx
–(3x2+2)
ex/2 dx

18. Formulas From Integration by Parts
Example: Find ∫(x3 + 2x) ex/2dx
x3+2x
differentiate
with alternate signs
6x –6
Set u = x3 + 2x, dv = ex/2dx
–(3x2+2) 0
ex/2 dx

19. Formulas From Integration by Parts
Example: Find ∫(x3 + 2x) ex/2dx
x3+2x
differentiate
with alternate signs
integrate
6x –6
Set u = x3 + 2x, dv = ex/2dx
–(3x2+2) 0
ex/2 dx

20. Formulas From Integration by Parts
Example: Find ∫(x3 + 2x) ex/2dx
x3+2x
differentiate
with alternate signs
integrate
6x –6
Set u = x3 + 2x, dv = ex/2dx
–(3x2+2) 0
2ex/2
ex/2 dx

21. Formulas From Integration by Parts
Example: Find ∫(x3 + 2x) ex/2dx
x3+2x
differentiate
with alternate signs
integrate
6x –6
Set u = x3 + 2x, dv = ex/2dx
–(3x2+2) 0
2ex/2
4ex/2ex/2 dx

22. Formulas From Integration by Parts
Example: Find ∫(x3 + 2x) ex/2dx
x3+2x
differentiate
with alternate signs
integrate
6x –6
Set u = x3 + 2x, dv = ex/2dx
–(3x2+2) 0
2ex/2
4ex/2 8ex/2 16ex/2
ex/2 dx

23. Formulas From Integration by Parts
Example: Find ∫(x3 + 2x) ex/2dx
x3+2x
differentiate
with alternate signs
integrate
6x –6
Set u = x3 + 2x, dv = ex/2dx
–(3x2+2) 0
2ex/2
4ex/2 8ex/2 16ex/2
Hence ∫(x3 + 2x) ex/2dx
=
ex/2 dx

24. Formulas From Integration by Parts
Example: Find ∫(x3 + 2x) ex/2dx
x3+2x
differentiate
with alternate signs
integrate
6x –6
Set u = x3 + 2x, dv = ex/2dx
–(3x2+2) 0
2ex/2
4ex/2 8ex/2 16ex/2
Hence ∫(x3 + 2x) ex/2dx
= 2ex/2(x3+2x)
ex/2 dx

25. Formulas From Integration by Parts
Example: Find ∫(x3 + 2x) ex/2dx
x3+2x
differentiate
with alternate signs
integrate
6x –6
Set u = x3 + 2x, dv = ex/2dx
–(3x2+2) 0
2ex/2
4ex/2 8ex/2 16ex/2
Hence ∫(x3 + 2x) ex/2dx
= 2ex/2(x3+2x) – 4ex/2(3x2+2)
ex/2 dx

26. Formulas From Integration by Parts
Example: Find ∫(x3 + 2x) ex/2dx
x3+2x
differentiate
with alternate signs
integrate
6x –6
Set u = x3 + 2x, dv = ex/2dx
–(3x2+2) 0
2ex/2
4ex/2 8ex/2 16ex/2
Hence ∫(x3 + 2x) ex/2dx
= 2ex/2(x3+2x) – 4ex/2(3x2+2) + 48xex/2 – 96ex/2 + c
ex/2 dx

27. Formulas From Integration by Parts
Example: Find ∫(x3 + 2x) ex/2dx
x3+2x
differentiate
with alternate signs
integrate
6x –6
Set u = x3 + 2x, dv = ex/2dx
–(3x2+2) 0
2ex/2
4ex/2 8ex/2 16ex/2
Hence ∫(x3 + 2x) ex/2dx
= 2ex/2(x3+2x) – 4ex/2(3x2+2) + 48xex/2 – 96ex/2 + c
= ex/2 [2(x3+2x) – 4(3x2+2) + 48x – 96] + c
ex/2 dx

28. Formulas From Integration by Parts
Example: Find ∫(x3 + 2x) ex/2dx
x3+2x
differentiate
with alternate signs
integrate
6x –6
Set u = x3 + 2x, dv = ex/2dx
–(3x2+2) 0
2ex/2
4ex/2 8ex/2 16ex/2
Hence ∫(x3 + 2x) ex/2dx
= 2ex/2(x3+2x) – 4ex/2(3x2+2) + 48xex/2 – 96ex/2 + c
= ex/2 [2(x3+2x) – 4(3x2+2) + 48x – 96] + c
= ex/2 [2x3 – 12x2 + 52x – 104] + c
ex/2 dx

29. Formulas From Integration by Parts
The tableau method is used when the
integrand is a product of a polynomial with
functions that can be integrated repeatedly
such as ex, sin(x), and cos(x).

30. Formulas From Integration by Parts
The tableau method is used when the
integrand is a product of polynomial with a
functions that can be integrated repeatedly
such as ex, sin(x), and cos(x).
Antiderivatives of trig-power

31. Formulas From Integration by Parts
The tableau method is used when the
integrand is a product of polynomial with a
functions that can be integrated repeatedly
such as ex, sin(x), and cos(x).
Antiderivatives of trig-power
From here on, we use s for sin(x), c for cos(x),
and t for tan(x).

32. Formulas From Integration by Parts
The tableau method is used when the
integrand is a product of polynomial with a
functions that can be integrated repeatedly
such as ex, sin(x), and cos(x).
Antiderivatives of trig-power
From here on, we use s for sin(x), c for cos(x),
and t for tan(x). Also we assume that n > 2.

33. Formulas From Integration by Parts
The tableau method is used when the
integrand is a product of polynomial with a
functions that can be integrated repeatedly
such as ex, sin(x), and cos(x).
Antiderivatives of trig-powers
∫ cosn(x) dx
From here on, we use s for sin(x), c for cos(x),
and t for tan(x). Also we assume that n > 2.

34. Formulas From Integration by Parts
The tableau method is used when the
integrand is a product of polynomial with a
functions that can be integrated repeatedly
such as ex, sin(x), and cos(x).
Antiderivatives of trig-powers
Write ∫cndx = ∫cn–1 cdx, use integration by parts.
∫ cosn(x) dx
From here on, we use s for sin(x), c for cos(x),
and t for tan(x). Also we assume that n > 2.

35. Formulas From Integration by Parts
The tableau method is used when the
integrand is a product of polynomial with a
functions that can be integrated repeatedly
such as ex, sin(x), and cos(x).
Antiderivatives of trig-powers
set u = cn–1 dv = cdx
Write ∫cndx = ∫cn–1 cdx, and use integration
by parts.
∫ cosn(x) dx
From here on, we use s for sin(x), c for cos(x),
and t for tan(x). Also we assume that n > 2.

36. Formulas From Integration by Parts
The tableau method is used when the
integrand is a product of polynomial with a
functions that can be integrated repeatedly
such as ex, sin(x), and cos(x).
Antiderivatives of trig-powers
set u = cn–1 dv = cdx
du = -(n – 1)cn–2sdx
Write ∫cndx = ∫cn–1 cdx, use integration by parts.
∫ cosn(x) dx
From here on, we use s for sin(x), c for cos(x),
and t for tan(x). Also we assume that n > 2.

37. Formulas From Integration by Parts
The tableau method is used when the
integrand is a product of polynomial with a
functions that can be integrated repeatedly
such as ex, sin(x), and cos(x).
Antiderivatives of trig-powers
set u = cn–1 dv = cdx
du = -(n – 1)cn–2sdx v = s
Write ∫cndx = ∫cn–1 cdx, use integration by parts.
∫ cosn(x) dx
From here on, we use s for sin(x), c for cos(x),
and t for tan(x). Also we assume that n > 2.

38. Formulas From Integration by Parts
∫cndx =
set u = cn–1 dv = cdx
du = -(n – 1)cn–2sdx v = s

39. Formulas From Integration by Parts
∫cndx = cn–1s
set u = cn–1 dv = cdx
du = -(n – 1)cn–2sdx v = s

40. Formulas From Integration by Parts
∫cndx = cn–1s + (n – 1)∫cn–2s2dx
set u = cn–1 dv = cdx
du = -(n – 1)cn–2sdx v = s

41. Formulas From Integration by Parts
∫cndx = cn–1s + (n – 1)∫cn–2s2dx
set u = cn–1 dv = cdx
du = -(n – 1)cn–2sdx v = s
∫cndx = cn–1s + (n – 1)∫cn–2(1 – c2)dx

42. Formulas From Integration by Parts
∫cndx = cn–1s + (n – 1)∫cn–2s2dx
set u = cn–1 dv = cdx
du = -(n – 1)cn–2sdx v = s
∫cndx = cn–1s + (n – 1)∫cn–2(1 – c2)dx
∫cndx = cn–1s + (n – 1)∫cn–2dx – (n – 1) ∫cndx

43. Formulas From Integration by Parts
∫cndx = cn–1s + (n – 1)∫cn–2s2dx
set u = cn–1 dv = cdx
du = -(n – 1)cn–2sdx v = s
∫cndx = cn–1s + (n – 1)∫cn–2(1 – c2)dx
∫cndx = cn–1s + (n – 1)∫cn–2dx – (n – 1) ∫cndx
combine like–terms

44. Formulas From Integration by Parts
∫cndx = cn–1s + (n – 1)∫cn–2s2dx
set u = cn–1 dv = cdx
du = -(n – 1)cn–2sdx v = s
∫cndx = cn–1s + (n – 1)∫cn–2(1 – c2)dx
∫cndx = cn–1s + (n – 1)∫cn–2dx – (n – 1) ∫cndx
n∫cndx = cn–1s + (n – 1)∫cn–2dx

45. Formulas From Integration by Parts
∫cndx = cn–1s + (n – 1)∫cn–2s2dx
set u = cn–1 dv = cdx
du = -(n – 1)cn–2sdx v = s
∫cndx = cn–1s + (n – 1)∫cn–2(1 – c2)dx
∫cndx = cn–1s + (n – 1)∫cn–2dx – (n – 1) ∫cndx
n∫cndx = cn–1s + (n – 1)∫cn–2dx
Hence, ∫cndx = + ∫cn–2dx
n
cn–1s
n
n–1

46. Formulas From Integration by Parts
∫cndx = + ∫cn–2dx
n
cn–1s
n
n–1
Using this formula repeatedly eventually
reduces the problem to integrating cosine to
the power 0 or 1, which terminates the
process.

47. Formulas From Integration by Parts
∫cndx = + ∫cn–2dx
n
cn–1s
n
n–1
Example: Find ∫ c5 dx
Using this formula repeatedly eventually
reduces the problem to integrating cosine to
the power 0 or 1, which terminates the
process.

48. Formulas From Integration by Parts
∫cndx = + ∫cn–2dx
n
cn–1s
n
n–1
Example: Find ∫ c5 dx
n = 5,
Using this formula repeatedly eventually
reduces the problem to integrating cosine to
the power 0 or 1, which terminates the
process.

49. Formulas From Integration by Parts
∫cndx = + ∫cn–2dx
n
cn–1s
n
n–1
Example: Find ∫ c5 dx
n = 5, ∫ c5 dx = +5
c4s
Using this formula repeatedly eventually
reduces the problem to integrating cosine to
the power 0 or 1, which terminates the
process.

50. Formulas From Integration by Parts
∫cndx = + ∫cn–2dx
n
cn–1s
n
n–1
Example: Find ∫ c5 dx
n = 5, ∫ c5 dx = + ∫c3dx5
c4s
5
4
Using this formula repeatedly eventually
reduces the problem to integrating cosine to
the power 0 or 1, which terminates the
process.

51. Formulas From Integration by Parts
∫cndx = + ∫cn–2dx
n
cn–1s
n
n–1
Example: Find ∫ c5 dx
n = 5, ∫ c5 dx = + ∫c3dx5
c4s
5
4
n = 3,
Using this formula repeatedly eventually
reduces the problem to integrating cosine to
the power 0 or 1, which terminates the
process.

52. Formulas From Integration by Parts
∫cndx = + ∫cn–2dx
n
cn–1s
n
n–1
Example: Find ∫ c5 dx
n = 5, ∫ c5 dx = + ∫c3dx5
c4s
5
4
n = 3, (5
c4s
5
4
= + 3
c2s + )
Using this formula repeatedly eventually
reduces the problem to integrating cosine to
the power 0 or 1, which terminates the
process.

53. Formulas From Integration by Parts
∫cndx = + ∫cn–2dx
n
cn–1s
n
n–1
Example: Find ∫ c5 dx
n = 5, ∫ c5 dx = + ∫c3dx5
c4s
5
4
n = 3, (5
c4s
5
4
= + 3
c2s + 3
2
∫ c dx )
Using this formula repeatedly eventually
reduces the problem to integrating cosine to
the power 0 or 1, which terminates the
process.

54. Formulas From Integration by Parts
∫cndx = + ∫cn–2dx
n
cn–1s
n
n–1
Example: Find ∫ c5 dx
n = 5, ∫ c5 dx = + ∫c3dx5
c4s
5
4
n = 3, (5
c4s
5
4
= + 3
c2s + 3
2
∫ c dx )
(5
c4s
5
4
= + 3
c2s + 3
2
s) + k
Using this formula repeatedly eventually
reduces the problem to integrating cosine to
the power 0 or 1, which terminates the
process.

55. Formulas From Integration by Parts
For ∫sndx
u = sn–1Set

56. Formulas From Integration by Parts
For ∫sndx
u = sn–1 dv = sdxSet

57. Formulas From Integration by Parts
For ∫sndx
u = sn–1 dv = sdx
du = (n – 1)sn–2cdx
Set

58. Formulas From Integration by Parts
For ∫sndx
u = sn–1 dv = sdx
du = (n – 1)sn–2cdx v = –c
Set

59. Formulas From Integration by Parts
∫sndx = + ∫sn–2dx
n
–sn–1c
n
n–1
For ∫sndx
u = sn–1 dv = sdx
du = (n – 1)sn–2cdx v = –c
Using integration by parts we get the formula:
Set

60. Formulas From Integration by Parts
∫sndx = + ∫sn–2dx
n
–sn–1c
n
n–1
For ∫sndx
u = sn–1 dv = sdx
du = (n – 1)sn–2cdx v = –c
Using integration by parts we get the formula:
For use the fact that∫secn(x)dx, ∫sec2(x)dx = t
Set

61. Formulas From Integration by Parts
∫sndx = + ∫sn–2dx
n
–sn–1c
n
n–1
For ∫sndx
u = sn–1 dv = sdx
du = (n – 1)sn–2cdx v = –c
Using integration by parts we get the formula:
For use the fact that∫secn(x)dx, ∫sec2(x)dx = t
Set u = secn–2(x)
Set

62. Formulas From Integration by Parts
∫sndx = + ∫sn–2dx
n
–sn–1c
n
n–1
For ∫sndx
u = sn–1 dv = sdx
du = (n – 1)sn–2cdx v = –c
Using integration by parts we get the formula:
For use the fact that∫secn(x)dx, ∫sec2(x)dx = t
Set u = secn–2(x) dv = sec2(x)dx
Set

63. Formulas From Integration by Parts
∫sndx = + ∫sn–2dx
n
–sn–1c
n
n–1
For ∫sndx
u = sn–1 dv = sdx
du = (n – 1)sn–2cdx v = –c
Using integration by parts we get the formula:
For use the fact that∫secn(x)dx, ∫sec2(x)dx = t
Set
du = (n – 2)secn–2(x)tdx
Set
u = secn–2(x) dv = sec2(x)dx

64. Formulas From Integration by Parts
∫sndx = + ∫sn–2dx
n
–sn–1c
n
n–1
For ∫sndx
u = sn–1 dv = sdx
du = (n – 1)sn–2cdx v = –c
Using integration by parts we get the formula:
For use the fact that∫secn(x)dx, ∫sec2(x)dx = t
Set
du = (n – 2)secn–2(x)tdx
Set
du = (n – 2)secn–2(x)tdx v = t
u = secn–2(x) dv = sec2(x)dx

65. Formulas From Integration by Parts
∫sndx = + ∫sn–2dx
n
–sn–1c
n
n–1
For ∫sndx
u = sn–1 dv = sdx
du = (n – 1)sn–2cdx v = –c
Using integration by parts we get the formula:
For use the fact that∫secn(x)dx,
∫secn(x)dx = + ∫secn–2(x)dxn – 1
secn–2(x)tan(x)
n–1
n–2
∫sec2(x)dx = t
Set
du = (n – 2)secn–2(x)tdx v = t
Using integration by parts we get the formula:
Set
u = secn–2(x) dv = sec2(x)dx

66. Formulas From Integration by Parts
For the antiderivatives of tn, use change of
variable via the formula t2 = sec2(x) – 1.

67. Formulas From Integration by Parts
For the antiderivatives of tn, use change of
variable via the formula t2 = sec2(x) – 1.
= ∫tn–2 t2 dx
∫tndx

68. Formulas From Integration by Parts
For the antiderivatives of tn, use change of
variable via the formula t2 = sec2(x) – 1.
= ∫tn–2 t2 dx
= ∫tn–2(sec2(x) –1)dx
∫tndx

69. Formulas From Integration by Parts
For the antiderivatives of tn, use change of
variable via the formula t2 = sec2(x) – 1.
= ∫tn–2 t2 dx
= ∫tn–2(sec2(x) –1)dx
∫tndx
= ∫ tn–2sec2(x) – tn–2dx

70. Formulas From Integration by Parts
For the antiderivatives of tn, use change of
variable via the formula t2 = sec2(x) – 1.
= ∫tn–2 t2 dx
= ∫tn–2(sec2(x) –1)dx
∫tndx
= ∫ tn–2sec2(x) – tn–2dx
Using substitution

71. Formulas From Integration by Parts
For the antiderivatives of tn, use change of
variable via the formula t2 = sec2(x) – 1.
= ∫tn–2 t2 dx
= ∫tn–2(sec2(x) –1)dx
∫tndx
= ∫ tn–2sec2(x) – tn–2dx
Using substitution
Set u = tan(x)

72. Formulas From Integration by Parts
For the antiderivatives of tn, use change of
variable via the formula t2 = sec2(x) – 1.
= ∫tn–2 t2 dx
= ∫tn–2(sec2(x) –1)dx
∫tndx
= ∫ tn–2sec2(x) – tn–2dx
Using substitution
Set u = tan(x) then du = sec2(x)dx

73. Formulas From Integration by Parts
For the antiderivatives of tn, use change of
variable via the formula t2 = sec2(x) – 1.
= ∫tn–2 t2 dx
= ∫tn–2(sec2(x) –1)dx
∫tndx
= ∫ tn–2sec2(x) – tn–2dx
Using substitution
Set u = tan(x) then du = sec2(x)dx
∫ tn–2 sec2(x) dx

74. Formulas From Integration by Parts
For the antiderivatives of tn, use change of
variable via the formula t2 = sec2(x) – 1.
= ∫tn–2 t2 dx
= ∫tn–2(sec2(x) –1)dx
∫tndx
= ∫ tn–2sec2(x) – tn–2dx
Using substitution
Set u = tan(x) then du = sec2(x)dx
∫ tn–2 sec2(x) dx = ∫ un–2 du =

75. Formulas From Integration by Parts
For the antiderivatives of tn, use change of
variable via the formula t2 = sec2(x) – 1.
= ∫tn–2 t2 dx
= ∫tn–2(sec2(x) –1)dx
∫tndx
= ∫ tn–2sec2(x) – tn–2dx
Using substitution
Set u = tan(x) then du = sec2(x)dx
∫ tn–2 sec2(x) dx = ∫ un–2 du = n – 1
un–1

76. Formulas From Integration by Parts
For the antiderivatives of tn, use change of
variable via the formula t2 = sec2(x) – 1.
= ∫tn–2 t2 dx
= ∫tn–2(sec2(x) –1)dx
∫tndx
= ∫ tn–2sec2(x) – tn–2dx
Using substitution
Set u = tan(x) then du = sec2(x)dx
∫ tn–2 sec2(x) dx = ∫ un–2 du = n – 1
un–1
= n – 1
tn–1

77. Formulas From Integration by Parts
For the antiderivatives of tn, use change of
variable via the formula t2 = sec2(x) – 1.
= ∫tn–2 t2 dx
= ∫tn–2(sec2(x) –1)dx
∫tndx
= ∫ tn–2sec2(x) – tn–2dx
Using substitution
Set u = tan(x) then du = sec2(x)dx
∫ tn–2 sec2(x) dx = ∫ un–2 du = n – 1
un–1
= n – 1
tn–1
So ∫tndx = – ∫tn–2dx
n – 1
tn–1

78. Formulas From Integration by Parts
∫sndx = + ∫sn–2dx
n
–sn–1c
n
n–1
We summarize the formulas below.
∫secn(x)dx = + ∫secn–2(x)dxn – 1
secn–2(x)tan(x)
n–1
n–2
∫cndx = + ∫cn–2dx
n
cn–1s
n
n–1
From integration by parts, we get:

79. Formulas From Integration by Parts
∫sndx = + ∫sn–2dx
n
–sn–1c
n
n–1
We summarize the formulas below.
∫tndx = – ∫tn–2dx
n – 1
tn–1
∫secn(x)dx = + ∫secn–2(x)dxn – 1
secn–2(x)tan(x)
n–1
n–2
∫cndx = + ∫cn–2dx
n
cn–1s
n
n–1
From integration by parts, we get:
From change of variable, we get:

80. Formulas From Integration by Parts
Exercise.
∫cscn(x)dx2. Find by integration by parts.
1. Finish the algebra for the antiderivatives
of sn, secn(x).
∫cotn(x)dx3. Find by change of variable.