The document discusses the LIATE rule for integration by parts. LIATE is an acronym that indicates the order of priority for choosing the u(x) function: logarithmic, inverse trigonometric, algebraic, trigonometric, exponential. Two examples are provided to illustrate applying the LIATE rule. In the first example, choosing u(x)=x for x sin x dx works according to LIATE. In the second example, choosing u(x)=sin x for ex sin x dx works according to LIATE after solving using integration by parts twice.

3. The LIATE Rule

4. The LIATE Rule
The difficulty of integration by parts is in choosing u(x) and v (x)
correctly.

5. The LIATE Rule
The difficulty of integration by parts is in choosing u(x) and v (x)
correctly.
The LIATE rule is a rule of thumb that tells you which function
you should choose as u(x):

6. The LIATE Rule
The difficulty of integration by parts is in choosing u(x) and v (x)
correctly.
The LIATE rule is a rule of thumb that tells you which function
you should choose as u(x):
LIATE

7. The LIATE Rule
The difficulty of integration by parts is in choosing u(x) and v (x)
correctly.
The LIATE rule is a rule of thumb that tells you which function
you should choose as u(x):
L
Log
IATE

8. The LIATE Rule
The difficulty of integration by parts is in choosing u(x) and v (x)
correctly.
The LIATE rule is a rule of thumb that tells you which function
you should choose as u(x):
L I
Inv. Trig
ATE

9. The LIATE Rule
The difficulty of integration by parts is in choosing u(x) and v (x)
correctly.
The LIATE rule is a rule of thumb that tells you which function
you should choose as u(x):
LI A
Algebraic
TE

10. The LIATE Rule
The difficulty of integration by parts is in choosing u(x) and v (x)
correctly.
The LIATE rule is a rule of thumb that tells you which function
you should choose as u(x):
LIA T
Trig
E

11. The LIATE Rule
The difficulty of integration by parts is in choosing u(x) and v (x)
correctly.
The LIATE rule is a rule of thumb that tells you which function
you should choose as u(x):
LIAT E
Exp

12. The LIATE Rule
The difficulty of integration by parts is in choosing u(x) and v (x)
correctly.
The LIATE rule is a rule of thumb that tells you which function
you should choose as u(x):
LIATE

13. The LIATE Rule
The difficulty of integration by parts is in choosing u(x) and v (x)
correctly.
The LIATE rule is a rule of thumb that tells you which function
you should choose as u(x):
LIATE
The word itself tells you in which order of priority you should use
u(x).

14. Example 1

15. Example 1
x sin xdx

16. Example 1
x sin xdx
Here we have an algebraic and a trigonometric function:

17. Example 1
x sin xdx
Here we have an algebraic and a trigonometric function:
x sin xdx

18. Example 1
x sin xdx
Here we have an algebraic and a trigonometric function:
A
x sin xdx

19. Example 1
x sin xdx
Here we have an algebraic and a trigonometric function:
x$$$XT
sin xdx

20. Example 1
x sin xdx
Here we have an algebraic and a trigonometric function:
x sin xdx

21. Example 1
x sin xdx
Here we have an algebraic and a trigonometric function:
x sin xdx
According to LIATE, the A comes before the T, so we choose
u(x) = x.

22. Example 2

23. Example 2
ex
sin xdx

24. Example 2
ex
sin xdx
In this case we have E and T:

25. Example 2
ex
sin xdx
In this case we have E and T:
b
E
ex
sin xdx

26. Example 2
ex
sin xdx
In this case we have E and T:
ex$$$XT
sin xdx

27. Example 2
ex
sin xdx
In this case we have E and T:
ex
sin xdx

28. Example 2
ex
sin xdx
In this case we have E and T:
ex
sin xdx
According to LIATE, the T comes before the E, so we choose
u(x) = sin x.

29. Example 2
ex
sin xdx
In this case we have E and T:
ex
sin xdx
According to LIATE, the T comes before the E, so we choose
u(x) = sin x.
Let’s solve this integral to show that this works:

30. Example 2
ex
sin xdx
In this case we have E and T:
ex
sin xdx
According to LIATE, the T comes before the E, so we choose
u(x) = sin x.
Let’s solve this integral to show that this works:
u(x) = sin x

31. Example 2
ex
sin xdx
In this case we have E and T:
ex
sin xdx
According to LIATE, the T comes before the E, so we choose
u(x) = sin x.
Let’s solve this integral to show that this works:
u(x) = sin x v (x) = ex

32. Example 2
ex
sin xdx
In this case we have E and T:
ex
sin xdx
According to LIATE, the T comes before the E, so we choose
u(x) = sin x.
Let’s solve this integral to show that this works:
u(x) = sin x v (x) = ex
u (x) = cos x

33. Example 2
ex
sin xdx
In this case we have E and T:
ex
sin xdx
According to LIATE, the T comes before the E, so we choose
u(x) = sin x.
Let’s solve this integral to show that this works:
u(x) = sin x v (x) = ex
u (x) = cos x v(x) = ex

34. Example 2
ex
sin xdx
In this case we have E and T:
ex
sin xdx
According to LIATE, the T comes before the E, so we choose
u(x) = sin x.
Let’s solve this integral to show that this works:
u(x) = sin x v (x) = ex
u (x) = cos x v(x) = ex
ex
sin xdx = ex
sin x − ex
cos xdx

35. Example 2
ex
sin xdx
In this case we have E and T:
ex
sin xdx
According to LIATE, the T comes before the E, so we choose
u(x) = sin x.
Let’s solve this integral to show that this works:
u(x) = sin x v (x) = ex
u (x) = cos x v(x) = ex
b
v (x)
ex
sin xdx = ex
sin x − ex
cos xdx

36. Example 2
ex
sin xdx
In this case we have E and T:
ex
sin xdx
According to LIATE, the T comes before the E, so we choose
u(x) = sin x.
Let’s solve this integral to show that this works:
u(x) = sin x v (x) = ex
u (x) = cos x v(x) = ex
ex$$$Xu(x)
sin xdx = ex
sin x − ex
cos xdx

37. Example 2
ex
sin xdx
In this case we have E and T:
ex
sin xdx
According to LIATE, the T comes before the E, so we choose
u(x) = sin x.
Let’s solve this integral to show that this works:
u(x) = sin x v (x) = ex
u (x) = cos x v(x) = ex
ex
sin xdx = b
v(x)
ex
sin x − ex
cos xdx

38. Example 2
ex
sin xdx
In this case we have E and T:
ex
sin xdx
According to LIATE, the T comes before the E, so we choose
u(x) = sin x.
Let’s solve this integral to show that this works:
u(x) = sin x v (x) = ex
u (x) = cos x v(x) = ex
ex
sin xdx = ex$$$Xu(x)
sin x − ex
cos xdx

39. Example 2
ex
sin xdx
In this case we have E and T:
ex
sin xdx
According to LIATE, the T comes before the E, so we choose
u(x) = sin x.
Let’s solve this integral to show that this works:
u(x) = sin x v (x) = ex
u (x) = cos x v(x) = ex
ex
sin xdx = ex
sin x − b
v(x)
ex
cos xdx

40. Example 2
ex
sin xdx
In this case we have E and T:
ex
sin xdx
According to LIATE, the T comes before the E, so we choose
u(x) = sin x.
Let’s solve this integral to show that this works:
u(x) = sin x v (x) = ex
u (x) = cos x v(x) = ex
ex
sin xdx = ex
sin x − ex
$$$Xu (x)
cos xdx

41. Example 2
ex
sin xdx
In this case we have E and T:
ex
sin xdx
According to LIATE, the T comes before the E, so we choose
u(x) = sin x.
Let’s solve this integral to show that this works:
u(x) = sin x v (x) = ex
u (x) = cos x v(x) = ex
ex
sin xdx = ex
sin x − ex
cos xdx

42. Example 2

43. Example 2
Now we need to solve that integral:

44. Example 2
Now we need to solve that integral:
ex
sin xdx = ex
sin x − ex
cos xdx

45. Example 2
Now we need to solve that integral:
ex
sin xdx = ex
sin x − ex
cos xdx
We need to use integration by parts again.

46. Example 2
Now we need to solve that integral:
ex
sin xdx = ex
sin x − ex
cos xdx
We need to use integration by parts again.
By using LIATE, we choose u(x) = cos x:

47. Example 2
Now we need to solve that integral:
ex
sin xdx = ex
sin x − ex
cos xdx
We need to use integration by parts again.
By using LIATE, we choose u(x) = cos x:
u(x) = cos x

48. Example 2
Now we need to solve that integral:
ex
sin xdx = ex
sin x − ex
cos xdx
We need to use integration by parts again.
By using LIATE, we choose u(x) = cos x:
u(x) = cos x v (x) = ex

49. Example 2
Now we need to solve that integral:
ex
sin xdx = ex
sin x − ex
cos xdx
We need to use integration by parts again.
By using LIATE, we choose u(x) = cos x:
u(x) = cos x v (x) = ex
u (x) = − sin x

50. Example 2
Now we need to solve that integral:
ex
sin xdx = ex
sin x − ex
cos xdx
We need to use integration by parts again.
By using LIATE, we choose u(x) = cos x:
u(x) = cos x v (x) = ex
u (x) = − sin x v(x) = ex

51. Example 2
Now we need to solve that integral:
ex
sin xdx = ex
sin x − ex
cos xdx
We need to use integration by parts again.
By using LIATE, we choose u(x) = cos x:
u(x) = cos x v (x) = ex
u (x) = − sin x v(x) = ex
ex
sin xdx = ex
sin x − ex
cos xdx

52. Example 2
Now we need to solve that integral:
ex
sin xdx = ex
sin x − ex
cos xdx
We need to use integration by parts again.
By using LIATE, we choose u(x) = cos x:
u(x) = cos x v (x) = ex
u (x) = − sin x v(x) = ex
ex
sin xdx = ex
sin x − b
v (x)
ex
cos xdx

53. Example 2
Now we need to solve that integral:
ex
sin xdx = ex
sin x − ex
cos xdx
We need to use integration by parts again.
By using LIATE, we choose u(x) = cos x:
u(x) = cos x v (x) = ex
u (x) = − sin x v(x) = ex
ex
sin xdx = ex
sin x − ex
$$$Xu(x)
cos xdx

54. Example 2
Now we need to solve that integral:
ex
sin xdx = ex
sin x − ex
cos xdx
We need to use integration by parts again.
By using LIATE, we choose u(x) = cos x:
u(x) = cos x v (x) = ex
u (x) = − sin x v(x) = ex
ex
sin xdx = ex
sin x − ex
cos xdx

55. Example 2
Now we need to solve that integral:
ex
sin xdx = ex
sin x − ex
cos xdx
We need to use integration by parts again.
By using LIATE, we choose u(x) = cos x:
u(x) = cos x v (x) = ex
u (x) = − sin x v(x) = ex
ex
sin xdx = ex
sin x − ex
cos xdx
= ex
sin x − ex
cos x − ex
(− sin x) dx

56. Example 2
Now we need to solve that integral:
ex
sin xdx = ex
sin x − ex
cos xdx
We need to use integration by parts again.
By using LIATE, we choose u(x) = cos x:
u(x) = cos x v (x) = ex
u (x) = − sin x v(x) = ex
ex
sin xdx = ex
sin x − ex
cos xdx
= ex
sin x − b
v(x)
ex
cos x − ex
(− sin x) dx

57. Example 2
Now we need to solve that integral:
ex
sin xdx = ex
sin x − ex
cos xdx
We need to use integration by parts again.
By using LIATE, we choose u(x) = cos x:
u(x) = cos x v (x) = ex
u (x) = − sin x v(x) = ex
ex
sin xdx = ex
sin x − ex
cos xdx
= ex
sin x − ex
$$$Xu(x)
cos x − ex
(− sin x) dx

58. Example 2
Now we need to solve that integral:
ex
sin xdx = ex
sin x − ex
cos xdx
We need to use integration by parts again.
By using LIATE, we choose u(x) = cos x:
u(x) = cos x v (x) = ex
u (x) = − sin x v(x) = ex
ex
sin xdx = ex
sin x − ex
cos xdx
= ex
sin x − ex
cos x − b
v(x)
ex
(− sin x) dx

59. Example 2
Now we need to solve that integral:
ex
sin xdx = ex
sin x − ex
cos xdx
We need to use integration by parts again.
By using LIATE, we choose u(x) = cos x:
u(x) = cos x v (x) = ex
u (x) = − sin x v(x) = ex
ex
sin xdx = ex
sin x − ex
cos xdx
= ex
sin x − ex
cos x − ex
$$$$$Xu (x)
(− sin x)dx

60. Example 2
Now we need to solve that integral:
ex
sin xdx = ex
sin x − ex
cos xdx
We need to use integration by parts again.
By using LIATE, we choose u(x) = cos x:
u(x) = cos x v (x) = ex
u (x) = − sin x v(x) = ex
ex
sin xdx = ex
sin x − ex
cos xdx
= ex
sin x − ex
cos x − ex
(− sin x) dx

61. Example 2

62. Example 2
So, we have that:

63. Example 2
So, we have that:
ex
sin xdx = ex
sin x − ex
cos x − ex
sin xdx

64. Example 2
So, we have that:
ex
sin xdx
!!
= ex
sin x − ex
cos x − ex
sin xdx
!!

65. Example 2
So, we have that:
ex
sin xdx
!!
= ex
sin x − ex
cos x − ex
sin xdx
!!
So, we add ex sin xdx to both sides of the equation:

66. Example 2
So, we have that:
ex
sin xdx
!!
= ex
sin x − ex
cos x − ex
sin xdx
!!
So, we add ex sin xdx to both sides of the equation:
2 ex
sin xdx = ex
(sin x − cos x)

67. Example 2
So, we have that:
ex
sin xdx
!!
= ex
sin x − ex
cos x − ex
sin xdx
!!
So, we add ex sin xdx to both sides of the equation:
2 ex
sin xdx = ex
(sin x − cos x)
ex
sin xdx =
ex
2
(sin x − cos x) + C

68. Example 2
So, we have that:
ex
sin xdx
!!
= ex
sin x − ex
cos x − ex
sin xdx
!!
So, we add ex sin xdx to both sides of the equation:
2 ex
sin xdx = ex
(sin x − cos x)
ex
sin xdx =
ex
2
(sin x − cos x) + C
This is in fact an example of Trick number 2.

69. Example 3

70. Example 3
Let’s do another example:

71. Example 3
Let’s do another example:
x2
ex
dx

72. Example 3
Let’s do another example:
x2
ex
dx
We have an A and an E:

73. Example 3
Let’s do another example:
x2
ex
dx
We have an A and an E:
A
x2
ex
dx

74. Example 3
Let’s do another example:
x2
ex
dx
We have an A and an E:
x2
b
E
ex
dx

75. Example 3
Let’s do another example:
x2
ex
dx
We have an A and an E:
x2
ex
dx

76. Example 3
Let’s do another example:
x2
ex
dx
We have an A and an E:
x2
ex
dx
LIATE says that the A comes before the E. So, we choose
u(x) = x2:

77. Example 3
Let’s do another example:
x2
ex
dx
We have an A and an E:
x2
ex
dx
LIATE says that the A comes before the E. So, we choose
u(x) = x2:
u(x) = x2

78. Example 3
Let’s do another example:
x2
ex
dx
We have an A and an E:
x2
ex
dx
LIATE says that the A comes before the E. So, we choose
u(x) = x2:
u(x) = x2 v (x) = ex

79. Example 3
Let’s do another example:
x2
ex
dx
We have an A and an E:
x2
ex
dx
LIATE says that the A comes before the E. So, we choose
u(x) = x2:
u(x) = x2 v (x) = ex
u (x) = 2x

80. Example 3
Let’s do another example:
x2
ex
dx
We have an A and an E:
x2
ex
dx
LIATE says that the A comes before the E. So, we choose
u(x) = x2:
u(x) = x2 v (x) = ex
u (x) = 2x v(x) = ex

81. Example 3
Let’s do another example:
x2
ex
dx
We have an A and an E:
x2
ex
dx
LIATE says that the A comes before the E. So, we choose
u(x) = x2:
u(x) = x2 v (x) = ex
u (x) = 2x v(x) = ex
x2
ex
dx =

82. Example 3
Let’s do another example:
x2
ex
dx
We have an A and an E:
x2
ex
dx
LIATE says that the A comes before the E. So, we choose
u(x) = x2:
u(x) = x2 v (x) = ex
u (x) = 2x v(x) = ex
u(x)
x2
ex
dx =

83. Example 3
Let’s do another example:
x2
ex
dx
We have an A and an E:
x2
ex
dx
LIATE says that the A comes before the E. So, we choose
u(x) = x2:
u(x) = x2 v (x) = ex
u (x) = 2x v(x) = ex
x2
b
v (x)
ex
dx =

84. Example 3
Let’s do another example:
x2
ex
dx
We have an A and an E:
x2
ex
dx
LIATE says that the A comes before the E. So, we choose
u(x) = x2:
u(x) = x2 v (x) = ex
u (x) = 2x v(x) = ex
x2
ex
dx = x2
ex
− 2xex
dx

85. Example 3
Let’s do another example:
x2
ex
dx
We have an A and an E:
x2
ex
dx
LIATE says that the A comes before the E. So, we choose
u(x) = x2:
u(x) = x2 v (x) = ex
u (x) = 2x v(x) = ex
x2
ex
dx =
u(x)
x2
ex
− 2xex
dx

86. Example 3
Let’s do another example:
x2
ex
dx
We have an A and an E:
x2
ex
dx
LIATE says that the A comes before the E. So, we choose
u(x) = x2:
u(x) = x2 v (x) = ex
u (x) = 2x v(x) = ex
x2
ex
dx = x2
b
v(x)
ex
− 2xex
dx

87. Example 3
Let’s do another example:
x2
ex
dx
We have an A and an E:
x2
ex
dx
LIATE says that the A comes before the E. So, we choose
u(x) = x2:
u(x) = x2 v (x) = ex
u (x) = 2x v(x) = ex
x2
ex
dx = x2
ex
− b
u (x)
2xex
dx

88. Example 3
Let’s do another example:
x2
ex
dx
We have an A and an E:
x2
ex
dx
LIATE says that the A comes before the E. So, we choose
u(x) = x2:
u(x) = x2 v (x) = ex
u (x) = 2x v(x) = ex
x2
ex
dx = x2
ex
− 2xb
v(x)
ex
dx

89. Example 3
Let’s do another example:
x2
ex
dx
We have an A and an E:
x2
ex
dx
LIATE says that the A comes before the E. So, we choose
u(x) = x2:
u(x) = x2 v (x) = ex
u (x) = 2x v(x) = ex
x2
ex
dx = x2
ex
− 2xex
dx

90. Example 3

91. Example 3
x2
ex
dx = x2
ex
− 2 xex
dx

92. Example 3
x2
ex
dx = x2
ex
− 2 xex
dx
Now we need to solve that integral by integration by parts again:

93. Example 3
x2
ex
dx = x2
ex
− 2 xex
dx
Now we need to solve that integral by integration by parts again:
u(x) = x

94. Example 3
x2
ex
dx = x2
ex
− 2 xex
dx
Now we need to solve that integral by integration by parts again:
u(x) = x v (x) = ex

95. Example 3
x2
ex
dx = x2
ex
− 2 xex
dx
Now we need to solve that integral by integration by parts again:
u(x) = x v (x) = ex
u (x) = 1

96. Example 3
x2
ex
dx = x2
ex
− 2 xex
dx
Now we need to solve that integral by integration by parts again:
u(x) = x v (x) = ex
u (x) = 1 v(x) = ex

97. Example 3
x2
ex
dx = x2
ex
− 2 xex
dx
Now we need to solve that integral by integration by parts again:
u(x) = x v (x) = ex
u (x) = 1 v(x) = ex
x2
ex
dx = x2
ex
− 2 xex
− 1.ex
dx

98. Example 3
x2
ex
dx = x2
ex
− 2 xex
dx
Now we need to solve that integral by integration by parts again:
u(x) = x v (x) = ex
u (x) = 1 v(x) = ex
x2
ex
dx = x2
ex
− 2 xex
− 1.ex
dx
= x2
ex
− 2xex
+ 2ex
= ex
x2
− 2x + 2

99. Example 3
x2
ex
dx = x2
ex
− 2 xex
dx
Now we need to solve that integral by integration by parts again:
u(x) = x v (x) = ex
u (x) = 1 v(x) = ex
x2
ex
dx = x2
ex
− 2 xex
− 1.ex
dx
= x2
ex
− 2xex
+ 2ex
= ex
x2
− 2x + 2