Integration by Parts, Part 3

In this video we talk about the LIATE rule, this is the secret for choosing u and v' correctly. Then we learn another useful trick for integration by parts.

Watch video: http://www.youtube.com/edit?ns=1&video_id=1SEUGvHTcQ0

For more lessons: http://www.intuitive-calculus.com/integration-by-parts.html

3. The LIATE Rule

4. The LIATE Rule
The diculty of integration by parts is in choosingu(x) andv
0
(x)
correctly.

5. The LIATE Rule
The diculty of integration by parts is in choosingu(x) andv
0
(x)
correctly.
The LIATE rule is a rule of thumb that tells you which function
you should choose asu(x):

6. The LIATE Rule
The diculty of integration by parts is in choosingu(x) andv
0
(x)
correctly.
The LIATE rule is a rule of thumb that tells you which function
you should choose asu(x):
LIATE

7. The LIATE Rule
The diculty of integration by parts is in choosingu(x) andv
0
(x)
correctly.
The LIATE rule is a rule of thumb that tells you which function
you should choose asu(x):
L|{z}
Log
IATE

8. The LIATE Rule
The diculty of integration by parts is in choosingu(x) andv
0
(x)
correctly.
The LIATE rule is a rule of thumb that tells you which function
you should choose asu(x):
LI|{z}
Inv. Trig
ATE

9. The LIATE Rule
The diculty of integration by parts is in choosingu(x) andv
0
(x)
correctly.
The LIATE rule is a rule of thumb that tells you which function
you should choose asu(x):
LIA|{z}
Algebraic
TE

10. The LIATE Rule
The diculty of integration by parts is in choosingu(x) andv
0
(x)
correctly.
The LIATE rule is a rule of thumb that tells you which function
you should choose asu(x):
LIAT|{z}
Trig
E

11. The LIATE Rule
The diculty of integration by parts is in choosingu(x) andv
0
(x)
correctly.
The LIATE rule is a rule of thumb that tells you which function
you should choose asu(x):
LIATE|{z}
Exp

12. The LIATE Rule
The diculty of integration by parts is in choosingu(x) andv
0
(x)
correctly.
The LIATE rule is a rule of thumb that tells you which function
you should choose asu(x):
LIATE

13. The LIATE Rule
The diculty of integration by parts is in choosingu(x) andv
0
(x)
correctly.
The LIATE rule is a rule of thumb that tells you which function
you should choose asu(x):
LIATE
The word itself tells you in which order of priority you should use
u(x).

14. Example 1

15. Example 1
Z
xsinxdx

16. Example 1
Z
xsinxdx
Here we have an algebraic and a trigonometric function:

17. Example 1
Z
xsinxdx
Here we have an algebraic and a trigonometric function:
Z
xsinxdx

18. Example 1
Z
xsinxdx
Here we have an algebraic and a trigonometric function:
Z

A
xsinxdx

19. Example 1
Z
xsinxdx
Here we have an algebraic and a trigonometric function:
Z
x
:T
sinxdx

20. Example 1
Z
xsinxdx
Here we have an algebraic and a trigonometric function:
Z
xsinxdx

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

22. Example 2

23. Example 2
Z
e
x
sinxdx

24. Example 2
Z
e
x
sinxdx
In this case we have E and T:

25. Example 2
Z
e
x
sinxdx
In this case we have E and T:
Z

>
E
e
x
sinxdx

26. Example 2
Z
e
x
sinxdx
In this case we have E and T:
Z
e
x

:T
sinxdx

27. Example 2
Z
e
x
sinxdx
In this case we have E and T:
Z
e
x
sinxdx

28. Example 2
Z
e
x
sinxdx
In this case we have E and T:
Z
e
x
sinxdx
According to LIATE, the T comes before the E, so we choose
u(x) = sinx.

29. Example 2
Z
e
x
sinxdx
In this case we have E and T:
Z
e
x
sinxdx
According to LIATE, the T comes before the E, so we choose
u(x) = sinx.
Let's solve this integral to show that this works:

30. Example 2
Z
e
x
sinxdx
In this case we have E and T:
Z
e
x
sinxdx
According to LIATE, the T comes before the E, so we choose
u(x) = sinx.
Let's solve this integral to show that this works:
u(x) = sinx

31. Example 2
Z
e
x
sinxdx
In this case we have E and T:
Z
e
x
sinxdx
According to LIATE, the T comes before the E, so we choose
u(x) = sinx.
Let's solve this integral to show that this works:
u(x) = sinx v
0
(x) =e
x

32. Example 2
Z
e
x
sinxdx
In this case we have E and T:
Z
e
x
sinxdx
According to LIATE, the T comes before the E, so we choose
u(x) = sinx.
Let's solve this integral to show that this works:
u(x) = sinx v
0
(x) =e
x
u
0
(x) = cosx

33. Example 2
Z
e
x
sinxdx
In this case we have E and T:
Z
e
x
sinxdx
According to LIATE, the T comes before the E, so we choose
u(x) = sinx.
Let's solve this integral to show that this works:
u(x) = sinx v
0
(x) =e
x
u
0
(x) = cosx v(x) =e
x

34. Example 2
Z
e
x
sinxdx
In this case we have E and T:
Z
e
x
sinxdx
According to LIATE, the T comes before the E, so we choose
u(x) = sinx.
Let's solve this integral to show that this works:
u(x) = sinx v
0
(x) =e
x
u
0
(x) = cosx v(x) =e
x
Z
e
x
sinxdx=e
x
sinx
Z
e
x
cosxdx

35. Example 2
Z
e
x
sinxdx
In this case we have E and T:
Z
e
x
sinxdx
According to LIATE, the T comes before the E, so we choose
u(x) = sinx.
Let's solve this integral to show that this works:
u(x) = sinx v
0
(x) =e
x
u
0
(x) = cosx v(x) =e
x
Z

>
v
0
(x)
e
x
sinxdx=e
x
sinx
Z
e
x
cosxdx

36. Example 2
Z
e
x
sinxdx
In this case we have E and T:
Z
e
x
sinxdx
According to LIATE, the T comes before the E, so we choose
u(x) = sinx.
Let's solve this integral to show that this works:
u(x) = sinx v
0
(x) =e
x
u
0
(x) = cosx v(x) =e
x
Z
e
x

:
u(x)
sinxdx=e
x
sinx
Z
e
x
cosxdx

37. Example 2
Z
e
x
sinxdx
In this case we have E and T:
Z
e
x
sinxdx
According to LIATE, the T comes before the E, so we choose
u(x) = sinx.
Let's solve this integral to show that this works:
u(x) = sinx v
0
(x) =e
x
u
0
(x) = cosx v(x) =e
x
Z
e
x
sinxdx=
>
v(x)
e
x
sinx
Z
e
x
cosxdx

38. Example 2
Z
e
x
sinxdx
In this case we have E and T:
Z
e
x
sinxdx
According to LIATE, the T comes before the E, so we choose
u(x) = sinx.
Let's solve this integral to show that this works:
u(x) = sinx v
0
(x) =e
x
u
0
(x) = cosx v(x) =e
x
Z
e
x
sinxdx=e
x

:
u(x)
sinx
Z
e
x
cosxdx

39. Example 2
Z
e
x
sinxdx
In this case we have E and T:
Z
e
x
sinxdx
According to LIATE, the T comes before the E, so we choose
u(x) = sinx.
Let's solve this integral to show that this works:
u(x) = sinx v
0
(x) =e
x
u
0
(x) = cosx v(x) =e
x
Z
e
x
sinxdx=e
x
sinx
Z

>
v(x)
e
x
cosxdx

40. Example 2
Z
e
x
sinxdx
In this case we have E and T:
Z
e
x
sinxdx
According to LIATE, the T comes before the E, so we choose
u(x) = sinx.
Let's solve this integral to show that this works:
u(x) = sinx v
0
(x) =e
x
u
0
(x) = cosx v(x) =e
x
Z
e
x
sinxdx=e
x
sinx
Z
e
x

:
u
0
(x)
cosxdx

41. Example 2
Z
e
x
sinxdx
In this case we have E and T:
Z
e
x
sinxdx
According to LIATE, the T comes before the E, so we choose
u(x) = sinx.
Let's solve this integral to show that this works:
u(x) = sinx v
0
(x) =e
x
u
0
(x) = cosx v(x) =e
x
Z
e
x
sinxdx=e
x
sinx
Z
e
x
cosxdx

42. Example 2

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

44. Example 2
Now we need to solve that integral:
Z
e
x
sinxdx=e
x
sinx
Z
e
x
cosxdx

45. Example 2
Now we need to solve that integral:
Z
e
x
sinxdx=e
x
sinx
Z
e
x
cosxdx
We need to use integration by parts again.

46. Example 2
Now we need to solve that integral:
Z
e
x
sinxdx=e
x
sinx
Z
e
x
cosxdx
We need to use integration by parts again.
By using LIATE, we chooseu(x) = cosx:

47. Example 2
Now we need to solve that integral:
Z
e
x
sinxdx=e
x
sinx
Z
e
x
cosxdx
We need to use integration by parts again.
By using LIATE, we chooseu(x) = cosx:
u(x) = cosx

48. Example 2
Now we need to solve that integral:
Z
e
x
sinxdx=e
x
sinx
Z
e
x
cosxdx
We need to use integration by parts again.
By using LIATE, we chooseu(x) = cosx:
u(x) = cosx v
0
(x) =e
x

49. Example 2
Now we need to solve that integral:
Z
e
x
sinxdx=e
x
sinx
Z
e
x
cosxdx
We need to use integration by parts again.
By using LIATE, we chooseu(x) = cosx:
u(x) = cosx v
0
(x) =e
x
u
0
(x) =sinx

50. Example 2
Now we need to solve that integral:
Z
e
x
sinxdx=e
x
sinx
Z
e
x
cosxdx
We need to use integration by parts again.
By using LIATE, we chooseu(x) = cosx:
u(x) = cosx v
0
(x) =e
x
u
0
(x) =sinx v(x) =e
x

51. Example 2
Now we need to solve that integral:
Z
e
x
sinxdx=e
x
sinx
Z
e
x
cosxdx
We need to use integration by parts again.
By using LIATE, we chooseu(x) = cosx:
u(x) = cosx v
0
(x) =e
x
u
0
(x) =sinx v(x) =e
x
Z
e
x
sinxdx=e
x
sinx
Z
e
x
cosxdx

52. Example 2
Now we need to solve that integral:
Z
e
x
sinxdx=e
x
sinx
Z
e
x
cosxdx
We need to use integration by parts again.
By using LIATE, we chooseu(x) = cosx:
u(x) = cosx v
0
(x) =e
x
u
0
(x) =sinx v(x) =e
x
Z
e
x
sinxdx=e
x
sinx
Z

>
v
0
(x)
e
x
cosxdx

53. Example 2
Now we need to solve that integral:
Z
e
x
sinxdx=e
x
sinx
Z
e
x
cosxdx
We need to use integration by parts again.
By using LIATE, we chooseu(x) = cosx:
u(x) = cosx v
0
(x) =e
x
u
0
(x) =sinx v(x) =e
x
Z
e
x
sinxdx=e
x
sinx
Z
e
x
:
u(x)
cosxdx

54. Example 2
Now we need to solve that integral:
Z
e
x
sinxdx=e
x
sinx
Z
e
x
cosxdx
We need to use integration by parts again.
By using LIATE, we chooseu(x) = cosx:
u(x) = cosx v
0
(x) =e
x
u
0
(x) =sinx v(x) =e
x
Z
e
x
sinxdx=e
x
sinx
Z
e
x
cosxdx

55. Example 2
Now we need to solve that integral:
Z
e
x
sinxdx=e
x
sinx
Z
e
x
cosxdx
We need to use integration by parts again.
By using LIATE, we chooseu(x) = cosx:
u(x) = cosx v
0
(x) =e
x
u
0
(x) =sinx v(x) =e
x
Z
e
x
sinxdx=e
x
sinx
Z
e
x
cosxdx
=e
x
sinx

e
x
cosx
Z
e
x
(sinx)dx

56. Example 2
Now we need to solve that integral:
Z
e
x
sinxdx=e
x
sinx
Z
e
x
cosxdx
We need to use integration by parts again.
By using LIATE, we chooseu(x) = cosx:
u(x) = cosx v
0
(x) =e
x
u
0
(x) =sinx v(x) =e
x
Z
e
x
sinxdx=e
x
sinx
Z
e
x
cosxdx
=e
x
sinx
"

>
v(x)
e
x
cosx
Z
e
x
(sinx)dx
#

57. Example 2
Now we need to solve that integral:
Z
e
x
sinxdx=e
x
sinx
Z
e
x
cosxdx
We need to use integration by parts again.
By using LIATE, we chooseu(x) = cosx:
u(x) = cosx v
0
(x) =e
x
u
0
(x) =sinx v(x) =e
x
Z
e
x
sinxdx=e
x
sinx
Z
e
x
cosxdx
=e
x
sinx

e
x
:
u(x)
cosx
Z
e
x
(sinx)dx

58. Example 2
Now we need to solve that integral:
Z
e
x
sinxdx=e
x
sinx
Z
e
x
cosxdx
We need to use integration by parts again.
By using LIATE, we chooseu(x) = cosx:
u(x) = cosx v
0
(x) =e
x
u
0
(x) =sinx v(x) =e
x
Z
e
x
sinxdx=e
x
sinx
Z
e
x
cosxdx
=e
x
sinx
"
e
x
cosx
Z

>
v(x)
e
x
(sinx)dx
#

59. Example 2
Now we need to solve that integral:
Z
e
x
sinxdx=e
x
sinx
Z
e
x
cosxdx
We need to use integration by parts again.
By using LIATE, we chooseu(x) = cosx:
u(x) = cosx v
0
(x) =e
x
u
0
(x) =sinx v(x) =e
x
Z
e
x
sinxdx=e
x
sinx
Z
e
x
cosxdx
=e
x
sinx
"
e
x
cosx
Z
e
x


:
u
0
(x)
(sinx)dx
#

60. Example 2
Now we need to solve that integral:
Z
e
x
sinxdx=e
x
sinx
Z
e
x
cosxdx
We need to use integration by parts again.
By using LIATE, we chooseu(x) = cosx:
u(x) = cosx v
0
(x) =e
x
u
0
(x) =sinx v(x) =e
x
Z
e
x
sinxdx=e
x
sinx
Z
e
x
cosxdx
=e
x
sinx

e
x
cosx
Z
e
x
(sinx)dx

61. Example 2

62. Example 2
So, we have that:

63. Example 2
So, we have that:
Z
e
x
sinxdx=e
x
sinxe
x
cosx
Z
e
x
sinxdx

64. Example 2
So, we have that:
Z
e
x
sinxdx
|{z}
!!
=e
x
sinxe
x
cosx
Z
e
x
sinxdx
|{z}
!!

65. Example 2
So, we have that:
Z
e
x
sinxdx
|{z}
!!
=e
x
sinxe
x
cosx
Z
e
x
sinxdx
|{z}
!!
So, we add
R
e
x
sinxdxto both sides of the equation:

66. Example 2
So, we have that:
Z
e
x
sinxdx
|{z}
!!
=e
x
sinxe
x
cosx
Z
e
x
sinxdx
|{z}
!!
So, we add
R
e
x
sinxdxto both sides of the equation:
2
Z
e
x
sinxdx=e
x
(sinxcosx)

67. Example 2
So, we have that:
Z
e
x
sinxdx
|{z}
!!
=e
x
sinxe
x
cosx
Z
e
x
sinxdx
|{z}
!!
So, we add
R
e
x
sinxdxto both sides of the equation:
2
Z
e
x
sinxdx=e
x
(sinxcosx)
Z
e
x
sinxdx=
e
x
2
(sinxcosx) +C

68. Example 2
So, we have that:
Z
e
x
sinxdx
|{z}
!!
=e
x
sinxe
x
cosx
Z
e
x
sinxdx
|{z}
!!
So, we add
R
e
x
sinxdxto both sides of the equation:
2
Z
e
x
sinxdx=e
x
(sinxcosx)
Z
e
x
sinxdx=
e
x
2
(sinxcosx) +C
This is in fact an example of

69. Example 3

70. Example 3
Let's do another example:

71. Example 3
Let's do another example:
Z
x
2
e
x
dx

72. Example 3
Let's do another example:
Z
x
2
e
x
dx
We have an A and an E:

73. Example 3
Let's do another example:
Z
x
2
e
x
dx
We have an A and an E:
Z


A
x
2
e
x
dx

74. Example 3
Let's do another example:
Z
x
2
e
x
dx
We have an A and an E:
Z
x
2

>
E
e
x
dx

75. Example 3
Let's do another example:
Z
x
2
e
x
dx
We have an A and an E:
Z
x
2
e
x
dx

76. Example 3
Let's do another example:
Z
x
2
e
x
dx
We have an A and an E:
Z
x
2
e
x
dx
LIATE says that the A comes before the E. So, we choose
u(x) =x
2
:

77. Example 3
Let's do another example:
Z
x
2
e
x
dx
We have an A and an E:
Z
x
2
e
x
dx
LIATE says that the A comes before the E. So, we choose
u(x) =x
2
:
u(x) =x
2

78. Example 3
Let's do another example:
Z
x
2
e
x
dx
We have an A and an E:
Z
x
2
e
x
dx
LIATE says that the A comes before the E. So, we choose
u(x) =x
2
:
u(x) =x
2
v
0
(x) =e
x

79. Example 3
Let's do another example:
Z
x
2
e
x
dx
We have an A and an E:
Z
x
2
e
x
dx
LIATE says that the A comes before the E. So, we choose
u(x) =x
2
:
u(x) =x
2
v
0
(x) =e
x
u
0
(x) = 2x

80. Example 3
Let's do another example:
Z
x
2
e
x
dx
We have an A and an E:
Z
x
2
e
x
dx
LIATE says that the A comes before the E. So, we choose
u(x) =x
2
:
u(x) =x
2
v
0
(x) =e
x
u
0
(x) = 2x v(x) =e
x

81. Example 3
Let's do another example:
Z
x
2
e
x
dx
We have an A and an E:
Z
x
2
e
x
dx
LIATE says that the A comes before the E. So, we choose
u(x) =x
2
:
u(x) =x
2
v
0
(x) =e
x
u
0
(x) = 2x v(x) =e
x
Z
x
2
e
x
dx=

82. Example 3
Let's do another example:
Z
x
2
e
x
dx
We have an A and an E:
Z
x
2
e
x
dx
LIATE says that the A comes before the E. So, we choose
u(x) =x
2
:
u(x) =x
2
v
0
(x) =e
x
u
0
(x) = 2x v(x) =e
x
Z


u(x)
x
2
e
x
dx=

83. Example 3
Let's do another example:
Z
x
2
e
x
dx
We have an A and an E:
Z
x
2
e
x
dx
LIATE says that the A comes before the E. So, we choose
u(x) =x
2
:
u(x) =x
2
v
0
(x) =e
x
u
0
(x) = 2x v(x) =e
x
Z
x
2

>
v
0
(x)
e
x
dx=

84. Example 3
Let's do another example:
Z
x
2
e
x
dx
We have an A and an E:
Z
x
2
e
x
dx
LIATE says that the A comes before the E. So, we choose
u(x) =x
2
:
u(x) =x
2
v
0
(x) =e
x
u
0
(x) = 2x v(x) =e
x
Z
x
2
e
x
dx=x
2
e
x

Z
2xe
x
dx

85. Example 3
Let's do another example:
Z
x
2
e
x
dx
We have an A and an E:
Z
x
2
e
x
dx
LIATE says that the A comes before the E. So, we choose
u(x) =x
2
:
u(x) =x
2
v
0
(x) =e
x
u
0
(x) = 2x v(x) =e
x
Z
x
2
e
x
dx=

u(x)
x
2
e
x

Z
2xe
x
dx

86. Example 3
Let's do another example:
Z
x
2
e
x
dx
We have an A and an E:
Z
x
2
e
x
dx
LIATE says that the A comes before the E. So, we choose
u(x) =x
2
:
u(x) =x
2
v
0
(x) =e
x
u
0
(x) = 2x v(x) =e
x
Z
x
2
e
x
dx=x
2

>
v(x)
e
x

Z
2xe
x
dx

87. Example 3
Let's do another example:
Z
x
2
e
x
dx
We have an A and an E:
Z
x
2
e
x
dx
LIATE says that the A comes before the E. So, we choose
u(x) =x
2
:
u(x) =x
2
v
0
(x) =e
x
u
0
(x) = 2x v(x) =e
x
Z
x
2
e
x
dx=x
2
e
x

Z

>
u
0
(x)
2xe
x
dx

88. Example 3
Let's do another example:
Z
x
2
e
x
dx
We have an A and an E:
Z
x
2
e
x
dx
LIATE says that the A comes before the E. So, we choose
u(x) =x
2
:
u(x) =x
2
v
0
(x) =e
x
u
0
(x) = 2x v(x) =e
x
Z
x
2
e
x
dx=x
2
e
x

Z
2x
>
v(x)
e
x
dx

89. Example 3
Let's do another example:
Z
x
2
e
x
dx
We have an A and an E:
Z
x
2
e
x
dx
LIATE says that the A comes before the E. So, we choose
u(x) =x
2
:
u(x) =x
2
v
0
(x) =e
x
u
0
(x) = 2x v(x) =e
x
Z
x
2
e
x
dx=x
2
e
x

Z
2xe
x
dx

90. Example 3

91. Example 3
Z
x
2
e
x
dx=x
2
e
x
2
Z
xe
x
dx

92. Example 3
Z
x
2
e
x
dx=x
2
e
x
2
Z
xe
x
dx
Now we need to solve that integral by integration by parts again:

93. Example 3
Z
x
2
e
x
dx=x
2
e
x
2
Z
xe
x
dx
Now we need to solve that integral by integration by parts again:
u(x) =x

94. Example 3
Z
x
2
e
x
dx=x
2
e
x
2
Z
xe
x
dx
Now we need to solve that integral by integration by parts again:
u(x) =x v
0
(x) =e
x

95. Example 3
Z
x
2
e
x
dx=x
2
e
x
2
Z
xe
x
dx
Now we need to solve that integral by integration by parts again:
u(x) =x v
0
(x) =e
x
u
0
(x) = 1

96. Example 3
Z
x
2
e
x
dx=x
2
e
x
2
Z
xe
x
dx
Now we need to solve that integral by integration by parts again:
u(x) =x v
0
(x) =e
x
u
0
(x) = 1v(x) =e
x

97. Example 3
Z
x
2
e
x
dx=x
2
e
x
2
Z
xe
x
dx
Now we need to solve that integral by integration by parts again:
u(x) =x v
0
(x) =e
x
u
0
(x) = 1v(x) =e
x
Z
x
2
e
x
dx=x
2
e
x
2

xe
x

Z
1:e
x
dx

98. Example 3
Z
x
2
e
x
dx=x
2
e
x
2
Z
xe
x
dx
Now we need to solve that integral by integration by parts again:
u(x) =x v
0
(x) =e
x
u
0
(x) = 1v(x) =e
x
Z
x
2
e
x
dx=x
2
e
x
2

xe
x

Z
1:e
x
dx

=x
2
e
x
2xe
x
+ 2e
x
=e
x

x
2
2x+ 2

99. Example 3
Z
x
2
e
x
dx=x
2
e
x
2
Z
xe
x
dx
Now we need to solve that integral by integration by parts again:
u(x) =x v
0
(x) =e
x
u
0
(x) = 1v(x) =e
x
Z
x
2
e
x
dx=x
2
e
x
2

xe
x

Z
1:e
x
dx

=x
2
e
x
2xe
x
+ 2e
x
=e
x

x
2
2x+ 2