integration by parts, part 3
DESCRIPTION
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.htmlTRANSCRIPT
The LIATE Rule
The LIATE Rule
The difficulty of integration by parts is in choosing u(x) and v ′(x)correctly.
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 functionyou should choose as u(x):
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 functionyou should choose as u(x):
LIATE
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 functionyou should choose as u(x):
L︸︷︷︸Log
IATE
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 functionyou should choose as u(x):
L I︸︷︷︸Inv. Trig
ATE
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 functionyou should choose as u(x):
LI A︸︷︷︸Algebraic
TE
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 functionyou should choose as u(x):
LIA T︸︷︷︸Trig
E
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 functionyou should choose as u(x):
LIAT E︸︷︷︸Exp
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 functionyou should choose as u(x):
LIATE
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 functionyou should choose as u(x):
LIATE
The word itself tells you in which order of priority you should useu(x).
Example 1
Example 1
∫x sin xdx
Example 1
∫x sin xdx
Here we have an algebraic and a trigonometric function:
Example 1
∫x sin xdx
Here we have an algebraic and a trigonometric function:∫x sin xdx
Example 1
∫x sin xdx
Here we have an algebraic and a trigonometric function:∫��A
x sin xdx
Example 1
∫x sin xdx
Here we have an algebraic and a trigonometric function:∫x���:
Tsin xdx
Example 1
∫x sin xdx
Here we have an algebraic and a trigonometric function:∫x sin xdx
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 chooseu(x) = x .
Example 2
Example 2
∫ex sin xdx
Example 2
∫ex sin xdx
In this case we have E and T:
Example 2
∫ex sin xdx
In this case we have E and T:∫��>
Eex sin xdx
Example 2
∫ex sin xdx
In this case we have E and T:∫ex���:
Tsin xdx
Example 2
∫ex sin xdx
In this case we have E and T:∫ex sin xdx
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 chooseu(x) = sin x .
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 chooseu(x) = sin x .Let’s solve this integral to show that this works:
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 chooseu(x) = sin x .Let’s solve this integral to show that this works:
u(x) = sin x
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 chooseu(x) = sin x .Let’s solve this integral to show that this works:
u(x) = sin x v ′(x) = ex
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 chooseu(x) = sin x .Let’s solve this integral to show that this works:
u(x) = sin x v ′(x) = ex
u′(x) = cos x
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 chooseu(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
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 chooseu(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
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 chooseu(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
∫��>
v ′(x)ex sin xdx = ex sin x −
∫ex cos xdx
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 chooseu(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���:
u(x)sin xdx = ex sin x −
∫ex cos xdx
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 chooseu(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 =��>
v(x)ex sin x −
∫ex cos xdx
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 chooseu(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���:
u(x)sin x −
∫ex cos xdx
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 chooseu(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 −
∫��>
v(x)ex cos xdx
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 chooseu(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���:
u′(x)cos xdx
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 chooseu(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
Example 2
Example 2
Now we need to solve that integral:
Example 2
Now we need to solve that integral:∫ex sin xdx = ex sin x −
∫ex cos xdx
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.
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 :
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
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
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
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
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
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 −
∫��>
v ′(x)ex cos xdx
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���:
u(x)cos xdx
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
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
]
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 −
[��>
v(x)ex cos x −
∫ex (− sin x) dx
]
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���:
u(x)cos x −
∫ex (− sin x) dx
]
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 −
∫��>
v(x)ex (− sin x) dx
]
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���
��:u′(x)(− sin x)dx
]
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
]
Example 2
Example 2
So, we have that:
Example 2
So, we have that:∫ex sin xdx = ex sin x − ex cos x −
∫ex sin xdx
Example 2
So, we have that:∫ex sin xdx︸ ︷︷ ︸
!!
= ex sin x − ex cos x −∫
ex sin xdx︸ ︷︷ ︸!!
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:
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)
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
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.
Example 3
Example 3
Let’s do another example:
Example 3
Let’s do another example: ∫x2exdx
Example 3
Let’s do another example: ∫x2exdx
We have an A and an E:
Example 3
Let’s do another example: ∫x2exdx
We have an A and an E: ∫���
Ax2exdx
Example 3
Let’s do another example: ∫x2exdx
We have an A and an E: ∫x2��>
Eexdx
Example 3
Let’s do another example: ∫x2exdx
We have an A and an E: ∫x2exdx
Example 3
Let’s do another example: ∫x2exdx
We have an A and an E: ∫x2exdx
LIATE says that the A comes before the E. So, we chooseu(x) = x2:
Example 3
Let’s do another example: ∫x2exdx
We have an A and an E: ∫x2exdx
LIATE says that the A comes before the E. So, we chooseu(x) = x2:
u(x) = x2
Example 3
Let’s do another example: ∫x2exdx
We have an A and an E: ∫x2exdx
LIATE says that the A comes before the E. So, we chooseu(x) = x2:
u(x) = x2 v ′(x) = ex
Example 3
Let’s do another example: ∫x2exdx
We have an A and an E: ∫x2exdx
LIATE says that the A comes before the E. So, we chooseu(x) = x2:
u(x) = x2 v ′(x) = ex
u′(x) = 2x
Example 3
Let’s do another example: ∫x2exdx
We have an A and an E: ∫x2exdx
LIATE says that the A comes before the E. So, we chooseu(x) = x2:
u(x) = x2 v ′(x) = ex
u′(x) = 2x v(x) = ex
Example 3
Let’s do another example: ∫x2exdx
We have an A and an E: ∫x2exdx
LIATE says that the A comes before the E. So, we chooseu(x) = x2:
u(x) = x2 v ′(x) = ex
u′(x) = 2x v(x) = ex∫x2exdx =
Example 3
Let’s do another example: ∫x2exdx
We have an A and an E: ∫x2exdx
LIATE says that the A comes before the E. So, we chooseu(x) = x2:
u(x) = x2 v ′(x) = ex
u′(x) = 2x v(x) = ex
∫���
u(x)
x2exdx =
Example 3
Let’s do another example: ∫x2exdx
We have an A and an E: ∫x2exdx
LIATE says that the A comes before the E. So, we chooseu(x) = x2:
u(x) = x2 v ′(x) = ex
u′(x) = 2x v(x) = ex
∫x2��>
v ′(x)exdx =
Example 3
Let’s do another example: ∫x2exdx
We have an A and an E: ∫x2exdx
LIATE says that the A comes before the E. So, we chooseu(x) = x2:
u(x) = x2 v ′(x) = ex
u′(x) = 2x v(x) = ex∫x2exdx = x2ex −
∫2xexdx
Example 3
Let’s do another example: ∫x2exdx
We have an A and an E: ∫x2exdx
LIATE says that the A comes before the E. So, we chooseu(x) = x2:
u(x) = x2 v ′(x) = ex
u′(x) = 2x v(x) = ex
∫x2exdx = ���
u(x)
x2ex −∫
2xexdx
Example 3
Let’s do another example: ∫x2exdx
We have an A and an E: ∫x2exdx
LIATE says that the A comes before the E. So, we chooseu(x) = x2:
u(x) = x2 v ′(x) = ex
u′(x) = 2x v(x) = ex
∫x2exdx = x2��>
v(x)ex −
∫2xexdx
Example 3
Let’s do another example: ∫x2exdx
We have an A and an E: ∫x2exdx
LIATE says that the A comes before the E. So, we chooseu(x) = x2:
u(x) = x2 v ′(x) = ex
u′(x) = 2x v(x) = ex
∫x2exdx = x2ex −
∫��>
u′(x)2xexdx
Example 3
Let’s do another example: ∫x2exdx
We have an A and an E: ∫x2exdx
LIATE says that the A comes before the E. So, we chooseu(x) = x2:
u(x) = x2 v ′(x) = ex
u′(x) = 2x v(x) = ex
∫x2exdx = x2ex −
∫2x��>
v(x)exdx
Example 3
Let’s do another example: ∫x2exdx
We have an A and an E: ∫x2exdx
LIATE says that the A comes before the E. So, we chooseu(x) = x2:
u(x) = x2 v ′(x) = ex
u′(x) = 2x v(x) = ex∫x2exdx = x2ex −
∫2xexdx
Example 3
Example 3
∫x2exdx = x2ex − 2
∫xexdx
Example 3
∫x2exdx = x2ex − 2
∫xexdx
Now we need to solve that integral by integration by parts again:
Example 3
∫x2exdx = x2ex − 2
∫xexdx
Now we need to solve that integral by integration by parts again:
u(x) = x
Example 3
∫x2exdx = x2ex − 2
∫xexdx
Now we need to solve that integral by integration by parts again:
u(x) = x v ′(x) = ex
Example 3
∫x2exdx = x2ex − 2
∫xexdx
Now we need to solve that integral by integration by parts again:
u(x) = x v ′(x) = ex
u′(x) = 1
Example 3
∫x2exdx = x2ex − 2
∫xexdx
Now we need to solve that integral by integration by parts again:
u(x) = x v ′(x) = ex
u′(x) = 1 v(x) = ex
Example 3
∫x2exdx = x2ex − 2
∫xexdx
Now we need to solve that integral by integration by parts again:
u(x) = x v ′(x) = ex
u′(x) = 1 v(x) = ex
∫x2exdx = x2ex − 2
[xex −
∫1.exdx
]
Example 3
∫x2exdx = x2ex − 2
∫xexdx
Now we need to solve that integral by integration by parts again:
u(x) = x v ′(x) = ex
u′(x) = 1 v(x) = ex
∫x2exdx = x2ex − 2
[xex −
∫1.exdx
]= x2ex − 2xex + 2ex = ex
(x2 − 2x + 2
)
Example 3
∫x2exdx = x2ex − 2
∫xexdx
Now we need to solve that integral by integration by parts again:
u(x) = x v ′(x) = ex
u′(x) = 1 v(x) = ex
∫x2exdx = x2ex − 2
[xex −
∫1.exdx
]= x2ex − 2xex + 2ex = ex
(x2 − 2x + 2
)
