Integration by parts to solve it clearly

Integration by Parts
dx
dv
u
dx
du
v
dx
dy


If ,
uv
y  where u and v are
both functions of x.
Substituting for y,
x
x
x
dx
x
x
d
cos
1
sin
)
sin
(




e.g. If ,
x
x
y sin

We develop the formula by considering how to
differentiate products.
dx
dv
u
dx
du
v
dx
uv
d



)
(

dx
x
x
dx
x
x
x

 
 cos
sin
sin
x
x
x
dx
x
x
d
cos
1
sin
)
sin
(




So,
Integrating this equation, we get
dx
x
x
dx
x
dx
dx
x
x
d


 
 cos
sin
)
sin
(
The l.h.s. is just the integral of a derivative, so,
since integration is the reverse of differentiation,
we get
Can you see what this has to do with integrating a
product?

dx
x
x
dx
x
x
x

 
 cos
sin
sin
Here’s the product . . .
if we rearrange, we get
dx
x
x
x
dx
x
x

 
 sin
sin
cos
The function in the integral on the l.h.s. . . .
. . . is a product, but the one on the r.h.s. . . .
is a simple function that we can integrate easily.

dx
x
x
dx
x
x
x

 
 cos
sin
sin
Here’s the product . . .
if we rearrange, we get
dx
x
x
x
dx
x
x

 
 sin
sin
cos
C
x
x
x 


 )
cos
(
sin
C
x
x
x 

 cos
sin
We need to turn this method ( called integration by
parts ) into a formula.
So, we’ve integrated !
x
x cos

Integrating:


 
 dx
dx
dv
u
dx
dx
du
v
dx
dx
uv
d )
(

 
 dx
dx
dv
u
dx
dx
du
v
uv

 
 dx
dx
du
v
uv
dx
dx
dv
u
dx
dv
u
dx
du
v
dx
uv
d


)
(
x
x
x
dx
x
x
d
cos
sin
)
sin
(


dx
x
x
dx
x
dx
dx
x
x
d


 
 cos
sin
)
sin
(
Rearranging:
Simplifying the l.h.s.:
 

 dx
x
x
dx
x
x
x cos
sin
sin

 
 dx
x
x
x
dx
x
x sin
sin
cos
Generalisation
Example

SUMMARY

 
 dx
dx
du
v
uv
dx
dx
dv
u
To integrate some products we can use the formula

dx
du
to get . . .

 
 dx
dx
du
v
uv
dx
dx
dv
u
Using this formula means that we differentiate one
factor, u
So,

So,

 
 dx
dx
du
v
uv
dx
dx
dv
u
and integrate the other ,
dx
dv
to get v
factor, u
dx
du
to get . . .

So,

 
 dx
dx
du
v
uv
dx
dx
dv
u
e.g. 1 Find
 dx
x
x sin
2
x
u 2
 x
dx
dv
sin

and
2

dx
du x
v cos


differentiate
integrate
and integrate the other ,
dx
dv
to get v
factor, u
dx
du
to get . . .
Having substituted in the formula, notice that the
1st term, uv, is completed but the 2nd term still
needs to be integrated.
( +C comes later )

)
cos
(
2 x
x     dx
x 2
)
cos
(
We can now substitute into the formula
So,
x
u 2

2

dx
du x
v cos


differentiate integrate
x
dx
dv
sin

and
u
dx
dv u v v
dx
du

 
 dx
dx
du
v
uv
dx
dx
dv
u

 dx
x
xsin
2

C
So,

 
 dx
dx
du
v
uv
dx
dx
dv
u
)
cos
(
2 x
x     dx
x 2
)
cos
(

x
sin
2
x
u 2

2

dx
du x
v cos


We can now substitute into the formula
x
dx
dv
sin

and

 dx
x
cos
2
x
x cos
2


The 2nd term needs integrating


 x
x cos
2

 dx
x
xsin
2


 
 dx
dx
du
v
uv
dx
dx
dv
u








2
)
(
2x
e
x   
 







dx
e x
1
2
2
x
u 
1

dx
du
x
e
dx
dv 2

and
dx
e
xe x
x



2
2
2
2
e.g. 2 Find
 dx
xe x
2
Solution:

v
2
2x
e
C
e
xe x
x



4
2
2
2

 dx
xe x
2
So,
This is a compound function,
so we must be careful.

Exercises
Find
 dx
xe x
1.
 dx
x
x 3
cos
2.
dx
e
xe x
x


C
e
xe x
x




 dx
x
x 3
cos
2. dx
x
x
x








3
3
sin
3
3
sin
)
(
C
x
x
x



9
3
cos
3
3
sin
1.
Solutions:

 dx
xe x

Definite Integration by Parts
With a definite integral it’s often easier to do the
indefinite integral and insert the limits at the end.


1
0
dx
xex
 dx
xe x
dx
e
xe x
x



 1
0
x
x
e
xe 

C
e
xe x
x



   
0
0
1
1
0
1 e
e
e
e 



   
1
0 

 1

We’ll use the question in the exercise you have just
done to illustrate.

Integration by parts cannot be used for every product.
Using Integration by Parts
It works if
 we can integrate one factor of the product,
 the integral on the r.h.s. is easier* than the
one we started with.
* There is an exception but you need to learn the
general rule.

Solution:
What’s a possible problem?
Can you see what to do?
If we let and , we will need to
differentiate and integrate x.
x
u ln
 x
dx
dv

x
ln
ANS: We can’t integrate .
x
ln
x
ln
Tip: Whenever appears in an integration by
parts we choose to let it equal u.

 
 dx
dx
du
v
uv
dx
dx
dv
u
e.g. 3 Find  dx
x
x ln


 
 dx
dx
du
v
uv
dx
dx
dv
u
So, x
u ln
 x
dx
dv

x
dx
du 1

2
2
x
v 

  dx
x
x ln x
x
ln
2
2

  dx
x
x 1
2
2
The r.h.s. integral still seems to be a product!
BUT . . .



 x
x
dx
x
x ln
2
ln
2
4
2
x
C

x cancels.
e.g. 3 Find  dx
x
x ln
So, 

 x
x
dx
x
x ln
2
ln
2
 dx
x
2

e.g. 4 dx
e
x x


2
Solution:

 
 dx
dx
du
v
uv
dx
dx
dv
u
Let 2
x
u 
x
e
dx
dv 

x
dx
du
2
 x
e
v 











 dx
xe
e
x
dx
e
x x
x
x
2
2
2








 dx
xe
e
x
dx
e
x x
x
x
2
2
2
2
2
1 I
e
x
I x


 
The integral on the r.h.s. is still a product but using
the method again will give us a simple function.
We write
and
1
I 2
I

e.g. 4 dx
e
x x


2
Solution:
2

dx
du x
e
v 


2
2
1 I
e
x
I x


 

2
I
So, 
  x
xe
2 

 dx
e x
2

  x
xe
2 

dx
e x
2

C
e
xe x
x


 

2
2

Substitute in ( 1 )
. . . . . ( 1 )
x
x
e
x
dx
e
x 




2
2
Let x
u 2
 x
e
dx
dv 

and


 dx
xe
I x
2
2
C
e
xe x
x


 

2
2


 
 dx
dx
du
v
uv
dx
dx
dv
u
Solution:
It doesn’t look as though integration by parts will
help since neither function in the product gets easier
when we differentiate it.
e.g. 5 Find  dx
x
e x
sin
However, there’s something special about the 2
functions that means the method does work.
Example


 
 dx
dx
du
v
uv
dx
dx
dv
u
x
e
u  x
dx
dv
sin

x
e
dx
du
 x
v cos


e.g. 5 Find  dx
x
e x
sin

  dx
x
e x
sin 
 x
e x
cos  dx
x
e x
cos


 x
e x
cos  dx
x
e x
cos
Solution:
We write this as: 2
1 cos I
x
e
I x





 
 dx
dx
du
v
uv
dx
dx
dv
u
x
e
u  x
dx
dv
cos

x
e
dx
du
 x
v sin

e.g. 5 Find  dx
x
e x
sin

x
e x
sin  dx
x
e x
sin

 dx
x
e
I x
sin
1
where
So, 2
1 cos I
x
e
I x



and 
 dx
x
e
I x
cos
2
We next use integration by parts for I2

 dx
x
e x
cos
1
2 sin I
x
e
I x





 
 dx
dx
du
v
uv
dx
dx
dv
u
x
e
u  x
dx
dv
cos

x
e
dx
du
 x
v sin

e.g. 5 Find  dx
x
e x
sin

 dx
x
e
I x
sin
1

x
e x
sin  dx
x
e x
sin
So,
where
2
1 cos I
x
e
I x



and 
 dx
x
e
I x
cos
2
We next use integration by parts for I2

 dx
x
e x
cos
1
2 sin I
x
e
I x





 
 dx
dx
du
v
uv
dx
dx
dv
u
e.g. 5 Find  dx
x
e x
sin
So, 2
1 cos I
x
e
I x



2 equations, 2 unknowns ( I1 and I2 ) !
Substituting for I2 in ( 1)
. . . . . ( 1 )
. . . . . ( 2 )


 x
e
I x
cos
1
2
I 1
sin I
x
e x




 
 dx
dx
du
v
uv
dx
dx
dv
u
e.g. 5 Find  dx
x
e x
sin
So, 2
1 cos I
x
e
I x



. . . . . ( 1 )
. . . . . ( 2 )


 x
e
I x
cos
1 1
sin I
x
e x

2
I 1
sin I
x
e x




 
 dx
dx
du
v
uv
dx
dx
dv
u
e.g. 5 Find  dx
x
e x
sin
So, 2
1 cos I
x
e
I x



. . . . . ( 1 )
. . . . . ( 2 )


 x
e
I x
cos
1
x
e
x
e
I x
x
sin
cos
2 1 



2
sin
cos
1
x
e
x
e
I
x
x



 C

1
sin I
x
e x

2
I 1
sin I
x
e x



Exercises
2.
dx
x
x sin
2

( Hint: Although 2. is not a product it can be turned
into one by writing the function as . )
x
ln
1
1. dx
x

2
1
ln

Solutions:
dx
x
x sin
2

1.
x
dx
du
2
 x
v cos


2
x
u  x
dx
dv
sin

and
Let
1
I



 dx
x
x
x
x
I cos
2
cos
2
1




 dx
x
x
x
x
I cos
2
cos
2
1
2
I
2

dx
du
x
v sin




 dx
x
x
x
I sin
2
sin
2
2
C
x
x
x 

 cos
2
sin
2
C
x
x
x
x
x
dx
x
x 




  cos
2
sin
2
cos
sin 2
2
x
u 2
 x
dx
dv
cos

and
Let
For I2:
. . . . . ( 1 )
Subs. in ( 1 )

2.
x
dx
du 1
 x
v 
x
u ln
 1

dx
dv
and
Let


  dx
x
ln
1 
x
x ln dx
x
x
 
1

 x
x ln dx
1
C
x
x
x 

 ln
This is an important
application of
integration by parts
dx
x

2
1
ln dx
x
 

2
1
ln
1
dx
x

2
1
ln
So,  2
1
ln x
x
x 

   
1
1
ln
1
2
2
ln
2 



1
2
ln
2 


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