Integration by part solving part by part

1. 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



)
(

2. Integration by Parts
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?

3. Integration by Parts
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.

4. Integration by Parts
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

5. Integration by Parts
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

6. Integration by Parts
SUMMARY

 
 dx
dx
du
v
uv
dx
dx
dv
u
To integrate some products we can use the formula
Integration by Parts

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

8. Integration by Parts
So,

 
 dx
dx
du
v
uv
dx
dx
dv
u
and integrate the other ,
dx
dv
to get v
Using this formula means that we differentiate one
factor, u
dx
du
to get . . .

9. Integration by Parts
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
Using this formula means that we differentiate one
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 )

10. Integration by Parts
)
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

11. Integration by Parts
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


differentiate integrate
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

12. Integration by Parts

 
 dx
dx
du
v
uv
dx
dx
dv
u








2
)
(
2x
e
x   
 







dx
e x
1
2
2
x
u 
1

dx
du
differentiate integrate
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.

13. Integration by Parts
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

14. Integration by Parts
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.

15. Integration by Parts
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.

16. Integration by Parts
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

17. Integration by Parts

 
 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
differentiate integrate
So, 

 x
x
dx
x
x ln
2
ln
2
 dx
x
2

18. Integration by Parts
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

19. Integration by Parts
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

20. Integration by Parts

 
 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

21. Integration by Parts

 
 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




22. Integration by Parts

 
 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




23. Integration by Parts

 
 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




24. Integration by Parts

 
 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



25. Integration by Parts

 
 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 1
sin I
x
e x

2
I 1
sin I
x
e x



26. Integration by Parts

 
 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
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



27. Integration by Parts
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

28. Integration by Parts
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 )

29. Integration by Parts
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 
