Integration by parts to solve it clearly

Integration by part solving part by part

1. Integration by Partsdx
dv
u
dx
du
v
dx
dy

If ,uvy where uand vare
both functions of x.
Substituting for y,xxx
dx
xxd
cos1sin
)sin(

e.g. If , xxy sin
We develop the formula by considering how to
differentiateproducts.dx
dv
u
dx
du
v
dx
uvd

)(

2. Integration by Partsdxxxdxxxx

 cossinsin xxx
dx
xxd
cos1sin
)sin(

So,
Integrating this equation, we getdxxxdxxdx
dx
xxd

 cossin
)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 Partsdxxxdxxxx

 cossinsin
Here’s the product . . .
if we rearrange, we getdxxxxdxxx

 sinsincos
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 Partsdxxxdxxxx

 cossinsin
Here’s the product . . .
if we rearrange, we getdxxxxdxxx

 sinsincos Cxxx  )cos(sin Cxxx  cossin
We need to turn this method ( called integration by
parts ) into a formula.
So, we’ve integrated ! xxcos

5. Integration by Parts
Integrating:
 dx
dx
dv
udx
dx
du
vdx
dx
uvd)( 
 dx
dx
dv
udx
dx
du
vuv 
 dx
dx
du
vuvdx
dx
dv
u dx
dv
u
dx
du
v
dx
uvd

)( xxx
dx
xxd
cossin
)sin(
 dxxxdxxdx
dx
xxd

 cossin
)sin(
Rearranging:
Simplifying the l.h.s.: 
 dxxxdxxxx cossinsin 
 dxxxxdxxx sinsincos
Generalisation
Example

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

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

8. Integration by Parts
So,
 dx
dx
du
vuvdx
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
vuvdx
dx
dv
u
e.g. 1 Find
dxxxsin2 xu2 x
dx
dv
sin
and2
dx
du xv 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
1
st
term, uv,is completed but the 2
nd
term still
needs to be integrated.
( +Ccomes later )

10. Integration by Parts)cos(2 xx  
 dxx2)cos(
We can now substitute into the formula
So,xu2 2
dx
du xv cos
differentiate integratex
dx
dv
sin
andu dx
dv u v v dx
du 
 dx
dx
du
vuvdx
dx
dv
u 
dxxxsin2

11. Integration by PartsC
So,
 dx
dx
du
vuvdx
dx
dv
u )cos(2 xx  
 dxx2)cos( xsin2 xu2 2
dx
du xv cos
differentiate integrate
We can now substitute into the formulax
dx
dv
sin
and
 dxxcos2 xxcos2
The 2
nd
term needs integrating xxcos2 
dxxxsin2

12. Integration by Parts
 dx
dx
du
vuvdx
dx
dv
u 







2
)(
2x
e
x  
 







dx
e
x
1
2
2 xu 1
dx
du
differentiate integratex
e
dx
dv
2

anddx
exe
xx


22
22
e.g. 2 Find
dxxe
x2
Solution:v 2
2x
e C
exe
xx

42
22 
dxxe
x2
So,
This is a compound function,
so we must be careful.

13. Integration by Parts
Exercises
Find
dxxe
x
1.
dxxx3cos 2.dxexe
xx

 Cexe
xx
 
dxxx3cos
2.dx
xx
x







3
3sin
3
3sin
)( C
xxx

9
3cos
3
3sin
1.
Solutions:
dxxe
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
dxxe
x 
dxxe
x dxexe
xx

  
1
0
xx
exe Cexe
xx
   
0011
01 eeee  10 1
We’ll use the question in the exercise you have just
done to illustrate.

15. Integration by Parts
Integration by parts cannotbe 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 integratex.xuln x
dx
dv
 xln
ANS: We can’t integrate .xln xln
Tip: Whenever appears in an integration by
parts we choose to let it equal u. 
 dx
dx
du
vuvdx
dx
dv
u
e.g. 3 Find 
dxxxln

17. Integration by Parts
 dx
dx
du
vuvdx
dx
dv
u
So,xuln x
dx
dv
 xdx
du1
 2
2
x
v 
dxxxln x
x
ln
2
2  
dx
x
x1
2
2
The r.h.s. integral still seems to be a product!
BUT . . . 

x
x
dxxx ln
2
ln
2 4
2
x C xcancels.
e.g. 3 Find 
dxxxln
differentiate
integrate
So,

x
x
dxxx ln
2
ln
2 
dx
x
2

18. Integration by Parts
e.g. 4dxex
x

2
Solution:
 dx
dx
du
vuvdx
dx
dv
u
Let2
xu x
e
dx
dv

 x
dx
du
2 x
ev

 

 dxxeexdxex
xxx
2
22 

 dxxeexdxex
xxx
2
22 2
2
1 IexI
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
and1I 2I

19. Integration by Parts
e.g. 4dxex
x

2
Solution:2
dx
du x
ev

 2
2
1 IexI
x

 
2I
So,
x
xe2 

 dxe
x
2 
x
xe2 

dxe
x
2  Cexe
xx


22 
Substitute in ( 1 )
. . . . . ( 1 )xx
exdxex


22
Letxu2 x
e
dx
dv

 and

 dxxeI
x
2
2 Cexe
xx


22

20. Integration by Parts
 dx
dx
du
vuvdx
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 
dxxe
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
vuvdx
dx
dv
u x
eu x
dx
dv
sin x
e
dx
du
 xvcos
e.g. 5 Find 
dxxe
x
sin 
dxxe
x
sin  xe
x
cos 
 dxxe
x
cos  xe
x
cos 
dxxe
x
cos
Solution:
We write this as:21 cosIxeI
x


22. Integration by Parts
 dx
dx
du
vuvdx
dx
dv
u x
eu x
dx
dv
cos x
e
dx
du
 xvsin
e.g. 5 Find 
dxxe
x
sin xe
x
sin 
dxxe
x
sin 
 dxxeI
x
sin
1
where
So,21 cosIxeI
x

and
 dxxeI
x
cos
2
We next use integration by parts for I
2
dxxe
x
cos 12 sinIxeI
x
 

23. Integration by Parts
 dx
dx
du
vuvdx
dx
dv
u x
eu x
dx
dv
cos x
e
dx
du
 xvsin
e.g. 5 Find 
dxxe
x
sin 
 dxxeI
x
sin
1 xe
x
sin 
dxxe
x
sin
So,
where21 cosIxeI
x
 and
 dxxeI
x
cos
2
We next use integration by parts for I
2
dxxe
x
cos 12 sinIxeI
x
 

24. Integration by Parts
 dx
dx
du
vuvdx
dx
dv
u
e.g. 5 Find 
dxxe
x
sin
So,21 cosIxeI
x

2equations, 2unknowns ( I
1
and I
2
) !
Substituting for I
2
in (1)
. . . . . ( 1 )
. . . . . ( 2 ) xeI
x
cos
1 2I 1sinIxe
x


25. Integration by Parts
 dx
dx
du
vuvdx
dx
dv
u
e.g. 5 Find 
dxxe
x
sin
So,21 cosIxeI
x

2equations, 2unknowns ( I
1
and I
2
) !
Substituting for I
2
in (1)
. . . . . ( 1 )
. . . . . ( 2 ) xeI
x
cos
1 1sin Ixe
x
 2I 1sin Ixe
x


26. Integration by Parts
 dx
dx
du
vuvdx
dx
dv
u
e.g. 5 Find 
dxxe
x
sin
So,21 cosIxeI
x

2equations, 2unknowns ( I
1
and I
2
) !
Substituting for I
2
in (1)
. . . . . ( 1 )
. . . . . ( 2 ) xeI
x
cos
1 xexeI
xx
sincos2
1  2
sincos
1
xexe
I
xx

 C 1sin Ixe
x
 2I 1sin Ixe
x


27. Integration by Parts
Exercises
2.dxxxsin
2

( Hint: Although 2. is not a product it can be turned
into one by writing the function as . )xln1
1.dxx
2
1
ln

28. Integration by Parts
Solutions:dxxxsin
2

1.x
dx
du
2 xvcos 2
xu x
dx
dv
sin andLet1I 
 dxxxxxI cos2cos
2
1 
 dxxxxxI cos2cos
2
1 2I 2
dx
du xvsin 
 dxxxxI sin2sin2
2 Cxxx  cos2sin2 Cxxxxxdxxx 
cos2sin2cossin
22 xu2 x
dx
dv
cos
andLetFor I
2
:
. . . . . ( 1 )
Subs. in ( 1 )

29. Integration by Parts
2.xdx
du1
 xv xuln 1
dx
dv
andLet
dxxln1 xxln dx
x
x


1  xxln dx

1 Cxxx ln
This is an important
application of
integration by partsdxx
2
1
ln dxx

2
1
ln1 dxx
2
1
ln
So, 
2
1
lnxxx   11ln122ln2  12ln2