Integration by Parts
We develop the formula by considering how to
differentiate products.
If
y  uv , dy
du
dv
v
 u
dx
dx
dx
Substituting for y, 
e.g. If
y  x sin x ,
where u and v are
both functions of x.
d ( uv )
du
dv
v
u
dx
dx
dx
d ( x sin x )
 sin x  1  x  cos x
dx
Integration by Parts
So,
d ( x sin x )
 sin x  1  x  cos x
dx
Integrating this equation, we get



x cos x dx


x cos x dx
d ( x sin x )
dx  sin x dx 
dx
The l.h.s. is just the integral of a derivative, so,
since integration is the reverse of differentiation,
we get
x sin x  sin x dx 
Can you see what this has to do with integrating a
product?
Integration by Parts

x sin x  sin x dx 

x cos x dx
Here’s the product . . .
if we rearrange, we get

x cos x dx  x sin x 

sin x dx
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.
Integration by Parts

x sin x  sin x dx 

x cos x dx
Here’s the product . . .
if we rearrange, we get

x cos x dx  x sin x 
So, we’ve

sin x dx
 x sin x  (  cos x )  C
 x sin x  cos x  C
integrated xcos x !
We need to turn this method ( called integration by
parts ) into a formula.
Integration by Parts
Example
Generalisation
d ( x sin x )
 sin x  x cos x
dx
d ( uv )
du
dv
v
 u
dx
dx
dx
Integrating:


d ( x sin x )
dx  sin x dx 
dx
Simplifying the l.h.s.:

x sinx  sinx dx 
Rearranging:




x cos x dx
x cos x dx



dv
u dx
dx
du
uv  v dx 
dx


dv
u dx
dx


d ( uv )
du
dx  v dx 
dx
dx
x cos x dx  x sinx  sinx dx
dv
du
u dx  uv  v dx
dx
dx
Integration by Parts
SUMMARY
Integration by Parts
To integrate some products we can use the formula


dv
du
u dx  uv  v dx
dx
dx
Integration by Parts
So,


dv
du
u dx  uv  v dx
dx
dx
Using this formula means that we differentiate one
factor, u to get du . . .
dx
Integration by Parts
So,


dv
du
u dx  uv  v dx
dx
dx
Using this formula means that we differentiate one
factor, u to get du . . .
dx
and integrate the other
dv ,
to get v
dx
Integration by Parts
So,


dv
du
u dx  uv  v dx
dx
dx
Using this formula means that we differentiate one
factor, u to get du . . .
dx
and integrate the other

dv ,
to get v
dx
e.g. 1 Find
2 x sin
dxformula, notice that the
Having substituted
in x
the
dv
st
 2 x andbut the
 sin
1 term, uv, is ucompleted
2nd xterm still
needs to be integrated.
differentiate
du
dx
2
dx
v   cos x
integrate
( +C comes later )
Integration by Parts
So,
dv
u  2 x and
 sin x
dx
du
v   cos x
2
dx
differentiate
integrate
We can now substitute into the formula


dv
du
u dx  uv  v dx
dx
dx
 2 x sin x dx  2 x( cos x )
u
dv
dx
u
v

 (  cos x ) 2 dx
v
du
dx
Integration by Parts
So,
dv
u  2 x and
 sin x
dx
du
v   cos x
2
dx
differentiate
integrate
We can now substitute into the formula


dv
du
u dx  uv  v dx
dx
dx
 2 x sin x dx  2 x( cos x )

2 cos x dx

 (  cos x ) 2 dx
 2 x cos x 
term
2 sinintegrating
x C
The
 22xndcos
x needs
Integration by Parts
e.g. 2 Find
2x
xe
dx

Solution:
differentiate
So,
 xe
2x

dv
u  x and
 e2x
dx
2x
du
e
1
v
dx
2
This is a compound
integrate
function,
 so we must
 e 2 x be
 careful.
 

 1 dx
 e2x
dx  ( x )
 2 


xe 2 x


2

dv
du
u dx  uv  v dx
dx
dx

e2x
dx
2
 
2 
xe 2 x e 2 x


C
2
4
Integration by Parts
Exercises
Find
1.
Solutions:
1.
 xe dx
2.
x
 xe
x
dx 
 x cos 3 x dx
xe   e dx
x
x
 xe  e  C
x
2.
x
sin
3
x
sin
3
x


dx
 x cos 3 x dx  ( x ) 3  
3
x sin 3 x cos 3 x


C
3
9

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.
We’ll use the question in the exercise you have just
done to illustrate.
x
xe
 dx

1
0
x
xe dx
 xe x   e x dx
 xe x  e x  C

 1e
 xe
1
x

x 1
e 0
e
1
  0e
 0   1  1
0
e
0

Integration by Parts
Using Integration by Parts
Integration by parts cannot be used for every product.
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.
Integration by Parts
e.g. 3 Find
Solution:
 x ln x dx


dv
du
u dx  uv  v dx
dx
dx
What’s a possible problem?
ANS: We can’t integrate
ln x .
Can you see what to do?
dv
If we let u  ln x and
 x , we will need to
dx
differentiate ln x and integrate x.
Tip: Whenever ln x appears in an integration by
parts we choose to let it equal u.
Integration by Parts
e.g. 3 Find
So,
differentiate
 x ln x dx
u  ln x
du 1

dx x
x2
ln x 
  x ln x dx 
2


dv
du
u dx  uv  v dx
dx
dx
dv
x
dx
integrate
x2
v
2

x2 1
 dx
2 x
The r.h.s. integral still seems to be a product!
BUT . . . x cancels.
2



So,

x
x
dx
x ln x dx 
ln x 
2
2
x2
x2
x ln x dx 
ln x 
C
2
4
Integration by Parts
e.g. 4
2 x
x
 e dx


dv
du
u dx  uv  v dx
dx
dx
Solution:
dv
Let
x
2

e
u  x and
dx
du
 2x
v  e  x
dx
  x 2 e  x dx   x 2 e  x    2 xe  x dx

2 x
2 x
x
e
dx


x
e


I1
x
2
xe
dx

I2
The integral on the r.h.s. is still a product but using
the method again will give us a simple function.
We write
I   x 2e  x  I
1
2
Integration by Parts
e.g. 4
Solution:
2 x
x
 e dx
I 2   2 xe  x dx
Let
I1   x 2e  x  I 2 . . . . . ( 1 )
u  2x
du
2
dx
So,
Substitute in ( 1 )
dv
 ex
dx
v  e  x
and
I 2   2 xe  x    2e  x dx
x
  2 xe  x 
2
e
dx

  2 xe  x  2e  x  C
 x e dx   x e
2 x
2 x
 2 xe
x
 2e
x
C
Integration by Parts
Example

e.g. 5 Find e sin x dx
Solution:
x


dv
du
u dx  uv  v dx
dx
dx
It doesn’t look as though integration by parts will
help since neither function in the product gets easier
when we differentiate it.
However, there’s something special about the 2
functions that means the method does work.
Integration by Parts


dv
du
e.g. 5 Find  e sin x dx
u dx  uv  v dx
dx
dx
Solution:
dv
x
 sin x
ue
dx
du
 ex
v  cos x
dx
  e x sin x dx   e x cos x    e x cos x dx
x
  e cos x 
x
We write this as:
x
e
 cos x dx
I1  e cos x  I 2
x
Integration by Parts


e.g. 5 Find e sin x dx
So,
where
x

dv
du
u dx  uv  v dx
dx
dx
I1  e x cos x  I 2
I 1   e x sin x dx and I 2  e x cos x dx

We next use integration by parts for I2
ue
dv
 cos x
dx
x
du
 ex
dx
v  sin x
x
x
e
cos
x
dx

e sinx 


x
e
 sin x dx
I 2  e sinx  I1
x
Integration by Parts


e.g. 5 Find e sin x dx
So,
where
x

dv
du
u dx  uv  v dx
dx
dx
I1  e x cos x  I 2
I 1   e x sin x dx and I 2  e x cos x dx

We next use integration by parts for I2
ue
dv
 cos x
dx
x
du
 ex
dx
v  sin x
x
x
e
cos
x
dx

e sinx 


x
e
 sin x dx
I 2  e sinx  I1
x
Integration by Parts

e.g. 5 Find e sin x dx
So,
x

I1  e x cos x  I 2 . . . . . ( 1 )
I 2  e x sinx  I1
.....(2)
2 equations, 2 unknowns ( I1 and I2 ) !
Substituting for I2 in ( 1 )
I1  e x cos x 

dv
du
u dx  uv  v dx
dx
dx
Integration by Parts

e.g. 5 Find e sin x dx
So,
x

I1  e x cos x  I 2 . . . . . ( 1 )
I 2  e x sin x  I 1

dv
du
u dx  uv  v dx
dx
dx
.....(2)
2 equations, 2 unknowns ( I1 and I2 ) !
Substituting for I2 in ( 1 )
I1  e x cos x  e x sin x  I 1
Integration by Parts

e.g. 5 Find e sin x dx
So,
x

I1  e x cos x  I 2 . . . . . ( 1 )
I 2  e x sin x  I 1

dv
du
u dx  uv  v dx
dx
dx
.....(2)
2 equations, 2 unknowns ( I1 and I2 ) !
Substituting for I2 in ( 1 )
I1  e x cos x  e x sin x  I 1
 2I1  e x cos x  e x sinx
 e x cos x  e x sin x
 I1 
C
2
Integration by Parts
Exercises
1.
x
2
sin x dx
2.
2
1 ln x dx
( Hint: Although 2. is not a product it can be turned
into one by writing the function as 1 ln x . )
Integration by Parts
Solutions:
1.
x
2
sin x dx Let u  x
2
and
du
 2x
dx
I1
dv
 sin x
dx
v  cos x
I 1   x 2 cos x    2 x cos x dx
 I 1   x cos x 
2
For I2:
Let
u  2 x and
du
2
dx
dv
2 x cos x dx . . . . . ( 1 )
I2
 cos x
dx
v  sin x
 I 2  2 x sin x   2 sin x dx
 2 x sin x  2 cos x  C
Subs. in ( 1 )

2
2
x
sin
x
dx


x
cos x  2 x sin x  2 cos x  C

Integration by Parts
2.
2
2
1 ln x dx  1 1 ln x dx
This is an important
application of
integration by parts
dv
u  ln x and
1
dx
du 1

vx
dx x
1
x
ln
x

  1 ln x dx 
x  dx
x
Let


So,
2
1 ln x dx

x ln x  1 dx

x ln x  x  C
  x ln x  x 
2
1
  2 ln2  2    1 ln1  1 
 2 ln 2  1
Integration by Parts