“Teach A Level Maths”
Vol. 2: A2 Core Modules
25: Integration by Parts
© Christine Crisp
Integration by Parts
Module C3
AQA
MEI/OCR
Module C4
Edexcel
OCR
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Integration by Parts
There is a formula for integrating some products.
I’m going to show you how we get the formula but
it is tricky so if you want to go directly to the
summary and examples click below.
Summary
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


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
The following exercises and examples are harder
so you may want to practice more of the
straightforward questions before you tackle them.
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
The next example is interesting but is not essential.
Click below if you want to miss it out.
Omit 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
Integration by Parts
The following slides contain repeats of
information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
For most purposes the slides can be printed
as “Handouts” with up to 6 slides per sheet.
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
e.g.
Find
2x
xe
dx

Solution:
u x
differentiate
So,
 xe
2x

xe 2 x


2
dv
 e2x
dx
2x
e
v
2
and
du
1
dx
 e2x
dx  ( x )
 2






e2x
dx
2


dv
du
u dx  uv  v dx
dx
dx

 e2x

 2

integrate

 1 dx


xe 2 x e 2 x


C
2
4
Integration by Parts
Using Integration by Parts
Integration by parts can’t 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
 x ln x dx


dv
du
u dx  uv  v dx
dx
dx
We can’t integrate ln x so,
u  ln x
differentiate
du 1

dx x
x2
ln x 
  x ln x dx 
2
dv
x
dx
x2
v
2

integrate
x2 1
 dx
2 x
The r.h.s. integral still seems to be a product!
BUT . . . x cancels.
2
2
x
x
So,  x ln x dx 
C
ln x 
2
4
Integration by Parts
2.
This is an important
application of
integration by parts
 ln x dx   1 ln x dx
Let
u  ln x
du 1

dx x
and
  1 ln x dx  x ln x 
 x ln x 
dv
1
dx
vx

1
x  dx
x

1 dx
 x ln x  x  C