integration by parts with examples

1. “Teach A Level Maths”
Vol. 2: A2 Core Modules
25: Integration by Parts
© Christine Crisp

2. Integration by Parts
Module C3 Module C4
AQA Edexcel
MEI/OCR OCR
"Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with
permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages"

3. 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

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

5. Integration by Parts
So, d ( x sin x )
 sin x  1  x  cos x
dx
Integrating this equation, we get
  
d ( x sin x )
dx  sin x dx  x cos 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 
 x cos x dx
Can you see what this has to do with integrating a
product?

6. 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.

7. 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
 x sin x  (  cos x )  C
 x sin x  cos x  C
So, we’ve integrated xcos x !
We need to turn this method ( called integration by
parts ) into a formula.

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

9. 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

10. 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

11. 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 dv ,
and integrate the other to get v
dx

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

13. Integration by Parts
So, dv
u  2 x and  sin x
dx
differentiate integrate
du
2 v   cos x
dx
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 u v

 (  cos x ) 2 dx
v du
dx dx

14. Integration by Parts
So, dv
u  2 x and  sin x
dx
differentiate integrate
du
2 v   cos x
dx
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 ) 
 (  cos x ) 2 dx
 2 x cos x 
 2 cos x dx
  2 x cos x  2 sin x  C
The 2nd term needs integrating

15. Integration by Parts
e.g. 2 Find
 xe 2 x dx

dv

du
u dx  uv  v dx
dx dx
Solution:
dv
u  x and  e2x
differentiate
dx 2x integrate
du e
1 v
dx 2
This is a compound function,
 e2x  so we mustx be careful.
 e2 
So,  xe dx  ( x )  
   1 dx
2x
 2  
  2 

xe 2 x e2x xe 2 x e 2 x

2

 2
dx 
2

4
C

16. Integration by Parts
Exercises
Find
1.
 xe dx 2.
 x cos 3 x dx
x
Solutions:
 xe dx  xe   e dx
x x x
1.
 xe  e  C
x x
 sin3 x 

sin3 x
2.
 x cos 3 x dx  ( x ) 3  
  3
dx
x sin 3 x cos 3 x
  C
3 9

17. 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.
 xe x dx  xe x   e x dx
 xe x  e x  C
 0
1 x
xe dx 
 xe x x 1
e 0 
 1e 1
e 1
  0e 0
e 0

 0   1  1

18. 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.

19. 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.

20. Integration by Parts
 
dv du
e.g. 3 Find  x ln x dx u dx  uv  v dx
dx dx
Solution:
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.

21. Integration by Parts
 
dv du
e.g. 3 Find  x ln x dx u dx  uv  v dx
dx dx
So, u  ln x
dv
x
dx integrate
differentiate du 1 x2
 v
dx x 2
x2 x2 1
  x ln x dx 
2
ln x 
  dx
2 x
The r.h.s. integral still seems to be a product!
BUT . . . x cancels. 2
 
x x
So, x ln x dx  ln x  dx
2 2
x2 x2

 x ln x dx 
2
ln x 
4
C

22. Integration by Parts
e.g. 4
 x 2 e  x dx
 
dv du
u dx  uv  v dx
dx dx
Solution:
Let dv
u  x and
2
 ex
dx
du
 2x v  e  x
dx
  x 2 e  x dx   x 2 e  x    2 xe  x dx
  x 2 e  x dx   x 2 e  x 
I1
 2 xe  x 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

23. Integration by Parts
e.g. 4
 x 2 e  x dx
Solution: I1   x 2e  x  I 2 . . . . . ( 1 )
I 2   2 xe  x dx
dv
Let u  2x and  ex
du
dx
2 v  e  x
dx
So, I 2   2 xe  x    2e  x dx
  2 xe  x 
 x
2e  x dx
  2 xe  x  2e  C
Substitute in ( 1 )
 x e dx   x e
2 x 2 x x x
 2 xe  2e C

24. Integration by Parts
The next example is interesting but is not essential.
Click below if you want to miss it out.
Omit Example
 
dv du

e.g. 5 Find e sin x dx
x
u dx  uv  v dx
dx dx
Solution:
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.

25. Integration by Parts
 
dv du
e.g. 5 Find  e sin x dx
x
u dx  uv  v dx
dx dx
Solution: dv
ue x
 sin x
dx
du
 ex v  cos x
dx
  e x sin x dx   e x cos x    e x cos x dx
  e cos x   e x cos x dx
x
We write this as: I1  e cos x  I 2
x

26. Integration by Parts
 
dv du

e.g. 5 Find e sin x dx
x
u dx  uv  v dx
dx dx
So, I1  e x cos x  I 2
I 1   e x sin x dx and I 2  e x cos x dx
where

We next use integration by parts for I2
dv
ue x
 cos x
du dx
 ex v  sin x
dx
 e x cos x dx  e x sinx   e x sin x dx
 I 2  e sinx  I1
x

27. Integration by Parts
 
dv du

e.g. 5 Find e sin x dx
x
u dx  uv  v dx
dx dx
So, I1  e x cos x  I 2
I 1   e x sin x dx and I 2  e x cos x dx
where

We next use integration by parts for I2
dv
ue x
 cos x
du dx
 ex v  sin x
dx
 e x cos x dx  e x sinx   e x sin x dx
 I 2  e sinx  I1
x

28. Integration by Parts
 
dv du

e.g. 5 Find e sin x dx
x
u dx  uv  v dx
dx dx
So, 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 

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

30. Integration by Parts
 
dv du

e.g. 5 Find e sin x dx
x
u dx  uv  v dx
dx dx
So, I1  e x cos x  I 2 . . . . . ( 1 )
I 2  e x sin x  I 1 .....(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

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

32. Integration by Parts
Solutions:
dv
1. x sin x dx Let u  x and  sin x
2 2
I1 du dx
 2x v  cos x
dx
I 1   x 2 cos x    2 x cos x dx
 I 1   x cos x 
dv
2
2 x cos x dx . . . . . ( 1 )
I2
For I2: Let u  2 x and  cos x
du dx
2 v  sin x
dx
 I 2  2 x sin x   2 sin x dx
 2 x sin x  2 cos x  C
Subs. in ( 1 )
  x 2 sin x dx   x 2 cos x  2 x sin x  2 cos x  C

33. Integration by Parts
2 2
2.
1 ln x dx  1 1 ln x dx This is an important
Let u  ln x and dv application of
1
du 1 dx integration by parts
 vx
dx x

1
  1 ln x dx  x ln x  x  dx
x


 C
x ln x  1 dx
x ln x  x
  x ln x  x 
2
1 ln x dx
2
So, 1
  2 ln2  2    1 ln1  1 
 2 ln 2  1

34. Integration by Parts

35. 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.

36. 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

37. Integration by Parts
e.g. Find
 xe 2 x dx

dv du
u dx  uv  v dx

dx dx
Solution:
dv
u x and  e2x
differentiate
dx 2x integrate
du e
1 v
dx 2
 e2x   e2x 
So,  xe dx  ( x )     1 dx
2x
 2  
 2 
   
xe 2 x e2x xe 2 x e 2 x

2

 2
dx 
2

4
C

38. 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.

39. Integration by Parts
 
dv du
e.g. 3 Find  x ln x dx u dx  uv  v dx
dx dx
We can’t integrate ln x so,
dv
u  ln x x
dx integrate
differentiate du 1 x2
 v
dx x 2
x2 x2 1
  x ln x dx 
2
ln x 
  dx
2 x
The r.h.s. integral still seems to be a product!
BUT . . . x cancels.
So,  x ln x dx 
x2 x2  C
ln x 
2 4

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