Mathematics Revision Guides – Integration by Parts
Author: Mark Kudlowski
Page 1 of 7
M.K. HOME TUITION
Mathematics Revision Guides
Level: AS / A Level
AQA : C3
Edexcel: C4
OCR: C4
OCR MEI: C3
INTEGRATION BY PARTS
Version : 2.4
Date: 12-01-2015
Examples 4- 6 are copyrighted to their respective owners and used with their permission.
Mathematics Revision Guides – Integration by Parts
Author: Mark Kudlowski
Page 2 of 7
INTEGRATION BY PARTS
This technique is a ‘reverse result’ of the product rule for differentiation.
Recall: If y = uv, where u and v are functions of x, then
dy d
dv
du
.
 (uv )  u  v
dx dx
dx
dx
Integrating with respect to x , we have
du
uv   u dv
dx dx   v dx dx .
Rearranging, we have the standard formula for integration by parts, i.e.
u
dv
dx
dx  uv  v du
dx dx , or, in function notation,
 f ( x) g ' ( x)dx  f ( x) g ( x)  g ( x) f ' ( x)dx .
dv
In order to be able to use this method, the function dx must be integrable.
Additionally, the product v du
dx should not be more difficult to integrate than the original.
Generally, if one of the factors of the product is a polynomial, then set u to be the polynomial, because
its degree will be reduced by 1 by each application of the process.
Example (1): Find i)
i) Take u = 3x
du
dx
 3x sin x dx ii) 
= 3.
 /6
0
3x sin x dx , leaving the answer as a surd in terms of .
dv
Take dx = sin x v = -cos x.
 3x sin x dx =  3x cos x   ( cos x  3) dx
=  3x cos x   3 cos x dx 

 3x cos x  3sin x  c
=
ii) The definite integral is worked out in the usual way:

 /6
0
3x sin x dx =  3x cos x  3sin x0
 /6
=
( ( 2 cos ( 6 ) )  3 sin ( 6 ) )  ( (0 cos (0) )  3 sin(0 ))
=
 3   3 
 0
 
 2
4


=
6  3
.
4
Mathematics Revision Guides – Integration by Parts
Author: Mark Kudlowski
Example (2): Find i)
 xe
du
 xe
=
1
2
xe 2 x   12 e 2 x dx 
=
1
2
xe 2 x  14 e 2 x  c
=
( 12 x  14 )e 2 x  c
1
2
ii)

2
1
xe 2 x dx , leaving the result in terms of e.
Let dx = e2x v = ½ e2x
xe 2 x   ( 12 e 2 x  1) dx

dx =
dx
dv
i) Let u = x dx = 1.

2x
2x
Page 3 of 7
(
ii) The definite integral is
1
2
x  14 )e 2 x

2
1
=
3 4 1 2 e 2 (3e 2  1)
.
e  e =
4
4
4
Sometimes the process might need to be carried out more than once.
Example (3): Find
x
2
cos x dx.
du
Let u = x2 dx = 2x.
dv
Let dx = cos x v = sin x.

x
2
cos x dx = x 2 sin x   2 x sin x dx
The integrand on the right-hand side needs another application of the process, but note how the degree
of the polynomial (the u function) has been reduced from 2 (quadratic) to 1 (linear).
du
Let u = 2x dx = 2.
dv
Let dx = sin x v = -cos x.

 2 x sin x dx =
 2 x cos x   2 cos x dx 
=  2 x cos x  2 sin x  c .
The two results need combining to give the final integral:
(watch out for the + and – signs !)
x
=
2
cos x dx = x 2 sin x   2 x cos x  2 sin x 
x 2 sin x  2 x cos x  2 sin x  c .
Mathematics Revision Guides – Integration by Parts
Author: Mark Kudlowski
Page 4 of 7
Example (4):
x e
2 x
i) Show that
dx =  e  x ( x 2  2 x  2)  c , given that  xe  x dx   e  x ( x  1)  c .
ii) Use the result from i) to find

2
0
x 3e  x dx , giving your answer in the form a – be-2 where a and b
are integer constants.
i) Two applications are needed here.
du
First application: Take u = x2 dx = 2x .

x e
2 x
dv
Take dx = e-x v = -e-x .
dx =  x 2e x    2( xe  x ) dx . Label the right-hand integrand I1.
Second application: This integrand I1 is (-2) times the one quoted in part i),
so the integral of I1 =
2e x ( x  1)  c .

x e
=
 e x ( x 2  2 x  2 )  c .
2 x
ii) To find
dx =  x2e x  2e x ( x  1)  c = e  x ( x 2  2 x  2 )  c

2
0
x 3e  x dx , we work out the indefinite integral first :
du
dv
Take u = x3 dx = 3x2.

x e
3 x
Take dx = e-x v = -e-x .
dx =  x3e x    3( x 2e x ) dx . Label the right-hand integrand I2 .
The integrand I2 on the RHS is (-3) times the integrand
x e
2 x
dx , so its integral is
3e x ( x2  2 x  2 )  c .
The complete integral
=
3 x
dx =  x3e x  3e x ( x2  2 x  2)  c
 x 3e  x  3e  x ( x 2  2 x  2)  c =
Hence

x e
 e

2
0
2

 e  x ( x 3  3x 2  6 x  6)  c .
x 3e  x dx =  e  x ( x 3  3x 2  6 x  6)

2
0
   e (0  0  0  6)  = 6 – 38e -2.
(8  12  12  6) -
0
Mathematics Revision Guides – Integration by Parts
Author: Mark Kudlowski
Page 5 of 7
Integrals involving products of ln x need to be handled a little differently, since ln x is not easy to
integrate, but it can be differentiated. In those cases, ln x should be taken as u.
Example (5): Find
x
du
Take u = ln x  dx =
3
ln x dx .
1
.
x
4
dv
Take dx = x3 v = x
4

 x4
3

=
ln
x

x
ln
x
dx

 4
=
1
4
x 4 ln x  14  x 3 dx
=
1
4
x 4 ln x  161 x 4  c
  x4 1 
      dx
  4 x
Sometimes an expression which does not look like a product can be defined as one, by treating 1 as a
term.
Example (6): Find
 ln x dx.
Treat the expression as ln x 1.
du
Thus u = ln x  dx =
1
.
x
dv
Take dx = 1 v = x
1
 ln x dx  (ln x)  x   ( x  x )dx
= x ln x   1dx = x ln x  x  c .

Mathematics Revision Guides – Integration by Parts
Author: Mark Kudlowski
Page 6 of 7
Example (7):
i) Evaluate

7.2
0
tan 1 x dx .
ii) The velocity v of a performance car (in metres per second) is modelled by the equation
v  84 tan 1 0.12t  where radian measure is used.
Use your knowledge of transformations to evaluate

60
0
84 tan 1 (0.12t ) dt , and hence calculate the
distance travelled in those 60 seconds.
Give your result in metres to 3 significant figures.
Treat the expression as tan-1 x 1.
du
Thus u = tan-1 x  dx =
1
.
1 x 2
dv
Take dx = 1 v = x

=

7.2
0


tan 1 x dx  x tan 1 x 0  
x tan
1
7.2

7.2
0

x
dx
1 x2

x  12 ln(1  x 2 ) 0 = 7.2 tan 1 7.2  12 ln(52.84)  0  0
7.2
= 10.316 – 1.984 = 8.332.
ii) The integrand

60
0
84 tan 1 (0.12t ) dt is obtained by transforming the integrand
by an x-stretch of scale factor

7.2
0
tan 1 x dx
1
followed by a y-stretch of scale factor 84, plus a trivial renaming
0.12
of the variable from x to t.
Now
1
 84  700 , so we multiply the result of part i) by 700 to obtain the distance travelled by
0.12
the car in those 60 seconds – namely 8.332 700, or 5830 metres to 3 significant figures.
Mathematics Revision Guides – Integration by Parts
Author: Mark Kudlowski
Example (8): Find
e
x
Page 7 of 7
sin x dx .
du
Here u = ex  dx = ex .
dv
Also dx = sin x v = -cos x.

e
x
sin x dx = e x  ( cos x)   ( cos x)  e x
 e x cos x   e x cos x dx
=
Applying integration by parts again we have
du
u = ex  dx = ex
dv
dx
= cos x v = sin x.
e
=
x
cos x dx  e x  (sin x)   (sin x)  e x
e x sin x   e x sin x dx
(1)
Combining the two results, we finally have
e
x
sin x dx =  e x cos x  e x sin x   e x sin x dx
The process appears to be going on for ever without simplifying the expression, but notice how the
original integrand now appears on both sides of the equation. Adding this integrand to both sides will
cancel it out from the RHS.
2 e x sin x dx =  e x cos x  e x sin x


e
x
sin x dx  12 (e x sin x  e x cos x)  c
(2)
From the above results in (1) and (2) we can also deduce that
 e cos x dx  e sin x  (e sin x  e cos x)
  e cos x dx  (e sin x  e cos x)  c
x
x
x
1
2
1
2
x
x
x
x