Study Advice Service
Mathematics
Worksheet
Integration 2
This is one of a series of worksheets designed to help you increase your confidence
in handling Mathematics. This worksheet contains both theory and exercises which
cover:1. Integration by Substitution
2. Integration by parts
3. Reduction methods
There are often different ways of doing things in Mathematics and the methods
suggested in the worksheets may not be the ones you were taught. If you are
successful and happy with the methods you use it may not be necessary for you to
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1. Integration by substitution (or change of variable)
Find the integral of ax  b n ie. find I    ax  b  dx
by the Fundamental Theorem of calculus this means that
dI
 ax  b n
dx
du
dI
 a and
 ax  b n  u n
Put u  ax  b then
dx
dx
n
By the Chain Rule
Giving
hence
integrating
dI dI
du


dx du dx
dI
un 
a
du
dI 1 n
 u
du a
1
I   u n du
a

1 u n 1
ax  b n 1
 
c 
c
a n 1
an  1
In practice, when doing the integration, we simplify this to
I =  ax  bn dx
)
u  ax  b gives du  a dx (i.e. dx  du
a
substituting
giving, I =
n
u
du 1 n
ax  bn 1  c
1 u n 1
  u du =

c 
a
a
a n 1
an  1
Examples
1. By substituting u  tan x, evaluate
I   tan5 x sec2 x dx
du
 sec 2 x  du  sec 2 xdx
Substituting u  tan x gives
dx

So  tan5 x sec 2 x dx   tan5 x sec 2 x dx

u6
  u du 
 c
6
5

1 tan6
6
x  c
1
2. By substituting u  3x 4  6 , evaluate the integral I  
Put u  3x 4  6 then
du
 12 x 3  du  12 x 3 dx
dx

x3
(3 x 4  6)5
dx

1
12 x 3
1
12 x 3 dx
dx



12 (3x 4  6) 5
12 (3x 4  6) 5
1 du
1
5
we have I 

 5
 u du
12 u
12
Writing I as
1 u 4
1


 c 
12 4
48u 4
1
 
 c
48(3 x 4  6)4
 c
3
3. Evaluate  04 tan 5 x sec 2 x dx
In example 1 we found, by substituting u  tan x ,  tan 5 x sec 2 x dx = 16 tan 6 x  c
Hence
3
4
0
5
2
tan x sec x dx 


1 tan 6
6
xc
 4   6

3
4
0

 1 tan 6 3  c  1 tan 6 0  c  1  16  c  0  c  1
6
6
6
Notice that the value of c is irrelevant as it cancels out. This is why we do not need
to put it in when doing definite integrals. In practice it is also easier to change the
limits when we do the substitution.
I=
3
4
0
tan 5 x sec 2 x dx .
du
 sec 2 x  du  sec 2 xdx
dx
the limits become tU  tan 3  1, t L  tan0  0
4
Substituting u  tan x gives
 
So
3
1
6
I   0 4 tan5 x sec 2 x dx   01 t 5 dt   61 t 6    61 1  0   61

0


 
Exercise 1
Find the following integrals using the given substitution
dx
1.  x 3x 2  1 dx, u  3x 2  1
2. 
u  4  3x
3


4

3
x
4
3.  5 x 3 x 4  3 dx, u  x 4  3
4.  sin3 x dx, u  cos x
e 3x
6.  x 3x  1 dx, u  3x 1
5. 
dx, u  e 3 x
3x
e 2


2
By using an appropriate substitution find the following integrals.
7.
10.
13.
t2
 te
dt
8.
x
a 2  x 2 dx

x
1  x2
cos v
 1  sin v dv
e 3 x 1
11.

14.
2
3
 1  3x 5  x  x dx
dx
1  e 3x

dx

5
p5
1
0
9.

12.
2
3  p6
t
3
3t 2  5
dp
dt

15.
5
02 cos x sin x dx
2 Integration by parts
It is not possible to integrate the function xe x by any of the methods outlined
2
previously. The more complicated looking function xex can be done by substituting
u  x 2 but xe x can only be integrated as a product.
du
etc., differentiating a product ( uv ) gives the rule
dx
d
uv   u ' v  uv'
dx
uv   u ' v dx   uv' dx
Integrating both sides we get
Writing u ' for
which gives the basic rule for integrating a product
 uv' dx  uv   u ' v dx
On the left hand side we have u and v ' and we need u ' and v on the right hand
side. This implies that we must be able to integrate one of the functions (easily) and
differentiate the other (always possible). In some cases there is no problem in
deciding which is which. For instance if one of the functions is ln x we need to
differentiate it as ln x cannot be integrated at sight.
If both functions can be integrated you should be able to decide which is which with
experience. If your first choice doesn’t work try again!!
Examples
1. Evaluate
 xe
x
dx
Both x and e x can be integrated so which do we choose?
First attempt:
Take u  x and v'  e x then u' 1 and v  e x
Using  uv' dx  uv   u ' v dx
we get
 xe
x
dx  xex  1 e x dx  xex  e x  c
3
Second attempt:
Take u  e x and v '  1x then u '  e x and v = 12 x 2
Using
 uv' dx  uv   u ' v dx
we get  xe x dx  e x

 x    e  x  dx
1 2
2
1 x 2e x
2
x 1 2
2
 21  x2e x dx
The last part  x 2e x dx is ‘worse’ than we started with as it involves the product of x 2
and e x whereas in the first attempt we could do the final integral immediately, in the
second attempt we’d have to try using integration by parts again (in practice you’d
come back to where you started!)
In general if both functions can be integrated easily and you have a power of x , then
differentiate the x -term.
Function
 x sin x dx
t
Differentiate
x
2 t
1
2. Evaluate
sin x
t2
e dt
 x tan x
Integrate
tan 1 x
dx
x
et
x
(as you can’t integrate tan1 x directly)
2
ln x dx
Both terms can be differentiated but ln x cannot be integrated immediately
1
x3
Take u  ln x and v'  x 2 then u ' and v 
x
3
Using  uv' dx  uv   u ' v dx and writing the function in the order uv' gives
 
x
2

ln
x
x
dx

ln
x

 3

3
3
  1  x dx
 x 3

x 3 ln x 1 2
x 3 ln x x 3

  x dx 

c
3
3
3
9
  
 1 x 3 3 ln x  1  c  1 x 3 ln x 3  1  c
9
9
Notice that the answer can be written in a number of different forms, all of which are
correct. Sometimes the question dictates which form to leave the answer in.
4
3. Evaluate

 0 x sin x dx
Take u  x and v'  sin x then u'  1 and v   cos x



 0 x sin x dx  x cos x 0   0 1   cos x dx
  x cos x  sin x  0

  cos   sin   0
   1  
4. Evaluate
x
2 3 x
e
dx
Take u  x 2 and v '  e 3 x
Giving I =
x
2 3 x
e
then u' 2 x and v   13 e 3 x




dx  x 2  1 e3x   2 x  1 e3x dx
3
3
2

3
x

3
x
 1 x e
 2 xe dx
3
3
3 x

we now need to integrate
 xe
dx by parts
It is important that this is not done separately but everything is kept together
otherwise some parts may get lost.
Take u  x and v '  e 3 x then u' 1 and v   13 e 3 x
We get
x
2 3 x
e
dx   13 x 2 e 3 x  23  xe3 x dx

x


1 e 3 x  1 e 3 x  c
3
9
  13 x 2 e 3 x  23 x  13 e 3 x    13 e 3 x dx
  13 x 2 e 3 x  23
2 e 3 x  c
  13 x 2 e 3 x  92 xe 3 x  27


1 e 3 x 9 x 2  6 x  2  c
  27
Exercise 2
Use the method of integration by parts to find the following
2.  x lnx  2  dx
1.  x sin x dx
3.  u 2 sin u du
ln t
4.  s 2 ln s ds
6.  sin 1 w dw
5. 
dt
3
t
7.  y 2 e 3 y dy
9.  x x  1 dx
8.  xx  14 dx
10.
13.
1
0
4
2
xx  15 dx
t  1ln2t  dt
11.
6
3
x
x2
dx
12.

14.
0
4
u cos 2u du
15.

2 3 s
s e
0
4 ln 
1  2
ds
d
5
3. Reduction methods
 cos
n
x dx can be evaluated, fairly easily, for some values of n, for instance.
a)  cos 2 x dx . From cos 2 x  2 cos2 x  1 we have cos2 x  1 cos 2 x  1


2
hence  cos 2 x dx  12  cos 2 x  1 dx  12 12 sin 2 x  x  c


b)  cos 3 x dx   1  sin2 x cos x dx using cos 2 x  1  sin 2 x
Substituting s  sin x,  ds  cos x dx giving


3
2
2
3
3
 cos x dx   1  sin x cos x dx   1  s ds  s  13 s  c  sin x  13 sin x  c
c)
 cos
4
x dx 
 12 1  cos 2 x  dx = 41  1  2 cos 2 x  cos
 1  1  2 cos 2 x  1 1  cos 4 x  dx
4
2
 1 x  sin 2 x   1 x  1 sin 4 x   c
4
8
4
2
2
2 x dx
What about  cos 20 x dx or  cos31 x dx ? In cases like this a reduction method,
based on integration by parts, is useful but not necessarily the only method.
Examples

1. If I n = cos n x dx (a) show that I n 
1
n 1
sin x cos n 1 x 
In2
n
n
(b) Find the values of (i) I 5 (ii) I 4
(a) Writing cos n x as cos x  cos n1 x we have I n   cos x cos n1 x dx
Using  uv' dx  uv   u ' v dx and
taking u  cos n1 x and v'  cos x
then u'  n  1 cos n2 x sin x  and v  sin x
then
I n  cos n 1 x sin x   ( n  1) cos n 2 x (  sin x )  sin x dx
 cos n 1 x sin x  ( n  1) cos n 2 x sin 2 x dx
 cos n 1 x sin x  ( n  1) cos n 2 x (1  cos 2 x ) dx
 cos n 1 x sin x  ( n  1) cos n 2 x  cos n x dx
 cos n 1 x sin x  ( n  1) cos n 2 x dx  ( n  1) cos n x dx
I n  cos n 1 x sin x  ( n  1)I n 2
 ( n  1)I n
I n ( n  1)I n  nI n  cos n 1 x sin x  ( n  1)I n 2
1
n 1
I n  sin x cos n 1 x 
I n2
n
n
6
(b) (i) From the formula I 5  1 sin x cos 4 x  4 I 3
5
5
but I 3  1 sin x cos 2 x  2 I 1 and I 1   cos x dx  sin x  C
3
3


hence I 5  1 sin x cos 4 x  4 1 sin x cos 2 x  2 sin x  C 
5
5 3
3
 1 sin x cos 4 x  4 sin x cos 2 x  8 sin x  K
5
15
15
(b) (ii) From the formula I 4  1 sin x cos 3 x  3 I 2
4
4
but I 2  1 sin x cos x  1 I 0 and I 0 
2
2

0
 cos x  dx   1 dx  x  C

hence I 4  1 sin x cos 3 x  3 1 sin x cos x  1 x  C 
4
4 2
2
 1 sin x cos 3 x  3 sin x cos x  3 x  K
4
8
2. Find a reduction formula for
x
8
n x
e dx
taking u  x n and v  e x then u = nx n 1 and v  e x
we get I n  x ne x   nx n 1e x dx  x ne x  n  x n 1e x dx  x ne x  nI n 1
Hence reduction formula is I n  x ne x  nI n 1
To evaluate
x e
5 x
dx we have
I 5  x5e x  5I 4
I 4  x 4 e x  4I 3
I3  x3e x  3I 2
I 2  x 2e x  2I1
I1  xex  I 0
I 0   e x dx  e x  c
Combining these (very carefully!!)
I1  xex  I 0  xex  e x  C


I 2  x2e x  2I1  x2e x  2( xex  e x  C )  x2  2 x  2 e x  2C


I3  x3e x  3I 2  x3e x  3 x 2  2 x  2 e x  2C


 x3  3x 2  6 x  6 e x  6C



I 4  x 4 e x  4 I 3  x 4 e x  4 x 3  3x 2  6 x  6 e x  6C



 x 4  4 x 3  12 x 2  24 x  24 e x  24C


I 5  x 5 e x  5 I 4  x 5 e x  5 x 4  4 x 3  12 x 2  24 x  24 e x  24C



 x 5  5 x 4  20 x 3  60 x 2  120 x  120 e x  K
7
Hence
x e
5 x


dx  x5  5 x 4  20 x3  60 x 2  120 x  120 e x  K
Exercise 3
Establish reduction formulas for the following
1.
2.
 sin x dx hence find  sin x dx,  sin
n
3
 xln x dx hence find  xln x dx
5
n
4
x dx
ANSWERS
Exercise 1
1.
3x 2  1
9
3
2
2.
c
4. 1 cos 3 x  cos x  c
3
2
7. 1 e t  c
2

10.  1 a 2  x 2
3

3
2
13.  1  x 2  c
c
1
64 x  32

c

3.
x 4  35  c
4
5
3
5. 1 ln e 3 x  2  c
6. 2 (3x  1) 2  2 (3x  1) 2  c
8.  ln1  sin v   c
9. 1 ln 4
3


45
6
27
3
7 
ln 1  e 3 x
11.
c
3e
12. 1 ln 22
6
14. 1 5  x  x 3  c
15. 1

6

6
6
8
Exercise 2
2


3. 2  u 2 cos u  2u sin u  c
5. 
1  2 ln t
4t
2




2. 1 x 2  4 lnx  2   1 x 2  4 x  c
1. sin x  x cos x  c

4

4. 1 s 3 ln s  1  c
3
3
6. w sin 1 w  1  w 2  c
c
 y2 2y 2 
 9y2  6y  2 
c
7. e 3 y 

   c  e3y 
 3



9 27 
27



8.
x  15 5 x  1  c
2x  1 2 3x  2 
9.
c
15
3
30
10.  1
11. 26
42
3
12. 2e 2  6
Exercise 3
n 1
1. I n 
I n  2  1 cos x sin n1 x
n
n

13. 4 ln 8  1

I 5   1 cos x 3 sin 4 x  4 sin3 x  8  c
15

14.   2
8
15. 3  2 ln 2
4

I 4   1 2 sin3 x cos x  3 sin x cos x  3x  c
8

2. I n  1 x 2 ln x n  nI n1
2



I 3  1 x 2 4ln x 3  6ln x 2  6 ln x  3  c
8
We would appreciate your comments on this worksheet, especially if
you’ve found any errors, so that we can improve it for future use.
Please contact the Maths tutor by email at studyadvice@hull.ac.uk
Updated 29th November 2004
The information in this leaflet can be made available in an
alternative format on request. Telephone 01482 466199
© 2009
9