LESSON 13 INTEGRALS WITH DISCONTINUOUS INTEGRANDS
Definition If the function f is continuous on the interval [ a , b ) and discontinuous
at x  b , then

b
f ( x ) dx  lim 
t  b
a

t
f ( x ) dx .
a
NOTE: Recall that a definite integral exists if the integrand is continuous on the
closed interval [ a , b ] of integration. Since the integrand f is continuous on the
interval [ a , b ) , then it is continuous on the closed interval [ a , t ] for all a  t  b .
Thus,

t
a
f ( x ) dx exists for all a  t  b . Of course, if t  a , then since the
function f is continuous at t  a , then the function f is defined at t  a and

a
a
f ( x ) dx  0 .
Definition If the function f is continuous on the interval ( a , b ] and discontinuous
at x  a , then

b
f ( x ) dx  lim 
t  a
a

b
f ( x ) dx .
t
Since the integrand f is continuous on the interval ( a , b ] , then it is continuous on
the closed interval [ t , b ] for all a  t  b . Thus,

b
t
f ( x ) dx exists for all
a  t  b . Of course, if t  a , then since the function f is continuous at t  a ,
then the function f is defined at t  a and
COMMENT: The integral

b
a

a
a
f ( x ) dx  0 .
f ( x ) dx in the definitions above are also called
improper integrals because of the integrand has a discontinuity on the closed
interval [ a , b ] . If the limit in the definition exists, we say that the improper
integral converges. If the limit does not exist, then we say that the improper
integral diverges.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Definition If the function f has a discontinuity at x  c in the open interval
( a , b ) but is continuous elsewhere in the closed interval [ a , b ] , then

b
a
f ( x ) dx =

c
a
f ( x ) dx +

b
c
f ( x ) dx =
lim 
r  c

r
a
f ( x ) dx + lim 
t  c

b
t
f ( x ) dx
provided that both of the improper integrals converge.
Examples Determine whether the following improper integrals converge or
diverge. If the integral converges, then give its value.
1.

8
0 3
3x
64  x 2
dx
NOTE: The integrand of f ( x ) 
3x
3
64  x 2
is continuous on its domain of
definition, which is the set of real numbers given by { x : x   8 and x  8} .
The interval of integration is the closed interval [ 0 , 8 ] . Thus, the integrand
has a discontinuity at x  8 , which is the upper limit of integration. Thus,
the Fundamental Theorem of Calculus can not be applied to this integral. We
will need to do the following.

8
0 3
3x
64  x 2
dx =
lim 
t  8

t
0 3
3x
64  x 2
dx
The integrand is continuous on the closed intervals [ 0 , t ] for all 0  t  8 .
Thus, the Fundamental Theorem of Calculus can be applied to these closed
intervals.

3x
3
64  x
2
dx
2
Let u  64  x
Then du   2 x dx
Copyrighted by James D. Anderson, The University of Toledo
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

3x
3
64  x
3
dx =
2
2

 2x
64  x
3
2
dx = 
3
2

du
3

=
3 u
2
u
 1/ 3
du =
3 3 2/3
9
 u  c =  ( 64  x 2 ) 2 / 3  c
2 2
4
Thus,

8
0 3
3x
64  x
2
dx =
lim
t  8

t
3x
t
0 3
64  x 2
9
dx = lim    ( 64  x 2 ) 2 / 3  =

t  8 
 4
0

9
9
lim  [ ( 64  t 2 ) 2 / 3  64 2 / 3 ] = 
lim  [ ( 64  t 2 ) 2 / 3  16 ] =
4 t8
4 t8

9
( 0  16 ) = 36
4
Answer: Converges; 36
2.

3/ 2
3
dy
9  y2
NOTE: The integrand of f ( y ) 
1
is continuous on its domain of
9  y2
definition, which is the interval (  3, 3 ) . The interval of integration is the

3 

3
,
 . Thus, the integrand has a discontinuity at
closed interval 
2 

y   3 , which is the lower limit of integration. Thus, the Fundamental
Theorem of Calculus can not be applied to this integral. We will need to do
the following.

3/ 2
3
dy
9  y2
=
lim
t   3

3/ 2
t
dy
9  y2
Copyrighted by James D. Anderson, The University of Toledo
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
3 
 for all
The integrand is continuous on the closed intervals  t ,
2 

3
3  t 
. Thus, the Fundamental Theorem of Calculus can be applied
2
to these closed intervals.
dy

9  y2
1
= sin
y
 c
3
Thus,

dy
3/ 2
9  y2
3
lim
t  3

=
lim
t   3

t
 1 1
t
 sin
 sin  1  =

3 
2

3/
dy
3/ 2
9  y2
lim
t  3

=
lim
t   3
 1 y 
 sin 3 
t
2
=

1 t 
  sin
 =
3
4
   

3
   =
 sin  1 (  1) =
4  2
4
4
Answer: Converges;
3.

4
6
3
4
5t 2
dt
t 3  64
5t 2
NOTE: The integrand of f ( t )  3
is continuous on its domain of
t  64
definition, which is the set of real numbers given by {t : t   4 } . The
interval of integration is the closed interval [  6 ,  4 ] . Thus, the integrand
has a discontinuity at of t   4 , which is the upper limit of integration.
Thus, the Fundamental Theorem of Calculus can not be applied to this
integral. We will need to do the following.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860

4
6
5t 2
dt =
t 3  64
lim
r   4

r
6
5t 2
dt
t 3  64
The integrand is continuous on the closed intervals [  6 , r ] for all
 6  r   4 . Thus, the Fundamental Theorem of Calculus can be applied to
these closed intervals.

5t 2
dt
t 3  64
3
Let u  t  64
2
Then du  3t dt

5t 2
5
3t 2
5 du
5
5
3
dt =  3
dt = 
= ln u  c = ln t  64  c
3
t  64
3 t  64
3 u
3
3
Thus,

4
6
5
3
5t 2
dt =
t 3  64
lim
r  4

lim
r   4

r
6
5t 2
dt =
t 3  64
( ln r 3  64  ln  216  64 ) =
  since
r
lim
r   4
5
3
lim
5

3
ln
t

64
=
 3

6
r  4

lim  ln r 3  64   
r  4
Answer: Diverges
4.

3
1
dx
x2
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( ln r 3  64  ln 152 ) =
1
is continuous on its domain of
x2
definition, which is the set of real numbers given by { x : x  0} . The interval
of integration is the closed interval [  1, 3 ] . Thus, the integrand has a
discontinuity at x  0 , which is in the interval of integration. Thus, the
Fundamental Theorem of Calculus can not be applied to this integral. We
will need to do the following.
NOTE: The integrand of f ( x ) 

3
1
dx
=
x2
0

1
dx
+
x2

3
0
dx
=
x2
lim 
r  0

r
1
dx
+
x2
lim 
t  0

3
t
dx
x2
The integrand is continuous on the closed intervals [  1, r ] for all
 1  r  0 and on the closed intervals [t , 3 ] for all 0  t  3 . Thus, the
Fundamental Theorem of Calculus can be applied to these closed intervals.
dx
=
x2


x
2
x1
1
 c =   c
dx =
1
x
Thus,

0
1
dx
=
x2
lim 
r  0
lim
r  0

r
1
dx
=
x2
r
 1
1

lim    
 lim    1    since
=
r  0 
r  0  r
x  1

1
 
r
dx
diverges, we do not have to
1 x2
3 dx
determine whether the second improper integral  0 2 converges or
x
3 dx
diverges. Even if it converges, the improper integral   1 2 can not
x
converge.
Since the first improper integral

0
Answer: Diverges
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
5.

1
4
11  3 x
dx
x2  4
11  3 x
is continuous on its domain of
x2  4
definition, which is the set of real numbers given by { x : x   2 and x  2 } .
The interval of integration is the closed interval [  4 , 1] . Thus, the integrand
has a discontinuity at x   2 , which is in the interval of integration. Thus,
the Fundamental Theorem of Calculus can not be applied to this integral. We
will need to do the following.
NOTE: The integrand of f ( x ) 

1
4
lim
11  3 x
dx =
x2  4
r   2

r
4

2
4
11  3 x
dx +
x2  4
11  3 x
dx +
x2  4
lim
t   2

1
t

1
2
11  3 x
dx =
x2  4
11  3 x
dx
x2  4
The integrand is continuous on the closed intervals [  4 , r ] for all
 4  r   2 and on the closed intervals [t , 1] for all  2  t  1 . Thus,
the Fundamental Theorem of Calculus can be applied to these closed
intervals.

11  3 x
dx
x2  4
We will find the partial fraction decomposition for
11  3 x
=
x2  4
11  3 x
. Thus,
( x  2)( x  2)
11  3 x
A
B

=
.
( x  2)( x  2)
x  2
x 2
Multiplying both sides of this equation by ( x  2 ) ( x  2 ) , we obtain the
following equation.
11  3 x  A ( x  2 )  B ( x  2 )
Copyrighted by James D. Anderson, The University of Toledo
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To solve for A, choose x   2 : 17   4 A  A  
5
4
To solve for B, choose x  2 : 5  4 B  B 
Thus,
17
4
11  3 x
A
B
11  3 x

=
=
=
( x  2)( x  2)
x  2
x 2
x2  4
17
5
1 5
17 
4  4


= 
x  2
x 2
4x  2
x  2 

Thus,

1
11  3 x
dx =
2
x 4
4

 5
17 

 dx =

x

2
x

2


1
1
( 5 ln x  2  17 ln x  2 )  c = ( ln x  2
4
4
5
 ln x  2
17
)  c =
1
( x  2) 5
ln
c
4
( x  2 ) 17
Thus,

1
4
1
4
2
4
11  3 x
dx =
x2  4
lim
r  2


r
4
11  3 x
dx =
x2  4
lim

(r  2) 5
(  6) 5 
1
 ln
 ln
17
17  =
( 2) 
4
 (r  2)
lim

(r  2) 5
2 53 5 
1

ln
 ln

17
2 17  = 4
 ( r  2)
r   2
r   2
r
lim
1
( x  2) 5 
 ln

( x  2 ) 17   4 =
4
lim
 ( r  2) 5
65 
 ln 17  =
 ln
17
(
r

2
)
2 

r   2
r   2
lim
r   2
 (r  2) 5
35 
 ln 12   
 ln
17
(
r

2
)
2 

Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
since
r
r
( r  2) 5
lim ln

  2
( r  2 ) 17
( r  2) 5
lim ln
  since
  2
( r  2 ) 17
lim
r  2

( r  2) 5
  since
( r  2 ) 17
lim
r   2
lim
r  2

( r  2) 5

( r  2 ) 17
1
   and
( r  2 ) 17
lim
r  2

( r  2 ) 5 is
finite (It is a finite negative number.)
Thus, the improper

1
4

2
4
11  3 x
dx diverges. Thus, the improper
x2  4
11  3 x
dx diverges.
x2  4
Answer: Diverges
6.


6
dx
x
x2  6
NOTE: The integrand of f ( x ) 
1
x
x2  6
is continuous on its domain of
definition, which is the interval (   ,  6 )  ( 6 ,  ) . The interval of
integration is the closed interval [ 6 ,  ) . Thus, the integrand has a
discontinuity at x  6 , which is the lower limit of integration. Thus, the
Fundamental Theorem of Calculus can not be applied to this integral. We
will need to do the following.
The integral


dx
x x2  6
and a discontinuity. However,
6
is improper with an infinite limit of integration
Copyrighted by James D. Anderson, The University of Toledo
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




 lim  lim 
t   r  ( 6)
x2  6

dx
6
x
dx
6

x2  6
x
lim
r  ( 6 )

r

 and
2
x  6 
dx
t
x

t
 lim 
 t   r x


2
x  6 
dx
because we want only one limit for the infinite limit of integration and one
limit for the discontinuity. In order to accomplish this, we will use the
following property of definite integrals.

b
a
f ( x ) dx =

c
a
f ( x ) dx +

b
c
f ( x ) dx ,
where c is a real number in the open interval ( a , b ) .
Since 3 is in the open interval ( 6 ,  ) , then we may write


dx
6
=
x2  6
x
lim
r  ( 6 )

dx
3
r

x
dx
3
6
x
+
x2  6
x2  6
3
dx
x
x2  6
=
dx
t
+ t lim
  3


x
x2  6
NOTE: You can use any real number in the open interval ( 6 ,  ) .
The integrand is continuous on the closed intervals [ r , 3 ] for all
6  r  3 and on the closed intervals [ 3 , t ] for all t  3 . Thus, the
Fundamental Theorem of Calculus can be applied to these closed intervals.

dx
x
x2  6
=
1
x
sec  1
c
6
6
Thus,
Copyrighted by James D. Anderson, The University of Toledo
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3

dx
3
6
1
6
1
6
=
x 6
2
x
lim
r  ( 6 )
lim
r  ( 6 )

r
x
x 6
x
x2  6

Now, consider

3
1
6
lim
r  ( 6 )
 1

1 x
sec

 =
6  r
 6


 sec  1 3  sec  1 1 =


6


1
3
sec  1
6
6
dx
3
6
=
2


 sec  1 3  sec  1 r  =

6
6 



 sec  1 3  0  =


6


Thus,

dx
3
converges to
1
3
sec  1
.
6
6
dx
x2  6
x
t


3
dx
x
= t lim
  3
x2  6

1
lim  sec  1
6 t   
t
lim sec  1

t  
6
Thus,
Thus,




3
t
3 
 sec  1
=
6
6 
1
6
 

  sec  1 3  since
 2
6 

converges to
1
6
 

  sec  1 3  .
 2
6 


2

x2  6
dx
x
x2  6
dx
3
6
x
x
dx
x
6
 1

1 x
lim
sec

 =
= t
6  3
x2  6
 6
dx
t
x2  6
+

converges and

3
dx
x
x2  6
=


dx
x2  6
=
1
3
sec  1
+
6
6
1
6
6
x
Copyrighted by James D. Anderson, The University of Toledo
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

  sec  1 3  =
 2
6 


2 6

=
6
12

Answer: Converges;
7.

6
12
dt
5
5
t
25  t 2
1
NOTE: The integrand of f ( t ) 
is continuous on its domain of
t 25  t 2
definition, which is the interval (  5 , 0 )  ( 0 , 5 ) . The interval of integration
is the closed interval [  5 , 5 ] . Thus, the integrand has discontinuities at
t   5 , which is the lower limit of integration, t  5 , which is the upper
limit of integration, and t  0 , which is in the interval. Thus, the
Fundamental Theorem of Calculus can not be applied to this integral. We
will need to do the following.
NOTE: We need one integral which will handle the discontinuity of t   5
as a lower limit of integration. We need two integrals that will handle the
discontinuity of t  0 as an upper limit of integration in one integral and as a
lower limit of integration in the other. We need one integral that will handle
the discontinuity of t  5 as an upper limit of integration. Thus, we need
four integrals. Since  1 is in the open interval (  5 , 0 ) and 1 is in the open
interval ( 0 , 5 ) , then we may write the following


dt
5
5
t
=
dt
5
1
25  t 2
t
25  t 2
=

1
5
lim
dt
r   5
+
25  t 2
t

1
r

dt
0
1
t
dt
t
25  t 2
+
+
25  t 2
lim 
u  0

Copyrighted by James D. Anderson, The University of Toledo
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0
t
25  t 2
dt
u
1

dt
1
t
25  t 2
+
+

lim 
v  0
dt
1
v
t
25  t 2
+
lim
w  5

dt
w
1
t
25  t 2
The integrand is continuous on the closed intervals [ r ,  1] for all
 5  r   1 , the closed intervals [  1, u ] for all  1  u  0 , the closed
intervals [v , 1] for all 0  v  1 , and the closed intervals [1, w ] for all
1  w  5 . Thus, the Fundamental Theorem of Calculus can be applied to
these closed intervals.

dt
t
25  t 2
BE CAREFUL: This integral is close to the form of the integral whose
dt
answer would involve the inverse secant function since 
=
t t 2  25
1
t
sec  1  c . We will need to use Trigonometric Substitution in order to
5
5
dt
evaluate 
t 25  t 2
Let t  5 sin  , where 


   . Then dt  5 cos  d .
2
2
2
2
2
2
Also, 25  t  25  25 sin  = 25 ( 1  sin  ) = 25 cos  .
25  t 2  25 cos 2   5 cos   5 cos  since cos   0






when
. Thus, we have that
2
2
Thus,

dt
t
25  t 2
=

5 cos  d
1
=
( 5 sin  ) ( 5 cos  )
5

d
1
=
sin 
5
1
ln csc   cot   c
5
Copyrighted by James D. Anderson, The University of Toledo
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 csc  d
=
Since t  5 sin  , then sin  
t
5
. Thus, csc  
Using right triangle
5
t
25  t 2
trigonometry, we have that cot  
.
t
5
t

25  t 2
Thus,
dt

t
1
5
ln

5
t
25  t
2
25  t 2
t
=
1
5
 csc  d
=
5
1
 c = ln
5
1
ln csc   cot   c =
5
25  t 2
t
c
Thus,

1
5
1
5
1
dt
5
t 25  t 2
lim
r   5

 ln ( 5 

lim  ln
r  5
5
=
lim
r   5

1
t 25  t 2
r
5
24 )  ln
25  r 2
r
dt

=
lim
r   5
1
5
 ln
 5
25  r 2 
 = 1 ln ( 5 
r
5

5  25  25
1
ln
5
5

Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
25  t
t
1
2
24 ) since
1
1
ln 1  ( 0 )  0
5
5



r
=

Thus, the improper integral
1
5
dt
t 25  t 2
Now, consider the improper integral

dt
0
1
t
25  t 2
=

lim 
u  0

5
1
lim   ln
5 u0 

25  u 2
u

5
1
lim   ln
5 u0 

25  u 2
u
lim 
5
u  0
25  u 2
u
=
1
Since
lim
u  0
25  t 2
.
1
5

lim
ln
= u  0 5

25  t
t
u
2

24 ) 

 ln ( 5 
5
25
0
0

0
 1

2  1/ 2

(
25

u
)
(

2
u
)
 2
 =
25  u 2
u
Thus, the improper integral
24 ) .
5  24 
 =
 ln
1

25  u 2
 0 , then
u

5
1
 ln
Thus, 5 u lim

 0

t
25  t 2
t
25  u 2
lim
= u lim
 0
u  0
u
u
0
lim 

0
2 1/ 2
u  0 ( 25  u )
5
5
1
1
ln ( 5 
5
dt
0
dt
u
5

converges to

u  0
 ln ( 5 
25  u 2
u

24 )    

dt
0
1
lim ln
5
t
25  t 2
diverges.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
  .


=

1

Since the improper integral
integral

5
dt
5
t 25  t 2
dt
0
1
t
25  t 2
diverges, then the improper
diverges.
Answer: Diverges
NOTE: If you want the practice, you can show that the improper integral
1
dt
 0 t 25  t 2   , which implies that it diverges, and the improper integral

8.

dt
5
1
t
 /4
  /2
25  t 2
1
converges to  ln ( 5 
5
24 ) .
tan  d

,
2
which is the lower limit of integration, and is continuous on the interval
  
  ,  . Thus, the Fundamental Theorem of Calculus can not be applied
 2 4
to this integral. We will need to do the following.
NOTE: The integrand of f (  )  tan  has a discontinuity at   

 /4
  /2
tan  d =
lim
t  (  / 2) 

 /4
t
tan  d
 
The integrand is continuous on the closed intervals  t ,  for all
 4


  t  . Thus, the Fundamental Theorem of Calculus can be applied to
2
4
these closed intervals.
 tan  d
= ln sec   c
Copyrighted by James D. Anderson, The University of Toledo
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Thus,

 /4
tan  d =
  /2
lim
t  (  /2) 
Since
lim
t  (  / 2) 

 /4
t
tan  d =



 ln sec t  =
 ln sec
4


lim
t  (  /2) 
sec t   , then
lim
t  (  /2) 
t  (   / 2) 
lim
t  (  /2) 
lim
( ln
 ln
sec 

 /4
t
=
2  ln sec t )   
ln sec t   .
Answer: Diverges
9.

0

2
sin  x
dx
7 x
NOTE: The integrand of f ( x ) 
sin
x
is continuous on its domain of
7 x
definition, which is the set of real numbers given by the open interval
(   , 0 ) . The interval of integration is the closed interval [   2 , 0 ] . Thus,
the integrand has a discontinuity at x  0 , which is the upper limit of
integration. Thus, the Fundamental Theorem of Calculus can not be applied
to this integral. We will need to do the following.

0

2
sin  x
dx =
7 x
lim 
t  0

t

2
sin  x
dx
7 x
2
The integrand is continuous on the closed intervals [   , t ] for all
  2  t  0 . Thus, the Fundamental Theorem of Calculus can be applied to
these closed intervals.

sin  x
dx
7 x
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Let u 
x
Then du  
1
dx
2 x
sin  x
1
dx =  2 
7
7 x


sin  x
2
dx = 
7
2 x

sin u du = 
2
(  cos u )  c =
7
2
2
cos u  c = cos  x  c
7
7
Thus,

0

2
sin  x
dx =
7 x
lim
t  0


sin  x
2

dx = lim   cos  x 
=
t  0 7
7 x
 2
t
t
2
2
2
lim  ( cos  t  cos  2 ) =
lim ( cos  t  cos  ) =
7 t0
7 t  0
2
2
2
4
lim  [ cos  t  (  1 ) ] =
lim  ( cos  t  1 ) = ( 1  1 ) =
7 t0
7 t0
7
7
cos  t  cos 0  1
since t lim
 0
Answer: Converges;
10.

3
0
4
7
ln y dy
NOTE: The integrand of f ( y )  ln y is continuous on its domain of
definition, which is the set of real numbers given by the open interval ( 0 ,  ) .
The interval of integration is the closed interval [ 0 , 3 ] . Thus, the integrand
has a discontinuity at x  0 , which is the lower limit of integration. Thus,
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
the Fundamental Theorem of Calculus can not be applied to this integral. We
will need to do the following.

3
0
ln y dy =
lim 
t  0

3
t
ln y dy
The integrand is continuous on the closed intervals [t , 3 ] for all 0  t  3 .
Thus, the Fundamental Theorem of Calculus can be applied to these closed
intervals.
 ln y dy
This integral can be evaluated using Integration by Parts.
u  ln y and
1
Then du  dy
y
Let
 ln y dy
=
 u dv
dv  dy
v  y
= uv 
 v du
= y ln y 
 dy
= y ln y  y  c
Thus,

3
0
ln y dy =
lim 
t  0

3
t
ln y dy = lim   y ln y  y  t =
t  0
lim [ 3 ln 3  3  ( t ln t  t ) ] =
t  0
3
lim ( 3 ln 3  3  t ln t  t )
t  0
lim t ln t  0  (   )
t  0
1
ln t
lim  t =  lim t   1 ( 0 )  0
lim  t ln t = t lim

=
1
1
 0
t  0
t  0
t  0
 2
t
t
( 3 ln 3  3  t ln t  t ) = 3 ln 3  3  0  0 = 3 ln 3  3
Thus, t lim
 0
Copyrighted by James D. Anderson, The University of Toledo
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Answer: Converges; 3 ln 3  3 or ln 27  3
11.

2 / 3
0
x csc 2 x dx
2
NOTE: The integrand of f ( x )  x csc x has a discontinuity at   0 ,
which is the lower limit of integration, and is continuous on the interval
2 

0,
. Thus, the Fundamental Theorem of Calculus can not be applied to
3 

this integral. We will need to do the following.

2 / 3
0
x csc 2 x dx =
lim 
t  0

2 / 3
t
x csc 2 x dx
 2 
The integrand is continuous on the closed intervals  t ,
for all
3 

2
0t 
. Thus, the Fundamental Theorem of Calculus can be applied to
3
these closed intervals.

x csc 2 x dx
This integral can be evaluated using Integration by Parts.
u  x
Let
Then du  dx

x csc 2 x dx =
dv  csc 2 x dx
v   cot x
and
 u dv
= uv 
 v du
=  x cot x 
 cot x dx
=
 x cot x  ln sin x  c = ln sin x  x cot x  c
Thus,

2 / 3
0
x csc 2 x dx =
lim 
t  0

2 / 3
t
x csc 2 x dx = lim   ln sin x  x cot x  t
t  0
Copyrighted by James D. Anderson, The University of Toledo
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2 / 3
=
2
2
2


lim   ln sin

cot
 ( ln sin t  t cot t )  =
t  0
3
3
3




3
3 
2 

lim  ln


 ln sin t  t cot t  =
t  0
2
3 
3 




3
2 3
lim   ln

 ln sin t  t cot t 

t  0 
2
9


lim t cot t  0  
t  0
lim t cot t =
t  0
lim 
t  0
t
=
tan t
lim 
t  0
1
1
1


1
sec 2 t sec 2 0 1
sin t  sin 0  0 , then lim  ln sin t    .
Since t lim
t  0
 0


3
2 3


lim
ln


ln
sin
t

t
cot
t
Thus, t  0  

2
9


Answer: Diverges
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860