X2 T05 04 reduction formula (2010)

1. Reduction Formula

2. Reduction Formula
Reduction (or recurrence) formulae expresses a given integral as
the sum of a function and a known integral.
Integration by parts often used to find the formula.

3. Reduction Formula
Reduction (or recurrence) formulae expresses a given integral as
the sum of a function and a known integral.
Integration by parts often used to find the formula.
e.g. (i) (1987)









 

2
0
5
2
2
0
cos evaluate hence,2 andinteger an is where

1
that prove ,cos Given that


xdx n n
I
n
n
I xdx I
n n
n
n

4. 

2
0
cos

xdx I
n
n

5. 

2
0
cos

xdx I
n
n



2
0
1
coscos

xdxx
n

6. 

2
0
cos

xdx I
n
n
x
u
n1
cos




xdxx ndu
n
sincos1
2




x
dxdv cos

x
vsin




2
0
1
coscos

xdxx
n

7. 

2
0
cos

xdx I
n
n
x
u
n1
cos





 
 
2
0
22
2
0
1
sincos1 sincos


xdxx nxx
n n


xdxx ndu
n
sincos1
2




x
dxdv cos

x
vsin




2
0
1
coscos

xdxx
n

8. 

2
0
cos

xdx I
n
n
x
u
n1
cos





 
 
2
0
22
2
0
1
sincos1 sincos


xdxx nxx
n n


xdxx ndu
n
sincos1
2




x
dxdv cos

x
vsin




2
0
1
coscos

xdxx
n



 






 
  
2
0
2 2 1 1
cos1cos10sin0cos
2
sin
2
cos


dxxx n
n n n

9. 

2
0
cos

xdx I
n
n
x
u
n1
cos





 
 
2
0
22
2
0
1
sincos1 sincos


xdxx nxx
n n


xdxx ndu
n
sincos1
2




x
dxdv cos

x
vsin




2
0
1
coscos

xdxx
n



 






 
  
2
0
2 2 1 1
cos1cos10sin0cos
2
sin
2
cos


dxxx n
n n n


 
 

2
0
2
0
2
cos1 cos1


xdx nxdx n
n n

10. 

2
0
cos

xdx I
n
n
x
u
n1
cos





 
 
2
0
22
2
0
1
sincos1 sincos


xdxx nxx
n n


xdxx ndu
n
sincos1
2




x
dxdv cos

x
vsin




2
0
1
coscos

xdxx
n



 






 
  
2
0
2 2 1 1
cos1cos10sin0cos
2
sin
2
cos


dxxx n
n n n


 
 

2
0
2
0
2
cos1 cos1


xdx nxdx n
n n




n n
I
n
I
n1 1
2






11. 

2
0
cos

xdx I
n
n
x
u
n1
cos





 
 
2
0
22
2
0
1
sincos1 sincos


xdxx nxx
n n


xdxx ndu
n
sincos1
2




x
dxdv cos

x
vsin


2
1



n n
I
nn
I



2
0
1
coscos

xdxx
n



 






 
  
2
0
2 2 1 1
cos1cos10sin0cos
2
sin
2
cos


dxxx n
n n n


 
 

2
0
2
0
2
cos1 cos1


xdx nxdx n
n n




n n
I
n
I
n1 1
2






12. 

2
0
cos

xdx I
n
n
x
u
n1
cos





 
 
2
0
22
2
0
1
sincos1 sincos


xdxx nxx
n n


xdxx ndu
n
sincos1
2




x
dxdv cos

x
vsin


2
1



n n
I
nn
I



2
0
1
coscos

xdxx
n



 






 
  
2
0
2 2 1 1
cos1cos10sin0cos
2
sin
2
cos


dxxx n
n n n


 
 

2
0
2
0
2
cos1 cos1


xdx nxdx n
n n




n n
I
n
I
n1 1
2





2
1






n n
I
n
n
I

13. 5
2
0
5
cosIxdx 



14. 5
2
0
5
cosIxdx 


3
5
4
I 

15. 5
2
0
5
cosIxdx 


3
5
4
I 
1
3
2
5
4
I

16. 5
2
0
5
cosIxdx 


3
5
4
I 
1
3
2
5
4
I


2
0
cos
15
8

xdx

17. 5
2
0
5
cosIxdx 


3
5
4
I 
1
3
2
5
4
I


2
0
cos
15
8

xdx

2
0
sin
15
8

x 

18. 5
2
0
5
cosIxdx 


3
5
4
I 
1
3
2
5
4
I


2
0
cos
15
8

xdx

2
0
sin
15
8

x 
15
8
0sin
2
sin
15
8







 


19. 


6
find ,cot Given that Ixdx I ii
n
n

20. 


6
find ,cot Given that Ixdx I ii
n
n

xdx I
n
n
cot

21. 


6
find ,cot Given that Ixdx I ii
n
n

xdx I
n
n
cot

xdxx
n22
cotcot

22. 


6
find ,cot Given that Ixdx I ii
n
n

xdx I
n
n
cot

xdxx
n22
cotcot 


 

dxxx
n
1cosec cot
2 2


 
 xdx xdxx
n n2 2 2
cot cosec cot

23. 


6
find ,cot Given that Ixdx I ii
n
n

xdx I
n
n
cot

xdxx
n22
cotcot 


 

dxxx
n
1cosec cot
2 2


 
 xdx xdxx
n n2 2 2
cot cosec cot
x
ucot

x
dx du
2
cosec



24. 


6
find ,cot Given that Ixdx I ii
n
n

xdx I
n
n
cot

xdxx
n22
cotcot 


 

dxxx
n
1cosec cot
2 2


 
 xdx xdxx
n n2 2 2
cot cosec cot
x
ucot

x
dx du
2
cosec


2
2




n
n
Iduu

25. 


6
find ,cot Given that Ixdx I ii
n
n

xdx I
n
n
cot

xdxx
n22
cotcot 


 

dxxx
n
1cosec cot
2 2


 
 xdx xdxx
n n2 2 2
cot cosec cot
x
ucot

x
dx du
2
cosec


2
2




n
n
Iduu2
1
1
1





n
n
Iu
n
2
1
cot
1
1





n
n
Ix
n

26. 6
6
cotIxdx 


27. 6
6
cotIxdx 

4
5
cot
5
1
Ix

28. 6
6
cotIxdx 

4
5
cot
5
1
Ix
2
3 5
cot
3
1
cot
5
1
Ixx 

29. 6
6
cotIxdx 

4
5
cot
5
1
Ix
2
3 5
cot
3
1
cot
5
1
Ixx 
0
3 5
cotcot
3
1
cot
5
1
Ixxx 

30. 6
6
cotIxdx 

4
5
cot
5
1
Ix
2
3 5
cot
3
1
cot
5
1
Ixx 
0
3 5
cotcot
3
1
cot
5
1
Ixxx 

dxxxx cotcot
3
1
cot
5
1
3 5

31. 6
6
cotIxdx 

4
5
cot
5
1
Ix
2
3 5
cot
3
1
cot
5
1
Ixx 
0
3 5
cotcot
3
1
cot
5
1
Ixxx 

dxxxx cotcot
3
1
cot
5
1
3 5
cxxxx  cotcot
3
1
cot
5
1
3 5

32. (iii) (2004 Question 8b)

4
2
0
Let tan and let 1 for 0,1,2,
n
n
nnn
IxdxJIn




2
1
) Show that
1
nn
aII
n




33. (iii) (2004 Question 8b)

4
2
0
Let tan and let 1 for 0,1,2,
n
n
nnn
IxdxJIn




2
1
) Show that
1
nn
aII
n



2 44
2
00
tan tan
nn
nn
I Idx dx




 


34. (iii) (2004 Question 8b)

4
2
0
Let tan and let 1 for 0,1,2,
n
n
nnn
IxdxJIn




2
1
) Show that
1
nn
aII
n



2 44
2
00
tan tan
nn
nn
I Idx dx




 


2 4
0
tan 1 tan
n
x
xdx



2 4
0
tan sec
n
x
xdx




35. (iii) (2004 Question 8b)

4
2
0
Let tan and let 1 for 0,1,2,
n
n
nnn
IxdxJIn




2
1
) Show that
1
nn
aII
n



2 44
2
00
tan tan
nn
nn
I Idx dx




 


2 4
0
tan 1 tan
n
x
xdx



2 4
0
tan sec
n
x
xdx



tan ux

2
sec du xdx


36. (iii) (2004 Question 8b)

4
2
0
Let tan and let 1 for 0,1,2,
n
n
nnn
IxdxJIn




2
1
) Show that
1
nn
aII
n



2 44
2
00
tan tan
nn
nn
I Idx dx




 


2 4
0
tan 1 tan
n
x
xdx



2 4
0
tan sec
n
x
xdx



tan ux

2
sec du xdx

when 0, 0
, 1
4
x
u
x
u






37. (iii) (2004 Question 8b)

4
2
0
Let tan and let 1 for 0,1,2,
n
n
nnn
IxdxJIn




2
1
) Show that
1
nn
aII
n



2 44
2
00
tan tan
nn
nn
I Idx dx




 


2 4
0
tan 1 tan
n
x
xdx



2 4
0
tan sec
n
x
xdx



tan ux

2
sec du xdx

1
0
n
udu 

when 0, 0
, 1
4
x
u
x
u






38. (iii) (2004 Question 8b)

4
2
0
Let tan and let 1 for 0,1,2,
n
n
nnn
IxdxJIn




2
1
) Show that
1
nn
aII
n



2 44
2
00
tan tan
nn
nn
I Idx dx




 


2 4
0
tan 1 tan
n
x
xdx



2 4
0
tan sec
n
x
xdx



tan
ux

2
sec du xdx

1
0
n
udu 

when 0, 0
, 1
4
x
u
x
u





1
1
0
1
n
u
n






39. (iii) (2004 Question 8b)

4
2
0
Let tan and let 1 for 0,1,2,
n
n
nnn
IxdxJIn




2
1
) Show that
1
nn
aII
n



2 44
2
00
tan tan
nn
nn
I Idx dx




 


2 4
0
tan 1 tan
n
x
xdx



2 4
0
tan sec
n
x
xdx



tan
ux

2
sec du xdx

1
0
n
udu 

when 0, 0
, 1
4
x
u
x
u





1
1
0
1
n
u
n





11
0
11 nn



40. 

1
1
) Deduce that for 1
21
n
nn
bJJ n
n

 



41. 

1
1
) Deduce that for 1
21
n
nn
bJJ n
n

 





1
12 22
11
nn
nnnn
JJ I I



 

42. 

1
1
) Deduce that for 1
21
n
nn
bJJ n
n

 





1
12 22
11
nn
nnnn
JJ I I



 



222
11
nn
nn
II

 

43. 

1
1
) Deduce that for 1
21
n
nn
bJJ n
n

 





1
12 22
11
nn
nnnn
JJ I I



 



222
11
nn
nn
II

 



222
1
n
nn
II

 

44. 

1
1
) Deduce that for 1
21
n
nn
bJJ n
n

 





1
12 22
11
nn
nnnn
JJ I I



 



222
11
nn
nn
II

 



222
1
n
nn
II

 

1
21
n
n




45. 

1
1
) Show that
421
n
m
m
n
cJ
n







46. 

1
1
) Show that
421
n
m
m
n
cJ
n








1
1
21
m
mm
JJ
m





47. 

1
1
) Show that
421
n
m
m
n
cJ
n








1
1
21
m
mm
JJ
m







1
2
11
2123
mm
m
J
mm



 


48. 




11
0
11 1
2123 1
mm
J
mm

 






1
1
) Show that
421
n
m
m
n
cJ
n








1
1
21
m
mm
JJ
m







1
2
11
2123
mm
m
J
mm



 


49. 




11
0
11 1
2123 1
mm
J
mm

 





4
0
1
1
21
n
m
n
dx
n









1
1
) Show that
421
n
m
m
n
cJ
n








1
1
21
m
mm
JJ
m







1
2
11
2123
mm
m
J
mm



 


50. 




11
0
11 1
2123 1
mm
J
mm

 





4
0
1
1
21
n
m
n
dx
n









4
0
1
1
21
n
m
n
x
n







1
1
214
n
m
n
n









1
1
) Show that
421
n
m
m
n
cJ
n








1
1
21
m
mm
JJ
m







1
2
11
2123
mm
m
J
mm



 


51. 1
2
0
) Use the substitution tan to show that
1
n
n
u
duxIdu
u




52. 1
2
0
) Use the substitution tan to show that
1
n
n
u
duxIdu
u



4
0
tan
n
n
Ixdx




53. 1
2
0
) Use the substitution tan to show that
1
n
n
u
duxIdu
u



4
0
tan
n
n
Ixdx



1
tan tan uxx u



2
1
du
dx
u


54. 1
2
0
) Use the substitution tan to show that
1
n
n
u
duxIdu
u



4
0
tan
n
n
Ixdx



1
tan tan uxx u



2
1
du
dx
u

when 0, 0
, 1
4
x
u
x
u






55. 1
2
0
) Use the substitution tan to show that
1
n
n
u
duxIdu
u



4
0
tan
n
n
Ixdx



1
tan tan uxx u



2
1
du
dx
u

when 0, 0
, 1
4
x
u
x
u





1
2
0
1
n
n
du
Iu
u



1
2
0
1
n
n
udu
I
u




56. 1
2
0
) Use the substitution tan to show that
1
n
n
u
duxIdu
u



4
0
tan
n
n
Ixdx



1
tan tan uxx u



2
1
du
dx
u

when 0, 0
, 1
4
x
u
x
u





1
2
0
1
n
n
du
Iu
u



1
2
0
1
n
n
udu
I
u



1
) Deduce that 0 and conclude that 0 as
1
nn
eI Jn
n
  


57. 1
2
0
) Use the substitution tan to show that
1
n
n
u
duxIdu
u



4
0
tan
n
n
Ixdx



1
tan tan uxx u



2
1
du
dx
u

when 0, 0
, 1
4
x
u
x
u





1
2
0
1
n
n
du
Iu
u



1
2
0
1
n
n
udu
I
u



1
) Deduce that 0 and conclude that 0 as
1
nn
eI Jn
n
  

2
0, for all 0
1
n
u
u
u



58. 1
2
0
) Use the substitution tan to show that
1
n
n
u
duxIdu
u



4
0
tan
n
n
Ixdx



1
tan tan uxx u



2
1
du
dx
u

when 0, 0
, 1
4
x
u
x
u





1
2
0
1
n
n
du
Iu
u



1
2
0
1
n
n
udu
I
u



1
) Deduce that 0 and conclude that 0 as
1
nn
eI Jn
n
  

2
0, for all 0
1
n
u
u
u


1
2
0
0, for all 0
1
n
n
u
Iduu
u





59. 2
1
1
nn
II
n




60. 2
1
1
nn
II
n



2
1
1
nn
II
n




61. 2
1
1
nn
II
n



2
1
1
nn
II
n



2
1
, as 0
1
nn
II
n





62. 2
1
1
nn
II
n



2
1
1
nn
II
n



1
0
1
n
I
n



2
1
, as 0
1
nn
II
n





63. 2
1
1
nn
II
n



2
1
1
nn
II
n



1
0
1
n
I
n



1
as , 0
1
n
n
 

2
1
, as 0
1
nn
II
n





64. 2
1
1
nn
II
n



2
1
1
nn
II
n



1
0
1
n
I
n



1
as , 0
1
n
n
 

0
n
I


2
1
, as 0
1
nn
II
n





65. 2
1
1
nn
II
n



2
1
1
nn
II
n



1
0
1
n
I
n



1
as , 0
1
n
n
 

0
n
I



2
10
n
nn
JI
 
2
1
, as 0
1
nn
II
n





66. 2
1
1
nn
II
n



2
1
1
nn
II
n



2
1
, as 0
1
nn
II
n




1
0
1
n
I
n



1
as , 0
1
n
n
 

0
n
I



2
10
n
nn
JI
 
Exercise 2D; 1, 2, 3, 6, 8,
9, 10, 12, 14