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) 
 n  1I
2
Given that I n   cos xdx, prove that I n  
n
 n2
0  n 

2
where n is an integer and n  2, hence evaluate  cos 5 xdx
0

4. 
2
I n   cos n xdx
0

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

6. 
2
I n   cos n xdx
0

2 u  cos n 1 x v  sin x
  cos n 1 x cos xdx du  n  1 cos n  2 x sin xdx dv  cos xdx
0

7. 
2
I n   cos n xdx
0

2 u  cos n 1 x v  sin x
  cos n 1 x cos xdx du  n  1 cos n  2 x sin xdx dv  cos xdx

0

 cos n 1 x sin x 0  n  1 cos n  2 x sin 2 xdx
2
2
0

8. 
2
I n   cos n xdx
0

2 u  cos n 1 x v  sin x
  cos n 1 x cos xdx du  n  1 cos n  2 x sin xdx dv  cos xdx

0

 cos n 1 x sin x 0  n  1 cos n  2 x sin 2 xdx
2
2
0 
cos n 1  sin  cos n 1 0 sin 0  n  1 cos n  2 x1  cos 2 x dx
2

 2 2
 
 
0

9. 
2
I n   cos n xdx
0

2 u  cos n 1 x v  sin x
  cos n 1 x cos xdx du  n  1 cos n  2 x sin xdx dv  cos xdx

0

 cos n 1 x sin x 0  n  1 cos n  2 x sin 2 xdx
2
2
0 
cos n 1  sin  cos n 1 0 sin 0  n  1 cos n  2 x1  cos 2 x dx
2

 2 2
 
 
0
 
2 2
 n  1 cos n  2 xdx  n  1 cos n xdx
0 0

10. 
2
I n   cos n xdx
0

2 u  cos n 1 x v  sin x
  cos n 1 x cos xdx du  n  1 cos n  2 x sin xdx dv  cos xdx

0

 cos n 1 x sin x 0  n  1 cos n  2 x sin 2 xdx
2
2
0 
cos n 1  sin  cos n 1 0 sin 0  n  1 cos n  2 x1  cos 2 x dx
2

 2 2
 
 
0
 
2 2
 n  1 cos n  2 xdx  n  1 cos n xdx
0 0
 n  1I n  2  n  1I n

11. 
2
I n   cos n xdx
0

2 u  cos n 1 x v  sin x
  cos n 1 x cos xdx du  n  1 cos n  2 x sin xdx dv  cos xdx

0

 cos n 1 x sin x 0  n  1 cos n  2 x sin 2 xdx
2
2
0 
cos n 1  sin  cos n 1 0 sin 0  n  1 cos n  2 x1  cos 2 x dx
2

 2 2
 
 
0
 
2 2
 n  1 cos n  2 xdx  n  1 cos n xdx
0 0
 n  1I n  2  n  1I n
 nI n  n  1I n  2

12. 
2
I n   cos n xdx
0

2 u  cos n 1 x v  sin x
  cos n 1 x cos xdx du  n  1 cos n  2 x sin xdx dv  cos xdx

0

 cos n 1 x sin x 0  n  1 cos n  2 x sin 2 xdx
2
2
0 
cos n 1  sin  cos n 1 0 sin 0  n  1 cos n  2 x1  cos 2 x dx
2

 2 2
 
 
0
 
2 2
 n  1 cos n  2 xdx  n  1 cos n xdx
0 0
 n  1I n  2  n  1I n
 nI n  n  1I n  2
In   n  1I
 n2
 n 

13. 
2

0
cos 5 xdx  I 5

14. 
2

0
cos 5 xdx  I 5
4
 I3
5

15. 
2

0
cos 5 xdx  I 5
4
 I3
5
4 2
  I1
5 3

16. 
2

0
cos 5 xdx  I 5
4
 I3
5
4 2
  I1
5 3

8 2
  cos xdx
15 0

17. 
2

0
cos 5 xdx  I 5
4
 I3
5
4 2
  I1
5 3

8 2
  cos xdx
15 0

8
 sin x 0
2
15

18. 
2

0
cos 5 xdx  I 5
4
 I3
5
4 2
  I1
5 3

8 2
  cos xdx
15 0

8
 sin x 0
2
15

  sin  sin 0 
8
 
15  2 
8

15

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

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

21. ii  Given that I n   cot n xdx, find I 6
I n   cot n xdx
  cot n  2 x cot 2 xdx

22. ii  Given that I n   cot n xdx, find I 6
I n   cot n xdx
  cot n  2 x cot 2 xdx
  cot n  2 xcosec 2 x  1dx
  cot n  2 xcosec 2 xdx   cot n  2 xdx

23. ii  Given that I n   cot n xdx, find I 6
I n   cot n xdx
  cot n  2 x cot 2 xdx
  cot n  2 xcosec 2 x  1dx
  cot n  2 xcosec 2 xdx   cot n  2 xdx u  cot x
du  cosec 2 xdx

24. ii  Given that I n   cot n xdx, find I 6
I n   cot n xdx
  cot n  2 x cot 2 xdx
  cot n  2 xcosec 2 x  1dx
  cot n  2 xcosec 2 xdx   cot n  2 xdx u  cot x
  u n2
du  I n  2 du  cosec 2 xdx

25. ii  Given that I n   cot n xdx, find I 6
I n   cot n xdx
  cot n  2 x cot 2 xdx
  cot n  2 xcosec 2 x  1dx
  cot n  2 xcosec 2 xdx   cot n  2 xdx u  cot x
  u n2
du  I n  2 du  cosec 2 xdx
1 n 1
 u  I n2
n 1
1
 cot n 1 x  I n  2
n 1

26.  cot 6 xdx  I 6

27.  cot 6 xdx  I 6
1 5
  cot x  I 4
5

28.  cot 6 xdx  I 6
1 5
  cot x  I 4
5
1 1
  cot 5 x  cot 3 x  I 2
5 3

29.  cot 6 xdx  I 6
1 5
  cot x  I 4
5
1 1
  cot 5 x  cot 3 x  I 2
5 3
1 5 1 3
  cot x  cot x  cot x  I 0
5 3

30.  cot 6 xdx  I 6
1 5
  cot x  I 4
5
1 1
  cot 5 x  cot 3 x  I 2
5 3
1 5 1 3
  cot x  cot x  cot x  I 0
5 3
1 5 1 3
  cot x  cot x  cot x   dx
5 3

31.  cot 6 xdx  I 6
1 5
  cot x  I 4
5
1 1
  cot 5 x  cot 3 x  I 2
5 3
1 5 1 3
  cot x  cot x  cot x  I 0
5 3
1 5 1 3
  cot x  cot x  cot x   dx
5 3
1 5 1 3
  cot x  cot x  cot x  x  c
5 3

32. (iii) (2004 Question 8b)

Let I n   4 tan n xdx and let J n   1 I 2 n for n  0,1, 2,
n
0
1
a ) Show that I n  I n2 
n 1

33. (iii) (2004 Question 8b)

Let I n   4 tan n xdx and let J n   1 I 2 n for n  0,1, 2,
n
0
1
a ) Show that I n  I n2 
n 1
 
I n  I n2   4 tan n dx   4 tan n2 dx
0 0

34. (iii) (2004 Question 8b)

Let I n   4 tan n xdx and let J n   1 I 2 n for n  0,1, 2,
n
0
1
a ) Show that I n  I n2 
n 1
 
I n  I n2   4 tan n dx   4 tan n2 dx
0 0

  4 tan n x 1  tan 2 x  dx
0

  4 tan n x sec 2 xdx
0

35. (iii) (2004 Question 8b)

Let I n   4 tan n xdx and let J n   1 I 2 n for n  0,1, 2,
n
0
1
a ) Show that I n  I n2 
n 1
 
I n  I n2   4 tan n dx   4 tan n2 dx
0 0

  4 tan n x 1  tan 2 x  dx
0

u  tan x
  4 tan n x sec 2 xdx du  sec 2 xdx
0

36. (iii) (2004 Question 8b)

Let I n   4 tan n xdx and let J n   1 I 2 n for n  0,1, 2,
n
0
1
a ) Show that I n  I n2 
n 1
 
I n  I n2   4 tan n dx   4 tan n2 dx
0 0

  4 tan n x 1  tan 2 x  dx
0

u  tan x
  4 tan n x sec 2 xdx du  sec 2 xdx
0
when x  0, u  0

x ,u  1
4

37. (iii) (2004 Question 8b)

Let I n   4 tan n xdx and let J n   1 I 2 n for n  0,1, 2,
n
0
1
a ) Show that I n  I n2 
n 1
 
I n  I n2   4 tan n dx   4 tan n2 dx
0 0

  4 tan n x 1  tan 2 x  dx
0

u  tan x
  4 tan n x sec 2 xdx du  sec 2 xdx
0
1
  u n du when x  0, u  0
0

x ,u  1
4

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

39. (iii) (2004 Question 8b)

Let I n   4 tan n xdx and let J n   1 I 2 n for n  0,1, 2,
n
0
1
a ) Show that I n  I n2 
n 1
 
I n  I n2   4 tan n dx   4 tan n2 dx
0 0

  4 tan n x 1  tan 2 x  dx
0

u  tan x
  4 tan n x sec 2 xdx du  sec 2 xdx
0
1
  u n du when x  0, u  0
0

u  n 1 1
x ,u  1
 4
 n  1 0

1 1
 0 
n 1 n 1

40.  1
n
b) Deduce that J n  J n1  for n  1
2n  1

41.  1
n
b) Deduce that J n  J n1  for n  1
2n  1
J n  J n1   1 I 2 n   1 I 2 n2
n n 1

42.  1
n
b) Deduce that J n  J n1  for n  1
2n  1
J n  J n1   1 I 2 n   1 I 2 n2
n n 1
  1 I 2 n   1 I 2 n2
n n

43.  1
n
b) Deduce that J n  J n1  for n  1
2n  1
J n  J n1   1 I 2 n   1 I 2 n2
n n 1
  1 I 2 n   1 I 2 n2
n n
  1  I 2 n  I 2 n2 
n

44.  1
n
b) Deduce that J n  J n1  for n  1
2n  1
J n  J n1   1 I 2 n   1 I 2 n2
n n 1
  1 I 2 n   1 I 2 n2
n n
  1  I 2 n  I 2 n2 
n
 1
n

2n  1

45.  1
n
 m
c) Show that J m  
4 n 1 2n  1

46.  1
n
 m
c) Show that J m  
4 n 1 2n  1
 1
m
Jm   J m1
2m  1

47.  1
n
 m
c) Show that J m  
4 n 1 2n  1
 1
m
Jm   J m1
2m  1
 1  1
m m 1
   J m2
2m  1 2m  3

48.  1
n
 m
c) Show that J m  
4 n 1 2n  1
 1
m
Jm   J m1
2m  1
 1  1
m m 1
   J m2
2m  1 2m  3
 1  1  1
m m 1 1
     J0
2m  1 2m  3 1

49.  1
n
 m
c) Show that J m  
4 n 1 2n  1
 1
m
Jm   J m1
2m  1
 1  1
m m 1
   J m2
2m  1 2m  3
 1  1  1
m m 1 1
     J0
2m  1 2m  3 1
 1 
n
m
   4 dx
n 1 2n  1 0

50.  1
n
 m
c) Show that J m  
4 n 1 2n  1
 1
m
Jm   J m1
2m  1
 1  1
m m 1
   J m2
2m  1 2m  3
 1  1  1
m m 1 1
     J0
2m  1 2m  3 1
 1 
n
m
   4 dx
n 1 2n  1 0
 1
n
m 
   x 04
n 1 2n  1
 1
n
m

 
n 1 2n  1 4

51. 1 un
d ) Use the substitution u  tan x to show that I n   du
0 1 u 2

52. 1 un
d ) Use the substitution u  tan x to show that I n   du
0 1 u 2

I n   4 tan n xdx
0

53. 1 un
d ) Use the substitution u  tan x to show that I n   du
0 1 u 2

I n   4 tan n xdx
0 u  tan x  x  tan 1 u
du
dx 
1 u2

54. 1 un
d ) Use the substitution u  tan x to show that I n   du
0 1 u 2

I n   4 tan n xdx
0 u  tan x  x  tan 1 u
du
dx 
1 u2
when x  0, u  0

x ,u  1
4

55. 1 un
d ) Use the substitution u  tan x to show that I n   du
0 1 u 2

I n   4 tan n xdx
0 u  tan x  x  tan 1 u
1 du du
In   un  dx 
0 1 u2 1 u2
1 u n du when x  0, u  0
In  
0 1 u2

x ,u  1
4

56. 1 un
d ) Use the substitution u  tan x to show that I n   du
0 1 u 2

I n   4 tan n xdx
0 u  tan x  x  tan 1 u
1 du du
In   un  dx 
0 1 u2 1 u2
1 u n du when x  0, u  0
In  
0 1 u2

x ,u  1
4
1
e) Deduce that 0  I n  and conclude that J n  0 as n  
n 1

57. 1 un
d ) Use the substitution u  tan x to show that I n   du
0 1 u 2

I n   4 tan n xdx
0 u  tan x  x  tan 1 u
1 du du
In   un  dx 
0 1 u2 1 u2
1 u n du when x  0, u  0
In  
0 1 u2

x ,u  1
4
1
e) Deduce that 0  I n  and conclude that J n  0 as n  
n 1
un
 0, for all u  0
1 u 2

58. 1 un
d ) Use the substitution u  tan x to show that I n   du
0 1 u 2

I n   4 tan n xdx
0 u  tan x  x  tan 1 u
1 du du
In   un  dx 
0 1 u2 1 u2
1 u n du when x  0, u  0
In  
0 1 u2

x ,u  1
4
1
e) Deduce that 0  I n  and conclude that J n  0 as n  
n 1
un
 0, for all u  0
1 u 2
n
1 u
 In   du  0, for all u  0
0 1 u2

59. 1
I n  I n 2 
n 1

60. 1
I n  I n 2 
n 1
1
In   I n 2
n 1

61. 1
I n  I n 2 
n 1
1
In   I n 2
n 1
1
 In  , as I n2  0
n 1

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

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

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

65. 1
I n  I n 2 
n 1
1
In   I n 2
n 1
1
 In  , as I n2  0
n 1
1
 0  In 
n 1
1
as n  , 0
n 1
 In  0
 J n   1 I 2 n  0
n

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