The document discusses reduction formulae, which expresses a given integral as the sum of a function and a known integral. It provides an example of using integration by parts to derive the reduction formula for the integral of cos(nx)dx and evaluates the integral of cos(5x)dx. It also shows how to find the reduction formula for the integral of cot(nx)dx and states that this formula can be used to find the value of I6, the integral of cot(6x)dx.

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