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
Given that I n cos n xdx, prove that I n
2
I n2
0 n
2
where n is an integer and n 2, hence evaluate cos 5 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 x1 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 x1 cos 2 x dx
2
2 2
0
2 2
n 1 cos n2
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 x1 cos 2 x dx
2
2 2
0
2 2
n 1 cos n2
xdx n 1 cos n xdx
0 0
n 1I n 2 n 1I 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 x1 cos 2 x dx
2
2 2
0
2 2
n 1 cos n2
xdx n 1 cos n xdx
0 0
n 1I n 2 n 1I n
nI n n 1I 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 x1 cos 2 x dx
2
2 2
0
2 2
n 1 cos n2
xdx n 1 cos n xdx
0 0
n 1I n 2 n 1I n
nI n n 1I n 2
In n 1I
n2
n
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 xcosec 2 x 1dx
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 xcosec 2 x 1dx
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 xcosec 2 x 1dx
cot n 2 xcosec 2 xdx cot n 2 xdx u cot x
u n2
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 xcosec 2 x 1dx
cot n 2 xcosec 2 xdx cot n 2 xdx u cot x
u n2
du I n 2 du cosec 2 xdx
1 n 1
u I n2
n 1
1
cot n 1 x I n 2
n 1
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 n2
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 n2
n 1
I n I n2 4 tan n dx 4 tan n2 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 n2
n 1
I n I n2 4 tan n dx 4 tan n2 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 n2
n 1
I n I n2 4 tan n dx 4 tan n2 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 n2
n 1
I n I n2 4 tan n dx 4 tan n2 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 n2
n 1
I n I n2 4 tan n dx 4 tan n2 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 n2
n 1
I n I n2 4 tan n dx 4 tan n2 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 n2
n 1
I n I n2 4 tan n dx 4 tan n2 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 n1 for n 1
2n 1
41. 1
n
b) Deduce that J n J n1 for n 1
2n 1
J n J n1 1 I 2 n 1 I 2 n2
n n 1
42. 1
n
b) Deduce that J n J n1 for n 1
2n 1
J n J n1 1 I 2 n 1 I 2 n2
n n 1
1 I 2 n 1 I 2 n2
n n
43. 1
n
b) Deduce that J n J n1 for n 1
2n 1
J n J n1 1 I 2 n 1 I 2 n2
n n 1
1 I 2 n 1 I 2 n2
n n
1 I 2 n I 2 n2
n
44. 1
n
b) Deduce that J n J n1 for n 1
2n 1
J n J n1 1 I 2 n 1 I 2 n2
n n 1
1 I 2 n 1 I 2 n2
n n
1 I 2 n I 2 n2
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 m1
2m 1
47. 1
n
m
c) Show that J m
4 n 1 2n 1
1
m
Jm J m1
2m 1
1 1
m m 1
J m2
2m 1 2m 3
48. 1
n
m
c) Show that J m
4 n 1 2n 1
1
m
Jm J m1
2m 1
1 1
m m 1
J m2
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 m1
2m 1
1 1
m m 1
J m2
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 m1
2m 1
1 1
m m 1
J m2
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 u2
52. 1 un
d ) Use the substitution u tan x to show that I n du
0 1 u2
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 u2
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 u2
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 u2
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 u2
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 u2
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 u2
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
61. 1
I n I n 2
n 1
1
In I n 2
n 1
1
In , as I n2 0
n 1
62. 1
I n I n 2
n 1
1
In I n 2
n 1
1
In , as I n2 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 n2 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 n2 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 n2 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 n2 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