X2 t04 04 reduction formula (2012)

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X2 t04 04 reduction formula (2012)

  1. 1. Reduction Formula
  2. 2. Reduction FormulaReduction (or recurrence) formulae expresses a given integral asthe sum of a function and a known integral.Integration by parts often used to find the formula.
  3. 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 n2 0  n   2 where n is an integer and n  2, hence evaluate  cos 5 xdx 0
  4. 4.  2I n   cos n xdx 0
  5. 5.  2I n   cos n xdx 0  2   cos n 1 x cos xdx 0
  6. 6.  2I 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. 7.  2I 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. 8.  2I 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. 9.  2I 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. 10.  2I 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. 11.  2I 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. 12.  2I 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. 13. 20 cos 5 xdx  I 5
  14. 14. 20 cos 5 xdx  I 5 4  I3 5
  15. 15. 20 cos 5 xdx  I 5 4  I3 5 4 2   I1 5 3
  16. 16. 20 cos 5 xdx  I 5 4  I3 5 4 2   I1 5 3  8 2   cos xdx 15 0
  17. 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. 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 8    sin  sin 0   15  2  8  15
  19. 19. ii  Given that I n   cot n xdx, find I 6
  20. 20. ii  Given that I n   cot n xdx, find I 6 I n   cot n xdx
  21. 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. 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. 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. 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. 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. 26.  cot 6 xdx  I 6
  27. 27.  cot 6 xdx  I 6 1 5   cot x  I 4 5
  28. 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. 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. 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. 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. 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 1a ) Show that I n  I n2  n 1
  33. 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 1a ) 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. 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 1a ) 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. 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 1a ) 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. 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 1a ) 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. 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 1a ) 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. 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 1a ) 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. 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 1a ) 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. 40.  1 nb) Deduce that J n  J n1  for n  1 2n  1
  41. 41.  1 nb) 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. 42.  1 nb) 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. 43.  1 nb) 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. 44.  1 nb) 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. 45.  1 n  mc) Show that J m   4 n 1 2n  1
  46. 46.  1 n  mc) Show that J m   4 n 1 2n  1  1 m Jm   J m1 2m  1
  47. 47.  1 n  mc) 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. 48.  1 n  mc) 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. 49.  1 n  mc) 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. 50.  1 n  mc) 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. 51. 1 und ) Use the substitution u  tan x to show that I n   du 0 1 u2
  52. 52. 1 und ) Use the substitution u  tan x to show that I n   du 0 1 u2  I n   4 tan n xdx 0
  53. 53. 1 und ) 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. 54. 1 und ) 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. 55. 1 und ) 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. 56. 1 und ) 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 1e) Deduce that 0  I n  and conclude that J n  0 as n   n 1
  57. 57. 1 und ) 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 1e) 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. 58. 1 und ) 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 1e) 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. 59. 1I n  I n 2  n 1
  60. 60. 1I n  I n 2  n 1 1In   I n 2 n 1
  61. 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. 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. 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. 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. 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. 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

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