x2 t04 04 reduction formula (2012)
TRANSCRIPT
Reduction Formula
Reduction FormulaReduction (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.
Reduction FormulaReduction (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
xdxnn
In
nIxdxI nnn
n
2
0
cos
xdxI nn
2
0
cos
xdxI nn
2
0
1 coscos
xdxxn
2
0
cos
xdxI nn
xu n 1cos xdxxndu n sincos1 2 xdxdv cos
xv sin
2
0
1 coscos
xdxxn
2
0
cos
xdxI nn
xu n 1cos
2
0
2220
1 sincos1sincos
xdxxnxx nn
xdxxndu n sincos1 2 xdxdv cos
xv sin
2
0
1 coscos
xdxxn
2
0
cos
xdxI nn
xu n 1cos
2
0
2220
1 sincos1sincos
xdxxnxx nn
xdxxndu n sincos1 2 xdxdv cos
xv sin
2
0
1 coscos
xdxxn
2
0
2211 cos1cos10sin0cos2
sin2
cos
dxxxn nnn
2
0
cos
xdxI nn
xu n 1cos
2
0
2220
1 sincos1sincos
xdxxnxx nn
xdxxndu n sincos1 2 xdxdv cos
xv sin
2
0
1 coscos
xdxxn
2
0
2211 cos1cos10sin0cos2
sin2
cos
dxxxn nnn
2
0
2
0
2 cos1cos1
xdxnxdxn nn
2
0
cos
xdxI nn
xu n 1cos
2
0
2220
1 sincos1sincos
xdxxnxx nn
xdxxndu n sincos1 2 xdxdv cos
xv sin
2
0
1 coscos
xdxxn
2
0
2211 cos1cos10sin0cos2
sin2
cos
dxxxn nnn
2
0
2
0
2 cos1cos1
xdxnxdxn nn
nn InIn 11 2
2
0
cos
xdxI nn
xu n 1cos
2
0
2220
1 sincos1sincos
xdxxnxx nn
xdxxndu n sincos1 2 xdxdv cos
xv sin
21 nn InnI
2
0
1 coscos
xdxxn
2
0
2211 cos1cos10sin0cos2
sin2
cos
dxxxn nnn
2
0
2
0
2 cos1cos1
xdxnxdxn nn
nn InIn 11 2
2
0
cos
xdxI nn
xu n 1cos
2
0
2220
1 sincos1sincos
xdxxnxx nn
xdxxndu n sincos1 2 xdxdv cos
xv sin
21 nn InnI
2
0
1 coscos
xdxxn
2
0
2211 cos1cos10sin0cos2
sin2
cos
dxxxn nnn
2
0
2
0
2 cos1cos1
xdxnxdxn nn
nn InIn 11 2
21
nn In
nI
5
2
0
5cos Ixdx
5
2
0
5cos Ixdx
354 I
5
2
0
5cos Ixdx
354 I
132
54 I
5
2
0
5cos Ixdx
354 I
132
54 I
2
0
cos158
xdx
5
2
0
5cos Ixdx
354 I
132
54 I
2
0
cos158
xdx
20sin
158
x
5
2
0
5cos Ixdx
354 I
132
54 I
2
0
cos158
xdx
20sin
158
x
158
0sin2
sin158
6 find ,cot Given that IxdxIii nn
6 find ,cot Given that IxdxIii nn
xdxI nn cot
6 find ,cot Given that IxdxIii nn
xdxI nn cot
xdxxn 22 cotcot
6 find ,cot Given that IxdxIii nn
xdxI nn cot
xdxxn 22 cotcot
dxxxn 1coseccot 22
xdxxdxx nn 222 cotcoseccot
6 find ,cot Given that IxdxIii nn
xdxI nn cot
xdxxn 22 cotcot
dxxxn 1coseccot 22
xdxxdxx nn 222 cotcoseccot xu cot
xdxdu 2cosec
6 find ,cot Given that IxdxIii nn
xdxI nn cot
xdxxn 22 cotcot
dxxxn 1coseccot 22
xdxxdxx nn 222 cotcoseccot xu cot
xdxdu 2cosec2
2
nn Iduu
6 find ,cot Given that IxdxIii nn
xdxI nn cot
xdxxn 22 cotcot
dxxxn 1coseccot 22
xdxxdxx nn 222 cotcoseccot xu cot
xdxdu 2cosec2
2
nn Iduu
21
11
n
n Iun
21cot
11
n
n Ixn
66cot Ixdx
66cot Ixdx
45cot
51 Ix
66cot Ixdx
45cot
51 Ix
235 cot
31cot
51 Ixx
66cot Ixdx
45cot
51 Ix
235 cot
31cot
51 Ixx
035 cotcot
31cot
51 Ixxx
66cot Ixdx
45cot
51 Ix
235 cot
31cot
51 Ixx
035 cotcot
31cot
51 Ixxx
dxxxx cotcot31cot
51 35
66cot Ixdx
45cot
51 Ix
235 cot
31cot
51 Ixx
035 cotcot
31cot
51 Ixxx
dxxxx cotcot31cot
51 35
cxxxx cotcot31cot
51 35
(iii) (2004 Question 8b)
420
Let tan and let 1 for 0,1,2,nnn n nI xdx J I n
21) Show that
1n na I In
(iii) (2004 Question 8b)
420
Let tan and let 1 for 0,1,2,nnn n nI xdx J I n
21) Show that
1n na I In
24 42 0 0
tan tann nn nI I dx dx
(iii) (2004 Question 8b)
420
Let tan and let 1 for 0,1,2,nnn n nI xdx J I n
21) Show that
1n na I In
24 42 0 0
tan tann nn nI I dx dx
24
0tan 1 tann x x dx
24
0tan secn x xdx
(iii) (2004 Question 8b)
420
Let tan and let 1 for 0,1,2,nnn n nI xdx J I n
21) Show that
1n na I In
24 42 0 0
tan tann nn nI I dx dx
24
0tan 1 tann x x dx
24
0tan secn x xdx
tanu x
2secdu xdx
(iii) (2004 Question 8b)
420
Let tan and let 1 for 0,1,2,nnn n nI xdx J I n
21) Show that
1n na I In
24 42 0 0
tan tann nn nI I dx dx
24
0tan 1 tann x x dx
24
0tan secn x xdx
tanu x
2secdu xdx
when 0, 0
, 14
x u
x u
(iii) (2004 Question 8b)
420
Let tan and let 1 for 0,1,2,nnn n nI xdx J I n
21) Show that
1n na I In
24 42 0 0
tan tann nn nI I dx dx
24
0tan 1 tann x x dx
24
0tan secn x xdx
tanu x
2secdu xdx1
0
nu du when 0, 0
, 14
x u
x u
(iii) (2004 Question 8b)
420
Let tan and let 1 for 0,1,2,nnn n nI xdx J I n
21) Show that
1n na I In
24 42 0 0
tan tann nn nI I dx dx
24
0tan 1 tann x x dx
24
0tan secn x xdx
tanu x
2secdu xdx1
0
nu du when 0, 0
, 14
x u
x u
11
01
nun
(iii) (2004 Question 8b)
420
Let tan and let 1 for 0,1,2,nnn n nI xdx J I n
21) Show that
1n na I In
24 42 0 0
tan tann nn nI I dx dx
24
0tan 1 tann x x dx
24
0tan secn x xdx
tanu x
2secdu xdx1
0
nu du when 0, 0
, 14
x u
x u
11
01
nun
1 10 1 1n n
1
1) Deduce that for 1
2 1
n
n nb J J nn
1
1) Deduce that for 1
2 1
n
n nb J J nn
1
1 2 2 21 1n nn n n nJ J I I
1
1) Deduce that for 1
2 1
n
n nb J J nn
1
1 2 2 21 1n nn n n nJ J I I
2 2 21 1n nn nI I
1
1) Deduce that for 1
2 1
n
n nb J J nn
1
1 2 2 21 1n nn n n nJ J I I
2 2 21 1n nn nI I
2 2 21 nn nI I
1
1) Deduce that for 1
2 1
n
n nb J J nn
1
1 2 2 21 1n nn n n nJ J I I
2 2 21 1n nn nI I
2 2 21 nn nI I
12 1
n
n
1
1) Show that
4 2 1
nm
mn
c Jn
1
1) Show that
4 2 1
nm
mn
c Jn
1
12 1
m
m mJ Jm
1
1) Show that
4 2 1
nm
mn
c Jn
1
12 1
m
m mJ Jm
1
2
1 12 1 2 3
m m
mJm m
1 1
0
1 1 12 1 2 3 1
m m
Jm m
1
1) Show that
4 2 1
nm
mn
c Jn
1
12 1
m
m mJ Jm
1
2
1 12 1 2 3
m m
mJm m
1 1
0
1 1 12 1 2 3 1
m m
Jm m
40
1
12 1
nm
n
dxn
1
1) Show that
4 2 1
nm
mn
c Jn
1
12 1
m
m mJ Jm
1
2
1 12 1 2 3
m m
mJm m
1 1
0
1 1 12 1 2 3 1
m m
Jm m
40
1
12 1
nm
n
dxn
4
01
12 1
nm
n
xn
1
12 1 4
nm
n n
1
1) Show that
4 2 1
nm
mn
c Jn
1
12 1
m
m mJ Jm
1
2
1 12 1 2 3
m m
mJm m
1
20) Use the substitution tan to show that
1
n
nud u x I du
u
1
20) Use the substitution tan to show that
1
n
nud u x I du
u
4
0tann
nI xdx
1
20) Use the substitution tan to show that
1
n
nud u x I du
u
4
0tann
nI xdx
1tan tanu x x u
21dudx
u
1
20) Use the substitution tan to show that
1
n
nud u x I du
u
4
0tann
nI xdx
1tan tanu x x u
21dudx
u
when 0, 0
, 14
x u
x u
1
20) Use the substitution tan to show that
1
n
nud u x I du
u
4
0tann
nI xdx
1tan tanu x x u
21dudx
u
when 0, 0
, 14
x u
x u
1
20 1n
nduI u
u
1
20 1
n
nu duI
u
1
20) Use the substitution tan to show that
1
n
nud u x I du
u
4
0tann
nI xdx
1tan tanu x x u
21dudx
u
when 0, 0
, 14
x u
x u
1
20 1n
nduI u
u
1
20 1
n
nu duI
u
1) Deduce that 0 and conclude that 0 as 1n ne I J n
n
1
20) Use the substitution tan to show that
1
n
nud u x I du
u
4
0tann
nI xdx
1tan tanu x x u
21dudx
u
when 0, 0
, 14
x u
x u
1
20 1n
nduI u
u
1
20 1
n
nu duI
u
1) Deduce that 0 and conclude that 0 as 1n ne I J n
n
2 0, for all 01
nu uu
1
20) Use the substitution tan to show that
1
n
nud u x I du
u
4
0tann
nI xdx
1tan tanu x x u
21dudx
u
when 0, 0
, 14
x u
x u
1
20 1n
nduI u
u
1
20 1
n
nu duI
u
1) Deduce that 0 and conclude that 0 as 1n ne I J n
n
2 0, for all 01
nu uu
1
200, for all 0
1
n
nuI du u
u
21
1n nI In
21
1n nI In
21
1n nI In
21
1n nI In
21
1n nI In
21 , as 0
1n nI In
21
1n nI In
21
1n nI In
101nI
n
21 , as 0
1n nI In
21
1n nI In
21
1n nI In
101nI
n
1 as , 0
1n
n
21 , as 0
1n nI In
21
1n nI In
21
1n nI In
101nI
n
1 as , 0
1n
n
0nI
21 , as 0
1n nI In
21
1n nI In
21
1n nI In
101nI
n
1 as , 0
1n
n
0nI
21 0nn nJ I
21 , as 0
1n nI In
21
1n nI In
21
1n nI In
21 , as 0
1n nI In
101nI
n
1 as , 0
1n
n
0nI
21 0nn nJ I
Exercise 2D; 1, 2, 3, 6, 8, 9, 10, 12, 14