05 substiution rule
DESCRIPTION
Advance Mathematics about Substitution Rule A Guide for BeginnersTRANSCRIPT
Substitution Rule
duufdxxuxgf
havewetervalinthatonThenxurangeoncontinuousisfIf
tervalinsomeonderivativecontinuoushavexuuLet
)()()((
:,)(
)(
The Differential
dxxfdyxfyLet
)(aldifferentiThe)(:
Examples
42)(5.
5sin2)(4.5tan2)(3.
52)(2.
52)(1.
:cases following theofeach for aldifferenti theFind
3
4
xxu
xxuxxuxxu
xxu
du
Solution
3 232
31
3
2
21
21
34
3
2324242)(5.
cos25sin2)(4.sec25tan2)(3.
5252)(2.
852)(1.
:cases following theofeach for aldifferenti theFind
x
dxdxxduxxxu
dxxduxxudxxduxxu
xdxdxxduxxxu
dxxduxxu
du
Substitution RuleBasic Problems
Example (1)
3 74
3
)52(
:
x
dxxI
Evaluate
duu
uxdux
x
dxxI
xdudxdxxdu
xuLet
x
dxxI
37
3 7
33
3 74
3
33
4
3 74
3
81
8
)52(
88
52:
)52(
cx
cu
cu
duu
3 44
3 4
34
37
)52(
1323
1323
)34(8
1
81
Example (2)
xdxx
I
Evaluate
7)52(1:
cxcu
duu
duxux
dxxx
I
duxx
dudxdxxdxxdu
xxuLet
dxxx
I
8)52(
8
1
)52(1
212
5252:
)52(1
88
7
7
7
21
21
21
21
7
Example (3)
3 7
2
)5tan2(
sec
:
x
dxxI
Evaluate
duu
u
xdux
x
dxxI
xdudxdxxdu
xuLetx
dxxI
37
3 7
22
3 7
2
22
3 7
2
21
sec2sec
)5tan2(
sec
sec2sec2
5tan2:)5tan2(
sec
cx
cu
cu
duu
3 4
3 4
34
37
)5tan2(
183
183
)34(2
1
21
Example (4)
3 7)5sin2(
cos:
x
dxxI
Evaluate
duu
ux
dux
xdxxI
xdudxdxxdu
xuLetx
dxxI
37
3 7
3 7
3 7
21
cos2cos
)5sin2(cos
cos2cos2
5sin2:)5sin2(
cos
cx
cu
cu
duu
3 4
3 4
34
37
)5sin2(
183
183
342
1
21
Example (5)
dxxxI
Evaluate
)42cos(
:87
cx
cu
duu
xduux
dxxxI
xdudxdxxdu
xuLet
dxxxI
)42sin(161
sin161
cos161
16cos
)42cos(
1616
42:
)42cos(
8
77
87
77
8
87
Example (6)
dxxxI
Evaluate
)42(sec
:827
cx
cu
duu
xduux
dxxxI
xdudxdxxdu
xuLet
dxxxI
)42tan(161
tan161
sec161
16sec
)42(sec
1616
42:
)42(sec
8
2
727
827
77
8
827
Example (7)
dxxxxI
Evaluate
)42tan()42sec(
:887
cx
cu
duuu
xduuux
dxxxxI
xdudxdxxdu
xuLet
dxxxxI
)42sec(161
sec161
tansec161
16tansec
)42tan()42sec(
1616
42:
)42tan()42sec(
8
77
887
77
8
887
Example (8)
3 2
33 )42tan()42sec(
:
x
dxxxI
Evaluate
cx
cu
duuux
duxuu
x
dxxxI
duxx
dudxdxxdu
xxuLet
x
dxxxI
)42sec(23
sec23
tansec232
3tansec
)42tan()42sec(
23
323
12
4242:
)42tan()42sec(
3
3 2
3 2
3 2
33
3 2
32
32
31
3
3 2
33
Example (9)
dxxxxI
Evaluate
)4sin2tan()4sin2sec(cos
:
cx
cu
duuu
xduuux
dxxxxI
xdudxdxxdu
xuLet
dxxxxI
)4cos2sec(21
sec21
tansec21
cos2tanseccos
)4sin2tan()4sin2sec(cos
cos2cos2
4sin2:
)4sin2tan()4sin2sec(cos
Substitution RuleDefinite Integral Case
Example (1)
1
02)12( x
dxI
Evaluate
3
12
3
12
1
02
1
02
212
)12(
31211100
22
12:)12(
udu
u
duxdxI
uxux
dudxdxdu
xuLetxdxI
31)
32(
21
)131(
21
121
121
21
21
3
1
3
1
1
3
1
2
3
12
uu
duu
udu
Example (2)
1
034
3
)12( xdxxI
Evaluate
3
13
3
13
33
1
034
3
33
4
1
034
3
818
)12(
31211100
88
12:
)12(
udu
uxdux
xdxxI
uxux
xdudxdxxdu
xuLet
xdxxI
181)
98(
161
)191(
161
1161
281
81
81
3
12
3
1
2
3
1
3
3
13
uu
duu
udu
Example (3)
dxxxI
Evaluate
cossin2
0
duu
xduxuI
ux
uxx
dudxdxxdu
xuLet
dxxxI
1
0
21
1
0
21
2
0
coscos
1)2
sin(2
00sin0cos
cos
sin:
cossin
32)01(
32
32
23
1
0
31
0
23
1
0
21
uu
duu
Substitution Rule
More Challenging Problems
Example (1)
25
:
xdxxI
Evaluate
Method 1
duuu
duu
uu
duuxdxxI
uxand
dudxdxdu
xuLetxdxxI
)2(251
225155
225
52
55
25:25
21
21
21
cxx
cuu
duuu
21
23
21
23
21
21
)25(254)25(
752
)
212
23(
251
)2(251
25
:
xdxxI
EvaluatetoMethodAnother
duu
u
duuuxdxx
I
uxand
duudxdxduu
xu
xuLetxdxxI
)2(252
52
52
25
52
52
52
25
25:25
2
2
2
2
cxx
cxx
cuu
212
3
212
3
3
)25(254
75)25(2
)25(23
)25(252
)23
(252
Note that the first method can be used to find the integral of any function of the form:f(x) = x(2n-1) (axn+b)k
for any positive integer n and any real number k (where k is not -1) as the following examples show:
Example (2)
dxxxI
x
dxxI
x
dxxI
dxxxI
dxxxI
EvaluateExamples
1235
3 22
3
3 2
12
12
)42()5(
)42()4(
)42()3(
)42()2(
)42()1(
::
In all of the first three examples, we let:u = 2x+ 4and so:du = 2dx → dx = du/2andx = (u - 4)/2
cxx
cuuduuu
duuuduuuI
uxand
dudxdxduxulet
Solution
xdxxI
dxxxI
dxxxI
IntegralsthreefirstThe
13)42(4
14)42(
41
)13
414
(41)4(
41
)4(41
2242
42
242
:
)42()3(
)42()2(
)42()1(
:
1314
13141213
12121
33
122
121
cxx
cuuduuu
duuuduuuI
uxand
dudxdxduxulet
Solution
xdxxI
dxxxI
dxxxI
)11()42(4
)10()42(
41
)11(4
)10(41)4(
41
)4(41
2242
42
242
:
)42()3(
)42()2(
)42()1(
1110
11101211
12122
33
122
121
cxx
cuuduuu
duuuduuuI
uxand
dudxdxduxulet
Solution
xdxxI
dxxxI
dxxxI
31
34
31
34
32
31
32
32
3
33
122
121
)42(12)42(43
41
314
344
1)4(41
)4(41
2242
42
242
:
)42()3(
)42()2(
)42()1(
In the fourth example, we let:u = 2x2+ 4and so:du = 4xdx → dx = du/4xandx2 = (u - 4)/2
beforeascontinueweThen
duuu
duuu
duuxu
xdux
I
uxand
xdudxdxxdu
xuLet
x
dxxI
)4(81
24
41
414
24
44
42:
)42()4(
32
32
32
2
32
3
4
2
2
3 22
3
4
In the fifth example, we let:u = 2x3+ 4and so:du = 6x2dx → dx = du/6x2
andx3 = (u - 4)/2
beforeascontinuethenand
duuu
duuu
duux
xduuxI
uxand
xdudxdxxdu
xuLet
dxxxI
)4(121
61
24
61
6
24
66
42:
)42()5(
12
12
123
2125
5
3
22
3
12355
Examples (3)
2
0
2
2
2
cos.
sin.
cos.
:
dxxIc
dxxIb
dxxIa
Evaluate
The double angle formulas can simplify these problems, by replacing cos2x by (1+cos2x)/2 and sin2x by (1- cos2x)/2
dxxIb
dxxIa
onesequivalentfollowingthetotransferedareproblemstheSo
22cos1.
22cos1.
:
cxx
Whycxx
dxxdx
dxx
dxxIa
2sin41
21
?22sin
21
21
2cos21
21
22cos1
cos. 2
cxx
Whycxx
dxxdx
dxx
dxxIa
2sin41
21
?22sin
21
21
2cos21
21
22cos1
sin. 2
4
0sin41)0(
21sin
41
221
2sin41
21
cos.
20
2
0
2
xx
dxxIc
Note: If the problems were what we have below, then his would be like the basic examples. Do them!
dxxxIb
dxxxIa
Evaluate
cossin.
sincos.
:
2
2
Other Problems having the same idea: Do them!
xxxxxHdxxx
If
dxxx
Ie
xxxxHdxxxId
xxxxxHdxxxIc
dxxx
Ib
dxxxIa
Evaluate
cos)sin1(coscoscos:int(cossin
3.
sincos
3.
cot)csccsccotcsc:int(cotcsc.
)tansecsectansec:int(tansec.
csccot
3.
sectan.
:
2233
5 2
5 2
21
23
3
99100100
2
5 2
2100