12 x1 t06 01 integration using substitution (2013)
TRANSCRIPT
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Integration Using Substitution
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Integration Using Substitution
Try to convert the integral into a “standard integral”
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Integration Using Substitution
Try to convert the integral into a “standard integral”STANDARD INTEGRALS
dxxn 0 if ,0;1 ,1
1 1
nxnxn
n
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Integration Using Substitution
Try to convert the integral into a “standard integral”STANDARD INTEGRALS
dxxn 0 if ,0;1 ,1
1 1
nxnxn
n
dxx1
0 ,ln xx
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Integration Using Substitution
Try to convert the integral into a “standard integral”STANDARD INTEGRALS
dxxn 0 if ,0;1 ,1
1 1
nxnxn
n
dxx1
0 ,ln xx
![Page 6: 12 x1 t06 01 integration using substitution (2013)](https://reader034.vdocuments.mx/reader034/viewer/2022042816/5591a1f01a28ab9f268b4738/html5/thumbnails/6.jpg)
Integration Using Substitution
Try to convert the integral into a “standard integral”STANDARD INTEGRALS
dxxn 0 if ,0;1 ,1
1 1
nxnxn
n
dxx1
0 ,ln xx
dxeax 0 ,1 ae
aax
![Page 7: 12 x1 t06 01 integration using substitution (2013)](https://reader034.vdocuments.mx/reader034/viewer/2022042816/5591a1f01a28ab9f268b4738/html5/thumbnails/7.jpg)
Integration Using Substitution
Try to convert the integral into a “standard integral”STANDARD INTEGRALS
dxxn 0 if ,0;1 ,1
1 1
nxnxn
n
dxx1
0 ,ln xx
dxeax 0 ,1 ae
aax
dxax cos 0 ,sin1 aax
a
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Integration Using Substitution
Try to convert the integral into a “standard integral”STANDARD INTEGRALS
dxxn 0 if ,0;1 ,1
1 1
nxnxn
n
dxx1
0 ,ln xx
dxeax 0 ,1 ae
aax
dxax cos 0 ,sin1 aax
a
dxax sin 0 ,cos1 aax
a
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dxax sec2 0 ,tan1 aax
a
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dxax sec2 0 ,tan1 aax
a
dxaxax tansec 0 ,sec1 aax
a
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dxax sec2 0 ,tan1 aax
a
dxaxax tansec 0 ,sec1 aax
a
dxxa
122 0 ,tan1 1 a
ax
a
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dxax sec2 0 ,tan1 aax
a
dxaxax tansec 0 ,sec1 aax
a
dxxa
122 0 ,tan1 1 a
ax
a
dxxa
122 axaa
ax
0, ,sin 1
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dxax sec2 0 ,tan1 aax
a
dxaxax tansec 0 ,sec1 aax
a
dxxa
122 0 ,tan1 1 a
ax
a
dxxa
122 axaa
ax
0, ,sin 1
dxax
122 0 ,ln 22 axaxx
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dxax sec2 0 ,tan1 aax
a
dxaxax tansec 0 ,sec1 aax
a
dxxa
122 0 ,tan1 1 a
ax
a
dxxa
122 axaa
ax
0, ,sin 1
dxax
122 0 ,ln 22 axaxx
dxax
122 22ln axx
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“NON” STANDARD INTEGRALSNot listed in the standard integrals, however will save time if you know
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“NON” STANDARD INTEGRALSNot listed in the standard integrals, however will save time if you know
duu 2 u
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“NON” STANDARD INTEGRALSNot listed in the standard integrals, however will save time if you know
duu 2 u
2
duu
1u
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“NON” STANDARD INTEGRALSNot listed in the standard integrals, however will save time if you know
ln xdx ln x x x
duu 2 u
2
duu
1u
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“NON” STANDARD INTEGRALSNot listed in the standard integrals, however will save time if you know
ln xdx ln x x x
duu 2 u
2
duu
1u
tan xdx logsec x
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“NON” STANDARD INTEGRALSNot listed in the standard integrals, however will save time if you know
ln xdx ln x x x
duu 2 u
2
duu
1u
complementary trig ratio complement of the answer
tan xdx logsec x
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“NON” STANDARD INTEGRALSNot listed in the standard integrals, however will save time if you know
ln xdx ln x x x
duu 2 u
2
duu
1u
complementary trig ratio complement of the answer
e.g.cot x dx
tan xdx logsec x
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“NON” STANDARD INTEGRALSNot listed in the standard integrals, however will save time if you know
ln xdx ln x x x
duu 2 u
2
duu
1u
complementary trig ratio complement of the answer
e.g.cot x dx
tan xdx logsec x
cot is the complement of tan
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“NON” STANDARD INTEGRALSNot listed in the standard integrals, however will save time if you know
ln xdx ln x x x
duu 2 u
2
duu
1u
complementary trig ratio complement of the answer
e.g.cot x dx log cosec x
tan xdx logsec x
cot is the complement of tan
so the answer is minus and the
complement of sec is cosec
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“NON” STANDARD INTEGRALSNot listed in the standard integrals, however will save time if you know
ln xdx ln x x x
duu 2 u
2
duu
1u
complementary trig ratio complement of the answer
e.g.cot x dx log cosec x
tan xdx logsec x
logsin xcot is the complement of tan
so the answer is minus and the
complement of sec is cosec
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e.g. (i) dxxx 42
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e.g. (i) dxxx 4242 xu
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e.g. (i) dxxx 4242 xu
u
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e.g. (i) dxxx 4242 xu
xdxdu 2u
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e.g. (i) dxxx 4242 xu
xdxdu 2udu
21
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e.g. (i) dxxx 4242 xu
duu
dxxx
21
2
21
4221
xdxdu 2udu
21
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e.g. (i) dxxx 4242 xu
duu
dxxx
21
2
21
4221
xdxdu 2udu
21
cu 23
32
21
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e.g. (i) dxxx 4242 xu
duu
dxxx
21
2
21
4221
xdxdu 2udu
21
cu 23
32
21
cu 23
31
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e.g. (i) dxxx 4242 xu
duu
dxxx
21
2
21
4221
xdxdu 2udu
21
cu 23
32
21
cu 23
31
cx 23
2 431
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e.g. (i) dxxx 4242 xu
duu
dxxx
21
2
21
4221
xdxdu 2udu
21
cu 23
32
21
cu 23
31
cx 23
2 431
cxx 4431 22
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dx
xxxii
7841
2
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dx
xxxii
7841
2
dxxduxxu
88784 2
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dx
xxxii
7841
2
dxxduxxu
88784 2
udu
81
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dx
xxxii
7841
2
dxxduxxu
88784 2
cu log81
udu
81
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dx
xxxii
7841
2
dxxduxxu
88784 2
cu log81
cxx 784log81 2
udu
81
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dx
xxxii
7841
2
dxxduxxu
88784 2
cu log81
cxx 784log81 2
udu
81
3log xx
dxiii
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dx
xxxii
7841
2
dxxduxxu
88784 2
cu log81
cxx 784log81 2
udu
81
3log xx
dxiii
xdxdu
xu
log
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dx
xxxii
7841
2
dxxduxxu
88784 2
cu log81
cxx 784log81 2
udu
81
3log xx
dxiii
xdxdu
xu
log
duuudu
3
3
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dx
xxxii
7841
2
dxxduxxu
88784 2
cu log81
cxx 784log81 2
udu
81
3log xx
dxiii
xdxdu
xu
log
duuudu
3
3
cu
cu
2
2
21
21
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dx
xxxii
7841
2
dxxduxxu
88784 2
cu log81
cxx 784log81 2
udu
81
3log xx
dxiii
xdxdu
xu
log
duuudu
3
3
cu
cu
2
2
21
21
c
x 2log2
1
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dx
xxiv
1
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dx
xxiv
1ududx
xuux2
11 2
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dx
xxiv
1ududx
xuux2
11 2
udu
uu 21 2
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dx
xxiv
1ududx
xuux2
11 2
duu 12 2
uduuu 21 2
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dx
xxiv
1ududx
xuux2
11 2
duu 12 2
cuu
3
312
uduuu 21 2
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dx
xxiv
1ududx
xuux2
11 2
duu 12 2
cuu
3
312
uduuu 21 2
cxx 12132 3
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dx
xxiv
1ududx
xuux2
11 2
duu 12 2
cuu
3
312
uduuu 21 2
cxx 12132 3
cxxx 121132
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dx
xxiv
1ududx
xuux2
11 2
duu 12 2
cuu
3
312
uduuu 21 2
2
3
5 cossin
xdxxv
cxx 12132 3
cxxx 121132
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dx
xxiv
1ududx
xuux2
11 2
duu 12 2
cuu
3
312
uduuu 21 2
2
3
5 cossin
xdxxvxdxduxu
cossin
cxx 12132 3
cxxx 121132
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dx
xxiv
1ududx
xuux2
11 2
duu 12 2
cuu
3
312
uduuu 21 2
2
3
5 cossin
xdxxvxdxduxu
cossin
cxx 12132 3
cxxx 121132
23
3sin,
3
u
ux
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dx
xxiv
1ududx
xuux2
11 2
duu 12 2
cuu
3
312
uduuu 21 2
2
3
5 cossin
xdxxvxdxduxu
cossin
cxx 12132 3
cxxx 121132
23
3sin,
3
u
ux
1 2
sin,2
u
ux
![Page 56: 12 x1 t06 01 integration using substitution (2013)](https://reader034.vdocuments.mx/reader034/viewer/2022042816/5591a1f01a28ab9f268b4738/html5/thumbnails/56.jpg)
dx
xxiv
1ududx
xuux2
11 2
duu 12 2
cuu
3
312
uduuu 21 2
2
3
5 cossin
xdxxvxdxduxu
cossin
1
23
5duu
cxx 12132 3
cxxx 121132
23
3sin,
3
u
ux
1 2
sin,2
u
ux
![Page 57: 12 x1 t06 01 integration using substitution (2013)](https://reader034.vdocuments.mx/reader034/viewer/2022042816/5591a1f01a28ab9f268b4738/html5/thumbnails/57.jpg)
dx
xxiv
1ududx
xuux2
11 2
duu 12 2
cuu
3
312
uduuu 21 2
2
3
5 cossin
xdxxvxdxduxu
cossin
1
23
5duu
123
6
61 u
cxx 12132 3
cxxx 121132
23
3sin,
3
u
ux
1 2
sin,2
u
ux
![Page 58: 12 x1 t06 01 integration using substitution (2013)](https://reader034.vdocuments.mx/reader034/viewer/2022042816/5591a1f01a28ab9f268b4738/html5/thumbnails/58.jpg)
dx
xxiv
1ududx
xuux2
11 2
duu 12 2
cuu
3
312
uduuu 21 2
2
3
5 cossin
xdxxvxdxduxu
cossin
1
23
5duu
123
6
61 u
38437
231
61
66
cxx 12132 3
cxxx 121132
23
3sin,
3
u
ux
1 2
sin,2
u
ux
![Page 59: 12 x1 t06 01 integration using substitution (2013)](https://reader034.vdocuments.mx/reader034/viewer/2022042816/5591a1f01a28ab9f268b4738/html5/thumbnails/59.jpg)
4
3225 x
xdxvi
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4
3225 x
xdxvixdxdu
xu2
25 2
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4
3225 x
xdxvixdxdu
xu2
25 2
9,416,3
uxux
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4
3225 x
xdxvixdxdu
xu2
25 2
16
9
21
9
16
21
21
duu
udu
9,416,3
uxux
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4
3225 x
xdxvixdxdu
xu2
25 2
16
9
21
9
16
21
21
duu
udu
16
9
21
u
9,416,3
uxux
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4
3225 x
xdxvixdxdu
xu2
25 2
16
9
21
9
16
21
21
duu
udu
16
9
21
u
1916
9,416,3
uxux
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5
0
225 dxxvii
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5
0
225 dxxvii
ududx
xuux
cos55
sinsin5 1
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5
0
225 dxxvii
ududx
xuux
cos55
sinsin5 1
2
1sin,50
0sin,0
1
1
u
uxuux
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5
0
225 dxxvii
ududx
xuux
cos55
sinsin5 1
2
0
2 cos5sin2525
uduu
2
1sin,50
0sin,0
1
1
u
uxuux
![Page 69: 12 x1 t06 01 integration using substitution (2013)](https://reader034.vdocuments.mx/reader034/viewer/2022042816/5591a1f01a28ab9f268b4738/html5/thumbnails/69.jpg)
5
0
225 dxxvii
ududx
xuux
cos55
sinsin5 1
2
0
2 cos5sin2525
uduu
2
0
2
2
0
2
cos25
cos5cos25
udu
uduu
2
1sin,50
0sin,0
1
1
u
uxuux
![Page 70: 12 x1 t06 01 integration using substitution (2013)](https://reader034.vdocuments.mx/reader034/viewer/2022042816/5591a1f01a28ab9f268b4738/html5/thumbnails/70.jpg)
5
0
225 dxxvii
ududx
xuux
cos55
sinsin5 1
2
0
2 cos5sin2525
uduu
2
0
2
2
0
2
cos25
cos5cos25
udu
uduu
2
0
2
0
2sin21
225
2cos1225
uu
duu2
1sin,50
0sin,0
1
1
u
uxuux
![Page 71: 12 x1 t06 01 integration using substitution (2013)](https://reader034.vdocuments.mx/reader034/viewer/2022042816/5591a1f01a28ab9f268b4738/html5/thumbnails/71.jpg)
5
0
225 dxxvii
ududx
xuux
cos55
sinsin5 1
2
0
2 cos5sin2525
uduu
2
0
2
2
0
2
cos25
cos5cos25
udu
uduu
2
0
2
0
2sin21
225
2cos1225
uu
duu2
1sin,50
0sin,0
1
1
u
uxuux
425
0sin21
2225
![Page 72: 12 x1 t06 01 integration using substitution (2013)](https://reader034.vdocuments.mx/reader034/viewer/2022042816/5591a1f01a28ab9f268b4738/html5/thumbnails/72.jpg)
5
0
225 dxxvii
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5
0
225 dxxviiy
x5
5
-5
225 xy
![Page 74: 12 x1 t06 01 integration using substitution (2013)](https://reader034.vdocuments.mx/reader034/viewer/2022042816/5591a1f01a28ab9f268b4738/html5/thumbnails/74.jpg)
5
0
225 dxxviiy
x5
5
-5
225 xy
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5
0
225 dxxviiy
x5
5
-5
225 xy
425
541 2
![Page 76: 12 x1 t06 01 integration using substitution (2013)](https://reader034.vdocuments.mx/reader034/viewer/2022042816/5591a1f01a28ab9f268b4738/html5/thumbnails/76.jpg)
Exercise 6C; 1, 2ace, 3, 4ace, 5ace etc, 6, 7 & 8ac, 11, 12Exercise 6D; 1, 2a, 3, 4b, 5ac, 6ace etc, 7ab(i,ii)
8ab (i,iii,vi), 9ab (ii,iv,vi), 11, 13
Patel Exercise 2A ace in all NOTE: substitution is not given in Extension 2
5
0
225 dxxviiy
x5
5
-5
225 xy
425
541 2