resolucion de integrales sesion 4 metodo algebraico
TRANSCRIPT
8/20/2019 Resolucion de Integrales Sesion 4 Metodo Algebraico
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INTEGRALES
1. ( )
( ) dt
t t t
t I .
23
23
1
2
2
∫ ++−
=
( )( ) ( )
dt t t t
t I .
12
23
1
2
∫ ++−
=
( )
( )( )
( )α ......
1212
2 2
+
+
+
+=
++
−
t
C
t
B
t
A
t t t
t
( ) ( ) ( )12* ++ t t t α
( ) ( )( ) ( ) ( )2..112.2 2 ++++++=− t t C t Bt t t At
( ) Ct Ct Bt Bt A At t At At .222..2 2222 +++++++=−
( ) ( ) ( ) At C B A At C B At 2.22.2 22 +++++++=−
( )
122
0.2.3
1.
=→==++−=++
A A
C B A
C B A
1
11
322
−=−=→=−
−=+ −=+
B
C C
C B
C B
Luego:
dt t
dt t
dt t
.1
1.
2
1.
1 3
1
3
1
3
1
∫ ∫ ∫ +−
+−
( ) ( )
3
1
3
1
3
1 )1(2 +−+− t Lnt Lnt Ln
( ) ( )( ) ( ) ( )( ) ( ) ( )( )243513 Ln Ln Ln Ln Ln Ln −−−+−
( ) )2(3
53 Ln Ln Ln −
−
10
9 Ln
Rpta: 11.0
8/20/2019 Resolucion de Integrales Sesion 4 Metodo Algebraico
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2. ( )
( ) ( )( ) dt
t t t
t t I .
321
731
0
2
∫ ++++
=
( )( ) ( ) ( )
( )α ......321321
73 2
++
++
+=
++++
t
C
t
B
t
A
t t t
t t
( ) ( ) ( ) ( )321* +++ t t t α
( ) ( ) ( )( ) ( ) ( )2.1.3132.73 2 ++++++++=+ t t C t t Bt t At t
( ) ( ) ( )23.3465.73 2222 ++++++++=+ t t C t t Bt t At t
( ) ( ) ( )C B At C B At C B At t
236345.73
22
++++++++=+
( )
0236
7.3.4.5
3.
=++=++
=++
B A
C B A
C B A
2
42
3
7
−==−
=++=++−
A
A
C B A
C B A
3
2
1734
5
==
=+=+
C
B
C B
C B
∫ ∫ ∫ ++
++
+−
1
0
1
0
1
0 3
33
2
12
1
12
t t dt
t
( ) ( ) 1
0
1
0
1
0)3(.32.212
+−+++− t Lnt Lnt Ln
( ) ( )( ) ( ) ( )( ) ( )( )3)4(.323.2122 Ln Ln Ln Ln Ln Ln −+−+−−
( )
+
+−
3
4.3
2
3.222 Ln Ln Ln
( )32
3
4
2
34
+
+ Ln Ln Ln
34 Ln
Rpta: 288.0
3. ( )
( ) dt t t t
t
I .342
33
2
23∫ ++−
=
8/20/2019 Resolucion de Integrales Sesion 4 Metodo Algebraico
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( )
( ) dt
t t t
t I .
342
33
2
23∫ ++−
=
( )( )( ) ( ) dt
t t t t I .
313
3
2∫ ++ −=
( )
( )( )( ) ( )α ......
3131
3
++
++=
++−
t
C
t
B
t
A
t t t
t
( ) ( ) ( ) ( )31* ++ t t t α
( ) ( ) ( ) ( ) ( ) ( )1..331.3 ++++++=− t t C t t Bt t At
( ) ( ) ( )t t C t t Bt t At ++++++=− 222
..334.3
( ) ( ) At C B At C B At 3.343 2 ++++++=−
( )
133
1.3.4
0.
=→=−=++
=++
A A
C B A
C B A
1
2
53
1
=−=
−=+−=+
C
B
C B
C B
dt t
dt t
dt t
.
3
1.
1
21 3
2
3
2
3
2
∫ ∫ ∫ +
+
+
−+
( ) ( ) 3
2
3
2
3
2 )3(1.2 +++− t Lnt Lnt Ln
( ) ( )( ) ( ) ( )( ) ( )( )5)6(34.223 Ln Ln Ln Ln Ln Ln −+−−−
+
−
5
6
3
4
2
3 2
Ln Ln Ln
80
81 Ln
Rpta: 012.0
4. ( )
( ) dt
t t
t I .
32
1
23∫ −−
=
( )
( )( ) dt t t
t
I .1
32
1
2
∫ +−
=
8/20/2019 Resolucion de Integrales Sesion 4 Metodo Algebraico
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( )
( )( ) ( )α ......
11
322 +++=
+−
t
C
t
B
t
A
t t
t
( ) ( )( )1* 2 +t t α
( )( ) ( )( ) ( )2.11.3 t C t t Bt t At ++++=−
( ) ( ) ( ) Bt B At C At ++++=− 2.3
( )
4
4
3
1
0.
−==−==+=+
C
A
B
B A
C A
dt t
dt t
dt t
.1
4.
34 2
1
2
1
2
2
1
∫ ∫ ∫ +−
+−
+
( ) ( ) 2
1
2
1
12
1)1(.4.3.4 +−+ −
t Lnt Lnt Ln
( ) ( )( ) ( )( )2)3(412
1.3124 Ln Ln Ln Ln −−
−+−
( ) 24)3(.42324 Ln Ln Ln +−−
2
3
3
2*24 −
Ln
Rpta:2
3
81
256−
Ln
5. ( )
( ) dt
t t
t I .
2
23
8
2∫ −
− −
+=
( )
( ) ( ) ( )α ......
222
222 −
+−
+=−
+
t
C
t
B
t
A
t t
t
( ) ( )( ) 2
2* −t t α
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( ) ( ) ( ) ( )t C t t Bt At .22.2 2 +−+−=−
( ) ( ) At C B At B At 4.24.2 2 +++−−+=−
( )
2
2
1
2
1
24
12.4
0.
=
=
=
==−−
=+
C
B
A
A
C B A
B A
( ) dt
t
dt
t
dt
t
.2
1.2.
2
1
2
11
2
1 3
82
3
8
3
8 ∫ ∫ ∫
−
−
−
−
−
− −+
−−
( ) ( ) 3
8
13
8
3
8 )2.(22.
2
1.
2
1 −
−−−
−
−
− −−−− t t Lnt Ln
( ) ( )( ) ( )( ) ( ) ( )[ ]11105.210)5(.
2
183
2
1 −− −−−−−−− Ln Ln Ln Ln
( ) ( )( )5
1)
2
1(.
2
183
2
1 +
−− Ln Ln Ln
( ) ( )5
1)
2
1(83
2
1+
−− Ln Ln Ln
5
1
2
1*8
3
2
1+
Ln
Rpta:5
1
4
3
2
1+
Ln
6.( ) ( )
dt t t
t I .212
94
0
2
2
∫ ++=
( ) ( ) ( ) ( )α ......
2212212
922
2
++
++
+=
++ t
C
t
B
t
A
t t
t
( ) ( ) ( ) 2212* ++ t t α
( ) ( )( ) ( )12.1222.9 22 ++++++= t C t t Bt At
( ) ( ) C B At t Bt t At +++++++++= 2425244.9 222
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( ) ( ) C B At C B At B At +++++++= 24254.29 22
( )
12
4
1
024
025.4
9.4
−===
=++=++
=+
C
B
A
C B A
C B A
B A
( ) dt
t
dt t
dt t
.2
1.12.
2
14
12
1
2
1 4
0
2
4
0
4
0
∫ ∫ ∫ +−
++
+
( ) ( ) 4
0
14
0
4
0)2.(122.412.
2
1 −++−+−+ t t Lnt Ln
( ) ( )( ) ( )( ) [ ]11 26.122)6(.4192
1 −− −+−+− Ln Ln Ln Ln
( )( ) ( )3
1.12)2()6(43 −−+ Ln Ln Ln
Rpta: ( )( ) 43.5 − Ln
7.( )
dt t t
t I .
24
2
23
3
∫ −−
=
( )( )
I
t t
t
t t
t
)1(
21
1
2
2
2
2
3
−−
+
−
−
Primera parte I
( ) ( )α ......
1)1(
222
2
−++=
−−
t
C
t
B
t
A
t t
t
( ) ( ) 22
.11.2 t C t Bt At t +−+−=−
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( ) ( ) Bt B At C At −−−+=− 22 2
( )
1
2
2
0
1
−===
=−=+
C
B
A
B A
C A
( ) ∫ ∫ ∫ ∫ −−++
4
2
4
2
2
4
2
4
2
.1
1.
1.2.
121 dt
t dt
t dt
t dt
( ) ( ) 4
2
4
2
14
2
4
21).(2.2 −−−+ −
t Lnt t Lnt
( ) ( )( ) ( ) )1()3(24.224.2)24( 11 Ln Ln Ln Ln −−−−−+− −−
( ) ( ) )3(4
1.2224.2)24( Ln Ln Ln −
−−−+−
−−
+−
4
1.2
3*2
4)24(
2
2
Ln
−−
+
2
1
3
4)2( Ln
Rpta:
+
25
34 Ln
8.( ) ( )
dt t t
t I .
12
35
0
2
2
∫ ++−
=
( )( ) ( ) ( )α ......
11212
322
2
++
++
+=
++−
t
C
t
B
t
A
t t
t
m.c.m (t+2).(t+1)2
( ) ( )( ) ( )2.211.3 22 ++++++=− t C t t Bt At
( ) ( ) C t C t t Bt t At 2.2312.3 222 +++++++=−
( ) ( ) C B At C B At B At 2232.3 22 +++++++=−
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( )
2
0
1
322
022
1
−==
=
−=++=++
=+
C
B
A
C B A
C B A
B A
( ) dt
t
dt t
dt t
.1
1.2.
1
10
2
1 5
0
2
5
0
5
0
∫ ∫ ∫ +−
++
+
( ) 14
0
5
0 )1.(2)1(.02 −+++++ t t Lnt Ln
Rpta: ( ) )16.(2)2(7 11 −− −+− Ln Ln
9.( ) ( )
dt t t
t I .
12
51
0
2∫ ++=
D.E.P.
( ) ( ) ( )( )( )
)........(1212
522
α
++
++
=++ t
C Bt
t
A
t t
t
( )( ) )(*12
2α
++ t t
( ) ( )( )21.5 2 ++++= t C Bt t At
C Ct t B Bt At At 2..2.5 22 +++++=
( ) ( ) C At C Bt B At t 2..20.5.0 22 +++++=++
B A B A −=→=+ 0 2−= A
555)2.(25.2 =→=+→=+ C C C C B 1=C
C BC BC A .2020.2 =→=+−→=+ 2= B
( )α En
dt t
t
t I .
1
1.2
2
21
0
2∫
++
++−
=
dt t
dt t
t dt
t I .
1
1.
1
.2.
2
1.2
1
0
2
1
0
2
1
0
∫ ∫ ∫ +
+
+
+
+
−=
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dt du
t u
=
+= 2
tdt du
t u
2
12
=+=
dt du
t u
=
=
( ) ( ) 1
0
1
0
21
0 12.2 arcTant t Lnt Ln I ++++−=
( ) ( )( ) ( ) ( )( ) ( ) ( )011223.2 arcTanarcTan Ln Ln Ln Ln I −+−+−−=
( )4
22
3.2
π ++
−= Ln Ln I
( )4
24
9 π ++
−= Ln Ln I
4
4
91
2
π
+
= Ln I
Rpta:49
8 π +
= Ln I
10. ( ) dt
t
t t I .
16
454
3
4
3
∫ −−
=
( )( )44
4522
3
+−−t t
t t
( )( )( )422
452
3
++−
−
t t t
t t
( )( )( ) ( ) ( ) )....(
422422
4522
3
α
++
++
+−
=++−
−t
DCt
t
B
t
A
t t t
t t
( ) ( ) ( )( ) ( ) ( )4.42.4.24.5 2223 −+++−+++=− t DCt t t Bt t At t
( ) ( ) B B At C B At D B At C B At t 488)444(224.5 233 −−+−+++−+++=−
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( )
0
3
1
1
0488
4444
022
5
====
=−−−=−+
=+−=++
D
C
B A
D D A
B A
D B A
C B A
dt t
dt t
t dt
t I .
4
1.3.
2
.2.
2
1 4
3
2
4
3
4
3
∫ ∫ ∫ ++
++
−=
( ) ( ) ( )4.2
322 24
3
4
3 ++++− t Lnt Lnt Ln
( ) ( )( ) ( ) ( )( ) ( )13)20(..2
35612 Ln Ln Ln Ln Ln Ln I −+−+−=
)13
20(..
2
3
5*1
6*2 Ln Ln +
Rpta: )13
20(.
2
3
5
12 Ln Ln +
11.( ) ( )
dt t t
t t I .
11
321
0
2
2
∫ ++++=
( ) ( ) ( ) )....(
1111
3222
2
α
++
++
=++++
t
D Bt
t
A
t t
t t
( ) ( ) ( )1.1.32 22 ++++=++ t C Bt t At t
( ) ( ) C At C Bt B At t +++++=++ ..32 22
( )
1
0
2
3
6222
3
1
2
===
=++=++
=+=+=+
C
B
A
C B A
C B A
C A
C B
B A
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dt t
dt t
I .1
1.
1
1.2
1
0
2
1
0
∫ ∫ ++
+=
( ) 1
0
1
0 )(1.2 t arcTant Ln ++
( ) ( )( ) )0(112.2 arcTanarcTan Ln Ln I −+−=
( )4
2.2 π
+= Ln I
Rpta: ( )4
4 π += Ln I
12.( )
dt t t
t I .
4
454
1
3
2
∫ ++
=
( ) dt
t t
t I .
4.
454
1
2
2
∫ ++=
( ) ( ) )....(
44.
4522
2
α
++
+=++
t
D Bt
t
A
t t
t
( ) ( ) t C Bt t At ..4.45 22 +++=+
( ) ( ) At C t B At .4..45 22 +++=+
( )
0
4
1
44
5
===
==+
C
B
A
A
B A
dt t
t dt
t I .
4
2
2
4.
1 4
1
2
4
1
∫ ∫ +
+=
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( ) ( ) 4
1
24
1 4.2 ++ t Lnt Ln
( ) ( )( ) ( ))5()20(.214 Ln Ln Ln Ln I −+−=
( ) 424 Ln Ln I +=
( )34 Ln I =
Rpta: ( )64 Ln I =
13.( ) ( )
dt t t
t I .
42
2162
022
2
∫ ++−
=
( ) ( ) ( ) ( ) )....(
42242
2162222
2
α
++
++
++
=++
−t
DCt
t
B
t
A
t t
t
( ) ( ) ( ) ( ) ( ) 2222 2.442.216 +++++++=− t DCt t Bt t At t
( ) ( ) ( )4.4.4824216 22232 +++++++++=− t t DCt B Bt t t t At t
( ) ( ) ( ) D B At DC At DC B At C At t 448.44442.216 232 +++++++++++=−
( )
0
23
1
23
0448
16444
242
0
=
−=
=
=
=++=++
=+++=+
D
C
B
A
D B A
DC A
DC B A
C A
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( ) ( )dt
t dt
t
dt t
I ∫ ∫ ∫ +
−+
++
+=
2
0
2
2
0
2
2
0 4
23
2
1.
2
23
( ) ( ) ( )
2
0
22
0
12
0 4.4
3
222
3
+−+++= −
t Lnt t Ln I
( ) ( )( ) ( ) ( ) ( )( )48.4
32424.
2
3 11 Ln Ln Ln Ln I −−−−−= −−
( ) 24
3
4
12
2
3 Ln Ln I −+=
Rpta: ( )4
12
4
3 += Ln I
14. ( )
( )( ) dt
t t
t t I .
11
2.41
0
2
2
∫ +++
=
D.F.P.
( )( )( ) ( ) ( ) )....(
111112.4 222
2
α
++++++=++ +t
D
t
C
t
B At
t t
t t
m.c.m. (t +1). (t + 1)2 ..x (α)
( ) ( ) ( ) ( ) ( )11.1.1..2.4 2222 +++++++=+ t t Dt C t B At t t
D Dt Dt C Ct B Bt Bt Bt At At t At t +++++++++++=+ 322232 22.2..2.4
( ) ( ) ( ) ( ) DC Bt D B At DC B At D At t +++++++++++=+ 2.2..2.4 232
02 =+→+ D D A 2−= D
0
22
4.2
=++=++
=+++
DC B
D B A
DC DB A
021
22
4.2
=−+==
C
B
A
1
1
2
===
C
B
A
Reemplazano:
( )( ) ( ) ( )
dt t t t
t I .
1
2
1
1
1
1.21
0
22∫
+−
+
+
+++
=
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( )
( )( )
( ) ( )
( )∫ ∫ ∫ ∫ +−++
++
+= −
1
0
1
0
21
0
2
1
0
2 1
12.1.
1
1.
1
.2dt
t dt t dt
t dt
t
t I
tdt du
t u
.2
12
=
+=dt du
t u
=
=dt du
t u
=
+= 1
dt du
t u
=
+= 1
( ) ( ) 1
0
1
0
1
0
1
0
2 121
1
11 +−
−+
+
++= t Ln
t t arcTant Ln I
( ) ( ) ( ) ( )[ ])1(221
101)1(2
1
0 Ln Ln
t arcTanarcTan Ln Ln I −−
−−−+−=
( ) ( )2.212
1
42 Ln Ln I −
−−+= π
Rpta: ( ) 212.2
4 +−= Ln I
π
15. ( )
( ) ( ) dt
t t
t t I .
93
62
022
2
∫ +−−
=
( )( ) ( ) ( ) ( )
)....(93393
62222
2
α
++
+−
+−
=+−
−t
DCt
t
B
t
A
t t
t t
( ) ( ) ( ) ( ) ( ) 2222 3.9.9.3.6 −+++++−=− t DCt t Bt t At t
( ) ( ) D Dt Dt Ct Ct Ct B Bt t t t At t 969692739.6 2232232 +−++−+++−−+=−
( ) ( ) ( ) D B At DC At DC A Bt C At t 992769963..6 232 ++−−+++−−++=−
09927
66.9.9
163
0
=++−−=−+=++−
=+
D B A
DC A
DC A B
C A
03
22.33
163
0
=++−−=−+=++−
=+
D B A
DC A
DC A B
C A
1
61
21
61
=
−=
−=
=
D
C
B
A
Reemplazano:
8/20/2019 Resolucion de Integrales Sesion 4 Metodo Algebraico
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( )( )
( ) ( )∫ ∫ ∫ ∫ ++
+−
−−
−=
2
0
22
2
0
2
2
0
2
2
0 3
1.
9
.2
2*6
1.
3
1.
3
1
6
1dt
t dt
t
t dt
t dt
t I
( ) ( ) ( ) 2
0
2
0
22
0
12
0
3
.
3
19.
12
13.
2
13
6
1
++−−+−= − t arcTant Lnt t Ln I
( )( ) ( ) ( )[ ] ( )[ ] ( )
−
+−−−−−+−−−= −−
03
2.
3
1)9(13.
12
131.
2
1)3(1
6
1 11arcTanarcTan Ln Ln Ln Ln I
( )
+
−
+−+
=
3
2.
3
1
9
13.
12
1
3
11.
2
1
3
1
6
1arcTan Ln Ln I
+
−
−+
=
3
2.
3
1
9
13.
12
1
3
2.
2
1
3
1
6
1arcTan Ln Ln I
Rpta:
+
−−
−=
3
2.
3
1
9
13.
12
13.
6
1
3
1arcTan Ln Ln I
16. ( ) dt
t t I .
12
12∫
+∞
∞− ++=
( ) dt
t I .
1112∫
+∞
∞− ++= dt du
t u
= += 1
( ) +∞
∞−+= 1t arcTan I
( ) ( )∞−−∞+= arcTanarcTan I
( ) ( )∞++∞+= arcTanarcTan I
22
π π += I
Rpta: π = I
17.( )
dt
t
t I .
11
22∫ +∞
+=
( ) dt t I .11
22∫ +∞
−+=
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( ) dt t t I .2.12
1
1
22∫ +∞
−+=
( ) +∞
−
−+=
1
12
1
1.
2
1 t I
( )
+∞
+
=1
21
1.
2
1
t I
−∞
=2
11.
2
1 I
Rpta:4
1= I
18. dt t Sent I ..2.2
0
3∫ =
π
dt t du
t u
2
3
.3=
=t Cosv
dt t Sendv
2.2
1
.2
−=
=
dt du
t u
=+= 1
dt du
t u
=+= 1
∫ ∫ −+−
==2
0
233 .2.2.
2.
2
32.
2..2.
π
dt t Sent t Sent
t Cost
dt t Sent I
t Cosv
dt t Sendv
dt du
t u
2.2
1
.2
−=
==
=
−+
−++−= ∫ dt t Cost Cos
t t Sen
t t Cost
t I .2.
2
12.
2.
2
32.
4
.32.
2
23
2
0
23
2.8
32.
2
32.
4
.32.
2
π
t Sent Cost
t Sent
t Cost t
I −++−=
Rpta: 760.0= I
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1!. dt et I t ..
4
1
0
42∫ =
dt t dut u
..2
2
== t
t
ev
dt edv
4
4
.4
1
.
=
=
41
0
442
...2
1.
4
−= ∫ dt et e
t I
t t
dt du
t u
== 2
t
t
ev
dt edv
4
4
.4
1
.
=
=
41
0
4442
.4
1.
4.
2
1.
4
−
−= ∫ dt eet
et
I t t t
41
0
4442
.16
1.
4.
2
1.
4
−
−= t t t
eet
et
I
41
0
4442
.32
1.
8.
4
−−= t t t
eet
et
I
41
0
24
32
1
84.
−−=
t t e I t
+−−
+−= 32
1
32
0
64
0
.32
1
32
1
64
1
. 0
ee I
32
1
64−= e
I
Rpta:64
2−= e
I
2". dt t Cost I .4.
2
0
2
∫ =
π
dt t du
t u
..2
2
==
t Senv
dt t Cost dv
4.4
1
.4
=
=
dt du
t u
=
=t Cosv
dt t Sendv
4.4
1
.4
−=
=
−−
−−= ∫ dt t Cost Cos
t t Sen
t I .4
4
14.
4.
2
14.
4
2
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4
0
2
44
1
4
14.
4.
2
14.
4
π
+
−−= t Sent Cos
t t Sen
t I
Rpta: 4
0
2
32
14.
8
4.
4
π
−+= t Cos
t t Sen
t I
21. dt t Sene I t
.4.
4
0
3
∫ =
π
dt edu
eu
t
t
..3 3
3
=
=t Cosv
dt t Sendv
4.4
1
.4
−=
=
dt edu
eu
t
t
..3 3
3
=
=
t Senv
dt t Cosdv
4.4
1
.4
=
=
−+== ∫ ∫ dt t Senet Senet Cosedt t Sene I
t t t t .4
4
34..
4
1.
4
34..
4
1.4.
3334
0
3
π
∫ ∫ −+== dt t Senet Senet Cosedt t Sene I t t t t
.416
94..
16
34..
4
1.4.
3334
0
3
π
∫ ∫ −+= dt t Senet Senet Cosedt t Sene
t t t t .494..34..4.4.16 3334
0
3
π
t Senet Cosedt t Sene t t t 4..34..4.4.25 33
4
0
3 +=∫
π
( )t Sent Cose
dt t Sene I
t
t 4.344.
25.4.
34
0
3 +== ∫
π
( )
+−+= 0304.(3.4251 043
senCoseSenCose I π π
π
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+= 4
25
14
3π
e I
Rpta: += 125
44
3π
e I
22. dt t Sene I t .3.3
0
2∫ =
π
dt t Cost Cose I t .3.
3
23.
3
1 2 ∫ +−=
dt edu
eu
t
t
..2 2
2
=
=t Cosv
dt t Sendv
3.3
1
.3
−=
=
dt edu
eu
t
t
.2 2
2
=
=
t Senv
dt t Cosdv
3.3
1
.3
=
=
..33
23..
3
1
3
23..
3
13.
2222
−+−== ∫ ∫ dt t Senet Senet Cosetdt Sene I
t t t t
dt t Senet Senet Cosetdt Sene I t t t t .3
9
43.
9
23..
3
13. 2222 ∫ ∫ −+−==
# !
dt t Senet Senet Cosetdt Sene I t t t t
.343..23..33..9 2222 ∫ ∫ −+−==
( )t Cost Senetdt Sene t t
3.3323..13 22 −=∫
( ) 3
0
223
0
3.332.13
13.
π
π
t Cost Senetdt Sene t t −=∫
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( ) ( )
−−−= 0.302180.31802.
13
1 032
CosSeneCosSene I π
( )
+= 33.
13
13
2π
e I
Rpta:
+= 1.
13
33
2π
e I
23. .1
1
02∫
−=
t
dt I
α
α α
α α
Cos R
d Cosdt
arcSent Sent
1
.1
1
==
=⇒=
α
α
α d
Cos
Cos I .
1
0
∫ =
α d I ∫ = .1
1
0α = I
01 arcSenarcSen I −=
Rpta:2
π = I
24. .2.
2
1222∫
−=
t t
dt I
α
α α
α α
Cos R
d Cosdt
t SenSent
2
.2
22
==
=⇒=
Reemplazano
8/20/2019 Resolucion de Integrales Sesion 4 Metodo Algebraico
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( ) .
2.2
.22∫ =
α α
α α
CosSen
d Cos I
( ) .
.4 2∫ =α
α
Sen
d
I
( )α α α Cot d Csc I −== ∫ .4
1..
4
1 2
2
1
24
4
1
−−=
t
t I
−−= 1
3
2
0
4
1 I
Rpta:4
3= I
25. .9.
.34
12∫ +
=t t
dt I
dt du
t u
==
.3.
.322∫
+=
t t
dt I
41
22
.33
3
1.3
++
−=
t
t Ln I
( ) ( )( )9132.1 ++−−= Ln Ln I
( ) 1032.1 +−−= Ln Ln I
( )( )( )
−+−
−=103103
103.2 Ln I
( )
−−
−=1
10.26 Ln I
Rpta: ( )610.2 −−= Ln I
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26.( )
.
9
4
0 2
32
∫ +
=t
dt I
( ).
9
4
032
∫ +
=t
dt I
( ) ( )
.
99
4
0
22∫ ++
=
t t
dt I
α
α α
α α
Sec R
d Secdt
t arcTanTant
.3
.3
33
2
==
=⇒=
( ) α
α α
α d
TanSec
Sec I .
99..3
.34
0
2
2
∫ +=
( ) α
α
α d
Tan
Sec I .
1.9 2∫ +
=
α
α
α d
Sec
Sec I .
..9 2∫ =
α
α
d Sec
I ...9
1
∫ =
α α d Cos I .9
1 ∫ =
α Sen I 9
1=
4
02 99
1
+=
t
t I
−= 05
4.
9
1 I
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Rpta:45
4= I
27. .82
6
42
∫ −−= t t
dt I
( ).
31
6
422∫
−−=
t
dt I
−=
−=
===−
=
3
1
3
1
.3
..3
31
.
t arcSec
t Sec
Tan R
d TanSecdt
Sect
d Secau
α
α
α
α α α
α
α α
Reemplazano:
.3
..3∫ =α
α α α
Tan
d TanSec I
.∫ = α α d Sec I
.)( α α TanSec Ln I +=
( ) 6
4
2
3
91
3
1
−−+
−= t t
Ln I
13 Ln Ln I −=
Rpta: 3 Ln I =