calculo de integrais_indefinidos_com_aplicacao_das_proprie
TRANSCRIPT
EXERCÍCIOS M1 Anatolie Sochirca ACM DEETC ISEL
1
Cálculo de integrais indefinidos com aplicação das propriedades e das
fórmulas da tabela dos integrais (integração imediata).
Para aplicar o método de integração imediata transformamos a expressão sob o sinal integral com o objectivo de obter um integral ou uma soma algébrica de integrais da tabela dos integrais.
Neste caso é útil a transformação: se )(xG é uma primitiva evidente da função )(xg e ))(()( xGuxf = então
( )∫∫∫ ⋅=⋅′⋅=⋅⋅ )())(()())(()()( xGdxGuxdxGxGuxdxgxf .
►1) =⋅
−−+⋅−=⋅−−+⋅−
∫∫ xdxx
x
x
x
x
xx
x
xxd
x
xxxxx33
5 3
333
3
3
5 33 223223
=⋅
−−+−=⋅
−−+⋅−= ∫∫−−−−+−
xdxxxxxxd
xx
x
x
x
x
xx
x
x 3
1
3
1
5
3
3
11
3
1
2
11
3
13
3
1
3
1
5
3
3
1
3
1
2
1
3
1
3
223223
=⋅
−−+−= ∫
−xdxxxxx 3
1
15
4
3
2
6
7
3
8
223
∫∫∫∫∫ =⋅−⋅−⋅+⋅−⋅=−
xdxxdxxdxxdxxdx 3
1
15
4
3
2
6
7
3
8
223
∫∫∫∫∫ =⋅⋅−⋅−⋅⋅+⋅⋅−⋅=−
xdxxdxxdxxdxxdx 3
1
15
4
3
2
6
7
3
8
223
=++−
⋅−+
−+
⋅++
⋅−+
=+−++++
Cxxxxx
13
12
115
41
3
22
16
73
13
8
13
11
15
41
3
21
6
71
3
8
=+⋅−−⋅+⋅−= Cxxxxx
3
22
15
19
3
52
6
133
3
11
3
2
15
19
3
5
6
13
3
11
=+⋅⋅−⋅−⋅⋅+⋅⋅−⋅= Cxxxxx 3
2
15
19
3
5
6
13
3
11
2
32
19
15
5
32
13
63
11
3
Cxxxxx +⋅−⋅−⋅+⋅−⋅= 3
2
15
19
3
5
6
13
3
11
319
15
5
6
13
18
11
3. ■
EXERCÍCIOS M1 Anatolie Sochirca ACM DEETC ISEL
2
►2) =⋅+
=⋅=⋅=⋅⋅ ∫∫∫∫ )(
5
1)(
)5(
1
)5(
1
)5(
1xnld
xnlnlxnld
xnlx
xd
xnlxd
xnlx
( ) ( ) CxnlnlCxnlnlnlxnlnldxnlnl
+=++=+⋅+
= ∫ )5(5)5(5
1. ■
Outro método:
► =⋅⋅⋅=
⋅⋅⋅=⋅
⋅⋅=⋅
⋅ ∫∫∫∫ x
xd
xnlx
xd
xnlxd
xnlxxd
xnlx 5
)5(
)5(
1
5
5
)5(
1
)5(5
5
)5(
1
( ) Cxnlnlxnldxnl
+=⋅= ∫ )5()5(()5(
1. ■
►3) ( )=⋅=+
⋅=⋅+ ∫∫∫ xarctgdxarctg
x
xdxarctgxd
x
xarctg 32
32
3
11
( )CxarctgC
xarctg +⋅=++
=+
413
4
1
13. ■
►4) =+++
⋅+⋅=⋅++
+⋅∫∫ 196
)3(2106
)3(222 xx
xdxarctgxd
xx
xarctg
( ) =+⋅+⋅=++
+⋅+⋅= ∫∫ )3()3(21)3(
)3()3(2
2xarctgdxarctg
x
xdxarctg
CxarctgCxarctg ++=+
++⋅=
+
)3(11
)3(2 2
11
. ■
►5) ( )( )
( ) ( ) =+
⋅−⋅=+
⋅−=⋅+
−∫∫∫ 222222
)2()2()2(
xsenxosc
xdxoscxdxsen
xsenxosc
xdxoscxsenxd
xsenxosc
xoscxsen
( ) ( ) =+
−−⋅⋅=+
⋅−⋅⋅⋅= ∫∫ 2222
)()(22
xsenxosc
senxdxoscdxosc
xsenxosc
xdxoscxdxoscxsen
( ) ( ) =+
+⋅⋅−=+
−⋅⋅−= ∫∫ 2222
)()(2)()(2
xsenxosc
senxdxoscdxosc
xsenxosc
senxdxoscdxosc
( ) ( ) =+
+−=+
+−= ∫∫ 22
2
22
2 )()()(
xsenxosc
senxxoscd
xsenxosc
senxdxoscd
EXERCÍCIOS M1 Anatolie Sochirca ACM DEETC ISEL
3
( ) ( ) =++−
+−=+⋅+−=+−
−
∫ Cxsenxosc
senxxoscdxsenxosc12
)(122
222
( )C
xsenxoscC
xsenxosc ++
=+−+−=
−
2
12 1
1. ■
►6) ( ) =−
⋅′⋅=
−
⋅⋅⋅=−
⋅⋅⋅=⋅
− ∫∫∫∫ 2
2
222 12
1
1
2
2
1
1
22
1
1 x
xdx
x
xdx
x
xdxxd
x
x
( )=
−
−⋅−=−
−⋅−=−
−−⋅=−
⋅= ∫∫∫∫2
12
2
2
2
2
2
2
2
1
)1(
2
1
1
)1(
2
1
1
)(
2
1
1
)(
2
1
x
xd
x
xd
x
xd
x
xd
( ) ( ) =++−
−⋅−=−⋅−⋅−=+−
−
∫ Cx
xdx1
2
11
2
1)1(1
2
11
2
12
22
12
( ) CxCx +−−=+−−= 22
12 11 . ■
►7) ( ) =−
⋅′⋅=
−
⋅⋅⋅=−
⋅⋅⋅=⋅
− ∫∫∫∫ 4
2
444 12
1
1
2
2
1
1
22
1
1 x
xdx
x
xdx
x
xdxxd
x
x
( ) ( )Cxarcsen
x
xd
x
xd +⋅=−
⋅=−
⋅= ∫∫ )(2
1
1
)(
2
1
1
)(
2
1 2
22
2
22
2
. ■
►8) =⋅
−+
−
⋅=⋅−
+⋅∫∫ xd
x
x
x
xarcsenxd
x
xxarcsen222 11
2
1
2
=⋅−
+⋅−
⋅⋅=⋅−
+⋅−
⋅= ∫∫∫∫4434421
)6
2222 11
12
11
2
exemplover
xdx
xxd
xxarcsenxd
x
xxd
x
xarcsen
( ) =+−−+⋅′⋅⋅= ∫ Cxxdxarcsenxarcsen 212
CxxarcsenCxxarcsen +−−=+−−⋅⋅= 2222 112
12 . ■
EXERCÍCIOS M1 Anatolie Sochirca ACM DEETC ISEL
4
►9) ( ) =+⋅⋅
⋅+⋅⋅=+⋅⋅⋅+⋅=⋅
+⋅⋅+⋅
∫∫∫ xxx
xdxdx
xxx
xdxxd
xxx
x
2
13
2
13
2
13
=+⋅
+
⋅
=+⋅
+⋅′
⋅⋅
=+⋅⋅
+⋅⋅= ∫∫∫xx
xdxd
xx
xdxdx
xxx
xdxdx
2
3
2
3
2
3
2
3
2
1
2
1
2
2
2
3
23
2
3
Cxxnl
xx
xxd
xx
xdxd
++⋅=+⋅
+⋅
=+⋅
+
⋅
= ∫∫ 2
3
2
3
2
3
2
3
2
3
2
2
2
2
2
. ■
►10) ( ) ( ) ( ) ( ) ( ) =⋅+=⋅′⋅+=⋅⋅+ ∫∫∫xxxxxx edexdeexdee
πππ222
( ) ( ) ( )C
eede
xxx +
++=+⋅+=
+
∫ 1
222
1
π
ππ
. ■
►11) ( ) ( )( ) ( ) =⋅⋅⋅
=⋅′
⋅⋅⋅
=⋅⋅=⋅⋅ ∫∫∫∫xxxxx ed
enlxde
enlxdexde 3
)3(
13
)3(
133
( ) ( ) ( )C
nl
eC
enlnl
eCe
enl
xxx +
+⋅=+
+⋅=+⋅⋅
⋅=
13
3
3
33
)3(
1. ■
►12) =⋅
−=⋅−=⋅=⋅ ∫∫∫∫ xd
xosc
xosc
xoscxd
xosc
xoscxd
xosc
xsenxdxtg
2
2
22
2
2
22 11
( ) =+−⋅′=⋅−⋅=⋅
−= ∫∫∫∫ Cxxdxtgxdxd
xoscxd
xosc1
11
122
( ) CxxtgCxxtgd +−=+−= ∫ . ■
►13) Ceedx
dexdx
exdx
e xxxxx
+−=
−=
⋅−=⋅′
⋅−=⋅ ∫∫∫∫1111
2
1
11. ■
EXERCÍCIOS M1 Anatolie Sochirca ACM DEETC ISEL
5
►14) ( ) ( ) =⋅⋅=⋅′⋅⋅=⋅⋅⋅⋅=⋅⋅ ∫∫∫∫222222
72
1
2
172
2
177 xdxdxxdxxdx xxxx
( ) ( ) Cnl
dnl
xdnlnl
xxx +
⋅=⋅⋅=⋅⋅⋅⋅= ∫∫ 72
77
7
1
2
177
7
1
2
12
222. ■
►15) =
−=⇔⋅−==⋅∫ 2
)2(121)2()6( 222 αααα osc
sensenoscxdxsen
=⋅−⋅=⋅
−=⋅−= ∫∫∫∫ xdxosc
xdxdxosc
xdxosc
2
)12(
2
1
2
)12(
2
1
2
)12(1
( ) =⋅′⋅⋅−⋅=⋅⋅−⋅= ∫∫∫∫ xdxsenxdxdxoscxd )12(12
1
2
1
2
1)12(
2
1
2
1
( ) Cxsenx
xsendxd +−=⋅−⋅= ∫∫ 24
)12(
2)12(
24
1
2
1. ■
►16) =⋅⋅+=⋅
⋅=
⋅ ∫∫∫ xdxoscxsen
xoscxsenxd
xoscxsenxoscxsen
xd 221
=⋅
+=⋅
⋅+
⋅= ∫∫ xd
xsen
xosc
xosc
xsenxd
xoscxsen
xosc
xoscxsen
xsen 22
( ) ( ) =⋅′
+⋅′
−=⋅+⋅= ∫∫∫∫ xdxsen
xsenxd
xosc
xoscxd
xsen
xoscxd
xosc
xsen
( ) ( ) ( ) ( ) =⋅′−⋅′=⋅′
−⋅′
= ∫∫∫∫ xdxoscnlxdxsennlxdxosc
xoscxd
xsen
xsen
CxtgnlCxosc
xsennlCxoscnlxsennl +=+=+−= . ■
►17) ( ) ( ) =+
⋅′+−=+
⋅′−=⋅
+ ∫∫∫ xosc
xdxosc
xosc
xdxoscxd
xosc
senx
5
5
55
( ) Cxoscnlxdxoscnl ++−=⋅′+−= ∫ 55 . ■
EXERCÍCIOS M1 Anatolie Sochirca ACM DEETC ISEL
6
►18) =⋅⋅=⋅−
⋅⋅⋅=⋅
−
⋅∫∫∫ xd
xosc
xsenxd
xsenxosc
xoscsenxxd
xsenxosc
xoscsenx
)2(
)2(
2
12
2
1
2222
( )( ) ( ) =⋅⋅−=⋅
′⋅−⋅= ∫∫
− )2()2(4
1
)2(
)2(2
1
2
12
1
xoscdxoscxdxosc
xosc
( ) ( ) ( ) CxoscCxosc
Cxosc +⋅−=+⋅−=++−
⋅−=+−
2
12
11
2
1
)2(2
1
2
1)2(
4
1
12
1)2(
4
1. ■
Outro método:
► ( ) ( )
=⋅−
⋅−⋅−⋅=
−
⋅′⋅⋅⋅=⋅
−
⋅∫∫∫
xsen
xsend
xsenxosc
xdsenxsenxxd
xsenxosc
xoscsenx2
2
2222 21
22
1
2
12
2
1
( ) ( ) ( ) =⋅−⋅⋅−⋅−=
⋅−
⋅−⋅−= ∫∫−
xsendxsenxsen
xsend 22
12
2
2
21214
1
21
21
4
1
( ) ( ) ( ) CxoscCxsenCxsen +⋅−=+⋅−⋅=+
+−
⋅−⋅−=+−
2
12
12
12
12
)2(2
121
2
1
12
121
4
1. ■
►19) =⋅⋅⋅⋅⋅+=⋅⋅⋅+ ∫∫ xdxoscxsenxoscxdxsenxosc 254)2(54 22
( ) ( ) =⋅′⋅⋅⋅⋅+−=⋅′−⋅⋅⋅⋅+= ∫∫ xdxoscxoscxoscxdxoscxoscxosc 254254 22
( ) ( ) =⋅⋅⋅+⋅−=⋅⋅+−= ∫∫ xoscdxoscxoscdxosc 2222 5545
154
( ) ( ) ( ) =++
⋅+⋅−=⋅+⋅⋅+⋅−=+
∫ Cxosc
xoscdxosc1
2
154
5
15454
5
11
2
12
22
12
( ) Cxosc +⋅+⋅−= 2
3254
15
2. ■