[RESOLUÇÃO]Calculo A - Diva Flemming - Cap6 Parte 1

Esta é uma pré-visualização de arquivo. Entre para ver o arquivo original

443 
CAPÍTULO 6 
 
6.2 – EXERCÍCIOS – pg. 246 
 
Nos exercícios de 1 a 10, calcular a integral e, em seguida, derivar as respostas para 
conferir os resultados. 
1. ∫ 3x
dx
 
32
2
2
3
1
2
1
2
1
2
xxdx
d
c
x
c
xdxx
=




 −
+
−
=+
−
=∫
−
−
 
2. d
t
t∫ 





+
3
2 19
 
.
19.
2
1
.2923
.
23
2
13
.9)9(
3
223
3
3
2
2
3
2
1
2
3
t
tttc
t
t
dt
d
c
t
tc
ttdttt
+=−−=





+−
+−=+
−
+=+
−
−
−
∫
 
 
3. ∫ ++ dxcbxax )3( 34 
.334
4
5
5
3
45
.3
45
343445
45
cbxaxcxbxaCcxxbxa
dx
d
Ccxxbxa
++=++=





+++
+++=
 
 
4. ∫ 






+ dxxx
x 3
1
 
 444 
3
1
.
2
5
.
15
2
2
12
15
22
.
15
22
2
5.3
1
2
13
1
2
3
2
1
2
5
2
52
5
2
1
2
3
2
1
xx
x
xxcxx
dx
d
cxxc
xxdxxx
+=+=





++
++=++=





+=
−
∫
−
 
 
5. ∫ − dxx
22 )32( 
( )
.91249125
5
494
5
4
.94
5
49
3
12
5
49124
242435
35
35
24
cxxcxxcxxx
dx
d
cxxxcx
xxdxxx
++−=++−=





++−
++−=++−=+−∫
 
 
6. ∫ xsen
dx
2 
( ) .1seccoscot
.cotseccos
2
2
2
2
xsen
xcxg
dx
d
cxgdxx
dxxsen
==+−
+−==
=
∫
∫
−
 
 
7. dy
y
y∫ 






−
2
12 
 445 
.
2
12
2
12
2
1
.
2
2
2
3
.
3
22
.
2
2
3
22
.1
3
22.
2
2
3
22
2
1.2
1
2
3
2
.
2
1
.2
2
1
2
1
2
1
2
1
2
1
2
3
2
1
2
3
2
1
2
3
2
1
2
1
y
y
yyyycyy
dx
d
cyycyy
c
yydyyy
−=
−=−=







+−
+





−=+−=
+−=





−=
−−
∫
−
 
 
8. cttgarc
t
dt
+=
+∫ 3
2
33
2
2 
.
33
2
1
1
.
3
2
3
2
22 +
=
+
=







+
tt
cttgarc
dt
d
 
 
9. ∫ dxxx
3
 
xxxcx
dx
d
c
xdxx
32
7
2
9
2
9
2
7
2
9
.
9
2
9
2
2
9
==





+
+=∫
 
10. ∫
−+ dx
x
xx
4
25 12
 
( )
.
1212
9
9.12
2
2
3
12
2
3
12
231
2
2
2
4
25
426
2
23
2
3
2312
42
x
xx
xx
x
x
x
x
x
c
xx
x
dx
d
c
xx
x
c
xxxdxxxx
−+
=−+=
−
+
−
−=





++−
++−=+
−
−
−
+=−+=
−−
−−
∫
 
 
 
 446 
 
 
Nos exercícios de 11 a 31, calcular as integrais indefinidas. 
11. ∫ +
dx
x
x
12
2
 
.
1
11 2 cxtgarcxdxx
+−=





+
−= ∫ 
12. ∫
+ dx
x
x
2
2 1
 
( ) .1
1
1
1
2 c
x
xc
x
xdxx +−=+
−
+=+∫
−
−
 
13. ∫ dxx
xsen
2cos
 
cxdxxxtgdx
xx
xsen
+=== ∫∫ secsec.cos
1
.
cos
 
14. ∫
−
dx
x 21
9
 
.3
1
3
2
cxsenarcdx
x
+=
−
= ∫ 
15. ∫
−
dx
xx 24
4
 
.sec2
1
2
2
cxarcdx
xx
+=
−
∫ 
16. ∫
+−+− dx
x
xxxx
2
234 12698
 
( )
c
x
xx
xx
dxxxxx
+−−+−=
+−+−= ∫
−−
1ln26
2
9
3
8
2698
23
212
 
 447 
17. ∫ 





++ dt
t
t
e t 1
2
 
.ln
3
2
2
1ln
2
32
1
2
32
3
cttect
t
e tt +++=+++= 
18. ∫ θθθ dtg.cos 
.cos
cos
.cos cdsendsen +−=== ∫∫ θθθθθ
θθ 
19. ∫
−
− dxee xx )( 
.cosh22 cxdxxhsen +== ∫ 
20. ∫ ++++ dtttttt )( 543 
ctttt
t
c
ttttt
+++++=
+++++=
5
6
3
4
2
3
5
65
43
4
2
3
6
5
5
4
4
3
3
2
2
5
6
4
5
3
4
2
32
4
52
2
 
21. ∫
−
−
dx
x
x 53/1
 
.||ln53||ln5
3
1
5
3
13
1
3
4
cxxcx
x
dx
x
x
+−−=+−
−
=






−=
−
−
∫
−
 
22. ∫ +− dtte
tt )cosh22 
.2
2ln
2
ctsenhet
t
++−= 
23. ∫ + dxxx )1(cossec 32 
 448 
.seccos.
cos
1 23
2 ctgxsenxdxxxx
++=





+= ∫ 
24. ∫ ≠+ ,0,)( 22 aaax
dx
 constante 
∫∫ +=+=+= cxtgarcaxa
dx
axa
dx
222222
1
)1( 
25. ∫ +
− dx
x
x
1
1
2
2
 
.2
1
21 2 cxtgarcxdxx
+−=





+
−= ∫ 
26. ∫ 





+− dttt3
3
6
2
1)2(8 
( )
.42
3
7
2
2
2
2
3
.
2
7
4
2
22
2
722424
2
12
2
1442
2
1)2(2
2
34234
23223
22
ctt
tt
ct
ttt
dttttdtttttt
dttttdttt
+++−=+





++−=






++−=





++−−+=






++−=





+−=
∫∫
∫∫
 
27. ∫ 





+− dt
t
tet 3
4 316 
.
2
3
5
8
2
3
4
52
2
2
4
54
5
cttec
tt
e tt +−−=+
−
+−= −
−
 
28. ∫ dxxx
x
2ln
ln
 
.||ln
2
1
ln2
ln
∫∫ +=== cxx
dxdx
xx
x
 
29. ∫ dxxecxtg
22 cos 
.sec
1
cos
2
22
2
∫∫ +=== cxtgdxxdxxsenx
xsen
 
 449 
30. ∫ +− dxxx
22 )1()1( 
( ) ( )
( )
( ) .
3
2
5
12
122422
1212
35
24
223234
22
cx
xxdxxx
dxxxxxxxxx
dxxxxx
++−=+−=
+++−−−++=
+++−=
∫
∫
∫
 
31. ∫ ∈






−
znonde
tn
dt
n
,
2
1
 
( ) cn
t
n
dtt
n
ct
t
dt
tn
dt
ctdt
n
n
n
+
−






−
=






−
≠
+==






−
=
+−=−=
−
−
∫
∫∫
∫
1
.
2
1
1
2
1
1
,1n Se
||ln22
2
1
,1n Se
22 0,n Se
1
 
32. Encontrar uma primitiva F , da função ,)( 3/2 xxxf += que satisfaça .1)1( =F 
( ) cxxdxxxxF ++=+= ∫ 2
3
5)(
23
5
3
2
 
1
2
1
5
3)1(
25
3)(
2
3
5
=++=
++=
cF
c
x
xxF
 
 
10
1
10
5610
2
1
5
31 −=−−=−−=c 
.
10
1
25
3)(
2
3
5
−+=
x
xxF 
 
33. Determinar a função )(xf tal que 
 450 
.222).2(
2
122cos
2
1
2cos
2
1)(
2
2
xsenxxsenxcxx
dx
d
cxxdxxf
−=−+=





++
++=∫
 
34. Encontrar uma primitiva da função 11)( 2 += xxf que se anule no ponto .2 x = 
( )
2
31)(
2
3
2
412
2
1
2
2
1)2(
1
1
111)(
1
2
2
−+−=
−
=
−
=−=
++−=
++−=++
−
=+=





+=
−
−
∫∫
x
x
xF
c
cF
cx
x
cx
xdxxdx
x
xF
 
35. Sabendo que a função )(xf satisfaz a igualdade. 
∫ +−−= ,2
1
cos)( 2 cxxxxsendxxf determinar ).4/(pif 
( )
.
8
22
2
12
.
4
1
2
2
4
1
444
)1(coscos
2
2
1)cos)((cos
2
1
cos 2
−
=
−
=







−=





−=





−=−=−−+=
−+−−=





+−−
pipipipipipi
senf
xsenxxxsenxxxxsenxx
xxxsenxxcxxxxsen
dx
d
 
36. Encontrar uma função f tal que .2)0(e0)( ==+′ fxsenxf 
0)( =+′ xsenxf 
.1cos)(
112
20cos)0(
cos)(
cos
)(
+=∴
=−=
=+=
++=
++=−
−=′
∫
xxf
c
cf
cxxf
cxdxxsen
senxxf