451
6.4 – EXERCÍCIOS – pg. 250
Calcular as integrais segu intes usando o método da subst ituição.
1.
+−+ dxxxx )12()322(
102
.
11
)322(
2
1
)12()322(
:
)12(2)24(
322
:
112
102
2
c
xx
dxxxx
Temos
dxxdxxdu
xxu
se
Fazendo
+
−+
=+−+
+=+=
−+=
−
2.
−dxxx
27/13
)2(
( ) ( )
.2
24
7
7
8
2
3
1
)2(
:
3
2
:
7
8
7
8
3
3
27/13
2
3
cxc
x
dxxx
Temos
dxxdu
xu
seFazendo
+−=+
−
=−
=
−=
−
3.
−
52
1x
dxx
( )
cxc
x
x
dxx
Temos
dxxdu
xu
seFazendo
dxxx
+−=+
−
=
−
=
−=
−
−
−
5
4
5
1
)1(
8
5
5
4
1
2
1
1
:
2
1
:
)1(
2
5/4
2
5 2
2
2
4.
−dxxx
2
345
452
(
)
( ) ( )
.34
9
5
2
3
34
6
1
.5345
:
6
34
:
345)34(5
2
3
2
3
2
1
2
1
2
2
2
2
22
cxc
x
dxxx
Temos
dxxdu
xu
seFazendo
dxxxdxxx
+−
−
=+
−−
=−
−=
−=
−
−=−=
5.
+dxxx
42
2
(
)
( )
( )
cx
c
x
dxxx
++=
+
+
=
+=
2
3
2
3
2
1
2
2
2
21
6
1
2
3
21
4
1
21
Fazendo: dxxdu
xu
4
21
2
=
+=
6.
+dtee
tt
22
3
1
)2(
( ) ( )
.2
8
3
3
4
2
2
1
)2(
:
2
2
:
3
4
3
1
2
3
4
2
22
2
2
cec
e
dtee
Temos
dtedu
eu
seFazendo
t
t
tt
t
t
++=+
+
=+
=
+=
−
7.
+
4
t
t
e
dte
. e 4 que sendo , 4ln dtedueuce
u
du
ttt
=+=++==
8.
+dx
x
e
x
2
/1
2
453
.
1
.
:se-doConsideran
.
2
1
.22
1
2
1
2
2
1
1
111
x
edu
eu
c
x
ec
x
edxxdx
x
e
x
x
xxx
−
=
=
+−−=+
−
+−=+=
−
−
9.
dxxxtg
2
sec
c
xtg +=
2
2
. considerando-se: dxxdu
x
tg
u
2
sec=
=
10.
dxxxsen cos
4
c
xsen +=
5
5
considerando-se:
dxxdu
x
sen
u
cos=
=
11.
dx
x
xsen
5
cos
cx
c
x
x
dxxsenx
+=
+=
−
−=
=
−
−
4
4
4
5
sec
4
1
cos4
1
4
cos
.cos
utiliz ando:
senxdx
du
x
u
−
=
=
cos
12.
−
dx
x
xxsen
cos
cos52
cxx
dx
x
xsen
+−−=
−=
5|cos|ln2
5
cos
2 utiliz ando:
senxdx
du
x
u
−
=
=
cos
13.
dxee
xx
2cos
.2
2
:se-doConsi deran
.2
2
1
dxedu
eu
cesen
x
x
x
=
=
+=
454
14.
dxx
x
2
cos
2
.2
:se-doConsideran
4
1
2
1
2
1
2
22
dxxdu
xu
cxsencxsen
=
=
+=+=
15.
−
θπθ
dsen )5(
( )
.
5
5
:se-doConsideran
.5cos
5
1
θ
πθ
πθ
d
du
u
c
=
−=
+−−=
16.
dy
y
ysenarc
−
2
12
(
)
( )
.
1
1
:se-doConsideran
.
4
1
22
1
2
2
2
dy
y
du
ysenarcu
cysenarcc
ysenarc
−
=
=
+=+=
17.
+
θ
θ
θ
d
tgba
2
sec2
Ctgba
b
++= ||ln
1
.2
θ
Considerando-se:
θθ
θ
dbdu
tgbau
2
sec.=
+
=
18.
+
2
16
x
dx
455
c
x
tgarcc
x
tgarc
x
dx +=+=
+
=
44
1
4
4
16
1
4
1
16
1
2
, utilizando:
dxdu
x
u
4
1
4
=
=
19.
+− 44
2
yy
dy
c
y
c
y
dyy
y
dy +
−
=+
−
−
=−=
−
=
−
−
2
1
1
)2(
)2(
)2(
1
2
2
, utilizando: dydu
yu
=
−
=
2
20.
θθθ
dsen cos
3
( )
.
4
3
3
4
)(
cos
3
4
3/4
3/1
csenc
sen
dsen +=+==
θ
θ
θθθ
21.
dx
x
x
2
ln
(
)
( ) ( )
.
22
ln
:se-doConsideran
.lnln4
4
1
)(ln
4
1
2
ln
2
1
2
2
22
22
2
2
dx
x
x
x
du
xu
cxcxcxc
x
==
=
+=+=+=+
22. dxee
axax 2
)(
−
+
( )
( )
.2
2
2
2
1
2
1
2
2
1
2
22
22
2
22
cx
a
axhsen
cxee
a
ce
a
xe
a
dxee
axax
axaxaxax
++=++−=
+−+=++=
−
−−
23.
+dttt
24
3
456
( ) ( )
(
)
( ) ( )
.13.
9
1
13.
2
3
.
6
1
2
3
13
6
1
1313
2
3
2
3
2
3
2
1
22
2
222
ctct
c
t
dtttdttt
++=++=
+
+
=+=+=
Considerando-se:
dttdu
tu
6
13
2
=
+= .
24.
++
34
20
4
4
2
x
x
dx
.
3
2
5
2
3
2
2
3
2
5
2
3
1
2
3
2
5
22
c
x
tgarcc
x
tgarc
x
dx +
+
=+
+
=
+
+
=
25.
+−
1
4
3
2
x
x
dx
( ) ( ) ( )
−
−
−=
−
−
−=
−−
=
3
2
1
3
1
3
3
2
3
3
3
3
32
3
222
x
dx
x
dx
x
dx
.
23
23
ln
2
3
3
2
1
3
2
1
ln
2
1
3c
x
x
c
x
x
+
−+
−+−
=+
−
−
−
+
−=
Considerando-se:
(
)
dxdu
x
u
x
u
3
1
3
2
3
2
2
2
=
−
=
−
=
Resposta alternativa:
457
.1
3
2
3
2
cot
1
3
2
3
2
>
−−
<
−−
x
se
x
hgarc
x
se
x
htgarc
26.
+
16
2
x
x
e
dxe
c
e
tgarc
x
+=
4
4
1
Considerando-se:
dx
e
du
eu
eu
x
x
x
2
22
=
=
=
27.
−
+dx
x
x
1
3
.
32
32
ln232
2
2
ln22
2
1
2
1
ln
2
1
2.22
2
1
22
44
4
4
82
4
42
4
4
12
4
2
2.
13
22
222
2
2
c
x
x
xc
u
u
u
c
u
u
u
u
du
u
u
du
u
u
du
udu
u
du
u
u
duu
u
u
+
+−
++
−+=+
−
+
−=
+
−
+
−=
−
−=
−
−
+=
−
+=
−
+=
−
=
−−
=
Considerando-se:
duudx
ux
xu
2
3
3
2
2
=
−=
+=
28.
xx
dx
3ln
3
2
458
( )
(
)
.
3ln
3
1
3ln
3
3
3ln
2
c
x
c
x
x
dx
x+
−
=+
−
==
−
Considerando-se:
dx
x
du
xu
3
3
3ln
=
=
29.
+dxxsen )2cos4(
π
( )
.2cos4cos
4
1
2cos4 cxxdxdxxsen ++−=+=
ππ
30.
+
dxx
x1
2
2
.
2ln
2
2ln
2
2
1
22
1
cc
xx
+=+=
+
Considerando-se:
dxxdu
xu
2
1
2
=
+=
31.
dxex
x
2
3
ce
x
+=
2
3
6
1
Considerando-se:
dxxd u
xu
6
3
2
=
=
32.
+
2
)2( t
dt
(
)
−
+=
2
2t c
t
c
t+
+
−
=+
−
+
=
−
2
1
1
)2(
1
.
Considerando-se:
dtdu
tu
=
+
=
2
459
33.
tt
dt
ln
.lnln ct +=
Considerando-se:
t
dt
du
tu
=
=
ln
34.
−dxxx
2
218
(
)
( )
.21
3
4
2
3
21
4
1
8
2
3
2
3
2
2
cxc
x+−
−
=+
−−
=
Considerando-se:
dxxdu
xu
4
21
2
−=
−=
35.
+dxee
xx
252
)2(
(
)
( )
.2
12
1
6
2
2
1
6
2
6
2
cec
e
x
x
++=+
+
=
Considerando-se:
dxedu
eu
x
x
2
2
2
2
=
+=
36.
+54
4
2
t
dtt
( )
dttt 454
2
1
2−
−=
(
)
.54
2
1
54
2
1
2
2
2
1
ctc
t++=+
+
=
Considerando-se:
dttdu
tu
8
54
2
=
+=
460
37.
−dx
xsen
x
3
cos
cxsen
+
−
−
=
|3|ln
Considerando-se:
dxxdu
x
sen
u
cos
3
−=
−
=
38.
+
5
)1( vv
dv
(
)
( )
c
v
c
v
+
+
−=
+
−
+
=
−
4
4
12
1
4
1
2
Considerando-se:
dv
v
du
vu
2
1
1
=
+=
39.
+dxxx 1
2
Considerando-se:
duudxux
ux
21
1
2
2
=−=
=+
(
)
(
)
( )
( ) ( ) ( )
( ) ( ) ( )
.11
3
2
11
5
4
11
7
2
1
3
2
1
5
4
1
7
2
3
2
5
4
7
2242
212211
23
357
357
246
224
2
22
cxxxxxx
cxxx
c
uuu
duuuu
duuuuduuuudxxx
++++++−++=
++++−+=
++−=+−=
+−=−=+
40.
−
dxex
x
5
4
Prévia do material em texto
451 6.4 – EXERCÍCIOS – pg. 250 Calcular as integrais seguintes usando o método da substituição. 1. ∫ +−+ dxxxx )12()322( 102 . 11 )322( 2 1)12()322( : )12(2)24( 322 : 112 102 2 c xxdxxxx Temos dxxdxxdu xxu seFazendo + −+ =+−+ +=+= −+= − ∫ 2. ∫ − dxxx 27/13 )2( ( ) ( ) .2 24 7 7 8 2 3 1)2( : 3 2 : 7 87 8 3 3 27/13 2 3 cxc xdxxx Temos dxxdu xu seFazendo +−=+ − =− = −= − ∫ 3. ∫ − 5 2 1x dxx ( ) cxc x x dxx Temos dxxdu xu seFazendo dxxx +−=+ − = − = −= − − ∫ ∫ − 5 4 5 1 )1( 8 5 5 4 1 2 1 1 : 2 1 : )1( 2 5/42 5 2 2 2 4. ∫ − dxxx 2345 452 ( ) ( ) ( ) .34 9 5 2 3 34 6 1 .5345 : 6 34 : 345)34(5 2 32 3 2 1 2 1 2 2 2 2 22 cxc xdxxx Temos dxxdu xu seFazendo dxxxdxxx +− − =+ −− =− −= −= − −=−= ∫ ∫∫ 5. ∫ + dxxx 42 2 ( ) ( ) ( ) cx c x dxxx ++= + + = += ∫ 2 3 2 3 2 1 2 2 2 21 6 1 2 3 21 4 1 21 Fazendo: dxxdu xu 4 21 2 = += 6. ∫ + dtee tt 22 31)2( ( ) ( ) .2 8 3 3 4 2 2 1)2( : 2 2 : 3 4 3 1 23 4 2 22 2 2 cec edtee Temos dtedu eu seFazendo t t tt t t ++=+ + =+ = += − ∫ 7. ∫ + 4t t e dte . e 4 que sendo , 4ln dtedueuce u du ttt =+=++== ∫ 8. ∫ + dx x e x 2 /1 2 453 . 1 . :se-doConsideran . 2 1 .221 2 1 2 2 1 1 111 x edu eu c x ec x edxxdx x e x x xxx − = = +−−=+ − +−=+= − − ∫∫ 9. ∫ dxxxtg 2sec c xtg += 2 2 . considerando-se: dxxdu xtgu 2sec= = 10. ∫ dxxxsen cos 4 c xsen += 5 5 considerando-se: dxxdu xsenu cos= = 11. ∫ dxx xsen 5cos cx c x x dxxsenx += += − −= = − − ∫ 4 4 4 5 sec 4 1 cos4 1 4 cos .cos utilizando: senxdxdu xu −= = cos 12. ∫ − dx x xxsen cos cos52 cxx dx x xsen +−−= −= ∫∫ 5|cos|ln2 5 cos 2 utilizando: senxdxdu xu −= = cos 13. ∫ dxee xx 2cos .2 2 :se-doConsideran .2 2 1 dxedu eu cesen x x x = = += 454 14. ∫ dxx x 2cos 2 .2 :se-doConsideran 4 1 2 1 2 1 2 22 dxxdu xu cxsencxsen = = +=+= 15. ∫ − θpiθ dsen )5( ( ) .5 5 :se-doConsideran .5cos 5 1 θ piθ piθ ddu u c = −= +−−= 16. dy y ysenarc ∫ − 212 ( ) ( ) . 1 1 :se-doConsideran . 4 1 22 1 2 2 2 dy y du ysenarcu cysenarccysenarc − = = +=+= 17. ∫ + θθ θ d tgba 2sec2 Ctgba b ++= ||ln1.2 θ Considerando-se: θθ θ dbdu tgbau 2sec.= += 18. ∫ + 216 x dx 455 c x tgarccxtgarc x dx +=+= + = ∫ 44 1 4 4 16 1 4 1 16 1 2 , utilizando: dxdu x u 4 1 4 = = 19. ∫ +− 442 yy dy c y c ydyy y dy + − =+ − − =−= − = − − ∫∫ 2 1 1 )2()2()2( 1 2 2 , utilizando: dydu yu = −= 2 20. ∫ θθθ dsen cos3 ( ) . 4 3 3 4 )( cos 3 4 3/4 3/1 csenc sendsen +=+== ∫ θ θθθθ 21. ∫ dxx x2ln ( ) ( ) ( ) . 22 ln :se-doConsideran .lnln4 4 1)(ln 4 1 2 ln 2 1 2 2 2222 22 dx xx xdu xu cxcxcxc x == = +=+=+=+ 22. dxee axax 2)( −+∫ ( ) ( ) .222 2 1 2 12 2 12 22 22222 cx a axhsen cxee a ce a xe a dxee axax axaxaxax ++=++−= +−+=++= − −− ∫ 23. ∫ + dttt 243 456 ( ) ( ) ( ) ( ) ( ) .13. 9 113. 2 3 . 6 1 2 3 13 6 11313 2 3 2 3 2 3 2 1 22 2 222 ctct c tdtttdttt ++=++= + + =+=+= ∫∫ Considerando-se: dttdu tu 6 13 2 = += . 24. ∫ ++ 34204 4 2 xx dx . 3 2 52 3 2 2 3 2 5 2 3 1 2 3 2 5 22 c x tgarcc x tgarc x dx + + =+ + = + + = ∫ 25. ∫ +− 14 3 2 xx dx ( ) ( ) ( )∫∫∫ − − −= − − −= −− = 3 213 13 3 2 3 3 33 32 3 222 x dx x dx x dx . 23 23ln 2 3 3 21 3 21 ln 2 13 c x x c x x + −+ −+− =+ − − − + −= Considerando-se: ( ) dxdu x u x u 3 1 3 2 3 2 22 = − = − = Resposta alternativa: 457 .1 3 2 3 2 cot 1 3 2 3 2 > −− < −− x se xhgarc x se xhtgarc 26. ∫ +162x x e dxe c e tgarc x += 44 1 Considerando-se: dxedu eu eu x x x 2 22 = = = 27. ∫ − + dx x x 1 3 . 32 32ln232 2 2ln22 2 1 2 1 ln 2 12.22 2 1 22 44 4 482 4 42 4 412 4 22. 13 22 222 2 2 c x x xc u u u c u u u u du u u du u u du udu u du u uduu u u + +− ++ −+=+ − + −= + − + −= − −= − − += − += − += − = −− = ∫∫ ∫∫∫∫ Considerando-se: duudx ux xu 2 3 3 2 2 = −= += 28. ∫ xx dx 3ln 3 2 458 ( ) ( ) . 3ln 3 1 3ln333ln 2 c x c x x dx x + − =+ − == ∫ − Considerando-se: dx x du xu 3 3 3ln = = 29. ∫ + dxxsen )2cos4( pi ( ) .2cos4cos 4 12cos4 cxxdxdxxsen ++−=+= ∫ ∫ pipi 30. ∫ + dxxx 1 2 2 . 2ln 2 2ln 2 2 1 22 1 cc xx +=+= + Considerando-se: dxxdu xu 2 12 = += 31. ∫ dxex x23 ce x += 23 6 1 Considerando-se: dxxdu xu 6 3 2 = = 32. ∫ + 2)2( t dt ( )∫ −+= 22 t ctc t + + − =+ − + = − 2 1 1 )2( 1 . Considerando-se: dtdu tu = += 2 459 33. ∫ tt dt ln .lnln ct += Considerando-se: t dtdu tu = = ln 34. ∫ − dxxx 2218 ( ) ( ) .21 3 4 2 3 21 4 18 2 32 3 2 2 cxc x +− − =+ −− = Considerando-se: dxxdu xu 4 21 2 −= −= 35. ∫ + dxee xx 252 )2( ( ) ( ) .2 12 1 6 2 2 1 62 62 cec e x x ++=+ + = Considerando-se: dxedu eu x x 2 2 2 2 = += 36. ∫ + 54 4 2t dtt ( ) dttt 454 212 −∫ −= ( ) .54 2 1 54 2 1 22 2 1 ctc t ++=+ + = Considerando-se: dttdu tu 8 54 2 = += 460 37. ∫ − dx xsen x 3 cos cxsen +−−= |3|ln Considerando-se: dxxdu xsenu cos 3 −= −= 38. ∫ + 5)1( vv dv ( ) ( ) cv c v + + −= + − + = − 4 4 12 1 4 12 Considerando-se: dv v du vu 2 1 1 = += 39. ∫ + dxxx 1 2 Considerando-se: duudxux ux 21 1 2 2 =⇒−= =+ ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) .11 3 211 5 411 7 2 13 21 5 41 7 2 3 2 5 4 7 2242 212211 23 357 357 246 224222 cxxxxxx cxxx c uuuduuuu duuuuduuuudxxx ++++++−++= ++++−+= ++−=+−= +−=−=+ ∫ ∫∫∫ 40. ∫ − dxex x 54 461 ce x + − = − 5 5 1 Considerando-se: 4 5 5xdu xu −= −= 41. ∫ dttt 2cos ctsen += 2 2 1 , utilizando: tdtdu tu 2 2 = = 42. ∫ + dxxx 568 32 ( ) ( ) ( ) .56 27 856 3 2 9 4 2 3 56 18 18 2 3 2 32 3 33 3 cxcxc x ++=++=+ + = Considerando-se: dxxdu xu 2 3 18 56 = += 43. ∫ θθθ dsen 2cos2 2/1 ( ) ( ) csencsen +=+= 2/32 3 1 2 3 2 2 1 2 3 θθ . Considerando-se: θθ θ ddu senu 2cos2 2 = = 44. ∫ + dxx )35(sec2 cxtg ++= )35( 5 1 . Considerando-se: 462 dxdu xu 5 35 = += 45. ∫ − 3)cos5( θ θθ dsen ( ) c+ − − = − 2 cos5 2θ . Considerando-se: θθ θ dsendu u = −= cos5 46. ∫ duugcot cusendu usen u +== ∫ ||ln cos Considerando-se: duudu usenu cos= = 47. ∫ >+ −− 0,)1( 2/3 adtee atat ( ) ( ) .1 5 2 2 5 11 252 5 ce a c e a at at ++−=+ +− = − − Considerando-se: ( )dtaedu eu at at −= += − −1 48. ∫ dx x xcos cxsen += 2 . Considerando-se: dx x du xu 2 1 = = 463 49. ∫ − dttt 4 ( ) ( ) ( ) ( ) ( ) ( ) ctttt ctt c uuduuuduuuu +−−+−−= +−+−= ++=+=+= ∫∫ 44 3 844 5 2 4 3 84 5 2 3 8 5 2822.4 2 35 35 242 Considerando-se: duudtut ut 24 4 2 2 =⇒+= =− 50. ∫ + dxxxsenx )42( 32 cxxc x x xdxxdxxsenx ++−=++−=+= ∫∫ 43 4 3332 2cos 6 142cos 6 142 , sendo que na primeira integral usamos: dxxdu xu 2 3 6 2 = =
