608 7.9 – EXERCÍCIOS – pg. 333 Nos exercícios de 1 a 14, calcular a integral indefinida. 1. ( ) ( ) dx x x sen x sen ∫ + + cos 1 1 Fazendo: 2 x tg t = Temos: ( ) ( ) ( ) ( ) ( ) ∫ ∫ ∫ ∫ ∫ + = + = + + + + = + - + + + + + + + = + - + + + + + = dt t t dt t t dt t t t t t t t t t t t dt t t t t t t t dt t t I 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 1 4 1 2 1 4 1 2 1 2 1 1 1 1 2 1 2 . 1 2 1 1 1 1 1 2 1 2 1 2 1 C x tg x tg x tg C t t t dt t t dt t t t + + + = + + + = + + = + + = ∫ ∫ 2 ln 2 1 2 4 2 ln 2 2 2 1 1 2 2 1 1 2 2 1 2 2 2 2. ∫ + + x x sen dx cos 1 Fazendo: 2 x tg t = Temos:
32
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608
7.9 – EXERCÍCIOS – pg. 333
Nos exercícios de 1 a 14, calcular a integral indefinida.
1. ( )
( )dx
xxsen
xsen∫ +
+
cos1
1
Fazendo:
2
xtgt =
Temos:
( )( )
( )
( ) ( )∫∫∫
∫∫
+=
+=
+
+
++
=
+
−++
+
++
++
=
+
−+
+
+
++
=
dtt
tdt
t
tdt
t
t
t
tt
t
tt
t
t
t
dt
t
tt
t
t
t
t
dt
t
t
I
22
22
22
2
2
22
2
22
2
2
2
2
22
1
2
1
4
12
1
4
1
212
1
11
1
21
2.
1
21
1
11
1
2
1
2
1
21
Cx
tgx
tg
xtg
Cttt
dtt
t
dtt
tt
+++=
+
++=
++=
++=
∫
∫
2ln
2
1
242
ln222
1
12
2
1
12
2
1
2
2
2
2. ∫ ++ xxsen
dx
cos1
Fazendo:
2
xtgt =
Temos:
609
Cx
tg
Ctt
dt
t
dt
t
t
t
t
t
dt
++=
++=+
=
+=
+
−+
++
+=
∫
∫∫
12
ln
1ln1
22
2
1
1
1
21
1
2
2
2
2
2
3. ∫ + xtgxsen
dx2
Fazendo:
2
xtgt =
Temos:
( ) ( ) ( )( )( )
( )( )( )
33
22
2
22
222
2
2
22
2
2222
11.
1
4
11
12121
4
1
1.
1
2
1
21
2.2
tttt
tt
t
dt
tt
ttttt
dt
t
t
t
t
t
t
t
dt
I
++−
−+
+=
−+
++−+
=
−
+
++
+
+=
∫
∫
∫
( )
C
xtg
xtg
Ct
t
dttt
dt
dtt
t
dtt
t
+−=
+−=
−=
−=
−=
∫ ∫
∫
∫
2
2
2ln
2ln
1
4
14
2
2
2
2
4. ∫ + x
dx
cos54
610
Fazendo:
2
xtgt =
Temos:
2
2
2
2
22
2
222
2
22
2
2
2
9
1.
1
2
1
9
1.
1
2
1
5544
1.
1
2
1
554
1.
1
2
1
1.54
1
2
t
t
t
dt
t
tt
dt
t
ttt
dt
t
tt
dt
t
t
t
dt
I
−
+
+=
+
−+=
+
−+++=
+
−+
+=
+
−+
+=
∫
∫
∫
∫∫
Cx
tg
xtg
Ct
t
Ct
t
t
dt
+
−
+
=
+−
+=
+−
+=
−= ∫
32
32ln
3
1
3
3ln
3
1
3
3ln
3.2
1.2
9
22
5. ∫ + x
dx
cos3
Fazendo:
2
xtgt =
Temos:
∫ ∫+
−+++=
+
−+
+=
2
222
2
2
2
1
133
1.
1
2
1
13
1
2
t
ttt
dt
t
t
t
dt
I
611
C
xtg
tgarc
Ct
tgarct
dt
+=
+=+
= ∫
2
22
2
22
2
24
22
6. ∫ − x
dx
cos1
Fazendo:
2
xtgt =
Temos:
Cx
tg
Ct
t
dt
t
dt
tt
t
t
dt
t
t
t
dt
I
+−
=
+−
=
==
+−+
+
+=
+
−−
+=
−
∫∫
∫∫
2
1
1
2
2
11
1.
1
2
1
11
1
2
1
22
22
2
2
2
2
2
7. ∫ −
+dx
xsen
x
1
cos1
Fazendo:
2
xtgt =
Temos:
612
( ) ( )∫
∫
∫
∫
+−=
++−=
+
+
−+
+
−++
=
+
+−
+
−+
=
22
22
2
2
2
2
22
2
2
2
2
11
4
1
2.
12
2
1
2.
1
21
1
11
1
2.
1
21
1
11
tt
dt
t
dt
tt
t
dt
t
tt
t
tt
t
dt
t
tt
t
I
( ) ( ) ( ) 11111
42222
+
++
−+
−=
+− t
DCt
t
B
t
A
tt
( ) ( ) ( ) ( ) ( )222 11114 −+++++−≡ tDCttBttA
2241 =∴=⇒= BBt
( ) ( ) ( )DDtDtCtCtCtBBtAAtAtAt
ttDCtBBttttA
+−++−+++−−+≡
+−++++−−+≡
22
1214
223223
2223
=++−
=−+
=+−+−
=+
4
02
02
0
DBA
DCA
DCBA
CA
0,2,2,2 ===−= DCBA
( )
( )
Cx
tgx
tg
xtg
Ctt
t
dtt
t
ttI
+++
−
−−−=
+++−
−+−−=
++
−+
−
−=
−
∫
12
ln
12
21
2ln2
1ln1
121ln2
1
2
1
2
1
2
2
2
1
22
613
8. ∫ + xsen
dx
23
Fazendo
dxdu
xu
2
2
=
=
e
2
utgt =
Temos:
∫
∫
∫∫
∫ ∫
+
+
=
+
+
=
++
=
++=
+
++
+=
++
+=+
=
C
t
tgarc
t
dt
tt
dt
tt
dt
t
tt
t
dt
t
tt
dt
usen
du
I
3
8
3
1
8
3.
3
1
9
8
3
13
1
13
23
1
323
1
2331
2
2
1
1
23
1
2
2
1
3
2
2
2
2
2
2
2
2
2
Cxtg
tgarc
C
utg
tgarc
Ct
tgarc
C
t
tgarc
++
=
+
+
=
++
=
+
+
=
8
13
8
1
8
12
3
8
1
8
13
8
1
8
3
13
8
1
9. ( )
( )∫ −−
−dt
t
t
12cos2
12cos
614
Fazendo:
dtdu
tu
2
12
=
−=
e
2
utgt =
Temos:
( )( )
( )
( )( ) ( )dt
tt
tdt
tt
t
dtt
t
t
t
t
tt
dtt
t
t
t
t
dt
t
t
du
u
uI
∫∫
∫∫
∫
++
−=
++
−=
+
+
+
−=
+
+−+
+
−
=
+
−−
++
−
=
−=
3
11
1
3
1
131
1
13
1.
1
1
1
122
1
12
2
1
1
12
1
2.
1
1
2
1
2.
cos2
cos
22
2
22
2
2
2
22
2
2
22
22
2
2
2
22
2
Usando:
( )( ) 3/113/11
12222
2
+
++
+
+=
++
−
t
DCt
t
BAt
tt
t
( )( ) ( ) ( )222 13/11 tDCttBAtt +++++≡−
2,0,3,0
13
1
03
1
1
0
==−==
=+
=+
−=+
=+
DCBA
DB
CA
DB
CA
615
.2
123
3
2
2
12
)3(3
2
3/1
2
1
3
3
122
Ct
tgarctgt
tgarctg
Ctarctgarctgt
dttt
I
+
−+
−−=
++−=
++
+
−= ∫
10. ∫ ++ tsent
dt
cos3
Fazendo :2
utgt =
∫∫
∫
∫
+
+
=++
=
++
+
+=
+
−+
++
+=
4
7
2
12
422
1.
1
2
1
1
1
23
1
2
22
2
2
2
2
2
2
2
u
du
uu
du
uu
u
u
du
u
u
u
u
u
du
I
C
ttg
tgarc
Cu
tgarc
C
u
tgarc
+
+
=
++
=
+
+
=
7
12
2
7
2
7
12
7
2
2
7
2
1
2
7
1
11. ∫−
xx
x
eesen
dxe
cos34
Fazendo
dxedu
eu
x
x
=
=
∫ −=
uusen
duI
cos34
616
Fazendo :2
utgt =
∫
∫ ∫
∫
−+
=
−+=
+
+−
+=
+
−−
+
+=
13
83
2
383
2
1
3381
2
1
1.3
1
2.4
1
2
2
2
2
2
2
2
2
2
2
tt
dt
tt
dt
t
ttt
t
t
t
t
tt
dt
I
Fazendo
3
131
3
83
2
2−
++
=
−+ t
B
t
A
tt
( ) ( )33
13
2++−≡ tBtA
51
31 =⇒= Bt
513 −=⇒−= At
Temos:
( )
Ce
tge
tg
Cu
tgu
tg
Ctt
dttt
I
xx
+−++−=
+−++−=
+−++−=
−+
+
−
= ∫
3
1
2ln
5
13
2ln
5
1
3
1
2ln
5
13
2ln
5
1
31ln
5
13ln
5
1
3/1
51
3
51
12. ∫ + θ
θθ
cos1
cos d
617
Fazendo :2
θtgt =
( )( ) ( )
( )dt
t
t
t
t
tt
t
dtt
t
tt
dt
t
t
2
1.
1
12
1
11
1
12
1
11
1
2.
1
1
2
22
2
2
22
22
2
2
2
22
2
+
+
−=
+
−++
−
−
=
+
−+
++
−
=
∫∫
∫
Ctgtgarctg
Cttgarct
dtt
dtt
t
+
+−=
++−=
+
−+−=
+
−=
∫
∫
22
2
2
1
21
1
1
2
2
2
θθ
13. ∫ + xxsen
dx
cos
Fazendo :2
xtgt =
∫
∫
∫
−−−=
+
−+
+=
+
−+
+
+=
12
2
1
12
1
2
1
1
1
2
1
2
2
2
2
2
2
2
2
2
tt
dt
t
tt
t
dt
t
t
t
t
t
dt
( ) ( )212112
12
+−+
−−=
−− t
B
t
A
tt
( ) ( )21211 −−++−≡ tBtA
618
22
1
22
1
−=
=
B
A
Cx
tgx
tg
Ctt
Ct
dt
t
dtI
++−+−−−=
++−+−−−=
+
+−−
−−−= ∫∫
212
ln2
121
2ln
2
1
21ln2
121ln
2
1
2122
1
2122
12
14. ∫ +− θθ
θ
cos4 sen
d
Fazendo :2
θtgt =
∫∫
∫
∫
∫
+
−
=
+−
=
+−=
−+−+
+
+=
+
−+
+−
+=
9
14
3
1
3
2
3
5
3
23
2
523
2
1244
1.
1
2
1
1
1
24
1
2
22
2
22
2
2
2
2
2
2
t
dt
tt
dt
tt
dt
ttt
t
t
dt
t
t
t
t
t
dt
I
C
tg
tgarcCt
tgarc
C
t
tgarc
+
−
=+−
=
+
−
=
14
12
3
14
2
14
13
14
2
3
14
3
1
3
14
1
3
2
θ
619
15. Calcular a área sob a curva xsen
y+
=2
1, de 0=x a
2
π=x .
A figura que segue mostra a região dada.
π/2
0.2
0.4
0.6
x
y
∫ +=
2
02
π
xsen
dxA
∫ +=
xsen
dxI
2
Fazendo :2
xtgt =
∫
∫∫
∫
+
+
=
++=
++=
+
++
+=
++
+=
4
3
2
1
1222
2
1
2221
2
1
22
1
2
2
22
2
2
2
2
2
t
dt
tt
dt
tt
dt
t
tt
t
dt
t
tt
dt
I
620
C
xtg
tgarc
Ct
tgarc
C
t
tgarc
+
+
=
++
=
+
+
=
3
12
2
3
2
3
12
3
2
2
3
2
1
2
3
1
..9
3
6
2
3
2
633
2
6.
3
2
3.
3
2
3
1
3
2
3
3
3
2
3
102
3
2
3
14
2
3
2
3
12
2
3
2
2
0
au
tgarctgarc
tgtgarc
tg
tgarc
xtg
tgarcA
πππ
ππ
ππ
π
π
=−
=
−=
−=
−=
+−
+
=
+
=
16. Calcule a área limitada pelas curvas x
ycos2
1
+= e
xy
cos2
1
−= entre
2π− e
2π
A figura que segue mostra a região dada.
621
-π/2 π/2
0.5
1
x
y
dxxx∫
−
+−
−
2
2
cos2
1
cos2
1π
π
dxxx
I ∫
+−
−=
cos2
1
cos2
1
Fazendo :2
xtgt =
∫∫
∫∫
∫
+−
+=
+
+
+−
+
+
+=
+
+
−+
−
+
−−
=
32
3/13
2
3
1.
1
2
13
1.
1
2
1
2
1
12
1
1
12
1
22
2
2
22
2
2
2
2
2
2
2
t
dt
t
dt
t
t
t
dt
t
t
t
dt
t
dt
t
t
t
tI
( )
Cx
tgtgarcx
tgtgarc
Ct
tgarcttgarc
+
−
=
+−+=
23
1
3
2
23
3
2
33
123
3
2
622
Portanto,
( )
.. 9
32
33
2
6.
3
2
3.
3
2
6.
3
2
3.
3
2
3
1
3
23
3
2
3
1
3
23
3
2
23
1
3
2
23
3
22
2
au
tgarctgarctgarctgarc
xtgtgarc
xtgtgarcA
ππ
ππππ
π
π
==
−+−=
−+−−−=
−
=
−
Nos exercícios de 17 a 33 calcular a integral indefinida: