Lista P - gabarito

6.4 – EXERCÍCIOS – pg. 250
Calcular as integrais seguintes usando o método da substituição.
1.
∫ (2 x
2
+ 2 x − 3)10 (2 x + 1) dx
Fazendo − se :
u = 2x 2 + 2x − 3
du = (4 x + 2)dx = 2(2 x + 1)dx
Temos :
1 (2 x 2 + 2 x − 3)11
+ c.
∫ (2 x + 2 x − 3) (2 x + 1) dx = 2
11
2
2.
∫ (x
3
10
− 2)1 / 7 x 2 dx
Fazendo − se :
u = x3 − 2
du = 3 x 2 dx
Temos :
∫ ( x − 2)
3
3.
∫
1/ 7
(
1 x3 − 2
x dx =
8
3
7
2
)
8
7
+c =
8
7 3
x − 2 7 + c.
24
(
)
x dx
5
x2 − 1
∫ (x
2
−1
− 1) 5 x dx
Fazendo − se :
u = x2 −1
du = 2 x dx
Temos :
∫
4.
5
∫ 5x
(
)
1 x2 −1
=
4
x2 −1 2
5
x dx
4/5
4
5
+ c = ( x 2 − 1) 5 + c
8
4 − 3 x 2 dx
451
1
(
1
)
= ∫ 5 x (4 − 3 x 2 ) 2 dx = ∫ 5 x 4 − 3 x 2 2 dx
Fazendo − se :
u = 4 − 3x 2
du = −6 x dx
Temos :
(
− 1 4 − 3x 2
5
x
4
−
3
x
dx
=
5
.
∫
3
6
2
2
5.
∫
3
2
)
+c=
3
−5
4 − 3 x 2 2 + c.
9
(
)
x 2 + 2 x 4 dx
(
)
1
= ∫ x 1 + 2 x 2 2 dx
(
1 1 + 2x 2
=
3
4
2
1
= 1 + 2x 2
6
(
6.
∫ (e
2t
)
)
3
2
3
2
Fazendo:
+c
u = 1 + 2x2
du = 4 x dx
+c
1
+ 2) 3 e 2t dt
Fazendo − se :
u = e2t + 2
du = 2e 2t dt
Temos :
(
1 e 2t + 2
(
e
+
2
)
e
dt
=
∫
4
2
3
2t
7.
3
2t
4
3
)
+c=
4
3 2t
e + 2 3 + c.
8
(
)
et dt
∫ et + 4
=∫
8.
1
du
= ln et + 4 + c , sendo que u = et + 4 e du = et dt.
u
e1 / x + 2
∫ x 2 dx
452
1
1
1
x −1
2
−2
x
dx
+
2
x
dx
=
−
e
+
2
.
+ c = −e x − + c.
2
∫
x
−1
x
Considerando - se :
1
= ∫ex
1
u = ex
1
du = e x .
9.
∫ tg x sec
x dx
u = tg x
du = sec 2 x dx
∫ sen x cos x dx
4
sen 5 x
+c
5
=
11.
2
tg 2 x
+ c . considerando-se:
2
=
10.
−1
.
x2
considerando-se:
u = sen x
du = cos x dx
sen x
dx
5
x
∫ cos
= ∫ cos −5 x . sen x dx
=−
cos − 4 x
1
=
+c
4 cos 4 x
−4
utilizando:
u = cos x
du = − senxdx
1
= sec 4 x + c
4
12.
∫
2 sen x − 5 cos x
dx
cos x
sen x
− 5 ∫ dx
cos x
= −2 ln | cos x | −5 x + c
= 2∫
13.
∫e
x
utilizando:
u = cos x
du = − senxdx
cos 2 e x dx
1
sen 2e x + c.
2
Considerando - se :
=
u = 2e x
du = 2e x dx.
453
14.
x
∫ 2 cos x
2
dx
11
1
sen x 2 + c = sen x 2 + c
22
4
Considerando - se :
=
u = x2
du = 2 x dx.
15.
∫ sen (5θ − π )dθ
1
= − cos (5θ − π ) + c.
5
Considerando - se :
u = 5θ − π
du = 5dθ .
16.
arc sen y
∫2
1 − y2
dy
2
1 (arc sen y )
1
2
+ c = (arc sen y ) + c.
2
2
4
Considerando - se :
=
u = arc sen y
du =
17.
1
1 − y2
dy.
2 sec 2 θ
∫ a + b tg θ dθ
1
= 2. ln | a + b tgθ | +C
b
Considerando-se:
u = a + b tgθ
du = b. sec 2 θ dθ
18.
dx
∫ 16 + x
2
454
=
19.
1
16 ∫
∫y
2
=∫
20.
∫
3
dx
 x
1+  
4
2
=
x
x
1
1
4 arc tg + c = arc tg + c , utilizando:
16
4
4
4
x
4
du =
1
dx
4
dy
− 4y + 4
u = y−2
dy
( y − 2) −1
1
−2
=
y
−
dy
=
+
c
=
+
c
,
utilizando:
(
2
)
2
∫
du = dy
−1
( y − 2)
2− y
sen θ cos θ dθ
= ∫ (senθ ) cos θ dθ =
1/ 3
21.
u=
4
( senθ ) 4 / 3
3
+ c = sen 3θ + c.
4
4
3
ln x 2
∫ x dx
(
)
2
1 ln x 2
1
1
2
2
+ c = (ln x 2 ) 2 + c = 4 (ln x ) + c = (ln x ) + c.
2
2
4
4
Considerando - se :
u = ln x 2
22.
du =
2x 2
= dx.
x2 x
∫ (e
+ e − ax ) 2 dx
ax
(
1 2 ax
1 − 2 ax
e + 2x −
e
+c
2a
2a
sen h 2ax
+ 2x + c =
+ 2 x + c.
a
)
2
= ∫ e 2 ax + 2 + e − 2 ax dx =
23.
=
1 2 ax
e − e − 2 ax
2a
∫
3t 4 + t 2 dt
(
)
455
(
3
)
1 3t 2 + 1 2
= ∫ t 3t + 1 dt = ∫ t 3t + 1 dt =
+c
3
6
2
3
3
1 3
1
= . . 3t 2 + 1 2 + c = . 3t 2 + 1 2 + c.
6 2
9
2
(
)
2
(
)
(
2
(
1
2
)
)
Considerando-se:
u = 3t 2 + 1
du = 6t dt
24.
∫ 4x
2
.
4 dx
+ 20 x + 34
5
5


2x + 
x+ 
dx
1
2
2
2
=∫
= arc tg 
+ c = arc tg 
+ c.
2
2
3
3
3
3
5 3

x+  + 
2
2
2 2

25.
∫x
2
3 dx
− 4x + 1
dx
3dx
1
dx
3
=∫
= −3 ∫
= −3 ∫
2
2
3
(x − 2) − 3
3 (x − 2)
(x − 2)2
−
1−
3
3
3
x−2
1
3 + c = − 3 ln x + 3 − 2 + c.
= − 3 ln
x
−
2
2 1−
2
3 +2− x
3
1+
Considerando-se:
u
2
2
(
x − 2)
=
3
x−2
u=
3
1
du =
dx
3
Resposta alternativa:
456
arc tg h
x−2
3
arc cot g h
26.
x−2
<1
3
se
x−2
3
se
x−2
> 1.
3
e x dx
∫ e2 x + 16
=
1
ex
arc tg + c
4
4
Considerando-se:
u 2 =e 2 x
u = ex
du =e 2 xdx
27.
∫
x+3
dx
x −1
=∫
u
2u 2
4 
du 


.
2
u
du
=
du = 2 ∫ 1 + 2
 du = 2 u + 4 ∫ 2
2
2
∫
u − 3 −1
u −4
u − 4 
 u −4

u
− du
1+
1
du
2 +c
= 2u + 8 ∫ 4 2 = 2u − 2 ∫
= 2u − 2 . 2 ln
2
u
4 u
2
u
 
1−
−
1−  
2
4 4
2
= 2u − 2 ln
2+u
2+ x+3
+ c = 2 x + 3 − 2 ln
+ c.
2−u
2− x+3
Considerando-se:
u2 = x + 3
x = u2 − 3
dx = 2u du
28.
3 dx
2
3x
∫ x ln
457
= ∫ (ln 3 x )
−2
3dx
(ln 3x ) + c = − 3 + c.
=3
x
−1
ln 3 x
Considerando-se:
u = ln 3 x
du =
29.
3
dx
3x
∫ ( sen 4 x + cos 2π )dx
= ∫ sen 4 x dx + cos 2π ∫ dx =
30.
∫2
x 2 +1
1
(− cos 4 x ) + x cos 2π + c.
4
x dx
2
2
1 2 x +1
2x
=
+c=
+ c.
2 ln 2
ln 2
Considerando-se:
u = x2 + 1
du = 2 x dx
31.
∫x e
=
3x2
dx
1 3x 2
e +c
6
Considerando-se:
u = 3x 2
du = 6 x dx
32.
dt
∫ (2 + t )
2
= ∫ (2 + t ) =
−2
(2 + t ) −1
−1
+c=
+c.
−1
2+t
Considerando-se:
u = 2+t
du = dt
458
33.
dt
∫ t ln t
= ln ln t + c.
u = ln t
Considerando-se:
34.
∫ 8x
du =
dt
t
1 − 2 x 2 dx
(
− 1 1 − 2x2
=8
3
4
2
3
2
)
+c =
3
−4
1 − 2 x 2 2 + c.
3
(
)
Considerando-se:
u = 1 − 2x2
du = −4 x dx
35.
∫ (e
2x
+ 2)5 e 2 x dx
(
)
6
6
1 e2 x + 2
1 2x
=
+c=
e + 2 + c.
2
6
12
(
)
Considerando-se:
u = e2 x + 2
du = 2e 2 x dx
36.
∫
4t dt
4t 2 + 5
(
= ∫ 4t 2 − 5
−
)
(
1
2
4t dt
1
)
1 4t 2 + 5 2
=
+ c = 4t 2 + 5 + c.
1
2
2
Considerando-se:
u = 4t 2 + 5
du = 8t dt
459
37.
cos x
∫ 3 − sen x dx
= − ln | 3 − sen x | + c
Considerando-se:
u = 3 − sen x
du = − cos x dx
38.
dv
v (1 + v )5
∫
(1 + v )
=2
−4
+c
−4
1
=−
(
2 1+ v
)
4
+c
Considerando-se:
u =1+ v
du =
39.
∫x
2
1
2 v
dv
1 + x dx
Considerando-se:
1 + x = u2
x = u 2 − 1 ⇒ dx = 2u du
∫x
2
(
)
2
(
)
1 + x dx = ∫ u 2 − 1 u 2u du = ∫ u 4 − 2u 2 + 1 2u 2 du
u7
u5
u3
−4 +2 +c
7
5
3
2
=
(1 + x )7 − 4 (1 + x )5 + 2 (1 + x )3 + c
7
5
3
2
4
2
3
2
= (1 + x ) 1 + x − (1 + x ) 1 + x + (1 + x ) 1 + x + c.
7
5
3
(
)
= ∫ 2u 6 − 4u 4 + 2u 2 du = 2
40.
∫x e
4 − x5
dx
460
=
− 1 − x5
e +c
5
Considerando-se:
u = − x5
du = −5x 4
41.
∫ t cos t dt
2
=
42.
1
u = t2
sen t 2 + c , utilizando:
2
du = 2tdt
∫ 8x
2
6 x 3 + 5 dx
3
(
)
3
3
1 6 x3 + 5 2
4 2
8
=8
+c=
6 x3 + 5 2 + c =
6 x 3 + 5 2 + c.
3
18
9 3
27
2
(
)
(
)
Considerando-se:
u = 6 x3 + 5
du = 18 x 2 dx
43.
∫ sen
1/ 2
2θ cos 2θ dθ
3
1 (sen 2θ )2
1
3/ 2
=
+ c = (sen 2θ ) + c .
3
2
3
2
Considerando-se:
u = sen 2θ
du = 2 cos 2θ dθ
44.
∫ sec (5x + 3)dx
2
1
= tg (5 x + 3) + c .
5
Considerando-se:
461
u = 5x + 3
du = 5 dx
45.
sen θ dθ
∫ (5 − cosθ )
=
3
(5 − cosθ )−2 + c .
−2
Considerando-se:
u = 5 − cos θ
du = sen θ dθ
46.
∫ cot g u du
=∫
cos u
du = ln | sen u | + c
sen u
Considerando-se:
47.
∫ (1 + e
− at 3 / 2
)
u = sen u
du = cos u du
e − at dt , a > 0
(
− 1 1 + e − at
=
5
a
2
5
2
)
+c=−
2
1 + e − at
5a
(
5
2
)
+ c.
Considerando-se:
u = 1 + e − at
du = e − at (− a ) dt
48.
∫
cos x
dx
x
= 2 sen x + c .
Considerando-se:
u= x
du =
1
2 x
dx
462
49.
∫t
t − 4 dt
(
)
(
)
= ∫ u 2 + 4 . u 2u du = ∫ 2u 4 + 8u 2 du = 2
u5
u3
+8 +c
5
3
2
(t − 4)5 + 8 (t − 4)3 + c
5
3
2
8
2
= (t − 4 ) t − 4 + (t − 4) t − 4 + c
5
3
=
Considerando-se:
t − 4 = u2
t = u 2 + 4 ⇒ dt = 2u du
50.
∫x
2
( sen 2 x 3 + 4 x) dx
= ∫ x 2 sen 2 x 3dx + ∫ 4 x 3dx =
−1
x4
−1
cos 2 x 3 + 4 + c =
cos 2 x 3 + x 4 + c ,
6
x
6
sendo que na primeira integral usamos:
u = 2 x3
du = 6 x 2 dx
463