Solução_Calculo_Stewart_6e

F.
326 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION
TX.10
27. Let u = e x .Thenx =lnu, dx = du/u, so
29.
e √
u2 − 1
2x
− 1 dx =
du = 41 u
u
2 − 1 − cos −1 (1/u)+C = e 2x − 1 − cos −1 (e −x )+C.
x 4
dx
√
x
10
− 2 =
x 4 dx
(x5 ) 2 − 2 = 1
5
31. Using cylindrical shells, we get
33. (a)
du
√
u2 − 2
u= x 5 ,
du= 5x 4 dx
43
= 1 5 ln u +
√
u2 − 2 + C =
1
5 ln x 5 + √ x 10 − 2 + C
V =2π
2
0
x · x 4 − x 2 dx =2π
2
=2π[(0 + 2 sin −1 1) − (0 + 2 sin −1 0] = 2π
0
x 2 4 − x 2 dx =2π
31 x
8 (2x2 − 4) 4 − x 2 + 16 x
8 sin−1
2 · π
2
=2π 2
d 1
a + bu − a2
du b 3 a + bu − 2a ln |a + bu| + C = 1
ba 2
b +
b 3 (a + bu) − 2ab
2 (a + bu)
= 1
b(a + bu) 2 + ba 2 − (a + bu)2ab
b 3 (a + bu) 2
= 1
b 3 u 2
=
b 3 (a + bu) 2
(b) Let t = a + bu ⇒ dt = bdu.Notethatu = t − a and du = 1 b
b dt.
u 2 du
(a + bu) = 1 (t − a)
2
dt = 1 t 2 − 2at + a 2
dt = 1
1 − 2a
2 b 3 t 2 b 3 t 2
b 3 t + a2
dt
t 2
= 1
t − 2a ln |t| − a2
+ C = 1
a + bu − a2
b 3 t b 3 a + bu − 2a ln |a + bu| + C
35. Maple and Mathematica both give sec 4 xdx= 2 3 tan x + 1 3 tan x sec2 x, while Derive gives the second
sin x
term as
3cos 3 x = 1 sin x 1
3 cos x cos 2 x = 1 3 tan x sec2 x. Using Formula 77, we get
sec 4 xdx= 1 tan x
3 sec2 x + 2 3 sec 2 xdx= 1 tan x 3 sec2 x + 2 3
tan x + C.
2
2
0
u 2
(a + bu) 2
37. Derive gives x 2 √ x 2 +4dx = 1 4 x(x2 +2) √ x 2 +4− 2ln √ x 2 +4+x .Maplegives
1
4 x(x2 +4) 3/2 − 1 x √ 2
x 2 +4− 2arcsinh 1
x 2
. Applying the command convert(%,ln); yields
1
4 x(x2 +4) 3/2 − 1 x √ x
2 2 +4− 2ln 1
x + 1 2 2
√
x2 +4 = 1 4 x(x2 +4) 1/2 (x 2 +4)− 2 − 2ln x + √ x 2 +4 /2
= 1 4 x(x2 +2) √ x 2 +4− 2ln √ x 2 +4+x +2ln2
Mathematica gives 1 4 x(2 + x2 ) √ 3+x 2 − 2arcsinh(x/2). Applying the TrigToExp and Simplify commands gives
1
4 x(2 + x 2 ) √ 4+x 2 − 8log √ 1
2 x + 4+x
2
= 1 4 x(x2 +2) √ x 2 +4− 2ln x + √ 4+x 2 +2ln2,soallare
equivalent (without constant).