Solução_Calculo_Stewart_6e

F.
320 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION
TX.10
23. Let u =1+ √ x.Thenx =(u − 1) 2 , dx =2(u − 1) du ⇒
1
1+ √ 8 2
x
0
dx =
1 u8 · 2(u − 1) du =2 2
1 (u9 − u 8 ) du = 1
5 u10 − 2 · 1 u9 2
= 1024 − 1024 − 1 + 2 = 4097 .
9 1 5 9 5 9 45
25.
3x 2 − 2
x 2 − 2x − 8 =3+ 6x +22
(x − 4)(x +2) =3+ A
x − 4 + B
x +2
⇒
6x +22=A(x +2)+B(x − 4). Setting
x =4gives 46 = 6A, soA = 23 .Settingx = −2 gives 10 = −6B, soB = − 5 .Now
3 3
3x 2
− 2
x 2 − 2x − 8 dx = 3+ 23/3
x − 4 − 5/3
dx =3x + 23 3
x +2
ln |x − 4| − 5 ln |x +2| + C.
3
27. Let u =1+e x ,sothatdu = e x dx =(u − 1) dx. Then
29.
1
u(u − 1) = A u +
Thus, I =
B
u − 1
−1
u + 1
u − 1
1
1
1+e dx = x u · du
u − 1 =
1
du = I. Now
u(u − 1)
⇒ 1=A(u − 1) + Bu. Setu =1to get 1=B. Setu =0to get 1=−A, soA = −1.
du = − ln |u| +ln|u − 1| + C = − ln(1 + e x )+lne x + C = x − ln(1 + e x )+C.
Another method: Multiply numerator and denominator by e −x and let u = e −x +1. This gives the answer in the
form − ln(e −x +1)+C.
5
3w − 1
5
w +2 dw = 3 − 7
5
dw = 3w − 7ln|w +2| =15− 7ln7+7ln2
w +2
0
0
0
=15+7(ln2− ln 7) = 15 + 7 ln 2 7
31. As in Example 5,
√ √
1+x 1+x 1+x
1 − x dx = √ · √ dx =
1 − x 1+x
Another method: Substitute u = (1 + x)/(1 − x).
1+x
√ dx = 1 − x
2
33. 3 − 2x − x 2 = −(x 2 +2x +1)+4=4− (x +1) 2 .Letx +1=2sinθ,
where − π ≤ θ ≤ π .Thendx =2cosθdθand
2 2
√
3 − 2x − x2 dx = 4 − (x +1) 2 dx =
4 − 4sin 2 θ 2cosθdθ
dx
√ + 1 − x
2
xdx
√
1 − x
2 =sin−1 x − 1 − x 2 + C.
=4 cos 2 θdθ=2 (1 + cos 2θ) dθ
=2θ +sin2θ + C =2θ +2sinθ cos θ + C
x +1
=2sin −1 +2· x +1
√
3 − 2x − x
2
·
+ C
2
2 2
x +1
=2sin −1 + x +1 √
3 − 2x − x2 + C
2 2
35. Because f(x) =x 8 sin x is the product of an even function and an odd function, it is odd.
Therefore, 1
−1 x8 sin xdx=0
[by (5.5.7)(b)].
37.
π/4
cos 2 θ tan 2 θdθ= π/4
sin 2 θdθ= π/4 1
(1 − cos 2θ) dθ = 1
θ − 1 sin 2θ π/4
= π
− 1
0 0 0 2 2 4 0 8 4 − (0 − 0) =
π
− 1 8 4