Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 7 REVIEW ¤ 347
49. Let u =2x +1.Then
∞
−∞
dx
∞
4x 2 +4x +5 =
−∞
1
du 2
u 2 +4 = 1 0
2 −∞
= 1 2
lim 1 tan−1 1
u 0
+ 1
t→−∞ 2 2 t 2
du
u 2 +4 + 1 ∞
2 0
du
u 2 +4
lim
1 tan−1 1
u t
= 1
t→∞ 2 2 0 4
0 − −
π
2 +
1 π − 0 4 2
= π . 4
51. We first make the substitution t = x +1,soln(x 2 +2x +2)=ln (x +1) 2 +1 =ln(t 2 +1). Then we use parts with
u =ln(t 2 +1), dv = dt:
ln(t 2 +1)dt = t ln(t 2 +1)−
t(2t) dt
t 2 +1 = t ln(t2 +1)− 2
t 2
dt
t 2 +1 = t ln(t2 +1)− 2 1 − 1
dt
t 2 +1
= t ln(t 2 +1)− 2t +2arctant + C
=(x +1)ln(x 2 +2x +2)− 2x + 2 arctan(x +1)+K, whereK = C − 2
[Alternatively, we could have integrated by parts immediately with
u =ln(x 2 +2x +2).] Notice from the graph that f =0where F has a
horizontal tangent. Also, F is always increasing, and f ≥ 0.
53. From the graph, it seems as though 2π
cos 2 x sin 3 xdxis equal to 0.
0
To evaluate the integral, we write the integral as
I = 2π
0
cos 2 x (1 − cos 2 x)sinxdxand let u =cosx ⇒
du = − sin xdx. Thus, I = 1
1 u2 (1 − u 2 )(−du) =0.
55.
√
4x2 − 4x − 3 dx = (2x − 1) 2 − 4 dx
39
= 1 2
u =2x − 1,
du =2dx
= √ u 2 − 2 2 1
2 du
u √
u2 − 2
2 2 − 22
2 ln
√ u + u2 − 2 2 + C = 1 u √ 4
u 2 − 4 − ln √ u + u2 − 4 + C
57. Let u =sinx,sothatdu =cosxdx.Then
59. (a)
= 1 4 (2x − 1) √ 4x 2 − 4x − 3 − ln 2x − 1+ √ 4x 2 − 4x − 3 + C
cos x 4+sin 2 xdx= √ 2 2 + u 2 du 21 = u √
22 + u
2 2 + 22
2 ln u + √ 2 2 + u 2 + C
= 1 sin x
4+sin 2 x +2ln sin x +
4+sin 2 x + C
2
d
− 1 √ u
a2 − u
du u
2 − sin −1 + C = 1 √ 1
a2 − u
a u 2 + √ 2 a2 − u − 1
2 1 − u2 /a · 1
2 a
= a 2 − u 2
−1/2 1 a 2 − u 2 √
a2 − u
+1− 1 =
2
u 2 u 2