Solução_Calculo_Stewart_6e

F.
322 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION
55. Let u = √
x,sothatx = u 2 and dx =2udu.Then
TX.10
dx
x + x √ x =
2udu
u 2 + u 2 · u =
2
du = I.
u(1 + u)
2
Now
u(1 + u) = A u + B ⇒ 2=A(1 + u)+Bu. Setu = −1 to get 2=−B,soB = −2. Setu =0to get 2=A.
1+u
2
Thus, I =
u − 2
du =2ln|u| − 2ln|1+u| + C =2ln √
x − 2ln 1+ √
x + C.
1+u
57. Let u = 3√ x + c. Thenx = u 3 − c ⇒
x
3 √ x + cdx= (u 3 − c)u · 3u 2 du =3 (u 6 − cu 3 ) du = 3 7 u7 − 3 4 cu4 + C = 3 7 (x + c)7/3 − 3 4 c(x + c)4/3 + C
59. Let u =sinx,sothatdu =cosxdx.Then
cos x cos 3 (sin x) dx = cos 3 udu= cos 2 u cos udu= (1 − sin 2 u)cosudu
= (cos u − sin 2 u cos u) du =sinu − 1 3 sin3 u + C = sin(sin x) − 1 3 sin3 (sin x)+C
61. Let y = √ x so that dy = 1
2 √ x dx ⇒ dx =2√ xdy =2ydy.Then
√xe √ x
dx =
ye y (2ydy)=
=2y 2 e y −
4ye y dy
2y 2 e y dy
U =4y,
dU =4dy
u =2y 2 ,
du =4ydy
dV = e y dy,
V = e y
dv = e y dy,
v = e y
=2y 2 e y − 4ye y − 4e y dy =2y 2 e y − 4ye y +4e y + C
=2(y 2 − 2y +2)e y + C =2 x − 2 √
x +2 e √x + C
63. Let u =cos 2 x,sothatdu =2cosx (− sin x) dx. Then
sin 2x
1+cos 4 x dx =
2sinx cos x
1+(cos 2 x) dx = 2
1
1+u 2 (−du) =− tan−1 u + C = − tan −1 (cos 2 x)+C.
65.
√ √
dx
1 x +1− x x
√
√ √ = √ √ · √ √ dx = x +1− x dx
x +1+ x x +1+ x x +1− x
= 2 3
(x +1) 3/2 − x 3/2
+ C
67. Let x =tanθ,sothatdx =sec 2 θdθ, x = √ 3 ⇒ θ = π 3 ,andx =1 ⇒ θ = π 4 .Then
√ 3
1
√ 1+x
2
π/3
sec θ
π/3
sec θ (tan 2 θ +1)
dx =
x 2 π/4 tan 2 θ sec2 θdθ=
dθ =
π/4 tan 2 θ
=
=
π/3
π/3
π/4
sec θ tan 2 θ
tan 2 θ
+ sec θ
dθ
tan 2 θ
π/3
(sec θ +cscθ cot θ) dθ = ln |sec θ +tanθ| − csc θ
π/4
π/4
ln √ 2+ 3 − √3
2
− ln √ 2+1 √ √ − 2 = 2 − √3
2
+ln 2+ √ 3 − ln 1+ √ 2