Solução_Calculo_Stewart_6e

F.
39. Let u =secθ, sothatdu =secθ tan θdθ.Then
1
u(u − 1) = A u +
Thus, I =
B
u − 1
−1
u + 1
u − 1
TX.10
SECTION 7.5 STRATEGY FOR INTEGRATION ¤ 321
sec θ tan θ
sec 2 θ − sec θ dθ = 1
u 2 − u du = 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|sec θ − 1| − ln |sec θ| + C [or ln |1 − cos θ| + C].
41. Let u = θ, dv =tan 2 θdθ= sec 2 θ − 1 dθ ⇒ du = dθ and v =tanθ − θ. So
θ tan 2 θdθ= θ(tan θ − θ) − (tan θ − θ) dθ = θ tan θ − θ 2 − ln |sec θ| + 1 2 θ2 + C
= θ tan θ − 1 2 θ2 − ln |sec θ| + C
43. Let u =1+e x ,sothatdu = e x dx. Then e x √ 1+e x dx = u 1/2 du = 2 3 u3/2 + C = 2 3 (1 + ex ) 3/2 + C.
Or:
Let u = √ 1+e x ,sothatu 2 =1+e x and 2udu = e x dx. Then
e
x √ 1+e x dx = u · 2udu= 2u 2 du = 2 3 u3 + C = 2 3 (1 + ex ) 3/2 + C.
45. Let t = x 3 .Thendt =3x 2 dx ⇒ I =
x 5 e −x3 dx = 1 3 te −t dt. Now integrate by parts with u = t, dv = e −t dt:
I = − 1 3 te−t + 1 3 e −t dt = − 1 3 te−t − 1 3 e−t + C = − 1 3 e−x3 (x 3 +1)+C.
47. Let u = x − 1,sothatdu = dx. Then
x 3 (x − 1) −4 dx = (u +1) 3 u −4 du = (u 3 +3u 2 +3u +1)u −4 du = (u −1 +3u −2 +3u −3 + u −4 ) du
=ln|u| − 3u −1 − 3 2 u−2 − 1 3 u−3 + C =ln|x − 1| − 3(x − 1) −1 − 3 2 (x − 1)−2 − 1 3 (x − 1)−3 + C
49. Let u = √ 4x +1 ⇒ u 2 =4x +1 ⇒ 2udu=4dx ⇒ dx = 1 2 udu.So
1
1
x √ 4x +1 dx = udu
2
1
4 (u2 − 1) u =2 √ =ln
4x +1− 1
√ + C
4x +1+1
du
u 2 − 1 =2 1
2
u − 1
ln u +1 + C [by Formula 19]
51. Let 2x =tanθ ⇒ x = 1 2 tan θ, dx = 1 2 sec2 θdθ, √ 4x 2 +1=secθ,so
53.
1
dx
x √ 4x 2 +1 = 2 sec2 θdθ sec θ
1
tan θ sec θ = tan θ dθ = csc θdθ
2
= − ln |csc θ +cotθ| + C [or ln |csc θ − cot θ| + C]
√ = − ln
4x2 +1
+ 1
√ 2x 2x
or + C ln
4x2 +1
− 1
2x 2x + C
x 2 sinh(mx)dx = 1 m x2 cosh(mx) − 2 m
x cosh(mx) dx
= 1 m x2 cosh(mx) − 2
1
m
m
x sinh(mx) − 1
sinh(mx) dx
m
= 1 m x2 cosh(mx) − 2 m 2 x sinh(mx)+ 2 m 3 cosh(mx)+C
u = x 2 ,
du =2xdx
U = x,
dU = dx
dv =sinh(mx) dx,
v = 1 m cosh(mx)
dV =cosh(mx) dx,
V = m 1 sinh(mx)