Solução_Calculo_Stewart_6e

F.
344 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION
TX.10
1.
3.
5
0
π/2
0
x
5
x +10 dx =
0
1 − 10
dx = x − 10 ln(x +10) 5
=5− 10 ln 15 + 10 ln 10
x +10
0
=5+10ln 10
15 =5+10ln2 3
cos θ
π/2
1+sinθ dθ = ln(1 + sin θ) =ln2− ln 1 = ln 2
0
π/2
5. sin 3 θ cos 2 θdθ= π/2
(1 − cos 2 θ)cos 2 θ sin θdθ= 0
(1 − 0 0 1 u2 )u 2 (−du)
= 1
0 (u2 − u 4 ) du = 1
3 u3 − 1 u5 1
= 1
− 1
5 0 3 5 − 0=
2
15
sin(ln t)
7. Let u =lnt, du = dt/t. Then
dt =
t
9.
4
1
x 3/2 ln xdx
11. Let x =secθ.Then
2
√
x2 − 1
dx =
x
1
u =lnx,
du = dx/x
π/3
0
dv = x 3/2 dx,
v = 2 5 x5/2
= 2 5
u =cosθ,
du = − sin θdθ
sin udu= − cos u + C = − cos(ln t)+C.
4
x 5/2 ln x − 2
1 5
4
1
x 3/2 dx = 2 5 (32 ln 4 − ln 1) − 2 5
= 2 4
128 124
(64 ln 2) − (32 − 1) = ln 2 −
5 25 5 25
tan θ
π/3
sec θ sec θ tan θdθ= tan 2 θdθ=
0
π/3
0
or
64
5
2
5 x5/2 4
1
124
ln 4 −
25
(sec 2 θ − 1) dθ = tan θ − θ π/3
0
= √ 3 − π 3 .
13. Let t = 3√ x.Thent 3 = x and 3t 2 dt = dx,so e 3√x dx = e t · 3t 2 dt =3I. ToevaluateI,letu = t 2 ,
dv = e t dt ⇒ du =2tdt, v = e t ,soI = t 2 e t dt = t 2 e t − 2te t dt. NowletU = t, dV = e t dt ⇒
dU = dt, V = e t .Thus,I = t 2 e t − 2 te t − e t dt = t 2 e t − 2te t +2e t + C 1 , and hence
3I =3e t (t 2 − 2t +2)+C =3e 3√x (x 2/3 − 2x 1/3 +2)+C.
15.
x − 1
x 2 +2x = x − 1
x(x +2) = A x + B
x +2
to get −1 =2A,soA = − 1 .Thus, 2
⇒ x − 1=A(x +2)+Bx. Setx = −2 to get −3 =−2B,soB = 3 .Setx =0
2
x − 1
−
1 3
x 2 +2x dx = 2
x + 2
dx = − 1 x +2 2 ln |x| + 3 ln |x +2| + C.
2
17. Integrate by parts with u = x, dv =secx tan xdx ⇒ du = dx, v =secx:
14
x sec x tan xdx= x sec x − sec xdx = x sec x − ln|sec x +tanx| + C.
19.
x +1
9x 2 +6x +5 dx =
x +1
(9x 2 +6x +1)+4 dx =
1
3
=
(u − 1)
+1 1
u 2 +4 3 du
= 1
u
9 u 2 +4 du + 1
9
x +1
(3x +1) 2 +4 dx
= 1 3 · 1 (u − 1) + 3
du
3 u 2 +4
u= 3x +1,
du= 3dx
2
u 2 +2 du = 1 2 9 · 1
2 ln(u2 +4)+ 2 9 · 1 1
2 tan−1 2 u + C
= 1
18 ln(9x2 +6x +5)+ 1 9 tan−1 1
2 (3x +1) + C