Solução_Calculo_Stewart_6e

F.
346 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION
35.
1
√ dx = x + x
3/2
dx
√ =
x (1 + x )
=4 √ u + C =4 1+ √ x + C
dx
√ x
1+
√ x
TX.10
⎡
⎣
u =1+ √ x,
du =
dx
2 √ x
⎤
⎦ =
2 du
√ = u
2u −1/2 du
37.
(cos x +sinx) 2 cos 2xdx= cos 2 x +2sinx cos x +sin 2 x cos 2xdx= (1 + sin 2x)cos2xdx
= cos 2xdx+ 1 2
sin 4xdx=
1
sin 2x − 1 cos 4x + C
2 8
Or: (cos x +sinx) 2 cos 2xdx= (cos x +sinx) 2 (cos 2 x − sin 2 x) dx
39. We’ll integrate I =
41.
and v = − 1 2 ·
1/2
Thus,
0
∞
1
43.
45.
∞
2
4
0
I = − 1 2 ·
1
1+2x ,so
= (cos x +sinx) 3 (cos x − sin x) dx = 1 4 (cos x +sinx)4 + C 1
xe 2x
(1 + 2x) dx by parts with u = dx
2 xe2x and dv =
(1 + 2x) .Thendu =(x · 2 2e2x + e 2x · 1) dx
xe 2x
1+2x −
xe 2x
1
(1 + 2x) dx = e 2x 2 4 −
1
t
3
dx = lim
(2x +1) t→∞
1
dx
x ln x
u =lnx,
du = dx/x
dx
x ln x = lim
t→∞
= − 1 4 lim
t→∞
t
2
=
du
u
− 1 2 · e2x (2x +1)
1+2x
dx = − xe2x
4x +2 + 1 2 · 1
2 e2x + C = e 2x 1
4 − x
+ C
4x +2
x 1/2 1
= e
4x +2
0
4 − 1 1
− 1
8
4 − 0 = 1 8 e − 1 4 .
t
1
dx = lim
(2x +1)
3 t→∞
1
1
(2t +1) 2 − 1
= − 1 9 4
1
(2x 2 +1)−3 2 dx = lim
t→∞
0 − 1
= 1
9 36
=ln|u| + C =ln|ln x| + C, so
t
1
−
4(2x +1) 2 1
dx
t
x ln x = lim ln |ln x| = lim [ln(ln t) − ln(ln 2)] = ∞, so the integral is divergent.
t→∞
2 t→∞
ln x
4
ln x
√ dx = lim √ dx = ∗ lim 2 √ x ln x − 4 √ 4
x
x t→0 + t x t→0 + t
= lim
t→0 +
(2 · 2ln4− 4 · 2) −
2
√
t ln t − 4
√
t
∗∗
=(4ln4− 8) − (0 − 0) = 4 ln 4 − 8
(∗) Letu =lnx, dv = 1 √
x
dx ⇒ du = 1 x dx, v =2√ x.Then
ln x
√ dx =2 √
x ln x − 2
x
dx
√
x
=2 √ x ln x − 4 √ x + C
(∗∗)
√ 2lnt
lim 2 t ln t = lim
t→0 + t→0 + t −1/2
= H 2/t √
lim = lim −4 t =0
t→0 + t−3/2 t→0 +
− 1 2
47.
1
0
x − 1
1
x√x
√ dx = lim − √ 1 dx = lim
x t→0 + t
x
= lim
t→0 + 2
3 − 2 −
1
t→0 + t
(x 1/2 − x −1/2 ) dx = lim
2
3 t3/2 − 2t 1/2
= − 4 3 − 0=− 4 3
2
t→0 +
3 x3/2 − 2x 1/2 1
t