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北大版高数第十一章习题解答


dx cos x ,
0
t=tan x 2dx = 1−a2 cos2 x
π 2
1 1+a cos x 1−a cos x · a cos x 1+a cos x 1 π 1−a cos x cos x 2 ] × (−1, 1) +∞ 2dt π √ . g (0) = 0, 1−a2 +t2 = 0 1−a2
√ x ′ x+3 √ x x+3 √ 3−x 2 x(x+3)2
.
(n+1)π nπ √ n−1 2 kπ (k+1)π +3 . k=0
√ x| cos x| dx x+3
º ¦
=
√ 2 nπ (n+1)π +3 .
þ
A 0
¨xû
cos xdx| ≤ 2,
î, ø Æ µ §Çß, ¦
√ x x+3
Ǒ lim
.
x→0
x+3
1 1 √ /√ 4 x +x 3 4 x
=
þ ,¦ þ
.
1 x
Ǒ lim
x→1
1 √ / √1 3 1−x4 3 1−x
= lim
Ù0,
(7)
(6)
¦ ¦
+∞ sin x x lim sin = x dx. x 0 →0 x A | 0 sin xdx| ≤ 2, +∞ α − x 2 x e dx (α > 0). 1
0 −t2 x2 0
+∞ −t2 x2 e dx, 0
(0 < t0 < t < +∞).
2 −t2 0x
.
+∞ −t2 x2 e 0 dx 0
§Çß,
(3)
+∞ −αx e sin xdx, 0
(i) (0 < α0 ≤ α ≤ +∞), (ii) (0 < α ≤ +∞). 3
µ.
¨x ∈ [0, +∞) 0 < α ≤ α ≤ +∞î, |e sin x| ≤ e , þ , M -§Çß, ¦ ¿ Ñ[α , +∞) èçþ . (ii) ¨α > 0î, | = . e sin xdx = à M Ó , α = A = 2kπ M < A < 2M î, | e sin xdx| > . ¦ ¿ Ñ(0, +∞) èçþ .
Ô ½ëè Ëõ ËÃ11.1 1. §Ç ¯¾þ ; þ ,
(1) (2) (3) (4) (5) (6) (8) (9) (10) (11) (12) 3. (1) (2) (3)
+∞ xe−x dx = 0 +∞ dx (x+1)(x+2) 0
.
1 √ σ 2π
u=x−x−1 +∞ du +∞ 1+x2 ∞ 1 π √ √ = arctan u|+ . −∞ = 1+x4 dx 0 −∞ u2 +2 = 2 2 +∞ ∞ x sin xdx = (−x cos x + sin x)|+ , . 0 0 √ t = x − 1 +∞ + ∞ + dx 2dt π √ √ √∞ = 4 3 t2 +1 = 2 arctan t| 3 = 3 . x x −1 +∞ ∞ dx dx = arctan(x + 1)|+ −∞ = π . −∞ x2 +2x+2 +∞ − x ∞ 1 −x e cos xdx = 2 e (sin x − cos x)|+ = 1 0 2. 0 π x =sin t 1 2 √ dx = dt = π . −π −1 1−x2 2
þ ,¦
2 dx √ 0 4x 1 √ 3 , 4
.
x→+∞ 2x+ √ 3
lim
+∞ dx x 2
þ . Ù ,¦
1 dx √ 0 3 1−x
þ
2 dx √ √ . 0 3 x+3 4 x+x3
Ǒ lim
(4) , (5)
¦
1 √ dx . 0 3 1−x4 1 sin x 0 x3/2 dx.
þ
π 2
= 1,
dx x2
0
,
.
4.
§
(1)
µ. ¨nǑêÃ
nπ 0 √
+∞ 0
√ x cos x x+3 dx.
¯
ð
þ î,
ð
þ
. ≥
√ (n+1)π nπ | cos x| (n+1)π +3 dx nπ √ A x| cos x| dx , x+3 0
x| cos x| dx x+3

, Ǒ( ) = +∞î, → 0, Õ Ǒ| Ò ,¦ þ .
2 2 2 ′
π 2
0
ln(a2 sin2 x + cos2 x)dx, a = 0.
′ t=tan x
0
π a+1

+1 . I (a) = π ln |a|2
+∞ t2 dt 2a (1+a2 t2 )(1+t2 ) a2 −1 0 π ′ a+1 a+1 2
2a sin2 xdx a2 sin2 x+cos2 x 1 a
sin x 1 3/2 / x1/2 x→0 x
= 1,
µ
1,
(8)
(9)
1 1 ln x ln x x lim 1 = lim − = −1, 1 . − x 0 1−x dx. x →1 x→1 1 1 ln xdx = (x ln x − x)|1 , . 0 = −1 0 π dx π 1 2 . 0 2 . lim 2 / 12 2 0 sin2 x cos2 x x→0 sin x cos x x
+∞ dx 3 2 1 x2 x3 b a+ε b a b a b a
π 2
(2)
+∞ cos(3x+2) √ √ dx. 1 x3 +1 3 x2 +1 cos(3x+2) √ | √x | ≤ 31 2 3 +1 3 x2 +1 x2 x3
k→0 0
dϕ 1−k2 sin2 ϕ
π 2
0 k→0 2
þ
Ǒ
þ ,¦ þ . Ǒ¨x → +∞î, ø §Çß, ¦ þ .
1 dx 0 x1/2 xα e−x
x2 e− 2 2
x→1
1 √ 3 1+x+x2 +x3
=
Æ þ Ù
,
.
x→+∞
lim
= 0,
+∞ − x 2 e 2 1
dx
Ù
ð
þ ¦ Ǒ
1
ð
þ
Ǒ lim
= 1,
ln x / ln x x→0 1−x
1 1−k2 sin2 ϕ
π 2
0
1 1−k2 sin2 ϕ π 2
π 2
1 1 2 2
π 2
2
k→1−0 0
2
2
π 2
π 2
0 k→1−0
2
2
π 2
0
1+α dx 1 1+x2 +α2 1+x2 +α2 α→0 0 1+α 1 dx 1+α π dx 4 1+x2 +α2 1 0 1+x2 1 π π 4 4 1 ex sin xy y +1 y →0 0 1 ex sin xy 0 y →0 y +1 a+ky a−ky cos y sin y
+∞ −α0 x e dx 0
(i)
0
−αx
−α0 x
2 = ln ln x|0 , 1 2
1
Ù
.
.
§
0
Ù
.
+∞ dx √ . 1 x2 + 3 x4 +3 +∞ √dx . 2 2x+ 3 x2 +1+6
¯
1 = − ln x |0 =
1 ln 2 .
Ǒ
x2 +
Ǒ
1 √ 3 4 x +3
<
1 x2 , 1 √ /1 3 2 x +1+6 x
+∞ dx x2 1 1 = 2 , 1 3,
x+1 +∞ = ln x = ln 2. +2 |0 √ 2 ( x − a ) +∞ − x−a= 2σt 1 +∞ −t2 √ 2σ2 dx e dt = −∞ e π −∞
(−xe
−x
−e
−x
∞ )|+ 0
= 1.
= 1 (σ > 0).
Ù
1 2
0
1 2
dx x ln x dx x ln2 x
+∞ 0
=
ËÃ11.3
1. (1)
µ. |
(2)
+∞ sin tx 1+x2 dx, 0 sin tx 1+x2 |
¯
¿ ù Ñ
èçþ þ
, ,
. M-

1 1+x2 ,
(−∞ < t < +∞).
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