高等数学第四章习题详细解答答案
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习 题 4-1
1.求下列不定积分: (1)解:C x x x x x
x x x x
+-=-=-⎰⎰
-
2
5232
122d )5(d )51(
(2)解:⎰+x x x
d )32(2
C x
x x ++⋅+=
3
ln 296ln 622ln 24 (3)略. (4) 解:⎰
⎰⎰
-+-=+-x x x x x x x d )1(csc d 1
1d )cot 1
1(
2222
=C x x x +--cot arcsin
(5) 解:⎰x x
x
d 2103 C x x x
x
x x +=
==⎰⎰80
ln 80d 80d 810 (6) 解:x x d 2sin
2
⎰
=C x x x x ++=-=⎰sin 2
1
21d )cos 1(21 (7)⎰
+x x x x d sin cos 2cos C x x x x x x x x x x +--=-=+-=⎰⎰cos sin d )sin (cos d sin cos sin cos 22 (8) 解:⎰x x x x d sin cos 2cos 22⎰⎰-=-=x x
x x x x x x d )cos 1
sin 1(d sin cos sin cos 222222 C x x +--=tan cot
(9) 解: ⎰⎰
⎰-=-x x x x x x x x x d tan sec d sec d )tan (sec sec 2
=C x x +-sec tan
(10) 解:},,1max{)(x x f =设⎪⎩
⎪
⎨⎧>≤≤--<-=1,11,11,)(x x x x x x f 则.
上连续在),()(+∞-∞x f ,
)(x F 则必存在原函数,⎪⎪⎩⎪
⎪⎨⎧>+≤≤-+-<+-=1,21
11,1,21)(32212
x C x x C x x C x x F 须处处连续,有又)(x F
)21(lim )(lim 12121C x C x x x +-=+-+-→-→ ,,2
1
112C C +-=+-即 )(lim )21(lim 21321C x C x x x +=+-+→→ ,,12
123C C +=+即
,1C C =联立并令.1,2
1
32C C C C +==+可得
.1,12111,211,21},1max{2
2
⎪⎪⎪⎩
⎪
⎪
⎪⎨⎧>++≤≤-++-<+-=⎰x C x x C x x C x dx x 故
2. 解:设所求曲线方程为)(x f y =,其上任一点),(y x 处切线的斜率为
3d d x x
y
=,从而 ⎰+=
=C x x x y 4
34
1d 由0)0(=y ,得0=C ,因此所求曲线方程为 44
1x y =. 3.解:因为 x x x cos sin sin 212
='⎪⎭⎫ ⎝⎛,x x x sin cos cos 212='
⎪⎭
⎫ ⎝⎛-
x x x x cos sin 2sin 212cos 41=='
⎪⎭
⎫
⎝⎛-
所以
x 2sin 21、 x 2cos 21-、 x 2cos 4
1
-都是x x cos sin 的原函数.
习 题 4-2
1.填空. (1)
2
1x
x d = d (x 1- + C) (2)x x d 1 = d (x ln + C) (3)x e x d = d (x
e + C) (4) x x d sec 2 = d (x tan + C)
(5)x x d sin = d (x cos -+ C) (6) x x d cos = d (x sin + C) (7)
x x d 112
- = d (x arcsin + C) (8)
x x x d 12
- = d (21x -+ C)
(9)x x x d sec tan = d (x sec + C) (10)
x x d 1
1
2
+ = d (x arctan + C) (11)x x
x d )1(1
+ = d (2x arctan + C) (12) x x d = d (22
x + C)
2.求下列不定积分: (1) 解:
⎰
+x x x d 4
2
)4d()4(21
)24d(4
1
221
222++=++=⎰⎰
-x x x x
=C x C x ++=++4)4(2212
(2) 解:x x x d ln 4⎰C x x x +==⎰5ln )d(ln ln 54
(3) 解:⎰x x
e x
d 21C
e x e x x +-=-=⎰1
1)1
d(
(4) 解:⎰++x e e e x
x x d )22(32C e e e e e e x x
x x x x +++=
++=⎰
22
131)d()22(4332 (5) 解:
⎰
-2
94d x x C x x
x x x +=-=
-=⎰
⎰
23arcsin 31)23(1)
23d(
31)2
3(
12d 2
2
(6) 解:
x x x x
d )ln (ln 12⎰+C x
x x x x x +-==⎰ln 1)ln d()ln (12
(7) 解:
x x x x d ln ln ln 1⎰C x x x x x x +===⎰⎰ln ln ln )ln d(ln ln ln 1
)d(ln ln ln ln 1
(8) 解:⎰-+x e e x x
d 1C
e e e x x x +=+=⎰arctan )d(1
1
2 (9) 解:⎰x x d cos 4
x x
x x x d 4
2cos 2cos 21d )22cos 1(
22⎰⎰++=+= x x
x d )42cos 22cos 41(2++
=⎰ ++=42sin x x 1cos4d 8
x x +ò 3sin 284
x x =++
sin 432x C +
(10) 解:
x x
x x
x d cos sin cos sin 3
⎰
-+C x x x x x
x +-=--=⎰3
2
3
)cos (sin 2)cos d(sin cos sin 1
(11) 解:⎰x x d cos 3
⎰=x x x d cos cos 2
)d(sin sin 12
⎰-=x x C x
x +-=3
sin sin 3 (12) 解:
x x
x d 1102
arccos ⎰
--=-=⎰)d(arccos 10
arccos x x
C x
+10
ln 10arccos