∫x^3 sin^2xdx
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x^3sin^2x
dx
=
∫
x^3
*
(1
-
cos(2x))/2
dx
=
(1/8)x^4
-
(1/2)K
where
K
=
∫
x^3cos(2x)
dx
令f
=
x^3
and
g
=
cos(2x),以下分别对f求导,对g求积分
f
=
x^3
g
=
cos2x
f'
=
3x^2
g(1)
=
(1/2)sin(2x)
f''
=
6x
g(2)
=
(-
1/4)cos(2x)
f'''
=
6
g(3)
=
(-
1/8)sin(2x)
f''''
=
0
g(4)
=
(1/16)cos(2x)、这个是分部积分法的速解法
于是∫
x^3cos(2x)
dx
=
[f
*
g(1)]
-
[f'
*
g(2)]
+
[f''
*
g(3)]
-
[f'''
*
g(4)]
=
[x^3
*
(1/2)sin(2x)]
-
[3x^2
*
(-
1/4)cos(2x)]
+
[6x
*
(-
1/8)sin(2x)]
-
[6
*
(1/16)cos(2x)]
=
(1/2)x^3sin(2x)
+
(3/4)x^2cos(2x)
-
(3/4)xsin(2x)
-
(3/8)cos(2x)
原式
=
(1/8)x^4
-
(1/4)x^3sin(2x)
-
(3/8)x^2cos(2x)
+
(3/8)xsin(2x)
+
(3/16)cos(2x)
+
C
印谈固肠:[答案] ∫ x^3sin^2x dx = ∫ x^3 * (1 - cos(2x))/2 dx = (1/8)x^4 - (1/2)K where K = ∫ x^3cos(2x) dx 令f = x^3 and g = cos(2x),以下分别对f求导,对g求积分 f = x^3 g = cos2x f' = 3x^2 g(1) = (1/2)sin(2x) f'' = 6x g(2) = (- 1/4)cos(2x) f''' = 6 g(3) = (- 1/8)sin(2x) f'''' = 0 g(4) ...
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