Question

∫sin2x * x dx 怎样求这个不定积分？？

Answer 1

2011-2-13
22:29
满意回答
解：∵∫x³cos(4x)dx=x³sin(4x)/4-3/4∫
x²sin(4x)dx
(应用分部积分法)
=x³sin(4x)/4+3x²cos(4x)/16-3/8∫xcos(4x)dx
(同上)
=x³sin(4x)/4+3x²cos(4x)/16-3xsin(4x)/32+3/32∫
sin(4x)dx
(同上)
=x³sin(4x)/4+3x²cos(4x)/16-3xsin(4x)/32-3cos(4x)/128+C1
同理可得∫x³cos(2x)dx=x³sin(2x)/2+3x²cos(2x)/4-3xsin(2x)/4-3cos(2x)/8+C2
∴∫
x³sin^4(x)dx=∫x³[3/8-cos(2x)/2+cos(4x)/8]dx
(应用半角公式)
=3/8∫x³dx-1/2∫x³cos(2x)dx+1/8∫x³cos(4x)dx
=3x^4/32-1/2[x³sin(2x)/2+3x²cos(2x)/4-3xsin(2x)/4-3cos(2x)/8]+1/8[x³sin(4x)/4+3x²cos(4x)/16-3xsin(4x)/32-3cos(4x)/128]+C
(C是积分常数)
=3x^4/32+[sin(4x)/32-sin(2x)/4]x³+3[cos(4x)/128-cos(2x)/8]x²+3[sin(2x)/8-sin(4x)/256]x+3[cos(2x)/16-cos(4x)/1024]+C。

Answer 2

∫
x
sin2x
dx
=
∫
x
d(-1/2
cos2x)
=
(-1/2)∫
x
d(cos2x)
=
(-1/2)x
cos2x
+
1/2
∫
cos2x
dx
=
(-1/2)x
cos2x
+
1/2
1/2
∫
cos2x
d(2x)
=
(-1/2)x
cos2x
+
(1/4)sin2x
+
C