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求不定积分[e的(-2x)次幂]*sin(x/2)
人气:401 ℃ 时间:2020-05-05 09:11:50
解答
I=∫e^(-2x)sin(x/2)dx
=2∫e^(-2x)dcos(x/2)
=2e^(-2x)cos(x/2)+4∫e^(-2x)cos(x/2)dx
=省略+8∫e^(-2x)dsin(x/2)
=省略+8e^(-2x)sin(x/2)+16∫e^(-2x)sin(x/2)dx
-17I=.
I= -(2/17)[4sin(x/2)+cos(x/2)]e^(-2x)
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