第5题，用分部积分法求不定积分，哪个大神会啊？
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∫e^(-2x)sin(x/2)dx=-1/2∫sin(x/2)de^(-2x)
=-1/2 e^(-2x)sin(x/2)+1/2 ∫e^(-2x)dsin(x/2)
= -1/2 e^(-2x)sin(x/2)+1/4 ∫e^(-2x)cos(x/2)dx
= -1/2 e^(-2x)sin(x/2)-1/8 ∫cos(x/2)de^(-2x)
= -1/2 e^(-2x)sin(x/2)-1/8 e^(-2x)cos(x/2) +1/8∫e^(-2x)dcos(x/2)
= -1/2 e^(-2x)sin(x/2)-1/8 e^(-2x)cos(x/2) -1/16∫e^(-2x)sin(x/2)dx
∫e^(-2x)sin(x/2)dx=16/17[-1/2 e^(-2x)sin(x/2)-1/8 e^(-2x)cos(x/2)]+C
=-1/2 e^(-2x)sin(x/2)+1/2 ∫e^(-2x)dsin(x/2)
= -1/2 e^(-2x)sin(x/2)+1/4 ∫e^(-2x)cos(x/2)dx
= -1/2 e^(-2x)sin(x/2)-1/8 ∫cos(x/2)de^(-2x)
= -1/2 e^(-2x)sin(x/2)-1/8 e^(-2x)cos(x/2) +1/8∫e^(-2x)dcos(x/2)
= -1/2 e^(-2x)sin(x/2)-1/8 e^(-2x)cos(x/2) -1/16∫e^(-2x)sin(x/2)dx
∫e^(-2x)sin(x/2)dx=16/17[-1/2 e^(-2x)sin(x/2)-1/8 e^(-2x)cos(x/2)]+C
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