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∫[e^(-2x)]sin(x/2)dx=-1/2∫sin(x/2)de^(-2x)
=-1/2e^(-2x)sin(x/2)+1/4∫[e^(-2x)]cos(x/2)dx
=-1/2e^(-2x)sin(x/2)-1/8∫cos(x/2)de^(-2x)
=-1/2e^(-2x)sin(x/2)-1/8e^(-2x)cos(x/2)dx-1/16∫e^(-2x)sin(x/2)dx
故∫e^(-2x)sin(x/2)dx
=16/17[-1/2e^(-2x)sin(x/2)-1/8e^(-2x)cos(x/2)dx]+C
=-1/2e^(-2x)sin(x/2)+1/4∫[e^(-2x)]cos(x/2)dx
=-1/2e^(-2x)sin(x/2)-1/8∫cos(x/2)de^(-2x)
=-1/2e^(-2x)sin(x/2)-1/8e^(-2x)cos(x/2)dx-1/16∫e^(-2x)sin(x/2)dx
故∫e^(-2x)sin(x/2)dx
=16/17[-1/2e^(-2x)sin(x/2)-1/8e^(-2x)cos(x/2)dx]+C
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