Question

用分步积分法求积分 e ^(-2x)sin(x/2)dx

Answer 1

M=∫e^(-2x)sin(x/2)dx

=(-1/2)∫sin(x/2)d[e^(-2x)]

=(-1/2)sin(x/2)e^(-2x)-(-1/2)∫e^(-2x)d[sin(x/2)]

=(-1/2)sin(x/2)e^(-2x)+(1/4)∫e^(-2x)cos(x/2)dx

=(-1/2)sin(x/2)e^(-2x)+(-1/8)∫cos(x/2)d[e^(-2x)]

=(-1/2)sin(x/2)e^(-2x)+(-1/8)cos(x/2)e^(-2x)+(1/8)∫e^(-2x)dcos(x/2)

=(-1/2)sin(x/2)e^(-2x)+(-1/8)cos(x/2)e^(-2x)+(-1/16)∫e^(-2x)sin(x/2)dx

=(-1/2)sin(x/2)e^(-2x)+(-1/8)cos(x/2)e^(-2x)+(-1/16)M

从而M=(16/17)*[(-1/2)sin(x/2)e^(-2x)+(-1/8)cos(x/2)e^(-2x)]+C

=(-2/17)e^(-2x)[ 4sin(x/2) + cos(x/2)] + C