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令z=e^(iX)=cosx+isinx,
则由sinx=(z-z^(-1))/(2i),cosx=(z+z^(-1))/2得,
∫&0,2π&[(sinx...
∫[(cosx)^4/(sinx)^3]dx
=∫[(cosx)^3/(sinx)^3]d(sinx)
=-1/2∫(cosx)^3d[1/(sinx)^2...
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