另x=2cosx dx=2cosx dx最终问题会归结到求解 ∫1/sinx dx而∫1/sinxdx=∫sinx/sin^2xdx=-∫dcosx/(1-cos^2x)=-∫dt/(1-t^2)[令t=cosx]=-1/2∫(1/(t+1)-1/(t-1))dt=-1/2(ln|t+1|-ln|t-1|)+C=-1/2ln|(cosx+1)/(cosx-1)|+C
