题目内容：

求标准正弦函数的倒数的不定积分

优质解答

∫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