∫[(1-sin(2x))/(sin^2(x)-x)]dx

∫[(1-sin(2x))/(sin^2(x)-x)]dx
=∫[(1-2sinxcosx)/(sin^2(x)-x)]dx
=∫1/[sin^2(x)-x]d[x-sin^2(x)]
=-∫1/[x-sin^2(x)]d[x-sin^2(x)]
=C-ln[x-sin^2(x)]

靠,很简单啊

1-sin(2x)可以放到dx里面去
即 原式等于∫1/(sin^2(x)-x)]d(x-sin^2(x))

接下来的该会算了吧 !!
比上面的简单多了