∫[(1-sin(2x))/(sin^2(x)-x)]dx 怎么解

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∫[(1-sin(2x))/(sin^2(x)-x)]dx 怎么解
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∫[(1-sin(2x))/(sin^2(x)-x)]dx 怎么解
∫[(1-sin(2x))/(sin^2(x)-x)]dx 怎么解

∫[(1-sin(2x))/(sin^2(x)-x)]dx 怎么解
∫[(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

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