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1/(1+sinx)的原函数
人气:464 ℃ 时间:2020-07-25 12:29:18
解答
∫1/﹙1﹢sinx﹚dx=∫1/﹙1﹢cos(90°-x﹚dx﹙令90°-x=t,则dx=-dt﹚=-∫1/﹙1+cost﹚dt=-tan﹙t/2﹚+C=-tan﹙﹙90°-x﹚/2﹚=-tan﹙45°﹣x/2﹚﹢C(C为任意实数﹚用积分法就是这么求解的,但是恩,就...
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