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数学
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1/sinx的原函数是多少
人气:231 ℃ 时间:2020-07-12 16:45:36
解答
∫1/sinx* dx=∫ sinx/sin^2 x* dx=-∫d(cosx)/(1-cos^2 x)=-0.5∫d(cosx)[1/(1-cosx)+1/(1+cosx)]=-0.5ln|(1+cosx)/(1-cosx)|+c=-ln|(1+cosx)/sinx|+c=-ln|cscx+cotx|+c
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