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∫Cos2x/Cosx^2sinx^2Dx不定积分多少

∫cos2x/cosx^2sinx^2dx =2∫cos2x/(sin2x)^2dx =∫1/(sin2x)^2dsin2x =-1/sin2x+C

我对于人性早有预备 还不算太黑

∫cos2x/(sinxcosx)^2dx = ∫√2cos2x/(2sinxcosx)^2dx =√2/2∫cot2xcsc2xd2x =-(√2/2)csc2x+c

∫sin²dx=∫1-cos²dx =∫1-(1+cos2x)/2dx =∫1/2-1/2cos2xdx =(1/2)x-(1/2)∫cos2xdx =(1/2)x-(1/4)sin2x+c(c为任意常数)

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