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∫Cos2x/sinxCosxDx

∫cos2x/(sinx+cosx)dx =∫(cos²x-sin²x)/(sinx+cosx)dx =∫(cosx-sinx)dx =sinx+cosx+C

=∫(1-2sin²x-2sinxcosx)/(cosx+sinx)dx =∫1/√2sin(x+π/4)-2sinxdx =-1/√2∫1/(1-cos²(x+π/4))dcos(x+π/4)+2cosx =-(1/2√2)ln(1+cos(x+π/4))/(1-cos(x+π/4))+2cosx+C

你好Rcos2x/(sinx+cosx)dx=∫[(cosx)^2-(sinx)^2]/(sinx+cosx)dx=∫(cosx-sinx)dx=sinx+cosx+c。经济数学团队帮你解答,请及时采纳。谢谢!

即 0.5∫x *sin2x dx 凑微分得到 = -0.25 ∫ x d(cos2x) 使用分部积分法 = -0.25 x *cos2x +0.25∫ cos2x dx = -0.25x *cos2x +0.125 sin2x +C,C为常数

∫(cos2x/[(cosx)^2.(sinx)^2] )dx =4∫[cos2x/(sin2x)^2]dx =2∫d(sin2x)/(sin2x)^2 =-2/(sin2x) + C

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

欢迎采纳,不要点错答案哦╮(╯◇╰)╭ 欢迎采纳,不要点错答案哦╮(╯◇╰)╭

(cos2x)/(cosx-sinx)=(cos²x-sin²x)/(cosx-sinx) =cosx+sinx =根号2(根号2/2cosx+根号2/2sinx) 根号2sin[x+(π/4)] 这样就可以画函数图像了。 采纳呀

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