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∫Cos2x/sinxCosxDx

=∫(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

cos2x=cosx^2-sinx^2=(cosx-sinx)(cosx+sinx) 所以上式化简为=cosx+sinx 所以原函数为 sinx-cosx

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

令u=3+sinxcosx, 则u'=cos²x-sin²x=cos2x ∴du=cos2x·dx 原式=∫du/u=lnu+C =ln(3+sinxcosx)+C

∫(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

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