∫cos2x/(cosx-sinx)*dx的值
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∫cos2x/(cosx-sinx)*dx的值∫cos2x/(cosx-sinx)*dx的值∫cos2x/(cosx-sinx)*dx的值∫cos2x/(cosx-sinx)*dx=∫(cosx^2-
∫cos2x/(cosx-sinx)*dx的值
∫cos2x/(cosx-sinx)*dx的值
∫cos2x/(cosx-sinx)*dx的值
∫cos2x/(cosx-sinx)*dx
=∫(cosx^2-sinx^2)/(cosx-sinx)*dx
=∫(cosx+sinx)dx
=sinx-cosx+C
∫cos2x/(sinx+cosx)dx
化简(cos2x/sinx+cosx)-(cos2x/sinx-cosx)
求不定积分∫(cos2x)/(sinx+cosx)dx
∫cos2x/(cosx-sinx)*dx的值
sinx×cos2x-sin2x×cosx
化简:sinx×cosx×cos2x
为什么cos2x=cosx*cosx-sinx*sinx
求不定积分∫cos2x/[(sinx)^2(cosx)^2] dx
求不定积分:∫(cos2x)/(sinX)^2.cosx^2
∫cos2x/(sinx^2*cosx^2)dx求积分
cos2x/sinx+cosx + 2sinx=
y=cos2x+sinx(sinx+cosx) 化简
f(sinx)=3-cos2x,则f(cosx)=A 3-cos2x B 3-sin2xC 3+cos2x D 3+sin2x
如果f(sinx)=cos2x,那么f(cosx)等于 A.-sin2x B.sin2x C.-cos2x D.cos2x
怎么证明cos2x/1-sin2x = cosx+sinx/cosx-sinx
为什么cos2X=(cosX)^2-(sinX)^2
cos2x/(cosx-sinx)的原函数是什么
cos2x+cosx=0,则sin2x+sinx