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计算不定积分∫ Cos2x/(Cosx%sinx)Dx

=∫(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

这个分子就是sin2x,然后就是相当于求cot2x的积分

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!

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

解:∫(0,π/2)cos^5xsin2xdx=∫(0,π/2)cos^5x2sinxcosxdx=2∫(0,π/2)cos^6xsinxdx=-2∫(0,π/2)cos^6xd(cosx)=-2×1/7cos^7x(0,π/2)=-2/7[cos^7(π/2)-cos^70)=-2/7(0-1)=2/7

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

∫cos2x/[(sinx)^2*(cosx)^2]dx =∫[(cosx)^2-(sinx)^2]/[(sinx)^2*(cosx)^2]dx =∫[1/(sinx)^2-1/(cosx)^2]dx =-cotx-tanx+c

∫ cos2x / (sin²x * cos²x) dx = ∫ cos2x / (1/2 * sin2x)² dx = 4∫ cos2x / (sin²2x) dx = 4∫ csc2x * cot2x dx = -2∫ csc2x * cot2x d(2x) = -2csc2x + C = -2/(sin2x) + C = -secx*cscx + C

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