仲宇铭
2021-04-21 08:27
f(x)=xcosx/[1+√(1-x^2)]f(-x)=-f(x)∫(-1->1)(2x^2+xcosx)/[1+√(1-x^2)]dx=∫(-1->1)2x^2/[1+√(1-x^2)]dx+∫(-开发者_JS百科1->1)xcosx/[1+√(1-x^2)]dx=∫(-1->1)2x^2/[1+√(1-x
仲宇铭
2021-04-21 08:27
f(x)=xcosx/[1+√(1-x^2)]f(-x)=-f(x)∫(-1->1)(2x^2+xcosx)/[1+√(1-x^2)]dx=∫(-1->1)2x^2/[1+√(1-x^2)]dx+∫(-
开发者_JS百科1->1)xcosx/[1+√(1-x^2)]dx=∫(-1->1)2x^2/[1+√(1-x^2)]dx+0=4∫(0->1)x^2/[1+√(1-x^2)]dx=2(2-π/2)=4-πletx=sinudx=cosudux=0,u=0x=1,u=π/2∫(0->1)x^2/[1+√(1-x^2)]dx=∫(0->1)[1-√(1-x^2)]dx=∫(0->π/2)(1-cosu)(cosudu)=(1/2)∫(0->π/2)(2cosu-1-cos2u)du=(1/2)[2sinu-u-(1/2)sin2u]|(0->π/2)=(1/2)(2-π/2)
张楠
2021-04-21 08:32
开发者_JAVA技巧
唐杰波
2021-04-21 08:36
∫(tanx)^2secxdx=∫[(secx)^2-1]secxdx=∫(secx)^3dx-∫secxdx=∫secxdtanx-∫secxdx=secxtanx-∫(tanx)^2secxdx-ln|secx+tanx|则2∫(tanx)^2secxdx=secxtanx-ln|secx+tanx|I=4π∫<0,√5>(tanx)^2sec
开发者_如何学Pythonxdx=2π[secxtanx-ln|secx+tanx|]<0,√5>=2π[sec√5tan√5-ln|sec√5+tan√5|]
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