∫0^1/2[arcsinx / (根号下1+x^2)]*dx 2.∫-1^1[xe^x2/2]*dx 3.∫0^a/2

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  • 1

    ∫(0^1/2)[arcsinx /√(1+x^2)]*dx

    这个没想好

    若∫(0^1/2)[arcsinx /√(1-x^2)]*dx可以做

    ∫(0^1/2)[arcsinx /√(1-x^2)]*dx

    =ʃ(0~1/2)arcsinxd(arcsinx)

    =1/2(arcsinx)²|(0~1/2

    =π²/72

    2.∫-1^1[1/2xe^x2]*dx

    =∫-1^1[e^(x^2)dx^2

    =e^(x^2))|(-1~1)

    =0

    3.∫0^a/2[xdx / √(a^2-x^2)] ( a>0)

    =1/2∫0^a/2[dx² / √(a^2-x^2)]

    设√(a²-x²)=t,

    那么a²-x²=t²

    ∴x²=a²-t²

    ∴dx²=-2tdt

    ∴ʃdx²/√(a²-x²)=ʃ-2dt=-2t=-2√(a²-x²)

    ∫0^a/2[xdx / √(a^2-x^2)] ( a>0)

    =1/2∫0^a/2[dx² / √(a^2-x^2)]

    =-√(a²-x²)|(0~a/2)

    =-√3/2a+a

    =(2-√3)a/2