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求x^2*arccosx的不定积分
人气:129 ℃ 时间:2020-10-02 03:41:55
解答
分部积分:
∫x^2*arccosxdx=1/3∫arccosxdx^3=1/3*x^3arccosx-1/3∫x^3darccosx
=1/3*x^3arccosx+1/3∫x^3/√(1-x^2)dx=1/3*x^3arccosx+1/6∫x^2/√(1-x^2)dx^2
=1/3*x^3arccosx+1/6∫{(x^2-1)/√(1-x^2)+1/√(1-x^2)}dx^2
=1/3*x^3arccosx+1/9(1-x^2)^(3/2)-1/3(1-x^2)^(1/2)+C
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