求解这道不定积分题
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利用分部积分与凑微分，过程与结果如图所示
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I = ∫x^2arccosxdx = (1/3)∫arccosxd(x^3)
= (1/3)x^3arccosx - (1/3)∫[-x^3/√(1-x^2)]dx
后者令 x = sint， 则
I2 = - (1/3)∫[-x^3/√(1-x^2)]dx = (1/3)∫[(sint)^3/cost]costdt
= (1/3)∫(sint)^3dt = (-1/3)∫[1-(cost)^2]dcost
= (-1/3)[cost-(1/3)(cost)^3]
= - (1/3)√(1-x^2) + (1/9)(1-x^2)^(3/2)
得 I = (1/3)x^3arccosx - (1/3)√(1-x^2) + (1/9)(1-x^2)^(3/2) + C
= (1/3)x^3arccosx - (1/3)∫[-x^3/√(1-x^2)]dx
后者令 x = sint， 则
I2 = - (1/3)∫[-x^3/√(1-x^2)]dx = (1/3)∫[(sint)^3/cost]costdt
= (1/3)∫(sint)^3dt = (-1/3)∫[1-(cost)^2]dcost
= (-1/3)[cost-(1/3)(cost)^3]
= - (1/3)√(1-x^2) + (1/9)(1-x^2)^(3/2)
得 I = (1/3)x^3arccosx - (1/3)√(1-x^2) + (1/9)(1-x^2)^(3/2) + C
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