积分题求解答?
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let
x=(3/2)sinu
dx=(3/2)cosu du
∫ (1-x)/√(9-4x^2) dx
=∫ dx/√(9-4x^2) -∫ x/√(9-4x^2) dx
=∫ dx/√(9-4x^2) +(1/8)∫ d(9-4x^2)/√(9-4x^2)
=∫ dx/√(9-4x^2) +(1/4)√(9-4x^2)
=(1/2)∫ du +(1/4)√(9-4x^2)
=(1/2)u+(1/4)√(9-4x^2)
=(1/2)arcsin(2x/3) +(1/4)√(9-4x^2) +C
x=(3/2)sinu
dx=(3/2)cosu du
∫ (1-x)/√(9-4x^2) dx
=∫ dx/√(9-4x^2) -∫ x/√(9-4x^2) dx
=∫ dx/√(9-4x^2) +(1/8)∫ d(9-4x^2)/√(9-4x^2)
=∫ dx/√(9-4x^2) +(1/4)√(9-4x^2)
=(1/2)∫ du +(1/4)√(9-4x^2)
=(1/2)u+(1/4)√(9-4x^2)
=(1/2)arcsin(2x/3) +(1/4)√(9-4x^2) +C
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谢谢
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