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求积分∫[arcsin√x/√(1-x)]dx
人气:333 ℃ 时间:2020-05-13 05:08:22
解答
令√x=u,则:x=u^2,dx=2udu.
∴∫[arcsin√x/√(1-x)]dx
=∫[arcsinu/√(1-u^2)]2udu
=-2∫arcsinu{-2u/[2√(1-u^2)]}du
=-2∫arcsinud[√(1-u^2)]
=-2[√(1-u^2)]arcsinu+2∫[√(1-u^2)]d(arcsinu)
=-2[√(1-u^2)]arcsinu+2∫[√(1-u^2)][1/√(1-u^2)]du
=-2[√(1-u^2)]arcsinu+2∫du
=2u-2[√(1-u^2)]arcsinu+C
=2√x-2[√(1-x)]arcsin√x+C
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