高等数学不定积分？

Question

如题所述

Answer 1

Answer 2

let
x= (sinu)^2
dx=2sinu.cosu du
x=3/4, u=π/3
x=1, u=π/2
∫(3/4->1) x/[√(1-x) -1] dx
=∫(π/3->π/2) [ (sinu)^2/(cosu -1) ] .[2sinu.cosu du]
=2∫(π/3->π/2) (sinu)^3. cosu/(cosu -1) du
=2∫(π/3->π/2) (sinu)^3. cosu( cosu +1) /[-(sinu)^2] du
=-2∫(π/3->π/2) sinu. cosu( cosu +1) du
=2∫(π/3->π/2) cosu( cosu +1) dcosu
=2 [ (1/3)(cosu)^3 + (1/2) (cosu)^2 ]|(π/3->π/2)
=-2[ (1/3)(√3/2)^3 +(1/2)(√3/2)^2 ]
=-2[ (1/8)√3 +(3/8)√3 ]
=-√3

Answer 3

令u=√(1-x),则x=1-u²,dx=-2u·du 积分限：u(½,0)
∫dx/[√(1-x)-1]
=∫-2udu/(u-1)
=-2∫[u/(u-1)]du
=-2∫[(u-1+1)/(u-1)]du
=-2∫[1+1/(u-1)]du
=-2u-2ln|u-1|+C
定积分=2u+2ln|u-1||(0,½)
=1+2ln½
=1-2ln2

Answer 4

详细过程如图rt所示

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