求算∫上限根号下(3)/2下限0arc cos x dx！
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令arccosx=u，则x=cosu
∫[0：√3/2]arccosxdx
=∫[π/2：π/6]ud(cosu)
=ucosu|[[π/2：π/6]-∫[π/2：π/6]cosudu
=(π/6)cos(π/6)-(π/2)cos(π/2)-sinu|[π/2：π/6]
=(π/6)(√3/2)-0-[sin(π/6)-sin(π/2)]
=√3π/12 - 1/2 +1
=(√3π+6)/12
∫[0：√3/2]arccosxdx
=∫[π/2：π/6]ud(cosu)
=ucosu|[[π/2：π/6]-∫[π/2：π/6]cosudu
=(π/6)cos(π/6)-(π/2)cos(π/2)-sinu|[π/2：π/6]
=(π/6)(√3/2)-0-[sin(π/6)-sin(π/2)]
=√3π/12 - 1/2 +1
=(√3π+6)/12
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