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請問一題,∫㏑(sinx)dx

也可以直接用極限來做:(以 99台中一中教甄題 為例 )
其中需用到公式:
sin2k+1sin22k+1sin32k+1sin2k2k+1=22k2k+1



0lnsinxdx=limnn(lnsinn+lnsinn2++lnsinn(n1)) 


=limnn1ln(sinnsinn2sinn(n1))

=limk12k+1ln(sin2k+1sin22k+1sin2k2k+1)


=limk12k+1ln(22k2k+1)


=limk2k+1ln(2k+1)ln22k


=limk2k+1ln(2k+1)2kln2

=limk2+1kkln(2k+1)2ln2


=202ln2


=ln2

[ 本帖最後由 Joy091 於 2011-2-14 11:39 AM 編輯 ]

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