解:设tan(x/2)=t,则dx=2dt/(1+t²),cosx=(1-t²)/(1+t²哪卜竖)
于是,F(x)=∫[(acosx-b)/(a²+b²-2abcosx)]dx
=∫弊腔{[(a-b)-(a+b)t²]/[(a-b)²+(a+b)²t²]}[2/李大(1+t²)]dt
=(1/b)∫{(a²-b²)/[(a-b)²+(a+b)²t²]-1/(1+t²)}dt
=(1/b)∫{[(a+b)/(a-b)]/[1+((a+b)t/(a-b))²]-1/(1+t²)}dt
=(1/b){arctan[(a+b)t/(a-b)]-arctant}
=(1/b)arctan{2bt/[(a-b)+(a+b)t²]}
=(1/b)arctan{2btan(x/2)/[(a-b)+(a+b)tan²(x/2)]}.
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