Here's an extrapolatory approach based on Richardson extrapolation of the integrals over each "period." (`Richardson[]` code from [Anton Antonov's answer here](http://mathematica.stackexchange.com/a/95166/4999)): Clear[Richardson] Richardson[A_, n_, N_] := Total@Table[(A[n + k]*(n + k)^N*If[OddQ[k + N], -1, 1])/(k! (N - k)!), {k, 0, N}] (* the integral from 2 n Pi to 2 (n+1) Pi *) Clear[aa]; aa[n_Integer /; n >= 0] := aa[n] = NIntegrate[ArcCot[x]*Sin[x]/(5/4 + Cos[x]), {x, 2 n Pi, 2 (n + 1) Pi}, WorkingPrecision -> 50]; We get around machine precision with 100 terms plus 6 extrapolation terms: sf[n_] := Sum[aa[i], {i, 0, n}]; (* partial sum = integral over {0, 2 (n+1) Pi} *) Richardson[sf, 100, 6] % - Pi*Log[3/2/(1 + 1/2*Exp[-1])] (* 0.743355751361227474780881457479018423835 -2.97535713733265290609713*10^-16 *) Without extrapolation: sf[100] % - Pi*Log[3/2/(1 + 1/2*Exp[-1])] (* 0.74207790286093203962531860751097243182349263171074 -0.00127784850029573269127658323333660172377741318898 *) ---- *Update* More extrapolatory terms yields an approximation accurate to machine precision with many fewer terms overall: Richardson[sf, 1, 16] % - Pi*Log[3/2/(1 + 1/2*Exp[-1])] (* 0.74335575136122767678505611095516483469029 -9.553153907978914419885698*10^-17 *)