$\displaystyle \int_0^\pi \! log(1-2tcos(x)+t^2) \, \mathrm{d} x$
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you can integrate by parts dv = dx , and $\displaystyle \log (1-2t\cos x + t^2 ) = u $ after that you can use the substitution $\displaystyle u = \tan \frac{x}{2} $ to the solve the resulting integral
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