$\displaystyle \int_0^\pi \! log(1-2tcos(x)+t^2) \, \mathrm{d} x$

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- May 9th 2012, 08:41 AMDanielMontesAnother "messy" integral
$\displaystyle \int_0^\pi \! log(1-2tcos(x)+t^2) \, \mathrm{d} x$

- May 12th 2012, 03:14 AMAmerRe: Another "messy" integral
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