# int(from (0) to (pi) Ln[(b-cosx) / (a-cosx)]dx a,b>0

Jitender Singh IIT Delhi
8 years ago
Ans:
$I = \int_{0}^{\pi}ln(\frac{b-cosx}{a-cosx})dx$…..............(1)
$I = \int_{0}^{\pi}ln(\frac{b-cos(\pi -x)}{a-cos(\pi -x)})dx$
$I = \int_{0}^{\pi}ln(\frac{b+cos(x)}{a+cos(x)})dx$.............(2)
(1) + (2)
$2I = \int_{0}^{\pi}ln(\frac{b^{2}-cos^{2}(x)}{a^{2}-cos^{2}(x)})dx$
Integration by Parts
$2I = (x.ln(\frac{b^{2}-cos^{2}(x)}{a^{2}-cos^{2}(x)}))_{0}^{\pi }-\int_{0}^{\pi}\frac{x.(a^{2}-cos^{2}x)}{b^{2}-cos^{2}x}dx$
$I = -\frac{1}{2}\int_{0}^{\pi}\frac{x.(a^{2}-cos^{2}x)}{b^{2}-cos^{2}x}dx$
$I = -\frac{1}{2}\int_{0}^{\pi}\frac{(\pi -x).(a^{2}-cos^{2}(\pi -x))}{b^{2}-cos^{2}(\pi -x)}dx$
$I = -\frac{1}{2}\int_{0}^{\pi}\frac{(\pi -x).(a^{2}-cos^{2}(x))}{b^{2}-cos^{2}(x)}dx$
$2I = -\frac{\pi}{2}\int_{0}^{\pi}\frac{(a^{2}-cos^{2}(x))}{b^{2}-cos^{2}(x)}dx$
$I = -\frac{\pi}{4}\int_{0}^{\pi}\frac{(a^{2}-cos^{2}(x))}{b^{2}-cos^{2}(x)}dx$
$I = -\frac{\pi}{4}\int_{0}^{\pi}\frac{(a^{2}-cos^{2}(x))}{b^{2}-cos^{2}(x)}.\frac{sec^{4}x}{sec^{4}x}dx$
$I = -\frac{\pi}{4}\int_{0}^{\pi}\frac{(a^{2}sec^{4}x-sec^{2}x)}{b^{2}sec^{4}x-sec^{2}x}dx$
$t = tanx$
$dt = sec^{2}x.dx$
$I = -\frac{\pi}{4}\int\frac{(a^{2}(t^{2}+1)-1)}{(t^{2}+1).(b^{2}(t^{2}+1)-1)}dt$
Simply using the partial fraction rule here, we have
$I = -\frac{\pi}{4}.(\frac{(a^{2}-b^{2})tan^{-1}(\frac{bt}{\sqrt{b^{2}-1}})+b\sqrt{b^{2}-1}tan^{-1}t}{b\sqrt{b^{2}-1}})$
$I = -\frac{\pi}{4}.(\frac{(a^{2}-b^{2})tan^{-1}(\frac{btanx}{\sqrt{b^{2}-1}})+b\sqrt{b^{2}-1}tan^{-1}tanx}{b\sqrt{b^{2}-1}})$
$I = -\frac{\pi}{4}.(\frac{(a^{2}-b^{2})tan^{-1}(\frac{btanx}{\sqrt{b^{2}-1}})}{b\sqrt{b^{2}-1}})+x)_{0}^{\pi}$
$I = -\frac{\pi }{4}.\pi$
$I = -\frac{\pi^{2} }{4}$
Thanks & Regards
Jitender Singh
IIT Delhi