An Integral

\texttt{putting x =} \cos \theta \\ \texttt{the integral becomes} \\ \int _0^\pi\ln (a^2+b^2-2ab\cos\theta)=\int_0^{\pi} \ln \left( (a^2)(1+(\frac{b}{a})^2-2\frac{b}{a}\cos\theta)\right)\mathrm{d\theta}
now we can use result

\int_{0}^{\pi}{\ln(1+a^2-2a\cos \theta)}\mathrm{d\theta}=2\pi \ln a \ \ (where \ a > 1)

hence answer is

2\pi \ln 3

@sonne
which property is this??

u may have a look at this

http://www.targetiit.com/iit-jee-forum/posts/prove-the-following-11283.html

I(\alpha)=\int_{0}^{\pi}{\ln(13-\alpha\cos \theta)}\mathrm{d\theta} \\ I'(\alpha)=\int_{0}^{\pi}{\frac{-\cos\theta}{13-\alpha\cos\theta}}\mathrm{d\theta}\\ I'(\alpha)=\frac{1}{\alpha}\int_{0}^{\pi}{\frac{13-\alpha\cos\theta-13}{13-\alpha\cos\theta}}\mathrm{d\theta}\\ I'(\alpha)=\frac{1}{\alpha}\left( \int_{0}^{\pi}1d\theta-13\int_{0}^{\pi}{\frac{1}{13-\alpha\cos\theta}}\mathrm{d\theta} \right) \\ I'(\alpha)=\frac{\pi}{\alpha}-\frac{26}{\alpha }\int_{0}^{\infty}{\frac{dt}{(13-\alpha)+(13+\alpha)t^2}} \\ I'(\alpha)=\frac{\pi}{\alpha}-\frac{26}{\alpha .\sqrt{169-\alpha^2} }\tan^{-1}{\frac{t}{\sqrt{(13-\alpha)}}}|_0^{\infty} \\ I'(\alpha)=\frac{\pi}{\alpha}-\frac{26}{\alpha .\sqrt{169-\alpha^2} }\frac{\pi}{2} \\ I(\alpha)=\pi\ln\alpha -13\pi \int \frac{1}{\alpha \sqrt{\alpha^2-169}} \\ \texttt{now putting }\alpha = 13\sin t \\ I(\alpha)=-\pi \ln( {{13+\sqrt{169-\alpha^2 }}})+C

sir is it right ?

I(0)=Ï€ln13

hence we can find C and then put α=6

really tiresome problem

anant sir has posted the answer here at

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=296&t=356856&start=0&hilit=kaymant

post no .2