Nice Integral-~~~~

\text{if we substituite } 1+a^2-2a\cos x =t \\ \text{ then the integral boils to }\\ \int_{a_1}^{a_2}{\frac{\log (t)dt}{\sqrt{t-a_1}\sqrt{t-a_2}}}\\ \text{now we can go by parts}\\ \text{here }a_1=(a-1)^2 \ and \ a_2=(a+1)^2
but how to go about by che's hint ?

well this was solved beautifully by nishant and theprophet sir

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

Consider integrating \,\phi(\alpha)=\int_0^\pi\,\ln(1-2\alpha\cos(x)+\alpha^2)\;dx\,
now,
\begin{align} \frac{d}{d\alpha}\,\phi(\alpha)\, &=\int_0^\pi \frac{-2\cos(x)+2\alpha}{1-2\alpha \cos(x)+\alpha^2}\;dx\, \\ &=\frac{1}{\alpha}\int_0^\pi\,\left(1-\frac{(1-\alpha)^2}{1-2\alpha \cos(x)+\alpha^2}\,\right)\,dx\, \\ &=\frac{\pi}{\alpha}-\frac{2}{\alpha}\left\{\,\arctan\left(\frac{1+\alpha}{1-\alpha}\cdot\tan\left(\frac{x}{2}\right)\right)\,\right\}\,\bigg|_0^\pi. \end{align}
As x varies from 0 to \pi, \left(\frac{1+\alpha}{1-\alpha}\cdot\tan\left(\frac{x}{2}\right)\right)\,varies through positive values from 0 to infinity when -1<\alpha <1 and \left(\frac{1+\alpha}{1-\alpha}\cdot\tan\left(\frac{x}{2}\right)\right)\, varies through negative values from 0 , -\infty when \alpha < -1,or,\alpha >1
Hence,
\\arctan\left(\frac{1+\alpha}{1-\alpha}\cdot\tan\left(\frac{x}{2}\right)\right)\,\bigg|_0^\pi=\frac{\pi}{2}\, when -1<\alpha <1
\\arctan\left(\frac{1+\alpha}{1-\alpha}\cdot\tan\left(\frac{x}{2}\right)\right)\,\bigg|_0^\pi=-\frac{\pi}{2}\, when \alpha < -1,or,\alpha >1
Therefore ,
\frac{d}{d\alpha}\,\phi(\alpha)\,=0\, when -1<\alpha <1
\frac{d}{d\alpha}\,\phi(\alpha)\,=\frac{2\pi}{\alpha}\, when \alpha < -1,or,\alpha >1
Upon integrating both sides with respect to wrt \alpha we get \phi (\alpha ) = C_{1} when -1<\alpha <1

and \phi (\alpha ) = 2\pi ln\left|\alpha \right|+C_{2} when \alpha < -1,or,\alpha >1
C₁ can be determined by setting \alpha = 0

we get C₁ = 0
to determine C₂ we substitute http://latex.codecogs.com/gif.latex?\alpha%20=\frac{1}{\beta%20} where -1<\beta <1
\begin{align} \phi(\alpha) &=\int_0^\pi\left(\ln(1-2\beta \cos(x)+\beta^2)-2\ln|\beta|\right)\;dx\, \\ &=0-2\pi\ln|\beta|\, \\ &=2\pi\ln|\alpha|\, \end{align}
hence we can conclude \phi (\alpha ) = 0,-1<\alpha <1 and
\phi (\alpha ) = 2\pi ln\left|\alpha \right| when,-1>\alpha ,\alpha >1

@soumik
I was wondering how you got that integral in the second step in terms of cos inverse.

sir i meant the bracket in second last line is closed after the integration is performed
sorry for that
actually these latex symbols made me so confused about placing
i hope its correct now

is this ok ?
following che's hint

For the 3rd sum, differentiate w.r.t a, then evaluate the integral, followed by an indefinite one w.r.t. a.

in the first question use partial fractions...

..
on putting the limits u will get..

(A/a + B/b)Ï€/2...

Sir can we derive this formula without using the hint given by che?

For 2) use the following general result:

\int_0^\pi \ln(1-2a\cos x+a^2)\ \mathrm dx =\left\{\begin{matrix} 0 & \text{ if } |a|\le 1\\[2ex] 2\pi\ln |a| & \text{ if } |a|>1 \end{matrix}}\right.
to get the required result as 2\pi \ln 3

substitute

P.S use of mathematica shud be avoided

@Tapas..can u plz explain ur method..

2 can be broken into integral of (below) and 2log2 parts;

no thats not a typo

In the 2nd one have you written 1 + 9 instead of 10 deliberately or is there a typo ?