TargetIIT

in the first question use partial fractions...
[image]43809400.jpg[/image]

[image]43809545.jpg[/image]..
on putting the limits u will get..

(A/a + B/b)π/2...

[image]43810969.jpg[/image]

substitute [im]http://alt2.mathlinks.ro/latexrender/pictures/d/d/2/dd22688bc564e64b5a901a619e3e8c4e42a9cc51.gif[/im]

P.S use of mathematica shud be avoided [im]http://www.mathlinks.ro/images/smiles/wink.gif[/im]

For 2) use the following general result:

[im]http://codecogs.izyba.com/gif.latex?\int_0^\pi%20\ln(1-2a\cos%20x+a^2)\%20\mathrm%20dx%20=\left\{\begin{matrix}%200%20&%20\text{%20if%20}%20|a|\le%201\\[2ex]%202\pi\ln%20|a|%20&%20\text{%20if%20}%20|a|%3E1%20\end{matrix}}\right.[/im]
to get the required result as [im]http://codecogs.izyba.com/gif.latex?2\pi%20\ln%203[/im]

Consider integrating [im]http://codecogs.izyba.com/gif.latex?\,\phi(\alpha)=\int_0^\pi\,\ln(1-2\alpha\cos(x)+\alpha^2)\;dx\, %20[/im]
now,
[im]http://codecogs.izyba.com/gif.latex?    \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} %20[/im]
As x varies from 0 to [im]http://codecogs.izyba.com/gif.latex?\pi[/im], [im]http://codecogs.izyba.com/gif.latex?\left(\frac{1+\alpha}{1-\alpha}\cdot\tan\left(\frac{x}{2}\right)\right)\, %20 [/im]varies through positive values from 0 to infinity when [im]http://codecogs.izyba.com/gif.latex?-1%3C\alpha%20%3C1[/im]   and [im]http://codecogs.izyba.com/gif.latex?\left(\frac{1+\alpha}{1-\alpha}\cdot\tan\left(\frac{x}{2}\right)\right)\, %20 [/im] varies through negative values from [im]http://codecogs.izyba.com/gif.latex?0%20,%20-\infty[/im] when [im]http://codecogs.izyba.com/gif.latex?\alpha%20%3C%20-1,or,\alpha%20%3E1[/im]
Hence,
[im]http://codecogs.izyba.com/gif.latex?\\arctan\left(\frac{1+\alpha}{1-\alpha}\cdot\tan\left(\frac{x}{2}\right)\right)\,\bigg|_0^\pi=\frac{\pi}{2}\,[/im] when [im]http://codecogs.izyba.com/gif.latex?-1%3C\alpha%20%3C1[/im]
[im]http://codecogs.izyba.com/gif.latex?\\arctan\left(\frac{1+\alpha}{1-\alpha}\cdot\tan\left(\frac{x}{2}\right)\right)\,\bigg|_0^\pi=-\frac{\pi}{2}\,[/im] when  [im]http://codecogs.izyba.com/gif.latex?\alpha%20%3C%20-1,or,\alpha%20%3E1[/im]
Therefore ,
[im]http://codecogs.izyba.com/gif.latex?\frac{d}{d\alpha}\,\phi(\alpha)\,=0\,[/im] when [im]http://codecogs.izyba.com/gif.latex?-1%3C\alpha%20%3C1[/im]
[im]http://codecogs.izyba.com/gif.latex?\frac{d}{d\alpha}\,\phi(\alpha)\,=\frac{2\pi}{\alpha}\,[/im] when [im]http://codecogs.izyba.com/gif.latex?\alpha%20%3C%20-1,or,\alpha%20%3E1[/im]
Upon integrating both sides with respect to wrt [im]http://codecogs.izyba.com/gif.latex?\alpha[/im] we get [im]http://codecogs.izyba.com/gif.latex?\phi%20(\alpha%20)%20=%20C_{1}[/im] when [im]http://codecogs.izyba.com/gif.latex?-1%3C\alpha%20%3C1[/im]

and [im]http://codecogs.izyba.com/gif.latex?\phi%20(\alpha%20)%20=%202\pi%20ln\left|\alpha%20\right|+C_{2}[/im] when [im]http://codecogs.izyba.com/gif.latex?\alpha%20%3C%20-1,or,\alpha%20%3E1[/im]
C[ss]1[/ss] can be determined by setting [im]http://codecogs.izyba.com/gif.latex?\alpha[/im] = 0

we get C[ss]1[/ss] = 0
to determine C[ss]2[/ss] we substitute http://codecogs.izyba.com/gif.latex?\alpha%20=\frac{1}{\beta%20} where [im]http://codecogs.izyba.com/gif.latex?-1%3C\beta%20%3C1[/im]
[im]http://codecogs.izyba.com/gif.latex?     \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} [/im]
hence we can conclude [im]http://codecogs.izyba.com/gif.latex?\phi%20(\alpha%20)%20=%200,-1%3C\alpha%20%3C1[/im] and
[im]http://codecogs.izyba.com/gif.latex?\phi%20(\alpha%20)%20=%202\pi%20ln\left|\alpha%20\right|%20when,-1%3E\alpha%20,\alpha%20%3E1[/im]

[im]http://codecogs.izyba.com/gif.latex?I=\int_{0}^{1}{\frac{x^{a-1}-x^{-a}}{(1+x)lnx}dx}%20\\=\int_{0}^{1}{\frac{x^{a-1}}{(1+x)lnx}dx}-\int_{0}^{1}{\frac{x^{-a}}{(1+x)lnx}dx}%20\\%20\\Substitute\:%20x=\frac{1}{t}%20\:%20in\:%20second\:%20integral%20\\we\:%20get:%20\\=\int_{0}^{1}{\frac{x^{a-1}}{(1+x)lnx}dx}+\int_{1}^{\infty}{\frac{t^{a-1}}{(1+t)lnt}dt}%20\\=\int_{0}^{\infty}{\frac{x^{a-1}}{(1+x)lnx}dx}%20\\Consider:\:%20f(a)=\int_{0}^{\infty}{\frac{x^{a-1}}{(1+x)lnx}dx}%20\\\frac{\partial%20f}{\partial%20a}=\int_{0}^{\infty}%20{\frac{x^{a-1}}{1+x}dx}%20\\Now\:%20substitute\:%20x=\frac{1}{z}-1%20\\You\:%20will\:%20get:%20\\\frac{\partial%20f}{\partial%20a}=%20\int_{0}^{1}{z^{-a}(1-z)^{1-a}}=\beta(1-a,a)=\pi cosec(a\pi)%20\\f(a)=-ln\left|cosec\pi%20a+cot\pi%20a%20\right|+c%20\\f\left(\frac{1}{2}%20\right)=0%20\\or,\:%20c-ln\left|cosec\frac{\pi}{2}%20+cot\frac{\pi}{2}%20\right|=0%20\\or,\:%20c=0%20\\Therefore:%20\:%20I=-ln\left|cosec\pi%20a+cot\pi%20a%20\right|[/im]

Beta function is not in JEE syllabus. I have used a property of beta function in my solution only because I was not able to evaluate the integral in any other way.

Do point out any mistake in my solution.