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![[Post New]](/templates/default/images/icon_minipost_new.gif) 14 Mar 2008 22:30:28 IST
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Find 
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Guide to latex:
http://www.goiit.com/posts/list/community-shelf-a-guide-to-latex-48056.htm
JEE and OLYMPIA INFINATUM
http://iit-redefined.theforum.name/index.php
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14 Mar 2008 22:36:06 IST
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We could set up a reduction formula by putting X^n = tany
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14 Mar 2008 22:37:11 IST
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do it. 
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Guide to latex:
http://www.goiit.com/posts/list/community-shelf-a-guide-to-latex-48056.htm
JEE and OLYMPIA INFINATUM
http://iit-redefined.theforum.name/index.php
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14 Mar 2008 22:39:27 IST
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is it tan^-1(x^n)
!!!!!!!!!!!
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Varsha
be cool,
tomorrow is what we make it today,
so if u can dream it , u can make it .
life is an ice cream ,eat it before it melts away!!!!!!!!!
    
 
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14 Mar 2008 22:49:39 IST
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nope
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Guide to latex:
http://www.goiit.com/posts/list/community-shelf-a-guide-to-latex-48056.htm
JEE and OLYMPIA INFINATUM
http://iit-redefined.theforum.name/index.php
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15 Mar 2008 15:25:42 IST
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hey i think i have an idea is it to do with complex numbers da!!! i remember in my iit class we used to do some special integrals with complex numbers??? is it do with x^2n =-1 roots and then substitute in the denominator and integrate???
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15 Mar 2008 17:52:30 IST
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well there is always more than one way to solve a problem. Do try! But no, the solution I have in mind doesnt not use complex numbers.
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Guide to latex:
http://www.goiit.com/posts/list/community-shelf-a-guide-to-latex-48056.htm
JEE and OLYMPIA INFINATUM
http://iit-redefined.theforum.name/index.php
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15 Mar 2008 19:33:24 IST
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any hint???? i'll try
!!!!!!!!!!
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Varsha
be cool,
tomorrow is what we make it today,
so if u can dream it , u can make it .
life is an ice cream ,eat it before it melts away!!!!!!!!!
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15 Mar 2008 20:27:18 IST
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u can use complex numbers
substitute x= iy
factorize the denominator
itll be sumthin like
i*[ ] [ ] dy/(y-e^ikpi/n) product (k runs frm 1 to 2n-1)
now its simple
use partial fractions and integrate
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15 Mar 2008 20:31:31 IST
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HEY BUT THAT WAS MY METHOD TOO!!!
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15 Mar 2008 20:52:27 IST
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I can give u an infinite series formula . Just expand ( 1/ ( 1 + x^2n ) ) binomially and integrate term by term . The condition is that x lies within its radius of convergence .
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16 Mar 2008 00:51:03 IST
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Let  be  .
Then  .
Hence,  .
Since 
Or 
 .
The remaining integral can be evaluated by a trigonometric substitution.
Let 
Thus, 
which simplifies to 
 .
Since  , and  , we can back substitute.
 .
Finally,  where  .
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Guide to latex:
http://www.goiit.com/posts/list/community-shelf-a-guide-to-latex-48056.htm
JEE and OLYMPIA INFINATUM
http://iit-redefined.theforum.name/index.php
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16 Mar 2008 00:54:01 IST
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was this question too tough? Extremely sorry if it was.
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Guide to latex:
http://www.goiit.com/posts/list/community-shelf-a-guide-to-latex-48056.htm
JEE and OLYMPIA INFINATUM
http://iit-redefined.theforum.name/index.php
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16 Mar 2008 09:32:31 IST
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no problem, we love challenges  |
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20 Mar 2008 07:38:25 IST
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