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![[Post New]](/templates/default/images/icon_minipost_new.gif) 20 Mar 2008 19:30:24 IST
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yeah 2tan....thing was a freaking question wasted half an hour on that so didnt check other qns......by the way the continuity sum b can take any value execpt 0 isnt it?
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its not just about hard work;its how smart you work. |
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20 Mar 2008 20:56:11 IST
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so till now no one has an answer to it kya??
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iitaspirant001@yahoo.com |
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20 Mar 2008 23:30:15 IST
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i got the ans to that question jus lemme know if i am rite integral will be represented as { 0 to 1{2[ tan-1 x ]^2dx [put x=tany] =2{y^2 sec^2 y dy =then integrate this by parts and the second part u will get 0 to pi/4 2{tany y dy here u can replace back y for tan-1 x and change the limits to 0 to 1 and then solve the integral again by parts and u will get the ans ...... please try it i am sure u will get please lemme now if i am correct....i cudnt write down the whole solution here but i have tried to be as clear as possible...please help.... i gues u will get this ans=(pi^2-8pi+16)/8
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20 Mar 2008 23:31:30 IST
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anyone with set 2 i need to discuss
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20 Mar 2008 23:33:27 IST
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@ nammi :HEY HAVENT YOU YET UNDERSTOOD THAT U ARE FACING A LOST BATTLE? THAT IS NOT SO EASILY INTEGRABLE AS QUOTED FROM THE WOLFRAM INTEGRATOR
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20 Mar 2008 23:38:26 IST
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i wasted 25 mins without any use......even if its jee level sum1 cud hav come up with a solution.......
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20 Mar 2008 23:59:05 IST
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nammi uve written the question wrong baba
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iitaspirant001@yahoo.com |
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21 Mar 2008 00:07:19 IST
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i know its squared but i have solved it correctly......jus scheck.....
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21 Mar 2008 10:28:08 IST
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Please tell me if u find ne mistake...
I =  2(tan-1x)2dx
Let tan-1x=y
x=tany
dx=sec2ydy
I=2 y2sec2ydy
= 2[ y2tany] -2 ytany dy
= 2[ y2tany] - y2tany + 1/2y2sec2y dy
= y2tany + y/2 + 1/2I
I = 2y2tany + y + C
Putting the limits
I = pi2/8 + pi/4
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21 Mar 2008 10:30:28 IST
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ohh no i made mistake
haha I = somethin + I
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iitaspirant001@yahoo.com |
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21 Mar 2008 10:31:44 IST
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= 2[ y2tany] -2 ytany dy
= 2[ y2tany] - y2tany + 2/2y2sec2y dy
u missed a 2 in d abv step d last part
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21 Mar 2008 10:35:47 IST
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i got pie^2/8 + pie log root 2
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21 Mar 2008 10:41:42 IST
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my dear fellow, that is the ans of the remaining part without integral log secx dx, the tough part lies therein only :D
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21 Mar 2008 10:45:38 IST
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no earlier it was pie^2 /8 - 4 int log sec x dx
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21 Mar 2008 11:02:48 IST
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Please don't refer to wolfram integrator for making important decisions like |
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