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![[Post New]](/templates/default/images/icon_minipost_new.gif) 17 Jun 2008 21:30:32 IST
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evaluate 
the box x is actually dx ..
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17 Jun 2008 21:38:39 IST
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Answer is 1,please tell me if i am correct,i will post solution then
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18 Jun 2008 07:23:16 IST
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no the answer is 0.....i solved it myself.....dont have the solution key..
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18 Jun 2008 07:32:14 IST
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u can post ur solution.........
plzz try others too
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18 Jun 2008 11:56:57 IST
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Hint: Try the substitution t = 
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18 Jun 2008 12:38:10 IST
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yup! i got it as 0. postin the soln
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FAILURE, THE FIRST STEP TO SUCCESS |
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18 Jun 2008 12:47:18 IST
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i put x=tan@ so dx=sec^2@d@
and hence the expression becomes as follows tan@ln(tan@)sec^2@/sec^4@
which finally gives as d@tan@ln(tan@)/1+tan^2@
now since x=tan@ the limits change and go from 0 to pie/2.
now apply that theorum of f(a+b-x)=f(x)
so the original expression changes to as follows
cot@ln(cot@)/1+cot^2@ and now let usince tan@ = 1/cot@ we use this in that result and we get -tan@ln(tan@)d@/1+tan^2@
so addin up both the integrals 2I = 0 and hence the ans is 0
try it outyourself. please correct me if i am wrong or nudge me if you have any doubts
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FAILURE, THE FIRST STEP TO SUCCESS |
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19 Jun 2008 10:45:02 IST
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another method;;
divide the integration from 0 to 1 and from 1 to inf
now for the second int. put x=1/t and the int. bcomes from 0 to 1
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20 Jun 2008 22:36:08 IST
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take x/(1+xsquare)square as v and logx as u.use integral of u*v form.
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