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
Moderation Team
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