1i think byparts will not work here,
my ans is comming
-(Î lna)/2a
7 Answers
62try integration by parts..
I think that should help..!
take logx as the first function
162hmm.. okie.. may be u are correct...!
i just saw the form and felt it should work...
anyways.. gr8 u solved it :)
did u use something like ln*(tanx) ??
1yeah pt x=atanθ.. dats r8...
i think something like this was asked in aits....
1this question is really easy..
put x=atanθ
then dx=asec2θdθ
θ varies from 0 to π/2
we get
π/2∫0ln(atanθ)dθ/a
=∫(lna/a)dθ+∫(lntanθ/a)dθ
but the second term is equaivalent to π/2∫0lnsin(π/2-θ)/cos(π/2-θ)/adθ
let I =∫(lntanθ/a)dθ
then I=π/2∫0lnsin(π/2-θ)/cos(π/2-θ)/adθ
adding both we get I=0
rest term left is π/2∫0ln(a)dθ/a=-πlna/2a