1@krish
i haven't got any soln yet
i only did what skygirl suggested and got back the question!!!!
and btw who told you this was "very very easy " ;p
23 Answers
1
1As an alternative to the differentiatiin under the integration, Use complex numbers. Write out the part inside the log as product of 1+me^(ix) and 1+me^(-ix), then split up into loga + log b form and integrate. This integration is easy.
62Krish bcos u are hell bent on gettin this solved ;)
I(m) = ∫(log(m2+2mcosx+1)dx
dI/dm=d/dm ∫(log(m2+2mcosx+1)dx
dI/dm= ∫d/dm(log(m2+2mcosx+1)dx
dI/dm= ∫(2m+2cosx)/(m2+2mcosx+1)dx
This I am sure you can solve..
The answer that you get for the RHS.. is Ix let
then I = ∫Ix. dm
62krish.. this is not in syllabus.. Yes poiinted that out very clearly..
If you really want the solution...
then do this..
f(m)= function that u defined..
take df/dm = derivative of the function
this given an integrable... function
now integrate it back with respect to m... you will get the answer :)
1just guessing that write 1 as sin2x+cos2x.then we get (cosx+m)2 +sin2x.
1just guessing that write 1 as sin2x+cos2x.then we get (cosx+m)2 +sin2x.
1@skygirl.
If you are getting the question again,then i think,it has been solved,Please post your solution because,I'm not getting !!!
1@yes, as far as i understand the question,,, m is a constant... so y differentiate wrt m?? n y did u take I as a function of m?? its rather a function of x ... all the rest are constants...
1@yes,,, expand your 'blah blah' a little....
@philip, u r r8... its a kinda reversible reaction :P..
very much like ∫tanθdθ widn limits 0 to pi/4 ..
1Even i know that its not so easy,But just to draw everyone's attention,i posted it like that!! :D
1take it equal to I(m)
differentiate it with respect to m ....and blah blah blah
dont worry guys, this is not in JEE syllabus
1@philip
I'm not getting it,Please post your solution upto that point!
1@ varun it happens to all of us sometimes
@skygirl the method you told leads back to the problem
meaning that when we integrate taking x as the first function
the first part gets cancelled out and x too vanishes giving us
∫log(1+2mcosx+m2)dx
1heyyy how can u dat??
log (a +b) ≠loga. log b
rather
loga +log b = log(ab) ...
u jus made a hasty mistake :) check it ... (fundamental :))
1take log(1+m2) out ? It is a constant right ?
log(1+m2)∫log(2mcosx)dx ?
1do it byparts..
take log(blah blah ) = first function..
then u will get a 'thing',,, take x=first func n the rest as second function...
u will get the ans i think.. (i din solve it fully.. :P)