**1**
Philip Calvert
**·**2008-12-08 06:28:19
@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

**1**
Aritra Das
**·**2015-06-11 23:02:36
As 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.

**62**
Lokesh Verma
**·**2008-12-08 19:29:55
Krish bcos u are hell bent on gettin this solved ;)

I(m) = âˆ«(log(m^{2}+2mcosx+1)dx

dI/dm=d/dm âˆ«(log(m^{2}+2mcosx+1)dx

dI/dm= âˆ«d/dm(log(m^{2}+2mcosx+1)dx

dI/dm= âˆ«(2m+2cosx)/(m^{2}+2mcosx+1)dx

This I am sure you can solve..

The answer that you get for the RHS.. is Ix let

then I = âˆ«Ix. dm

**62**
Lokesh Verma
**·**2008-12-08 18:24:18
krish.. 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 :)

**1**
krish1092
**·**2008-12-08 18:20:14
Someone please **solve** this!!!!!!!!!!

**1**
prasanna
**·**2008-12-08 07:40:24
just guessing that write 1 as sin^{2}x+cos^{2}x.then we get (cosx+m)^{2} +sin^{2}x.

**1**
prasanna
**·**2008-12-08 07:38:36
just guessing that write 1 as sin^{2}x+cos^{2}x.then we get (cosx+m)^{2} +sin^{2}x.

**1**
krish1092
**·**2008-12-08 07:06:03
@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**
skygirl
**·**2008-12-08 07:00:00
@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**
skygirl
**·**2008-12-08 06:58:14
@yes,,, expand your 'blah blah' a little....

@philip, u r r8... its a kinda reversible reaction :P..

very much like âˆ«tanÎ¸dÎ¸ widn limits 0 to pi/4 ..

**1**
krish1092
**·**2008-12-08 06:31:27
Even i know that its not so easy,But just to draw everyone's attention,i posted it like that!! :D

**1**
yes no
**·**2008-12-08 06:29:41
take 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**
skygirl
**·**2008-12-08 05:58:34
THIS IS A WRONG QUESTION!

wat should be taken?? ---- dm or dx ?? wat?

**1**
krish1092
**·**2008-12-08 06:24:18
@philip

I'm not getting it,Please post your solution upto that point!

**1**
skygirl
**·**2008-12-08 06:22:08
well is it?

i din solve it actually... ok i will check it...

**1**
Philip Calvert
**·**2008-12-08 06:19:40
@ 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

âˆ«log(1+2mcosx+m^{2})dx

**1**
varun
**·**2008-12-08 06:16:46
omg lol ... I most of the times make that mistake ...

Sorry :(

**1**
skygirl
**·**2008-12-08 06:15:14
heyyy how can u dat??

log (a +b) â‰ loga. log b

rather

loga +log b = log(ab) ...

u jus made a hasty mistake :) check it ... (fundamental :))

**1**
varun
**·**2008-12-08 06:12:11
take log(1+m^{2}) out ? It is a constant right ?

log(1+m^{2})âˆ«log(2mcosx)dx ?

**1**
skygirl
**·**2008-12-08 06:09:08
do 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)

**1**
skygirl
**·**2008-12-08 05:59:21
can guess dat it is dx...

but should be mentioned in the question...