# Integration-very very easy!!

1
Philip Calvert ·

@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 ·

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 ·

Krish bcos u are hell bent on gettin this solved ;)

I(m) = âˆ«(log(m2+2mcosx+1)dx

dI/dm=d/dm âˆ«(log(m2+2mcosx+1)dx

dI/dm= âˆ«d/dm(log(m2+2mcosx+1)dx

dI/dm= âˆ«(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 = âˆ«Ix. dm

62
Lokesh Verma ·

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 ·

1
prasanna ·

just guessing that write 1 as sin2x+cos2x.then we get (cosx+m)2 +sin2x.

1
prasanna ·

just guessing that write 1 as sin2x+cos2x.then we get (cosx+m)2 +sin2x.

1
krish1092 ·

@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 ·

@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 ·

@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 ·

Even i know that its not so easy,But just to draw everyone's attention,i posted it like that!! :D

1
yes no ·

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 ·

THIS IS A WRONG QUESTION!

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

1
krish1092 ·

@philip

1
skygirl ·

well is it?
i din solve it actually... ok i will check it...

1
Philip Calvert ·

@ 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+m2)dx

1
varun ·

omg lol ... I most of the times make that mistake ...

Sorry :(

1
skygirl ·

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 ·

take log(1+m2) out ? It is a constant right ?

log(1+m2)âˆ«log(2mcosx)dx ?

1
skygirl ·

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
krish1092 ·

edited!

62
Lokesh Verma ·

u r right sky..

1
skygirl ·

can guess dat it is dx...
but should be mentioned in the question...