Homework Statement
If the derivative of f(x) w.r.t x is (0.5- sin^2x)/f(x) then fundamental period of f(x) is
2. The attempt at a solution
I wrote 1/2-sin^2x as cos2x/2. since f'(x)=cos2x/2f(x) its integral will be f(x), I would be grateful is someone could provide intuition on how to proceed further
What is "fundamental period"? Can you please provide the problem statement exactly as written down (or at least translated as good as you can), since it does not give information about ##f##.
What is "fundamental period"? Can you please provide the problem statement exactly as written down (or at least translated as good as you can), since it does not give information about ##f##.
You're interpreting the question correctly, this is a MCQ type question and I have written the statement unaltered- all the data is mentioned. Fundamental period here means the period of the function (e.g 2pi for sinx) etc.Additionally 4 options are provided (but I should be able to get the answer regardless of whether there are options or not) the options are- 1)pi 2)2pi 3)pi/2 4)3pi/2I suspect that the problem should have said "A periodic function ##f(x)## satisfies the equation ##f'(x) = (0.5 - \sin^2 x)/f(x).## What is the period of ##f##?"
F(x) will be sin(2x)/2, I am trying to equate ∫2f(x)f'(x) with this now to find f(x)...So: ##2 f(x) f'(x) = \cos(2x),## and you should recognize that ##2 f(x) f'(x)## is the derivative of some function ##F(x)## that is related to ##f(x)## in some way. That allows you to find ##F(x)## and then, from that, to find ##f(x)##, up to a constant of integration (which will not affect the period, if any).
Got it! I integrated by parts and found f(x)^2=sin2x/2, so the answer is pi. Thanks a lot! :)F(x) will be sin(2x)/2, I am trying to equate ∫2f(x)f'(x) with this now to find f(x)...