hiiiiiiii liku

 

this is typically a very different kind of question ... when u read the solution u may wonder that what is going on .. but just relax .... look at question .. the first idea that shud strike u is that what shud i use ..

 

well clearly no property seems to be applicable here ... but there is some hope ..  there is an arbitary m in the expression to be integrated ..

 

this gives an idea .. we can use the reduction formula ..

 

so let us try our hand on it and see what happens ..

 

so let us call I_m = ₀  sin²mx/sin²x dx

 

 

 

 

 

 

similarily I_m - I_m-1        = ₀   sin(m-1/2)t/sin(t/2) dt 

 

so (I_m+1 - I_m) - (I_m - I_m-1) = ₀  [sin(m+1/2)t - sin(m-1/2)t]/sin(t/2)dt

 

so (I_m+1 - I_m) - (I_m - I_m-1) = 0

or, 2.I_m = I_m+1 + I_m-1

or, I_m-1,I_m,I_m+1 are in A.P.

 

Now, I_m = ₀  sin²mx/sin²x dx

 

               = 0

 

and,  I₁ = ₀  sin²1x/sin²x dx   

               = ₀  dx =        

                                                

So common difference of A.P is  

 

and hence I_m = I₀ + n.

                    = n.       ... ans

 

I hope u got this tough one .. :)

 

cheers

Most welome ... LIKU ..

 

However there is no need to say "SIR" ... i am just a kid like u ..

 

cheers