its a gd one really

Here there are infinite terms so if we add 1 more sin term it should make no difference.
This is only possible if sin x=x
therefor req. soln. =∫x dx=x^2/2+c

^^^^^^^^^^
I don't think so,

this one is a beauty -1≤ sinx ≤+1 now sin of +1 is smthing around .08 .. and sin of that is even smaller thus both the value tends towards 0. thus there for ∫0=c and should b the answer . i think!!!

You are right. This is fixed point iteration. The thing u've to integrate is equal to the solution of sin x = x , i.e 0. Try the same for cosine. ;) is fixed point iteration. The thing u've to integrate is equal to the solution of sin x = x , i.e 0. Try the same for cosine. ;)