I'm not convinced... i dont get it really. Do you mead x = a(-cos(yt/y)) ?
How do i differentiate that one and get the same thing in the brickets sin(y).
Ganesh... maybe i didn't make my self clear that y is a function of t.?? Or is it
me that have misunderstood you? Coud you show me how you diferentiate the
expression?

Forgive me if I am wrong, but it may simply be impossible without knowing the function y(t).

dx = a sin(y(t)) dt

Let us just call the funtion "ƒ" instead of y:

dx = a sin(ƒ(t)) dt

If, for example, there were no sine or factor applied, we could have:

dx = ƒ(t) dt

Which is just a general statement.

So, yes, what is y(t) ?

*** This is my interpretation of the problem:
x=x(t)
y=y(t)

dx=a.sin(y(t))dt

I want to solve this for x.

dx/dt = a.sin( y(t) )
∫ (dx/dt)dt = ∫a.sin( y(t) )dt
∫ dx = ∫a.sin( y(t) )dt
x(t)+C=∫a.sin( y(t) )dt

*** if..
y(t)=a.t
      then
dy/dt=a

*** therefore...
∫a.sin( y(t) )dt = ∫a.sin( a.t) )dt
Let z=a.t => dz/dt = a
∫a.sin( a.t) )dt = ∫(dz/dt)sin(z)dt = ∫sin(z)dz = -cos(z)+C = -cos(a.t)+C

*** putting it all together...
x(t) = -cos(a.t)+C
y(t)= a.t

dx/dt= a.sin(a.t)
dx= a.sin(a.t) dt
dx=a sin(y(t)) dt <=> dx=a sin(y) dt

***I think this is a possible solution for your problem