I am attempting to integrate
$$ \int _{0}^{t} \sin(s) dW_s $$
whereas $W_s$ is brownian motion, in some sense a normal random variable with mean 0 and variance $s$.
I looked around in stack exchange and I couldn't really find a clear solution.
One idea I had was using Ito's formula (Which really I want to believe is overkill given it appears much later in the text). If I can convert $\sin(s) dW_s$ to $dE(s,W_s)$ then I just integrate to yield $E(s,W_s)$ (and then something has to be set to $t$ and something else to $0$ and subtracted but i'll defer that part of the problem to when I get there).
Ito's formula gives:
$$ df(s,W_s) = \frac{\partial f}{\partial W_s} dW_s +\frac{1}{2} \frac{\partial ^2 f}{\partial W_s ^2 } dS +\frac{\partial f}{\partial s} dS $$
So if $df(s,W_s) = \sin(s) dW_s$ then it must be the case that
$$\frac{1}{2} \frac{\partial ^2 f}{\partial W_s ^2 } dS = -\frac{\partial f}{\partial s} dS $$
And
$$ \frac{\partial f}{\partial W_s} dW_s = \sin(s) dW_s$$
I notice this implies
$$ \frac{\partial f}{\partial W_s} = \sin(s) \rightarrow \frac{\partial^2 f}{\partial W_s^2} = 0 \rightarrow f = \sin(s)W_s + C(s)$$
But we have that $\frac{\partial f}{\partial s} = 0 \rightarrow C(s) = K + \cos(s)W_s$
and this is where things fall apart.
So I conclude that Ito's was not only overkill, but at least the way I approached, a dumb strategy overall. So how should I tackle this alternatively?