for the life of me I can't seem to figure out how to solve $\int_{0}^{t}W_{s}dW_{s}$. I know an Ito process $X$ is given by: $$ X(t) = X(t_{0}) + \int_{0}^{t}bdt + \int_{0}^{t}\sigma dW_{t} $$ and Ito's formula states that given a function $f(t, X)$, we have that: $$ df(X_{t}) = (\partial _{t}f + \partial_{x}fb + \frac{1}{2}\partial_{xx}f\sigma)dt + \partial_{x}f\sigma dW_{t} $$
In a book I'm reading it says suppose that $f(x) = x^{2}$ and apply Ito's formula to $f(W_{t})$, so we have that: $$ \partial_{t}f = 0, \quad \partial_{x}f = 2x, \quad \partial_{xx}f = 2 $$ However, what are $b$ and $\sigma$? Through reading other stack exchange questions, other people define $b = 0$ and $\sigma = 1$, however, if this is the case, would it not be the case that $X_{t} = dW_{t}$? Also what is the solution to the above integral?
Yep, when you take $b=0$ and $\sigma=1$ you do indeed get $X_t = \int_0^t 1\,dW_t = W_t$ (not $dW_t$, that wouldn't make sense). And applying Ito's formula to this with $f(x) = x^2$ tells you that $$W_t^2 = 2 \int_0^t W_s\,dW_s + t$$ which rearranges to $$\int_0^t W_s\,dW_s = \frac{1}{2}W_t^2 - \frac{1}{2}t$$ which is is presumably the answer you are looking for.