Let stochastic processes $S^a=(S^a_t)_{t\geq 0}$ and $S^b=(S^b_t)_{t\geq 0}$ solve the following two SDEs $$dS^a_t=S^a_tdB^a_t$$ and $$dS^b_t=S^b_tdB^b_t$$ respectively, with $S_0^a>0$ and $S_0^b>0$ and Brownian motions $B^a$ and $B^b$ are independent. Let $W^a_t=\frac{S^a_t}{S^a_t+S^b_t}$. Then how do we obtain explicit expression for $\mathbb{E}[W^a_t]$?
My attempt: Since they are both geometric Brownian motions, so we have $S^k_t=S_0^ke^{-\frac{1}{2}t+B^k_t}$ and $\mathbb{E}[S^k_t]=S_0^k$ for $k\in\{a,b\}$. By Ito lemma, we have $$dW^a_t=W^a_t(1-W^a_t)(1-2W^a_t)dt+W^a_t(1-W^a_t)dB^a_t-W^a_t(1-W^a_t)dB^b_t.$$ It follows by taking the expectation on both sides and we have $$\mathbb{E}[W^a_t]=W_0^a+\int_0^t\mathbb{E}\bigg[W^a_s(1-W^a_s)(1-2W^a_s)\bigg]ds.$$
Then I don't know how to proceed from here.
We can actually compute $\mathbb{E}[W_t^a]$ without needing to go through Ito's formula. First, note that since $S_t^a \stackrel{(d)}{=} S_t^b$ we have
$$1-W_t^a = 1-\frac{S_t^a}{S_t^a+S_t^b} = \frac{S_t^b}{S_t^a+S_t^b} \stackrel{(d)}{=} W_t^a.$$
Now we can compute
$$ 1 = \mathbb{E}[W_t^a + (1-W_t^a)] = \mathbb{E}[W_t^a] + \mathbb{E}[(1-W_t^a)] = 2 \mathbb{E}[W_t^a]$$
so $\mathbb{E}[W_t^a] = \frac 12$.