Find variance of $x(t,\omega)$ given: $$dx(t) = \beta(t)\cos{t}\text{ dt} + \sin{t}\text{ d}\beta(t)$$
where,
$\beta(t)$ is a scalar Brownian motion process with statistics: $$E[\beta(t)] = 0\text{ ; } E[\beta^2(t)] = t$$
Attempt at a solution:
From given we can arrive at: $$ x(t) = x(t_0) + \int_{t_0}^{t}\beta(\tau)\cos{\tau}\text{ d}\tau + \int_{t_0}^t\sin{\tau}\text{ d}\beta(\tau) \\ \implies x^2(t) = x^2(t_0) + \int_{t_0}^t\beta^2(\tau)\cos^2{\tau}\text{ d}\tau^2 + \int_{t_0}^t\sin^2{\tau}\text{ d}\beta^2(\tau) + 2x(t_0)\int_{t_0}^t\beta(\tau)\cos{\tau}\text{ d}\tau + \\2x(t_0)\int_{t_0}^t\sin{\tau}\text{ d}\beta(\tau) + 2\int_{t_0}^t\beta(\tau)\cos{\tau}\sin\tau\text{ d}\tau\text{ d}\beta(\tau) $$
Taking the expectation, $\text{E}(\cdot)$ of both sides: $$\begin{equation} \begin{split} \text{E}[x^2(t)] = \text{E}[x^2(t_0)] + \int_{t_0}^t\text{E}[\beta^2(\tau)]\cos^2\tau\text{ d}\tau^2 + \int_{t_0}^t(\sin^2\tau)\text{ E}[\text{d}\beta^2(\tau)] + \\ 2x(t_0)\int_{t_0}^t\text{E}[\beta(\tau)]\cos\tau\text{ d}\tau + 2x(t_0)\int_{t_0}^t(\sin\tau)\text{E}[\text{d}\beta(\tau)] + \\ 2\int_{t_0}^t\text{E}[\beta(\tau)\text{d}\beta(\tau)]\sin\tau\cos\tau\text{ d}\tau \\ \end{split} \end{equation}$$
Now from the definition of Brownian motion: $\beta(t_0) = 0, \text{E}[d\beta(\tau)] = 0$ and $\text{E}[d\beta^2(\tau)] = q\text{d}\tau$, where $q$ is the diffusion of the Brownian motion. Using this fact and the given statistics of $\beta(t)$:
$$\text{E}[x^2(t)] = \text{E}[x^2(t_0)] + \int_{t_0}^t\tau\cos^2\tau\text{ d}\tau^2 + \int_{t_0}^tq\sin^2\tau\text{ d}\tau + \int_{t_0}^t\text{E}[(\beta(\tau) - \beta(t_0))d\beta(\tau)]\sin\tau\cos\tau\text{ d}\tau$$
The expectation in the last term can be viewed as the expectation of a product of two independent increments of $\beta(t)$ and so we can evaluate it to be zero. Is this approach correct?
You can use (stochastic) integration by parts to get
$$ x(t) = \sin(t) \beta(t) = \int_0^t \cos(s) \beta(s) d s + \int_0^t \sin(s) d \beta(s). $$ Then apply the usual properties of mean and variance and what you already know about $\beta(t)$.