I want to evaluate $$\int_{0}^{T} W(t) \sin(t) \mathrm{d} t$$ where $W(t)$ is a Wiener process.
Context: I want to simulate paths of Karhunen Loeve expansion and compare them to a given path of brownian motion. To do so I need to evaluate the given integral. I've tried with Ito formula and got $$ \frac{1}{6}W^3(t)\cdot \sin(t)=\frac{1}{2}\int_{0}^{t} W^2(s) \cdot \sin(s)\mathrm{d}W(s)-\frac{1}{6} \int_{0}^{t}W^3(s)\cdot \cos(s)\mathrm{d}s +\frac{1}{2} \int_{0}^{t}W(s)\cdot \sin(s)\mathrm{d}s$$ which doesn't seem to make it any easier. My alternative is to solve it numerical but I'd prefer a closed form.
$\textbf{Edit}$ For more context see Pathwise Limit of Karhunen-Loeve Expansion
You can partially integrate to obtain
$$ \int_0^T \sin(t)W_t dt = \int_0^T \cos(t)dW_t - \cos(T)W_T $$
This is a centered Gaussian random variable. It's variance is given by
$$ \operatorname{Var}\left(\int_0^T \cos(t)dW_t\right) + 2 \operatorname{Cov}\left(\int_0^T \cos(t)dW_t, -\cos(T)W_T\right) + \operatorname{Var}\left(\cos(T)W_T\right) =: A+2B+C $$
The terms above can be calculated explicitly:
$$ \begin{aligned} A &= \int_0^T \cos^2(t)dt = \frac 12 (T + \sin(T)\cos(T))\\ B &= -\cos(T) \int_0^T \cos(t) dt = -\cos(T)\sin(T)\\ C &= T \cos^2(T)\\ \end{aligned} $$