calculate ${\int}_{0}^{\pi}{\int}_{0}^{x}\mathrm{log}(\mathrm{sin}(x-y))dydx$

I was asked to find the integral ${\iint}_{A}\mathrm{log}(\mathrm{sin}(x-y))dxdy$ where $A$ is the triangle $y=0,x=\pi ,y=x$ in the first quadrant.

I was given a hint: evaluate ${\int}_{0}^{\pi}\mathrm{log}(\mathrm{sin}(t))dt$ using symmetry.

What I did:

I inferred from the hint that the variable change $t=x-y$ is the way to go, so $t=x-y$, and since we integrate by $y$ first, then $x$ is a "constant" and $dt=-dy$, and since $y$ transitions from $0$ to $x$, then $t$ transitions from $x$ to $0$, so we can rewrite the integral:

${\int}_{0}^{\pi}{\int}_{0}^{x}\mathrm{log}(\mathrm{sin}(x-y))dydx={\int}_{0}^{\pi}{\int}_{x}^{0}-\mathrm{log}(\mathrm{sin}(t))dtdx={\int}_{0}^{\pi}{\int}_{0}^{x}\mathrm{log}(\mathrm{sin}(t))dtdx$

And here I am stuck. Firstly, I don't know how ${\int}_{0}^{\pi}\mathrm{log}(\mathrm{sin}(t))dt$ is related to the question, since in the question the limits are $0$ and $x$. not $0$ and $\pi $. They are not the same thing (even though $x$ transitions from $0$ to $\pi $).

But even if I did, how would I evaluate ${\int}_{0}^{\pi}\mathrm{log}(\mathrm{sin}(t))dt$??