The inner product is defined as the integral from $a$ to $b$ of $f(x)g(x)dx$. In this case, $f(x) = x$ and $g(x) = \sin 2x$, and the interval is $[-\pi, \pi]$. So, we have:
$$\langle f, g\rangle = \int_{-\pi}^{\pi} x \sin 2x dx$$
We can solve this integral using
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