$X$ is a random variable with probability density function:
$$f(x)=\{\begin{array}{ll}\frac{1}{2\pi }& -\pi \le x\le \pi \\ 0& \text{elsewhere}\end{array}$$
Then $E[\sin x]$ is:

The expectation $E[g(X)]$ is given by ${\int }_{-\infty }^{\infty }g(x)f(x)dx$.
$$E[\sin x]={\int }_{-\pi }^{\pi }(\sin x)\cdot \frac{1}{2\pi }dx$$
$$E[\sin x]=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }\sin xdx$$
Since $\sin x$ is an odd function and the interval $[-\pi ,\pi ]$ is symmetric about the origin, the integral is zero.
$${\int }_{-\pi }^{\pi }\sin xdx=[-\cos x{]}_{-\pi }^{\pi }=-(\cos \pi -\cos (-\pi ))=-(-1-(-1))=0$$
Thus, $E[\sin x]=0$.