Setting up the expected value for $x_t=\sin(2\pi U t)$. We have the series $x_t=\sin(2\pi U t)$ where $t=1,2,3,\ldots$ and $U$ is uniform on the interval $(0,1)$. I have to find the expected value of $x_t$. I always thought that if $X$ is a continuous random variable then $$E(X)=\int_{-\infty}^{\infty} xf_x(x) \, dx.$$ In this case I was thinking it be $$E(x_t)=\int_0^1 u\sin(2\pi U t) \, du$$ rather than $$E(x_t)=\int_0^1 \sin(2\pi U t) \, du.$$ I think the first way is correct, but I don't see why the second way would be correct since we are trying to vary $U$. I am not sure. Thanks

U is uniformly distributed hence $f_u \,du = 1 \, du$. The function you are taking the expected value of is $x_t (U) = \sin(2\pi Ut)$. Thus, you want $\Bbb E_U[x_t]$ = $\int_0^1 x_t f_u \,du$ = $\int_0^1 \sin(2\pi Ut)\cdot 1\, du$ 
For further information, look up the box muller method for simulating normal random variables. It's very similar.
