Expected value of $\sin(x)$ I have a uniformly distributed random variable $ \omega $ in the range $[\frac\pi2, \frac\pi{-2}]$. Then I have the function $ s = \sin(\omega) $  I want to calculate the expected value of this function $ s $. 
So far I know that the uniformly distributed random variable can be written as $$ \omega = \frac1{\frac\pi2 - - \frac\pi2} = \frac 1\pi $$
Then I don't know if the correct way of calculating the expected value is $$ E = \int_{-\frac\pi2}^{\frac\pi2} \frac1\pi \sin(x) dx $$
or if I'm completely off. 

The expected value of any random variable $s(\omega)$ where $\omega$ is having the probability distribution function $f(\omega)$ is given by:
$$ E(s(\omega)) = \int_{-\infty}^{\infty} s(\omega)f(\omega)d\omega$$
since $\omega$ is distributed uniformly in the interval $[-\pi/2,\pi/2]$ we have $$f(\omega) = \frac{1}{(\pi/2-(-\pi/2))} = \frac{1}{\pi}$$
Now, $$E(s(\omega))=\int_{-\infty}^{\infty} \sin(\omega)f(\omega)d\omega$$
or, $$E(s(\omega))=\int_{-\infty}^{\infty} \sin(\omega)\frac{1}{\pi}d\omega$$
The limits for $\omega$ is from $[-\pi/2,\pi/2]$ so the integral is 
$$E(s(\omega))=\int_{-\pi/2}^{\pi/2} \sin(\omega)\frac{1}{\pi}d\omega$$
The answer of this integral is $0$
