What would you like to understand?

The sine term is eliminated because the integrand $\sin(\omega t)$ is an odd function with respect to $t$, and the integration interval $[-1, 1]$ is symmetric about the origin. The definite integral of an odd function over a symmetric interval is always zero.

Conditions: The integration interval is $[-1, 1]$.; The function being integrated includes $\sin(\omega t)$.; $\omega$ is a constant with respect to the integration variable $t$.

The sine term is eliminated because the integrand $\sin(\omega t)$ is an odd function with respect to $t$, and the integration interval $[-1, 1]$ is symmetric about the origin. The definite integral of an odd function over a symmetric interval is always zero.

Conditions: The integration interval is $[-1, 1]$.; The function being integrated includes $\sin(\omega t)$.; $\omega$ is a constant with respect to the integration variable $t$.