Evaluate the integral for output current over half a cycle:

The terms $V_m$, $R_D$, and $R_L$ are constants with respect to the integration variable $\omega t$. We can pull them out of the integral.

$$\frac{V_m}{R_D + R_L} \int_{0}^{\pi} \sin(\omega t) d(\omega t)$$

The integral of $\sin(x)$ with respect to $x$ is $-\cos(x)$.

$$\int_{0}^{\pi} \sin(\omega t) d(\omega t) = [-\cos(\omega t)]_{0}^{\pi}$$

Substitute the upper limit ($\pi$) and the lower limit (0) into the result of the integration.

$$[-\cos(\omega t)]_{0}^{\pi} = (-\cos(\pi)) - (-\cos(0))$$

We know that $\cos(\pi) = -1$ and $\cos(0) = 1$.

$$= (-(-1)) - (-1)$$ $$= 1 + 1$$ $$= 2$$

Multiply the constant term by the result of the definite integral.

$$\frac{V_m}{R_D + R_L} \times 2$$ $$= \frac{2 V_m}{R_D + R_L}$$

The value of the integral representing the output current over half a cycle is $\boxed{\frac{2 V_m}{R_D + R_L}}$.