Full-wave controlled bridge rectifier and the RMS voltage derivation?
Following the assumption that there is a purely resistive load: by the definition of the root mean square, you can easily get to:

$$V_{rms} = \sqrt{\frac{1}{\pi} \int_\alpha^\pi V_m^2 \sin^2{\omega t} \text{ d}(\omega t)}$$

without any trouble, for any signal. The rest is trig identities and calculus.

$$V_{rms} = V_m\sqrt{\frac{1}{\pi}\left( \int_\alpha^\pi \frac{1}{2} -\frac{\cos{\omega t}}{2} \text{ d}(\omega t)\right)}$$

$$V_{rms} = V_m\sqrt{\frac{1}{\pi} \left(\left[\frac{\omega t}{2}\right]_\alpha^\pi - \left[\frac{\sin{2\omega t}}{4}\right]_\alpha^\pi\right)}$$

$$V_{rms} = V_m\sqrt{\frac{1}{\pi} \left(\frac{\pi}{2} - \frac{\alpha}{2} +\frac{\sin{\alpha}}{4}\right)}$$

$$\boxed{V_{rms} = V_m \sqrt{\frac{1}{2}-\frac{\alpha}{2\pi}+\frac{\sin{2\alpha}}{4\pi}}}$$