Efficiency of Half wave Rectifier and input RMS value
The "DC output power" is not really calculated on only one cycle: it is computed on the whole \$2\pi\$ cycle, but only one half cycle contributes to it (indeed, because the diode is conducting). If it were actually computed on only one half cycle, it would have the same power of a full-wave rectified sinusoid, which it doesn't.

Consider the RMS formula: \$V_{rms}=\sqrt{\frac{1}{T}\int_0^{T}v^2(t)\mathrm{dt}}\$ where \$v(t) = V\sin(\omega t)\$ is \$T\$-periodic (\$V\$ is the maximum value of the voltage). Obviously, \$v^2(t) = V^2\sin^2(\omega t)\$ is only \$T/2\$-periodic; thus \$V_{rms}=\sqrt{\frac{1}{T}\int_0^{T}\sin^2(\omega t)\mathrm{dt}} = \sqrt{\frac{1}{T/2}\int_0^{T/2}\sin^2(\omega t)\mathrm{dt}}\$ doesn't depend on its calculation being made in a half or full cycle.

However, what is of interest in evaluating the efficiency is not \$V_{rms}\$, but \$P_{rms}\$. The latter is obtained by calculating the root mean square of the absorbed power \$p(t) = v(t)i(t)\$, which is \$0\$ during the half cycle in which the half wave rectifier doesn't conduct; it is still evaluated considering a whole cycle, but only half a cycle contributes to it:
\begin{equation}
P_{rms} = \sqrt{\frac{1}{T}\int_0^{T}\left[VI\sin^2(\omega t)\right]^2\mathrm{dt}} = \sqrt{\frac{1}{T}\int_0^{T/2}\left[VI\sin^2(\omega t)\right]^2\mathrm{dt} + \frac{1}{T}\int_{T/2}^{T}\left[V\sin(\omega t)\cdot0\right]^2\mathrm{dt}} = \sqrt{\frac{1}{T}\int_0^{T/2}\left[VI\sin^2(\omega t)\right]^2\mathrm{dt}+0} 
\end{equation}