In the following picture,
I'm trying to compute the average value of the blue wave which represents the output of a half-wave rectifier circuit. In the book, it is $V_o \approx (1/\pi)V_s - V_D/2$ where the black wave is a sine wave. I'm not sure how the average is computed.
Note that $v_0(t) = \max(v_s(t)-v_D,0)$.
In the above case, the signal $v_s$ is a $T$-periodic $\sin $ wave, you want to compute ${1 \over T} \int_0^T v_0(t) dt$.
Since $v_s$ is $T$-periodic, it follows that $v_0$ is too, we can shift the time origin to suit the computation.
Take $v_s(t) = V_s \sin (2 \pi {t \over T})$, note that if $V_s \le V_D$, then $v_0 = 0$, so we will assume that $V_S > V_D$.
Let $t^*$ be the smallest positive time such that $v_s(t^*) = V_D$, that is $t^* = {T \over 2 \pi } \arcsin {V_D \over V_s}$.
Then \begin{eqnarray} {1 \over T} \int_0^T v_0(t) dt &=& {1 \over T} \int_0^{T\over 2} v_0(t) dt \\ &=& {2 \over T} \int_0^{T \over 4} v_0(t) dt \\ &=& {2 \over T} \int_{t^*}^{T \over 4} v_0(t) dt \\ &=& {2 \over T} \int_{t^*}^{T \over 4} (v_s(t)-V_D) dt \\ &=& {2 \over T} \int_{t^*}^{T \over 4} (V_s \sin (2 \pi {t \over T})-V_D) dt \\ &=& {2 \over T} \left [ -{T \over 2 \pi} V_s \cos (2 \pi {t \over T})-V_D t \right ]_{t^*}^{T \over 4} \\ &=& {1 \over \pi} V_S \cos (2 \pi {t^* \over T}) - {1 \over 2} V_D + 2 V_D {t^* \over T} \\ &=& {1 \over \pi} V_S \sqrt{1 - ({V_D \over V_S})^2 } - {1 \over 2} V_D + {1 \over \pi} V_D \arcsin {V_D \over V_s} \\ &=& {1 \over \pi} V_S \left[ \sqrt{1 - ({V_D \over V_S})^2 } + {V_D \over V_S}\arcsin {V_D \over V_s}\right ] - {1 \over 2} V_D \end{eqnarray} If we expand this expression in a Taylor series in terms of $V_D$, then the first two terms are ${1 \over \pi} V_S - {1 \over 2} V_D$.
Note, from a less accurate and less computationally intensive perspective, one could approximate the output as $v_0(t) = V_s \sin (2 \pi {t \over T}) -V_D$ for the first half cycle and $v_0(t) = 0$ for the second. This will yield the expression ${1 \over \pi} V_S - {1 \over 2} V_D$.
