Derivation for average value input current in buck-boost converter
$$ \langle i_g(t)\rangle_{T_s}=\frac{1}{T_s} \int_{0}^{T_s} i_g(t) dt =\frac{1}{T_s} \int_{0}^{dT_s} \langle i(t)\rangle_{T_s}  dt = d(t)\langle i(t)\rangle_{T_s} $$

Lets consider i(t).  Taking the integral of i(t) with respect to t will result in ti(t).  I am assuming also Ts is a subscript in this derivation.  The steps missing would be the following:

$$ \frac{dT_s}{T_s}i(t)_{Ts} - \frac{0}{T_s}i(t)_{Ts} = d(t)i(t)_{Ts} $$

d(t) needs to be taken with respect to t which is why it appears in the final derivation.  The computation of the integral is really the only step missing in this case.