I've got a book (Power Electronics: Converters, Applications, and Design, Mohan, 2003) which gives the following equation:
$$L_s \frac{di_u}{dt} = \frac{{v_a}_n - {v_c}_n}{2} $$
It then goes on to say we multiply both sides by $\omega$ (rad/sec) and integrate, giving
$$\omega L_s \int_0^{I_d} di_u=\int_0^u \frac{{v_a}_n - {v_c}_n}{2} d(\omega t)$$
What I can't figure out is, after multiplying by $\omega$ on the right-hand side, why does $\omega$ get incorporated into the variable of integration? The variable was $dt$, and now we're saying it's $d(\omega t)$. Am I missing some mathematical property that would allow this?
I think I see what is happening, assuming $\omega$ is constant. Let me walk through the steps and see if it looks right. I am going to use $\omega = \frac{dr}{dt}$ (where $r$ is radians). This also means that $dr = \omega\,dt$.
Therefore: $$ L_s \frac{di_u}{dt} = \frac{{v_a}_n - {v_c}_n}{2} ~~\text{(starting equation)} \\ dr\,L_s \frac{di_u}{dt} = \frac{{v_a}_n - {v_c}_n}{2}\,dr ~~\text{(multiply both sides by $dr$)} \\ \frac{dr}{dt} L_s\, di_u = \frac{{v_a}_n - {v_c}_n}{2}\,dr ~~\text{(rearrange)} \\ \omega\, L_s\, di_u = \frac{{v_a}_n - {v_c}_n}{2}\omega\,dt ~~\text{(substitute)} \\ \omega\, L_s\, di_u = \frac{{v_a}_n - {v_c}_n}{2}d(\omega\,t) ~~\text{(constant multiplier rule)} \\ \int \omega\, L_s\, di_u = \int \frac{{v_a}_n - {v_c}_n}{2}d(\omega\,t) ~~\text{(integrate both sides)} \\ \omega\, L_s \int di_u = \int \frac{{v_a}_n - {v_c}_n}{2}d(\omega\,t) ~~\text{(constant rule)} $$