Integrate both side limits Given the inductor equation
$$
L\frac{\mathrm{d}i(t)}{\mathrm{d}t} = \nu_L(t),
$$
an online lecture I found said that, by variable separation, we can integrate over a switching period (T_s) to obtain
$$
\int_{\color{red}t}^{\color{red}{t + T_s}} \mathrm{d}i = \frac{1}{L}\int_t^{t + T_s} \nu L(\tau) \mathrm{d}\tau
$$
Shouldn't it be instead
$$
\int_{\color{green}{i(t)}}^{\color{green}{i(t + T_s)}} \mathrm{d}i = \frac{1}{L}\int_t^{t + T_s} \nu L(\tau) \mathrm{d}\tau
$$
?
 A: Notice, and yes you're right:


*

*$$\text{V}_{\text{L}}(t)=\text{L}\frac{\partial\text{I}_{\text{L}}(t)}{\partial t}=\text{L}\text{I}'_{\text{L}}(t)$$


So:
$$\text{V}_{\text{L}}(t)=\text{L}\text{I}'_{\text{L}}(t)\Longleftrightarrow$$
$$\int\text{V}_{\text{L}}(t)\space\text{d}t=\int\text{L}\text{I}'_{\text{L}}(t)\space\text{d}t\Longleftrightarrow$$
$$\int\text{V}_{\text{L}}(t)\space\text{d}t=\text{L}\int\text{I}'_{\text{L}}(t)\space\text{d}t\Longleftrightarrow$$
$$\int\text{V}_{\text{L}}(t)\space\text{d}t=\text{L}\text{I}_{\text{L}}(t)\Longleftrightarrow$$
$$\frac{1}{\text{L}}\int\text{V}_{\text{L}}(t)\space\text{d}t=\text{I}_{\text{L}}(t)\Longleftrightarrow$$
$$\text{I}_{\text{L}}(t)=\frac{1}{\text{L}}\int\text{V}_{\text{L}}(t)\space\text{d}t$$
Now, for example the Voltage across the inductor is $\text{V}_{\text{L}}(t)=230\sqrt{2}\sin(100\pi t)$ and the inductor is $\frac{1}{2}\space\text{H}$:
$$\text{I}_{\text{L}}(t)=\frac{1}{\frac{1}{2}}\int230\sqrt{2}\sin(100\pi t)\space\text{d}t\Longleftrightarrow$$
$$\text{I}_{\text{L}}(t)=2\cdot230\sqrt{2}\int\sin(100\pi t)\space\text{d}t\Longleftrightarrow$$
$$\text{I}_{\text{L}}(t)=460\sqrt{2}\int\sin(100\pi t)\space\text{d}t\Longleftrightarrow$$
$$\text{I}_{\text{L}}(t)=460\sqrt{2}\int\sin(100\pi t)\space\text{d}t\Longleftrightarrow$$

Subsitute $u=100\pi t$ and $\text{d}u=100\pi\space\text{d}t$:

$$\text{I}_{\text{L}}(t)=\frac{460\sqrt{2}}{100\pi}\int\sin(u)\space\text{d}u\Longleftrightarrow$$
$$\text{I}_{\text{L}}(t)=\frac{23\sqrt{2}}{5\pi}\int\sin(u)\space\text{d}u\Longleftrightarrow$$
$$\text{I}_{\text{L}}(t)=-\frac{23\sqrt{2}\cos(u)}{5\pi}+\text{C}\Longleftrightarrow$$
$$\text{I}_{\text{L}}(t)=-\frac{23\sqrt{2}\cos(100\pi t)}{5\pi}+\text{C}$$
Notice, now that $\text{C}$ represent the initial voltage
