RMS is, by definition, about power. For current or voltage it is about finding the equivalent that would dissipate the same power in a resistor if administered as a DC level. Since power is proportional to the square of current and/or the square of voltage, you can expect that this will involve squares of functions. Also note that the square of a value, whether positive or negative, is always positive. And that's what we would expect for power, regardless of whether the current or voltage was negative or positive, itself.
In general form, we are interested in the average power given any arbitrary period:
$$\begin{align*}\overline{P}&=\frac1{t_1-t_0}\int_{t=t_0}^{t_1}P_t\:\text{d}t\end{align*}$$
However, since we are just interested in the simpler case where a resistance is assumed there are two possible cases:
$$\begin{align*}\overline{P}&=\frac1{t_1-t_0}\int_{t=t_0}^{t_1}P_t\:\text{d}t&\overline{P}&=\frac1{t_1-t_0}\int_{t=t_0}^{t_1}P_t\:\text{d}t\\\\&=\frac1{t_1-t_0}\int_{t=t_0}^{t_1}V_t\,I_t\:\text{d}t &&=\frac1{t_1-t_0}\int_{t=t_0}^{t_1}V_t\,I_t\:\text{d}t\\\\&=\frac{1}{t_1-t_0}R\int_{t=t_0}^{t_1}V_t^2\:\text{d}t &&=\frac1{t_1-t_0}\frac1{R}\int_{t=t_0}^{t_1}I_t^2\:\text{d}t\\\\\frac{\overline{P}}{R}&=\frac{1}{t_1-t_0}\int_{t=t_0}^{t_1}V_t^2\:\text{d}t &R\cdot\overline{P}&=\frac1{t_1-t_0}\int_{t=t_0}^{t_1}I_t^2\:\text{d}t\\\\V_{_\text{RMS}}&=\sqrt{\frac{1}{t_1-t_0}\int_{t=t_0}^{t_1}V_t^2\:\text{d}t} &I_{_\text{RMS}}&=\sqrt{\frac1{t_1-t_0}\int_{t=t_0}^{t_1}I_t^2\:\text{d}t}\end{align*}$$
That summarizes the reasoning behind the term root-mean-square and why current and/or voltage is squared in the process of finding this mean DC-equivalent value from varying voltages and/or currents.
Of course, if either voltage or current is zero over the entire period from \$t_0\$ to \$t_1\$ then the result is zero, as well.
So you can segment the analysis (piecewise.) Analyze the part that isn't everywhere zero from the part that is. Set \$t_0=0\$ at the start of the actively changing period, \$t_1\$ at the end of that actively changing period and the start of the inactive period, and \$T_s\$ is the full period width. Then:
$$\begin{align*}I_{_\text{RMS}}&=\sqrt{\frac1{T_s}\left[\int_{0}^{t_1}I_t^2\:\text{d}t + \int_{t_1}^{t_2=T_s}0\:\text{d}t\right]}\\\\&=\sqrt{\frac1{T_s}\int_{0}^{t_1}I_t^2\:\text{d}t}\\\\&=\sqrt{\frac1{T_s}\int_{0}^{t_1}I_p^2\sin^2\left(\omega\,t\right)\:\text{d}t}\\\\&=I_p\sqrt{\frac1{T_s}\int_{0}^{t_1}\sin^2\left(\omega\,t\right)\:\text{d}t}\end{align*}$$
And that conforms to one of your equations above.
Continuing, you know (at least in the cases where you show pictures) that \$\omega\,t_1=\pi\$:
$$\begin{align*}I_{_\text{RMS}}&=I_p\sqrt{\frac1{T_s}\int_{0}^{t_1}\sin^2\left(\omega\,t\right)\:\text{d}t}\\\\&=I_p\sqrt{\frac1{T_s}\left[\int_{0}^{t_1}\frac12\:\text{d}t-\int_{0}^{t_1}\cos\left(2\,\omega\,t\right)\:\text{d}t\right]}\\\\&=I_p\sqrt{\frac1{T_s}\left[\frac{t_1}{2}-\int_{0}^{t_1}\cos\left(2\,\omega\,t\right)\:\text{d}t\right]}\\\\&=I_p\sqrt{\frac1{T_s}\left[\frac{t_1}{2}-\frac{t_1}{\pi}\int_{0}^{\pi}\cos\left(2\,\phi\right)\:\text{d}\phi\right]}\\\\&=I_p\sqrt{\frac{t_1}{T_s}\left[\frac{1}{2}-\frac{1}{\pi}\int_{0}^{\pi}\cos\left(2\,\phi\right)\:\text{d}\phi\right]}\\\\&=I_p\sqrt{\frac{t_1}{2\,T_s}}\end{align*}$$
If \$\delta=\frac{t_1}{T_s}\$ then \$I_{_\text{RMS}}=I_p\sqrt{\frac{\delta}{2}}\$.
Which is another of your results.
So, in the first case with \$t_1=100\:\mu\text{s}\$ and \$T_s=\frac1{f_s=500\:\text{Hz}}=2\:\text{ms}\$ it follows that \$\delta=0.05\$.
Does any of this help to clarify?