\[ \begin{aligned}W&=\int_0^TP\text{ d}t\\&=\int_0^{T} i^2R \text{ d} t\\&=RI_m^2 \int_0^{T} \sin^2(\omega t) \text{ d}t\\\\\text{RHS}&=RI_m^2(\frac{t}{2}-\frac{\sin\left(2\omega t\right)}{4\omega})\,\bigg|_0^{T}\\\\&=\frac{RI_m^2}{4\omega}(2\omega t-\sin(2\omega t))\,\bigg|_0^{T}\\\\\therefore \text{RHS}&=F(T)-F(0)\\&=\frac{1}{2}I_m^2RT\\\\\because\text{LHS}&=I^2RT\\&=\text{RHS}\\\\\therefore I^2&=\frac12 I_m^2\\\Leftrightarrow I&=\frac{\sqrt2}{2}I_m\end{aligned} \]

来源：https://blog.jiejiss.com/