Hi,
$$ P = \frac{\Delta W}{\Delta T} = \frac{\int_0^{\Delta T} p(t) dt}{\Delta T} = \frac{i_0^2 R\int_0^{\Delta T} \sin^2(\omega t) dt} {\Delta T} = \frac{i_0^2 R \left( \frac{\Delta T}{2} - \frac{\sin(2 \omega \Delta T)}{4 \omega} \right) }{\Delta T} $$
Wegen $$ \omega = \frac{ 2 \pi }{\Delta T} $$ folgt $$ \sin(2 \omega \Delta T) = 0 $$
Also insgesamt $$ P = \frac{1}{2}i_0^2 R $$