In this case,

`T=(2pi)/omega`

So

`i_("rms")=sqrt(1/T int_0^Ti^2dt)`

`=sqrt(omega/(2pi)int_0^((2pi)/omega)(i_0\ sin\ omegat)^2dt)`

`=sqrt((omega(i _0)^2)/(2pi)int_0^((2pi)/omega)sin^2omegat\ dt)`

Let's just take the integral part first, and put

`sin^2omegat=(1-cos\ 2 omegat)/2`

`int_0^(2pi//omega)sin^2omegat\ dt =1/2int_0^(2pi//omega)(1-cos 2 omegat)dt`

`=1/2[t-(sin\ 2omega t)/(2omega)]_0^(2pi//omega)`

`=1/2[((2pi)/omega-0)-(0-0)]`

`=pi/omega`

So

`i_("rms")=sqrt((omega(i_0)^2)/(2pi)int_0^((2pi)/omega)sin^2omegat\ dt)`

`=sqrt{(omega (i_0)^2)/(2pi)pi/omega}`

`=sqrt(((i_0)^2)/2)`

`=(i_0)/sqrt2`

This is what the question required us to show.

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