In this case, T = 2π.

So

`i_("rms") =sqrt(1/Tint_0^Ti^2dt)`

`=sqrt(1/(2pi)int_0^(2pi)(3+2\ cos\ t)^2dt)`

Now

`(3+2\ cos\ t)^2=9+12\ cos\ t+4\ cos^2 t`

Since `cos\ 2t = 2 cos^2 t - 1`, it follows that `cos^2t=(cos 2t+1)/2`.

So

`(3+2\ cos\ t)^2 =9+12 cos\ t+4((cos\ 2t+1)/2)`

`=9+12 cos\ t+2(cos\ 2t+1)`

`=11+12\ cos\ t+2\ cos\ 2t`

So

`int_0^Ti^2dt =int_0^(2pi)(11+12\ cos\ t+2\ cos\ 2t)dt`

`=[11t+12\ sin\ t+sin\ 2t]_0^(2pi)`

`=22pi`

And, finally:

`i_("rms") =sqrt(1/Tint_0^Ti^2dt) `

`=sqrt((22pi)/(2pi))`

`=sqrt11`

`=3.317`

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