Calculate the root, mean, square of v(t) where v(t) = 0 0 < wt <= theta v(t) = Vm*sin(wt) theta < wt <= pi() v(t) = 0 pi() < wt <= pi()+theta v(t) = Vm*sin(wt) pi()+theta < wt <= 2*pi() In integral form, I am confident that it looks like this. Vrms = SQRT(1/T*INTEGRAL(Vm^2*sin^2(wt)dt)) from theta/w to pi/w AND pi+theta/w to 2*pi/w I believe it integrates like this. My thought was that I could integrate only one of the ranges and then double my answer due to symmetry in the sinusoidal function. Vrms = Vm*SQRT(w/2*pi*(t/2-sin(2wt)/4w)) evaluated from theta/w to pi/w After substitions and simplification (which I have been over about 10-15 times) I get this. Vrms = Vm - Vm*SQRT(1/pi*(theta-sin(theta)*cos(theta))) I know this is wrong because when I evaluate this equation at theta = pi, the value is zero. The correct value is Vrms = Vm/SQRT(2) when theta = pi. I have either defined the problem wrong somehow or I have performed the integral wrong. Many thanks for any help.