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Math noob here, I work mainly on electronics and math is not my strongest suit.

I am trying to find out the type function of total energy transfered throughout a AC half wave. If any one of you is in electronics, it's for trying to linearize the output from an thyristor/TRIAC AC chopper to a purely resistive load.

As I am just trying to find out what kind of function it is, other nitty gritty details doesnt really matter. Lets set all amplitudes to 1, phase difference to 0, and initial condition as 0 to simplify things.

Instantaneous power, P is a product of voltage of current at any point of. Both voltage and current are sinusoidal functions.

Voltage, $V(t)=\sin(\omega t)$

Current, $I(t)=\sin(\omega t)$

Instantaneous power, $P(t) = V(t)I(t) = (\sin(\omega t))^2$

The accumulative energy transferd, $E$, will be the time integral of instantaneous power

$E(t) = \int_{0}^{t} P dt = \int_{0}^{t} (\sin(\omega t))^2dt$

The circuit is off at $t \lt 0$ so let the initial condition $c=0$

And I need only information for the positive half of the result, so $t \leqslant \frac T 2$, where $T$ is the period for the voltage and current functions $V(t)$ and $I(t)$.

This is where I start to face problems. I expected that I will get some sort of trigonometric function out of this, but when I run the calculation through a spreadsheet (using trapezoidal approximation to calculate the integral), what I get is a sigmoid function that I had a hard time fitting into any of the curves of fundamental trigo functions.

I had tried $\frac{1-\cos(\omega t)}{2}$, $\frac{1-\cos^2(\omega t)}{2}$, and took a shot with $\arctan()$ due to it's sigmoid-like shape. No luck.

What exactly am I dealing with here?

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    Hint: $\cos 2x = 1 -2 \sin^2 x$ – Damien Feb 02 '22 at 07:45
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    As well as the first comment, you could also note that $sin(x) = \frac{e^{jx}-e^{-jx}}{2j}$ and express your integral in terms of complex exponentials. – Paul Feb 02 '22 at 08:39

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Thanks guys I got it. $$E(t)=\frac{1}{2}(t-\frac{1}{2\omega}sin(2\omega t)$$ Should have trusted calculus. Never thought that time itself can be part of the energy, mind bogling when you think in voltage and currents.