Jtdylktuy

I have to solve an integral. It's the equation to calculate the rectified value of the output voltage generated by a controlled half-wave rectifier (with a thyristor).
$$U_{d}(alpha) = f(U, alpha) $$
$$U_{d} = frac{1}{2 cdot pi} cdot int_{alpha}^{pi}{hat{U} cdot sin( omega t) cdot d omega t}$$

Because I didn't know what to that $d omega t$ meant and because I knew this equation a bit different I was nasty and wrote it as:
$$U_{d} = frac{1}{2 cdot pi} cdot int_{alpha}^{pi}{hat{U} cdot sin( omega t) cdot d t}$$
$$U_d = bigg[ frac{hat{U} cdot cos(omega alpha) - cos(omega pi)}{2cdot pi cdot omega} bigg]^{pi}_{0,698132}$$
When I compare this for $hat{U} = 40V, alpha = 40°, omega = 2cdot pi cdot 50Hz$ with what I should get, it's obvious that there must be something wrong. (I'd expect $~23.25V$ and got something way below)

The first issue seems to be obvious $pi$ and $alpha$ aren't temporal sizes. My idea was that this was meant as a multiple integral but then I had expected something like $dt domega$ and not $domega t$ and I'd expect also two integral signs with a limit. I never saw something like $domega t$ and don't know how to handle this.

I believe that $omega t$ is just the variable name. When you replace it with a single letter like $tau$, you get:

$U_d = frac{1}{2pi} int_{alpha}^{pi} hat{U}sin(tau)dtau$.

With your given values, this gives $U_d approx 11.2$ V. Try making the substitution $tau = omega t$, with $omega$ constant to understand this notation. I believe $tau$ is the phase.

I think it's meant to be $mathrm{d}(wt)$, i.e. integral with respect to $omega t$. If this is the case, then your resuls has an extra $omega^{-1}$ factor in it.