Need verification for this integral

Question

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: $$U_{d} = \frac{1}{2 \cdot \pi} \cdot \int_{\alpha}^{\pi}{\hat{U} \cdot sin( \omega t) \cdot d t}$$ $$U_d = \bigg[ \frac{\hat{U} \cdot cos(\omega \alpha) - cos(\omega \pi)}{2\cdot \pi \cdot \omega} \bigg]^{\pi}_{0,698132}$$ When I compare this for $\hat{U} = 40V, \alpha = 40°, \omega = 2\cdot \pi \cdot 50Hz$ 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 $dt d\omega$ and not $d\omega t$ and I'd expect also two integral signs with a limit. I never saw something like $d\omega t$ and don't know how to handle this.

Answer 1

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

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

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.

Answer 2

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.