Load RMS voltage of a bridge rectifier

I was given a homework assignment shown below and I have a few quick questions that I appear to be stuck on. Any help would be appreciated.

I can get Part i) and the answer is 168.64 degrees.

Part ii) I can't seem to get correct. My professor gave us the equation:

$$\displaystyle V_{L_{rms}} = \frac{1}{\pi} \int_{Q_{1}}^{Q_{2}} (V_p\sin(\theta)-1.4)^2 d\theta$$

with Q1&2 being the two points at the which diode conducts (5.68 and 174.32 degrees respectively). When I integrate this is, I get huge numbers like 249Vrms which doesn't seem be correct. Does anyone have any guidance on what I could possibly be doing wrong?

Thanks

• Are you correctly converting θ to radians where required? Dec 9, 2019 at 16:04
• I was just about to comment on the same thing Dave just said. You seem to have $\pi$ in your integral formula but you're also talking about degrees and not radians.
– user103380
Dec 9, 2019 at 16:06
• Yes I tried with both radians and degrees, I edited the post to add my workings, Thanks to both of ye in advance.
– S_G
Dec 9, 2019 at 16:36