Constant angular acceleration suvat-related equations: OCR Further Maths A Level "Focus on Proof" page $130$ Angular displacement is represented by the symbol $\theta.$ Angular velocity is the rate of change of angle and is denoted by $\omega$ or $\overset{.}{\theta}. $ Angular acceleration, which is the rate of change of angular velocity, is denoted by $\overset{..}{\theta},\ $ or $\ \alpha,$ as we now consider angular acceleration to be constant.
My book then derives/obtains the following equation: $\omega_1 = \omega_0 + \alpha t,\ $ and I am vaguely OK with the derivation, although I personally find it non-rigorous and handwavy.
Then it asks to:

Show by integration, that:
$$ \theta = \omega_0 t + \frac{1}{2} \alpha t^2.$$

My attempt:
$$\int_{0}^{t}\omega_1\ dt = \int_{0}^{t} \omega_0 + \alpha t\ dt \implies \omega_1 t= \omega_0 t + \frac{1}{2} \alpha t^2. $$
This suggests that $\omega_1 t = \theta,\ $ which doesn't make sense to me. What is going on? I think there is some speed = distance / time thing but I'm confused as to how it is being applied here.
 A: I will address why you have to integrate $\omega_1$ with respect to $t$:
$\omega_1$ is a function of $t$; and $$\omega_1(t)=\dot\theta=\frac{d\theta}{dt}$$ If $\omega_1$ were constant, then your reasoning of the integral $\displaystyle\int_0^t \omega_1(t)dt$ being equal to $\omega_1t$ would be correct. However, that is not the case, as $\omega_1$ clearly varies with $t$.
Thus we separate the differential equation and write : $$\int\omega_1(t)dt=\int d\theta$$ so that we get $$\theta=\int( \omega_0+\alpha t )dt$$$$=\omega_0 t+\frac12 \alpha t^2+C.$$ Now we have to find $C$ so we use the initial values. In the problem it should be given that $\theta$ at time $t=0$ is equal to $0$ (which you seem to have assumed). Otherwise, for a more general case, if we take the initial angular displacement at time $t=0$ to be $\theta_0$, we get $$\theta= \theta_0+\omega_0 t+\frac12 \alpha t^2.$$

POSTSCRIPT: If you want to write the integral $\displaystyle\int_0^t\omega_1 t dt$ as $c\cdot t$, where $c$ is some constant, then $c$ must be the time average of $\boldsymbol\omega_1$ during the time interval $0-t$, i.e, $\displaystyle\int_0^t \omega_1(t)dt=\langle\omega_1(t)\rangle t$, where $\langle\omega_1(t)\rangle$ is the time average.
