Answer
$ (\frac{1}{5}t^6-\frac{1}{3}t^4+2t+1)\ rad$
Work Step by Step
The angular position $\theta_o =1\ rad$
Also, we have:
Angular speed $\omega =(\frac{6}{5}t^5-\frac{4}{3}t^3+2)$
Since Angular speed is the rate of change of angular displacement;
$\omega = \frac{d\theta}{dt}$
$d\theta = \omega dt$
Integrate with respect to its limits:
$\int_{\theta_o}^{\theta}d\theta =\int_{0}^{t}\omega dt$
$\theta -\theta_o = \int_{0}^{t}(\frac{6}{5}t^5-\frac{4}{3}t^3+2)dt$
$\theta -1\ rad = \frac{6}{5}(\frac{t^6}{6})-\frac{4}{3}(\frac{t^4}{4})+2t$
$\theta -1\ rad = \frac{1}{5}t^6-\frac{1}{3}t^4+2t$
$\theta = (\frac{1}{5}t^6-\frac{1}{3}t^4+2t+1)\ rad$
