The range, $r$, of a cannon projectile under the influence of gravity, $g$. It is fired with a velocity, $v$. Its range is determined by the angle that it's barrel makes with the ground, theta. Given $g = 9.8 m/s$ and $v = \sqrt{980} m/s$. Theta is a random variable uniformly distributed between $0$ and $\frac{\pi}{2}$.
I know that the range can be found using $r = V^2 . \frac{sin(2\theta)}{g}$ which is $r=100sin(2\theta)$ once simplify. So my question is how would I find the mean and Taylor Series expansion of the range and theta?
Edited: more context added.
Hint 1: use the Taylor Series $$ \sin(x)=\sum_{k=0}^\infty(-1)^k\frac{x^{2k+1}}{(2k+1)!} $$ Hint 2: the mean of $\sin(2\theta)$ over $\left[0,\frac\pi2\right]$ is $$ \begin{align} \frac2\pi\int_0^{\pi/2}\sin(2\theta)\,\mathrm{d\theta} &=\left.-\frac1\pi\cos(2\theta)\,\right|_0^{\pi/2}\\[6pt] &=\frac2\pi \end{align} $$