A projectile is launched at an angle $θ$ with respect to the surface with velocity $v_0$ (deterministic). If the angle of inclination is a uniform random variable on $[0,π/2]$, calculate the function distribution of the variable $R$ defined as the point of impact of the projectile on the ground, measured from the origin. Also calculate the expectation.
$\textbf{Hint}$: Remember that the equations of motion are given by: $$x(t)=x_{0}+\left(v_{0} \cos \theta\right) t, \quad y(t)=y_{0}+\left(v_{0} \operatorname{sen} \theta\right) t-\frac{1}{2} g t^{2} .$$
The first thing I did was determine the distance between the origin and the drop point, which is $R(\theta)=v_{0}\cos(\theta)t=(2/g)v_{0}^{2}\sin(\theta)\cos(\theta)=(v_{0}^{2}/g)\sin(2\theta)$. The latter, since at the point of impact the height is $0$, then $y_{0}+\left(v_{0} \operatorname{sen} \theta\right) t-\frac{1}{2} g t^{2}=0$. Then the greatest distance is reached in $\theta=\pi/4$, so $R(\theta)\in [0,v_{0}^2/g]$. Thus obtaining that the distribution function of $R$ is: $$F_{R}(r)=\mathbb{P}(R\leq r)=\mathbb{P}\left(\frac{v_{0}^2\sin(2\theta)}{g}\leq r\right)=\mathbb{P}\left(\theta\leq \frac{\sin^{-1}\left(\frac{gr}{v_{0}^2}\right)}{2}\right)=\frac{2}{\pi}\cdot\frac{\sin^{-1}\left(\frac{gr}{v_{0}^2}\right)}{2}=\frac{\sin^{-1}\left(\frac{gr}{v_{0}^2}\right)}{\pi}$$ That is, the density function is: $$f_{R}(r)=\frac{dF_{R}(r)}{dr}=\frac{g}{\pi\sqrt{v_{0}^4-g^{2}r^{2}}}, r\in [0,v_{0}^2/g]$$ But when I integrate the density function I don't get 1. What part is wrong?
$\sin(2\theta)$ is not one-to-one over $[0, \pi/2]$. So $P[\frac{v_0^2 \sin(2\theta)}{g} \leq r] \neq P[\theta \leq \frac{\arcsin(\frac{gr}{v_0^2})}{2}]$. Instead \begin{align*} P\Big[\frac{v_0^2 \sin(2\theta)}{g} \leq r\Big] &= P\Big[\theta \leq \frac{\arcsin(\frac{gr}{v_0^2})}{2}\Big] + P\Big[\frac{\arcsin(\pi - \frac{gr}{v_0^2})}{2} \leq \theta\Big] \\ &= 2 \times \frac{2}{\pi}\frac{\arcsin(\frac{gr}{v_0^2})}{2} = \frac{2\arcsin(\frac{gr}{v_0^2})}{\pi} \end{align*} for $r \in [0, v_0^2/g]$.