A projectile is fired at an angle $\theta$ above the earth with a velocity $V$. Assuming that $\theta$ is an RV with PDF A projectile is fired at an angle $\theta$ above the earth with a velocity $V$.  Assuming that $\theta$ is an RV with PDF
Let $f(\theta)= \begin{cases} \frac{12}{\pi} ; \frac{\pi}{6} < \theta < \frac{\pi}{4} \\ 0 ; o.w. \end{cases}$
find the PDF of the range R of the projectile, where $R = V^2 sin2\theta /g$, g being
the gravitational constant.
I was thinking of usinf the formula but i don't think   able to continue. I dont need the complete answer but some hints regarding the computation will be helpful
 A: When you have a random variable $X$ with know pdf $\rho_X(x)$ and another variable of the form $Y = f(X)$ whose pdf you want to compute, the easiest way to perform the computation is usually to use the following lemma :

The function $\rho_Y(y)$ is the pdf of $Y$ if and only if for any continuous bounded function $\phi$, we have :
$$\mathbb E[\phi(Y)] = \int  \phi(y)\rho_Y(y) \text d y \tag 1$$

To actually compute $\rho_Y(y)$, we take an arbitrary continuous bounded function $\phi$ and we write :
$$\mathbb E[\phi(Y)] = \mathbb E[\phi\circ f(X)] = \int \phi\circ f(x)\rho_X(x)\text dx$$
Then, a change of variable should bring you to an integral as in $(1)$, where you can identify $\rho_Y$.
Edit We can do basically the same thing using another formula :
$$\mathbb P(Y\in[a,b]) = \int_{a}^b \rho_Y(y)\text dy = \mathbb P(X \in f^{-1}([a,b])) = \int_{f^{-1}([a,b])} \rho_X(x)\text dx$$To compute the last integral, determine explicitely $f^{-1}([a,b])$ and do a change of variable to get back to an integral between $a$ and $b$.
