probability distribution of a random variable that is uniformly distributed in [-1, 1] If $X$ is a random variable that is uniformly distributed between $-1$ and $1$, find the PDF of $\sqrt{\vert X\vert}$ and the PDF of $-\ln\vert X\vert$.
Solution or an explantion of approach wooulf work.
thank you in advance for your help

$X \sim U[-1, 1]$, $Y = \sqrt{|X|} \Rightarrow$
$$
F_Y(y) = \text{Pr}\left(Y \leq y\right), \quad y \geq 0, 
$$
is a cumulative distribution function (c.d.f.) of $Y$ (by definition).
$$
\begin{aligned}
\text{Pr}\left(Y \leq y\right) &= \text{Pr}\left(\sqrt{|X|} \leq y\right) = \text{Pr}\left(|X| \leq y^2\right) = \text{Pr}\left(-y^2 \leq X \leq y^2\right) = \\
&= \left|\begin{aligned}
&\text{since }X\sim U[-1, 1] \Leftrightarrow \text{c.d.f. of } X\text{ is }\\
&F_X(x) = 
\left\{
\begin{array}{rl}
0, & x \leq -1 \\
\frac{x-(-1)}{1-(-1)} = \frac{x+1}{2}, & -1 < x \leq 1 \\
1, & x > 1 
\end{array}
\right.
\end{aligned}\right| = \\
&= F_X(y^2) - F_X(-y^2) = 
\left\{
\begin{array}{rl}
\frac{y^2+1}{2}-\frac{-y^2+1}{2} = y^2, & 0 \leq y^2 \leq 1 \Leftrightarrow 0 \leq y \leq 1\\
1, & y^2 > 1  \Leftrightarrow y > 1
\end{array}
\right.
\end{aligned}
$$
So, the c.d.f. of $Y$, $F_Y(y) = \text{Pr}(Y \leq y)$, $\quad y \geq 0$, is equal to
$$
F_Y(y) = 
\left\{
\begin{array}{rl}
y^2, & 0 \leq y \leq 1\\
1, & y > 1
\end{array}
\right.
$$
Then, the probability density function (p.d.f.) of $Y$ equals to
$$
f_Y(y) = F'_Y(y) = 
\left\{
\begin{array}{rl}
2y, & 0 \leq y \leq 1\\
0, & y > 1
\end{array}
\right.
$$
