Let $X$ be a continuous random variable with density function
$f_{X}(x)=\frac{3}{8}(x+1)^{2}, \; -1 < x < 1$.
Find the density function of $Y=1-X^{2}$. The answer given is
$f_{X}(x)=\frac{3}{8}(1-y)^{-1/2} + \frac{3}{8}(1-y)^{1/2}, \; 0 < y < 1$.
can someone help me
until now i thought it was something like that
$f_{Y}(y)=\frac{3}{16}(1-\sqrt{1-y})^{2}(1-y)^{-1/2} + \frac{3}{16}(1+\sqrt{1-y})^{2}(1-y)^{-1/2}, \; 0 < y < 1$.