Find PDF of random variable Y=X^2 using CDF of random variable X

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Random variable $X$ - length of square's side. find PDF $f(x)$ of square's area.

$$X: f(x) = \begin{cases} \frac{1}{2}\sin(x), \ x\in[0, \pi] \\ 0, \ x\in[\pi,2\pi] \end{cases}$$

That's what I have come up with:

$f(x)$ doesn't have pdf because $\int_{-\infty}^{\infty}f^{'}(x) \ne 1$

I need to find $Y=X^2$

$$g_1^{-1}(x)=\sqrt{x}$$

$$g_2^{-1}(x)=-\sqrt{x}$$

cdf: $f_Y(x)=f_X(g_1^{-1}(x)\lvert\frac{x}{dx}g_1^{-1}(x)\rvert + f_X(g_2^{-1}(x)\lvert\frac{x}{dx}g_2^{-1}(x)\rvert \ \Rightarrow f_Y(x)=\frac{\sin(\sqrt{x})}{2\sqrt{x}}$

pdf: $f_Y(x) = cdf(f_Y(x))$