Given the probability density of random variable $X$: $$f(x) = \begin{cases} \frac{2x}{\pi^2}, & {0 \lt x\lt \pi,} \\[2ex] 0, & {others.} \end{cases}$$
Find the probability density of $Y = \sin X$.
Here is my work: $$ h(y): X=\arcsin Y $$ $$f_Y(y) = \begin{cases} f_X[h(y)]|h^\acute ' (y)|, & {0 \lt y \lt 1,} \\[2ex] 0, & {others.} \end{cases}$$
I got: $$f(y) = \begin{cases} \frac{2\arcsin y}{\pi^2 \sqrt {1-y^2}}, & {0 \lt y\lt 1,} \\[2ex] 0, & {others.} \end{cases}$$
But the answer is: $$f(y) = \begin{cases} \frac{2}{\pi \sqrt {1-y^2}}, & {0 \lt y\lt 1,} \\[2ex] 0, & {others.} \end{cases}$$
Frist of of all calculate
$$F_X(x)=\int_{0}^{x}\frac{2}{\pi^2}t dt=\frac{x^2}{\pi^2}$$
then look at the following drawing
By definition you have
$$\mathbb{P}[Y>y]=F_X(\pi-\alpha)-F_X(\alpha)=\frac{(\pi-\alpha)^2}{\pi^2}-\frac{\alpha^2}{\pi^2}=1-\frac{2\alpha}{\pi}$$
Thus
$$F_Y(Y)=1-\mathbb{P}[Y>y]=\frac{2\alpha}{\pi}$$
derivating you have your correct solution, having set $\alpha=\arcsin(y)$