Given that the random variable $X$ is having pdf given by: $$f_X(x)= \begin{cases}\frac{2}{3\pi} & \text { for } \frac{-\pi}{2} \leq x \leq \pi, \\ 0 & \text { else } \end{cases}$$ Find the CDF of the random variable $Y=\sin X$ without finding the pdf.
My try:
Case $1.$ When $y \leq -1$, we have $F_Y(y)=0$
Case $2.$ When $y \in (-1,0)$, we have $$F_Y(y)=P(-1<\sin X \leq y)=P(-\frac{\pi}{2}<\sin^{-1}(\sin X)\leq \sin^{-1}(y))=P(\frac{-\pi}{2}<X \leq \sin^{-1}(y))$$ So we get $$F_Y(y)=\int_{\frac{-\pi}{2}}^{\sin^{-1}(y)}\frac{2}{3\pi}{dy}=\frac{2}{3\pi}\left(\sin^{-1}y+\frac{\pi}{2}\right),\:-1<y<0$$
But i am unable to do for $0<y<1$