Evaluate conditional expectation of $X^2$ given $\sin(X)$ and uniform distribution of $X$ over $[0,\pi]$ I want to find the following conditional expectation: $\mathbb{E}(X^2\mid\sin(X))$ where $X$ is distributed uniformly on $[0, \pi]$.


I guess if $\sin(X)=y$, then $X^2$ can be either $\arcsin(y)$ or $\pi-\arcsin(y)$. And maybe the formal proof should be based on this?
Let $X$ have the pdf $\rho_X$. Then
$$\rho_{g(X)}(y) =
\int_{\mathbb R} \delta(g(\tau) - y) \rho_X(\tau) d\tau =
\sum_{\tau: \,g(\tau) = y} \mu(\tau),
\quad \mu(\tau) = \frac {\rho_X(\tau)} {|g'(\tau)|},\\
\rho_{X, g(X)}(x, y) =
\int_{\mathbb R} \delta(\tau - x) \delta(g(\tau) - y) \rho_X(\tau) d\tau =
\sum_{\tau: \,g(\tau) = y} \mu(\tau) \delta(\tau - x), \\
\mathbb E(f(X) \mid g(X) = y) =
\frac {\int_{\mathbb R} f(x) \rho_{X, g(X)}(x, y) dx} {\rho_{g(X)}(y)} =
\frac
 {\sum_{\tau: \,g(\tau) = y} \mu(\tau) f(\tau)}
 {\sum_{\tau: \,g(\tau) = y} \mu(\tau)}, \\
\mathbb E(f(X) \mid g(X)) =
\frac
 {\sum_{\tau: \, g(\tau) = g(X)} \mu(\tau) f(\tau)}
 {\sum_{\tau: \, g(\tau) = g(X)} \mu(\tau)}.$$
If $\rho_X(x) = [0 &lt; x &lt; \pi]/\pi$, $g(x) = \sin(x)$ and $X$ takes a value in $(0, \pi)$, the sums have two non-zero terms, corresponding to $\tau = X$ and $\tau = \pi - X$.
Continue with your approach and note $y=\sin x$ has two solutions, i.e. over either [0, $\pi/2$] or [$\pi/2, \pi]$. The density function for $y$ is then,

$$ f(y) = \frac2\pi \frac{dx}{dy} 
= \frac2\pi \frac1{\sqrt{1-y^2}}$$

Thus, 

$$E[X^2|\sin(X)]= \int_0^1 \arcsin^2(y)f(y)dy
= \frac2\pi \int_0^1 \frac{\arcsin^2(y)}{\sqrt{1-y^2}}dt=\frac{\pi^2}{12}$$