# Given a function $f(x)$ where $x$ is uniformly distributed between $a$ and $b$, how do I find the probability density function of $f$?

For example, if $f(x) = \sin x$ and $x$ is uniformly distributed on $[0, \pi]$, how is the equation found that satisfies the probability distribution function of $f(x)$? I imagine the distribution function will be greater when the derivative of $f(x)$ is closer to zero, but this is just a guess.

I apologize if this question is vague or not advanced enough, but I can't find the answer anywhere.

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Note that $\sin(x)$ increases from $x = 0$ to $x = {\pi \over 2}$, then decreases from ${\pi \over 2}$ to $\pi$, in a way symmetric about ${\pi \over 2}$. So for a given $0 \leq \alpha \leq 1$, the $x \in [0,\pi]$ for which $\sin(x) \leq \alpha$ consists of two segments, $[0,\beta]$ and $[\pi - \beta, \pi]$, where $\beta$ is the number for which $\sin(\beta) = \alpha$. In other words $\beta = \arcsin(\alpha)$.
Since $x$ is uniformly distributed on $[0,\pi]$, the probability $x$ is in $[0,\beta]$ is ${\beta \over \pi}$, and the probability $x$ is in $[\pi - \beta, \pi]$ is also ${\beta \over \pi}$. So the chance that $x$ is in one of these two segments is $2{\beta \over \pi}$. This means the probability $\sin(x) \leq \alpha$ is $2{\beta \over \pi}$, or ${2 \over \pi} \arcsin(\alpha)$. Thus this gives the distribution function of $\sin(x)$. The density function is obtained by differentiating with respect to $\alpha$; the result is ${2 \over \pi \sqrt{1 - \alpha^2}}$.
Note that if $0 \leq \alpha \leq 1$, then $0 \leq (2/\pi)\arcsin(\alpha) \leq 1$. –  Shai Covo Dec 30 '10 at 8:28