Probability of a vector lying in an interval

Suppose we have a vector $vec{v}$ with constant length with direction specified by $theta$ in a random direction w.r.t the $x$ axis.

How can I compute the probability of the $x$ component of $vec{v}$ lying in the range $v_{x}$ to $v_{x} + mathop{dx}$


By working and looking online, I have been able to figure out the following:

Let $|v|$ denote the magnitude of the vector $v$. Then,

$v_{x}$ = $|v|cos(theta)$

We have $P(v_{x} leq u) = P(vcos(theta) leq u) = P(vcos(theta) leq u, pi leq theta leq 2pi) + P(vcos(theta) leq u, pi leq theta leq 2pi)$.

Let the variable $w = u/v$. Then, our probability $P(v_{x} leq u)$ equals

$P(theta geq arccos(w), 0 leq theta leq pi) + P(theta leq arccos(w), pi leq theta leq 2pi) = frac{|pi – arccos(w)|}{pi} $


However, I’m not sure how I’m supposed to proceed. I don’t see how this relates, since it doesn’t introduce $mathop{dx}$? Could someone please help me from here?

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