Find the pdf of the area of an isosceles triangle given the pdf of its vertex angle This question is a follow-up to a discussion on MESE.  On the linked post, the OP asked about the following question (which I will paraphrase slightly):

Suppose $\theta$ is a random variable which takes its values in $[0, \pi/2]$ and with a pdf proportional to $\theta(\pi-\theta)$.  Let $\triangle ABC$ be a triangle with congruent legs of length $\ell$ and vertex angle $\theta$.  Find the pdf for the area of $\triangle ABC$.

On the original question, the OP wanted to know whether this question was fair, reasonable, too difficult, etc.  I have a different question: Is my solution to the problem (below) correct?
Solution.  First, let $f_\theta = k\theta(\pi-\theta)$ be the pdf for $\theta$.  Then
$$1 = \int_0^{\pi/2} f(\theta) \, d\theta = k \int_0^{\pi/2} \theta(\pi-\theta) \, d\theta = k\frac{\pi^3}{12},$$
so $k = \frac{12}{\pi^3}$.  Then the cdf for $\theta$ is
$$F_\theta(x) = \frac{6x^2}{\pi^2} - \frac{4x^3}{\pi^3}$$
Now, the area of the triangle is related to the angle by the equation $A = \frac{\ell^2}2 \sin\theta$, or equivalently $\theta = \arcsin\left(\frac{2A}{\ell^2}\right)$.  Therefore, if we let $F_A$ be the cdf of $A$, then:
$$\begin{aligned} F_A(x) &= P(A \le x) \\ &=P\left(\theta \le \arcsin \left( \frac{2x}{\ell^2} \right) \right) \\ &=F_\theta\left(\arcsin\left(\frac{2x}{\ell^2} \right) \right) \\
&=\frac{6 \left(\arcsin\left(\frac{2x}{\ell^2} \right) \right)^2}{\pi^2} - \frac{4 \left(\arcsin\left(\frac{2x}{\ell^2} \right) \right)^3}{\pi^3} \\
\end{aligned}$$
and the pdf of $A$ is therefore
$$f_A(x) = \frac{d}{dx} F_A(x)$$
which -- assuming I haven't made any errors in differentiation -- works out to be
$$f_A(x) = \frac{24 \arcsin\left(\frac{2x}{l^2}\right) \left(\pi - \arcsin\left(\frac{2x}{l^2}\right) \right)}{\pi^3 \sqrt{l^4 - 4x^2}}$$
Are there any errors in the above reasoning, or in the work leading to the solution?
Can this result be simplified any further?  Is there an easier approach to this problem?
 A: Thank you for the reference to my test question, mweiss. There are several approaches one can take with the question. I have a very long solution (3 pages) for the benefit of the students that breaks it down in exhaustive detail. Here is what I can tell you so far (before others make any comments) and before I tell you if the answer is correct.
A proportional probability density function to $x (\pi -x)$ (I notice you used theta, which is nice) at each point within $x\in{(0,\displaystyle\frac{\pi}{2}})$
is represented by:
$f(x)=\begin{cases} kx(\pi-x) & \text{if}& x\in [0,\pi/2]\\c & \text{if}& x\not\in [0,\pi/2]\end{cases}$
To find k, impose the following condition:
$1=\displaystyle\int_{-\infty}^{\infty}f(x)dx=\displaystyle\int_{0}^{\pi/2}x(\pi-x)dx$
The area is given by:
$A(x)=\dfrac{1}{2}L^2sin(x)$ or your equivalent (you list $\theta = \arcsin\left(\frac{2A}{\ell^2}\right)$ and that is fine too (I used that form in one of my alternative solutions).
Now, here you can also use the "Change of Variable" Theorem used in Statistics. Are you familiar with it?
We can transform the variable to:
$g(a)=f(A^{-1}(a))\dfrac{1}{A'(A^{-1}(a))}$
$a\in [0,L^2/2]$
I am awful when it comes to Latex and I hardly know it, so I hope it comes out correctly.
Another way you can do it: Sorry, had to do it with an image because I do not feel completely confident in Latex.
solution
