Spread in the Sengupta-Mitra distribution $p(\lambda) = C\frac{\sqrt{(\lambda_{\max}-\lambda)(\lambda-\lambda_{\min})}}{\lambda}$ is defined in the interval $[\lambda_{\min},\lambda_{\max}]$ and zero outside the interval.
By spread, I assume, we mean the variance $\big(\operatorname{Var}(\lambda)\big)^2 = \langle \lambda ^2 \rangle - \langle \lambda \rangle ^2$.
Now, I've tried computing the integral $\displaystyle \langle \lambda ^2 \rangle = \int_{\lambda_{\min}}^{\lambda_{\max}}\lambda^2p(\lambda) \, d\lambda$. But, it doesn't solve that easily for me.
Any guess on an alternative way of finding the so-called spread?
Context.
$p$ describes the distribution of eigenvalues of the Wishart matrix.
 A: First let us complete the square:
\begin{align}
& (\lambda_\max-\lambda)(\lambda - \lambda_\min) \\[10pt] = {} & -\Big(\lambda^2 -(\lambda_\min+\lambda_\max) \lambda\Big) - \lambda_\min \lambda_\max \\[10pt]
= {} & -\left(\lambda^2 -(\lambda_\min+\lambda_\max) \lambda + \left(\frac{\lambda_\min+\lambda_\max} 2 \right)^2 \right) - \lambda_\min \lambda_\max + \left(\frac{\lambda_\min+\lambda_\max} 2 \right)^2 \\[10pt]
= {} & -\left( \lambda- \frac{\lambda_\min + \lambda_\max} 2 \right)^2 + \left( \frac{\lambda_\max - \lambda_\min} 2 \right)^2
\end{align}
This suggests a substitution:
\begin{align}
\left( \frac{\lambda_\max - \lambda_\min} 2 \right) \sin\theta & = \lambda - \frac{\lambda_\min+\lambda_\max} 2 \\[10pt]
\left( \frac{\lambda_\max - \lambda_\min} 2 \right) \cos\theta \, d\theta & = d\lambda
\end{align}
The integral then becomes
$$
\int_{\lambda_\min}^{\lambda_\max} \lambda^2 p(\lambda) \, d\lambda
$$
$$
= \int_{-\pi/2}^{\pi/2} \Big( (\lambda_\min+\lambda_\max) + (\lambda_\max-\lambda_\min)\sin\theta) \Big) (\cos\theta) (\lambda_\max - \lambda_\min) (\cos\theta \, d\theta).
$$
