Bessel again

$\text {We begin by utilizing the following real-valued variation of the Jacobi-Anger expansion:}$

$\\[3mm]$ $\cos(z\cos (\theta ))=J_{0}(z)+2\sum_{n=1}^{\infty}(-1)^nJ_{2n}(z)\cos(2n\theta)$

$\\[3mm]$ $\text{Taking the second partial derivative of both sides with respect to}$ $\theta$ , $\text{ and dividing by 2 on both sides, we get:}$

$\\[3mm]$ $\frac{\partial^2 }{\partial \theta^2}(\cos(z\cos (\theta)))=2\sum_{n=1}^{\infty}(-1)^n(2n)^2J_{2n}(z)(-\cos(2n\theta))$

$\\[3mm]$ $\Rightarrow \frac{z [\cos(\theta) \sin(z \cos(\theta)) - z \sin^2(\theta) \cos(z \cos(\theta))]}{2}=\sum_{n=1}^{\infty}(-1)^n(2n)^2J_{2n}(z)(-\cos(2n\theta))$

$\\[3mm]$ $\text{It should be apparent that the right hand side of the previous step generates the numerator of the expression in our original integral when } \\ \theta=k\pi \text{, where } k\in \mathbb{Z}. \text{ Plugging this in for }\theta,\text{ we get:}$

$\\[3mm]$ $-\sum_{n=1}^{\infty}(-1)^n(2n)^2J_{2n}(z)=\frac{z\sin(z)}{2}$

$\\[3mm]$ $\text{Similarly, for the denominator, we utilize the following real-valued variation of the Jacobi-Anger expansion:}$

$\\[3mm]$ $\sin(z\cos (\theta ))=-2\sum_{n=1}^{\infty}(-1)^nJ_{2n-1}(z)\cos[(2n-1)\theta)]$

$\\[3mm]$ $\text{Taking the second partial derivative with respect to } \theta \text{, and dividing by 2 on both sides, we get:}$ $\\[3mm]$ $\frac{\partial^2 }{\partial \theta^2}(\sin(z\cos (\theta ))=-2\sum_{n=1}^{\infty}(-1)^n(2n-1)^2J_{2n-1}(z)(-\cos[(2n-1)\theta)])$ $\\[3mm]$ $\Rightarrow \frac{-z [\cos(\theta) \cos(z \cos(\theta)) + z \sin^2(\theta) \sin(z \cos(\theta))]}{2}=- \sum_{n=1}^{\infty}(-1)^n(2n-1)^2J_{2n-1}(z)(-\cos[(2n-1)\theta)])$ $\\[3mm]$ $\text{The expression on the right hand side of the previous step, is equal to the denominator of the expression in our original integral when } \\ \theta=k\pi \text{, where } k\in \mathbb{Z}. \text{ Plugging this in for } \theta, \text{we get:}$

$\\[3mm]$ $\frac{z\cos(z)}{2}= -\sum_{n=1}^{\infty}(-1)^n(2n-1)^2J_{2n-1}(z)$

$\\[3mm]$ $\text{Our integral has now become the following}:$

$\\[3mm]$ $\int_{0}^{\frac{\pi}{2}}\frac{\frac{z\sin(z)}{2}}{\frac{z\cos(z)}{2}}^{2/3} dz=\int_{0}^{\frac{\pi}{2}}\tan(z)^{2/3}dz$ $\\[3mm]$ $\text{Evaluating this integral, we finally arrive at our answer:}$ $\\[3mm]$ $\int_{0}^{\frac{\pi}{2}}\tan(z)^{2/3}dz\approx\fbox{3.14158}$