I'm studying alternative methods for elliptic boundary conditions. I picked the formula of Bessel function from this site https://dlmf.nist.gov/10.9 I'm looking for any available approach to solve this apparently simple equivalence. Any hint or suggestion for me would be useful. I think I should be able to answer to this question but I still don't get it.
$$ J_0(x)=\frac{1}{\pi}\int_{0}^{\pi}\cos(x\sin(\theta))d\theta= \frac{1}{\pi}\int_{0}^{\pi}\cos(x\cos(\theta))d\theta $$
With $x$ fixed, the function $\cos(x \sin \theta)$ is a continuous, $\pi$-periodic function in the real variable $\theta$.
Therefore, the value of $ \frac{1}{\pi} \int \cos(x \sin \theta) \, \mathrm d \theta$ over any finite interval of length $\pi$ is the same.
In particular, $$\frac{1}{\pi} \int _{0}^{\pi} \cos(x \sin \theta) \, \mathrm d \theta = \frac{1}{\pi} \int_{-\pi/2}^{\pi/2} \cos(x \sin \theta) \, \mathrm d \theta. $$
Now make the substitution $u = \theta + \pi/2$ to get $$\frac{1}{\pi} \int_{-\pi/2}^{\pi/2} \cos(x \sin \theta) \, \mathrm d \theta = \frac{1}{\pi}\int_{0}^{\pi} \cos(-x \cos u) \, \mathrm du = \frac{1}{\pi}\int_{0}^{\pi} \cos(x \cos u) \, \mathrm du.$$