I got curious about this integral because we have the following identity:
$$\frac{2}{\pi}\int_0^{\pi/2} \cos (x\cos t) dt=J_0(x)$$
So we have an interesting (if useless) symmetry:
$$\int_0^{\pi/2} J_0 (\cos x) dx=\frac{2}{\pi}\int_0^{\pi/2}\int_0^{\pi/2} \cos (\cos x \cos t) ~dt~dx=\int_0^{\pi/2} J_0 (\cos t) dt$$
Wolfram Alpha doesn't show a closed form for this integral. However, if we use the series for the Bessel function:
$$J_0(x)=\sum_{k=0}^\infty \frac{(-1)^k x^{2k}}{4^kk!^2}$$
And the known closed form for the family of integrals:
$$\int_0^{\pi/2} \cos^{2k} (x) dx=\frac{\sqrt{\pi}}{2}\frac{\Gamma \left(k+\frac{1}{2} \right)}{k!}$$
We obtain:
$$\int_0^{\pi/2} J_0 (\cos x) dx=\frac{\sqrt{\pi}}{2} \sum_{k=0}^\infty \frac{(-1)^k \Gamma \left(k+\frac{1}{2} \right)}{4^kk!^3}$$
Wolfram Alpha evaluates this series, giving a closed form for the integral:
$$\int_0^{\pi/2} J_0 (\cos x) dx=\frac{\pi}{2} \left(J_0 \left(\frac{1}{2} \right)\right)^2=1.3834405\dots$$
This agrees with the numerical value. However, I have not been able to show this myself.
How do we prove this closed form? Is there a more general case, involving Bessel functions of order $n$?