How to show the integral equality $\int_0^1J_0\left(a\sqrt{1-x^2}\right)\cos(bx) dx = \frac{\sin\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}}$.

Question

I have a rather simple integral equality

$$ \int_{0}^1 dx~J_{0}\left(a\sqrt{1-x^2}\right)\cos(bx) = \frac{\sin\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}}, $$ where $J_0$ is a Bessel function, $a>0$ and $b$ is real.

How can I derive this, and show it holds?

How about if I add $e^{c x^2}$ to the integrand $$ \int_{0}^1 dx~J_{0}\left(a\sqrt{1-x^2}\right)\cos(bx) e^{c x^2} ~?$$

Edit I

One solution to the first part was proposed here. Unfortunately, this form was not very telling how to extend it for the second integral.