How to evaluate the integral $$ \int_0^{\infty}\cos bx\frac{\sin t\sqrt{x^2-a^2}}{\sqrt{x^2-a^2}}\,dx \quad (t>|b|), $$ knowing that $$ \frac{\sin ut}{u}=\sqrt{\frac{\pi t}{2u}}J_{1/2}(ut) =\frac{\sqrt{\pi t}}{2}\frac{1}{2\pi i}\int_{C_1} \frac{e^{z-u^2t^2/(4z)}}{z^{3/2}}\,dz, $$ where $C_1=\{z:\Re z=\alpha\} \; (\alpha>0)$ ? The hint being to write down the consecutive integrals, and then change the order of integration.
The latter equality is easily established using the expansion $$ J_{\nu}(z)=\sum_{k=0}^{\infty}\frac{(-1)^k}{k!\Gamma(k+\nu+1)} \left(\frac{z}{2}\right)^{\nu+2k}, $$ and the sin series.
One apparent thing is to evaluate the integral with $\cos bx$ replaced through $e^{ibx}$ (or $e^{-ibx}$), but I don't see what to do next.
The integral representation should be written as \begin{equation} \frac{\sin ut}{u}=\sqrt{\frac{\pi t}{2u}}J_{1/2}(ut) =\frac{\sqrt{\pi }t}{2}\frac{1}{2\pi i}\int_{C_1} \frac{e^{z-u^2t^2/(4z)}}{z^{3/2}}\,dz \end{equation} (prefactor of the integral contains $t$ and not $\sqrt{t}$). Introducing this integral representation, it comes \begin{equation} I=\frac{\sqrt{\pi} t}{2}\frac{1}{2i\pi}\int_{C_1} \frac{e^{z+\frac{a^2t^2}{4z}}}{z^{3/2}}\,dz\int_0^\infty\cos bx \,e^{-\frac{t^2x^2}{4z}}\,dx \end{equation} Using the Fourier transform of the Gaussian (with $\Re z>0$), \begin{equation} \int_0^\infty\cos bx e^{-\frac{t^2x^2}{4z}}\,dz=\frac{\sqrt{\pi z}}{t}e^{-\frac{b^2z}{t^2}} \end{equation} it comes \begin{equation} I=\frac{\pi}{2}\frac{1}{2i\pi}\int_{C_1}e^{z(1-\frac{b^2}{t^2})+\frac{a^2t^2}{4z}}\frac{dz}{z} \end{equation} As $1-\frac{b^2}{t^2}>0$, it can be written as \begin{equation} I=\frac{\pi}{2}\frac{1}{2i\pi}\int_{C_1}e^{s+\frac{a^2\left( t^2-b^2 \right)}{4s}}\frac{ds}{s} \end{equation} We recognize the Schläfli representation for $J_0\left( ia\sqrt{t^2-b^2} \right)$: \begin{align} I&=\frac{\pi}{2}J_0\left( ia\sqrt{t^2-b^2} \right)\\ &=\frac{\pi}{2}I_0\left( a\sqrt{t^2-b^2} \right) \end{align}