Consider,
$$ \frac{2}{\pi}\int_0^{\infty } \frac{1}{t}\sin \left(\frac{c t}{2}\right) \cos (t \sin (\phi )) \, dt $$
Numerical integral shows that the result is 1 when $c>2|\sin(\phi)|$ and 0 otherwise, Mathematica also says the same thing.
How do I see this analytically?
This can be found from the product-to-sum trig identity $$ 2\sin a \cos b = \sin(a + b) + \sin(a-b) $$ and the well-known integral $$ \int_0^\infty\sin(k x)\frac{dx}{x} = \frac{\pi}{2}\operatorname{sgn}(k), $$ where $\operatorname{sgn}$ is $1$ for positive $k$, $-1$ for negative $k$, and $0$ for $k = 0$. Using those in the original integral gives \begin{eqnarray} \frac{2}{\pi}\int_0^{\infty }\sin \left(\frac{c t}{2}\right) \cos \left[t \sin (\phi )\right]\frac{dt}{t} &=& \frac{1}{\pi}\left[\int_0^\infty \sin\left(\left[\frac{c}{2}+\sin\phi\right]t\right)\frac{dt}{ t}+ \int_0^\infty \sin\left(\left[\frac{c}{2}-\sin\phi\right]t\right)\frac{dt}{t}\right] \\ &=& \frac{1}{2}\left[\operatorname{sgn}\left(\frac{c}{2}+\sin\phi\right)+\operatorname{sgn}\left(\frac{c}{2}-\sin\phi\right)\right]. \end{eqnarray} You can then work out the cases to get $$ \frac{2}{\pi}\int_0^{\infty }\sin \left(\frac{c t}{2}\right) \cos \left[t \sin (\phi )\right]\frac{dt}{t} = \operatorname{sgn} c\cdot\begin{cases}1 & |c| > 2|\sin\phi|\\ \frac{1}{2} & |c| = 2|\sin\phi|\\0 & |c| < 2|\sin\phi|\end{cases}. $$