By chance, I see this formula
$\int_0^1 T_{2n+1}(x)\sin(ax) { dx \over \sqrt{1-x^2}}=(-1)^n\frac{\pi}{2}J_{2n+1}(a)$ but what is the closed form if we have
$\int_0^1 T_{2n}(x)\sin(ax) { dx \over \sqrt{1-x^2}}=?$
Here $T_n$ is Chebyshev polynomials of first kind and $J_n$ is Bessel function of first kind