<s>[INST] Evaluate a definite integrals given in term of Bessel functions The problem is to solve the following integrals. The parameters are all positives.
\begin{align}
& \int \sqrt{ a^2 - x^2} \,  J_1\left(r \sqrt{a^2-x^2}\right) \cos (x z) dx
\\
& \int   x    J_0\left(r \sqrt{a^2-x^2}\right)  \sin (x z) dx
\end{align}
I try using trigonometric variable change and Euler formula. I think it makes them more complex to solve Here. I show how the integrals are transformed.
\begin{align}
        & x=a sin\theta ,  \,\,\,\,\, dx=a\cos\theta d \theta\\ \\
         & e^{ i a \sin \theta z}= \cos (a \sin \theta z)  + i\sin (a \sin \theta z)
         \\
         \\ 
           & a^2 \int   \cos^2\theta  J_1\left(r  a \cos\theta \right)     \cos (a \sin \theta z)   
            d\theta \\ &  a  \int  \sin \theta  J_0\left(r a \cos \theta\right) \sin (a \sin \theta z)  
         d\theta
       \end{align}
The goal is to obtain close solutions for both integrals.
 [/INST] You can evaluate them on $(0,a)$ in terms of spherical Bessel functions (which are expressible in terms of trigonometric functions) as follows:
$$
\int_0^a {\sqrt {a^2  - x^2 } J_1 \left( {r\sqrt {a^2  - x^2 } } \right)\cos (xz)dx}  = a^2 r\frac{{\mathsf{j}_1 (a\sqrt {r^2  + z^2 } )}}{{\sqrt {r^2  + z^2 } }} \\ = r\frac{{\sin (a\sqrt {r^2  + z^2 } )}}{{r^2  + z^2}} - ar\frac{{\cos (a\sqrt {r^2  + z^2 } )}}{{\sqrt {r^2  + z^2 } }}
$$
and
$$
\int_0^a {xJ_0 \left( {r\sqrt {a^2  - x^2 } } \right)\sin (xz)dx}  = a^2 z\frac{{\mathsf{j}_1 (a\sqrt {r^2  + z^2 } )}}{{\sqrt {r^2  + z^2 } }} \\ = z\frac{{\sin (a\sqrt {r^2  + z^2 } )}}{{r^2  + z^2 }} - az\frac{{\cos (a\sqrt {r^2  + z^2 } )}}{{\sqrt {r^2  + z^2 } }}.
$$
See (1.13.50) and (2.13.50) in A. Erdélyi, W. Magnus, F. Oberhettinger, F. G. Tricomi, Tables of Integral Transforms. Vol. I., McGraw-Hill Book Company, Inc., New York-Toronto-London, 1954.
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