We can generalize the integral by manipulating the Laplace transform of $J_{n}(bx)$, namely $$ \int_{0}^{\infty} J_{n}(bx) e^{-sx} \, dx = \frac{(\sqrt{s^{2}+b^{2}}-s)^{n}}{b^{n}\sqrt{s^{2}+b^{2}}}\ , \quad \ (n \in \mathbb{Z}_{\ge 0} \, , \text{Re}(s) >0 , \, b >0 )\tag{1}. $$
(See this question for a derivation of $(1)$ using contour integration.)
First let $s=p+ia$, where $p,a >0$.
A slight modification of the answer here shows that $\int_{0}^{\infty} J_{n}(bx) e^{-(p+ia)x} \, dx $ converges uniformly for all $p \in [0, \infty$).
This allows us to conclude that $$\begin{align} \int_{0}^{\infty} J_{n}(bx) e^{-iax} \, dx &= \lim_{p \downarrow 0}\int_{0}^{\infty} J_{n}(bx) e^{-(p+ia)x} \, dx \\ &= \lim_{p \downarrow 0} \frac{\left(\sqrt{(-p+ia)^2+b^{2}}-p-ia\right)^{n}}{b^{n}\sqrt{(p+ia)^2+b^{2}}} \\ &= \frac{\left(\sqrt{b^{2}-a^{2}}-ia\right)^{n}}{b^{n}\sqrt{b^{2}-a^{2}}}. \end{align}$$
So if $ a < b$, $$ \begin{align} \int_{0}^{\infty} J_{n}(bx) e^{-iax} \, dx &= \frac{\left(\sqrt{b^{2}-a^{2}+a^{2}} e^{-i \arcsin \left(\frac{a}{b}\right)}\right)^{n}}{b^{n} \sqrt{b^{2}-a^{2}}} \\ &= \frac{e^{-in \arcsin \left(\frac{a}{b}\right)}}{\sqrt{b^{2}-a^{2}}} .\end{align}$$
And if $a >b$, $$ \begin{align} \int_{0}^{\infty} J_{n}(bx) e^{-iax} \, dx &= \frac{\left(i\sqrt{a^{2}-b^{2}}-ia \right)^{n}}{b^{n}i \sqrt{a^{2}-b^{2}}} \\ &= \frac{-i e^{i \pi n /2} \left(\sqrt{a^{2}-b^{2}}-a \right)^{n}}{b^{n} \sqrt{a^{2}-b^{2}}}. \end{align}$$
Therefore,
$$\int_{0}^{\infty} J_{n}(bx) \sin(ax) \, dx = \begin{cases}
\frac{\sin \left(n \arcsin \left(\frac{a}{b} \right) \right)}{\sqrt{b^{2}-a^{2}}} \, & \quad 0 < a < b \\
\frac{\cos \left(\frac{\pi n}{2} \right) \left(\sqrt{a^{2}-b^{2}} -a \right)^{n}}{b^{n} \sqrt{a^{2}-b^{2}}} & \quad a > b >0
\end{cases} $$
Best Answer
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ First step: Multiply numerator and denominator by
$\ds{\sec^{2m + 2n}\pars{\theta} = \sec^{2m - 1}\pars{\theta}\ \sec^{2n - 1}\pars{\theta}\ \sec^{2}\pars{\theta}}$. \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\pi/2} {\sin^{2m - 1}\pars{\theta}\cos^{2n - 1}\pars{\theta} \over \bracks{a\sin^{2}\pars{\theta} + b\cos^{2}\pars{\theta}}^{m + n}} \,\dd\theta} \\[5mm] = & \int_{0}^{\pi/2} {\tan^{2m - 1}\pars{\theta} \over \bracks{a\tan^{2}\pars{\theta} + b}^{m + n}} \sec^{2}\pars{\theta}\,\dd\theta \\[5mm] \stackrel{x\ =\ \tan\pars{\theta}}{=}\,\,\,& \int_{0}^{\infty}{x^{2m - 1} \over \pars{ax^{2} + b}^{m + n}}\,\dd x \\[5mm] \stackrel{x^{2}\ \mapsto\ x}{=}\,\,\,& {1 \over 2}\int_{0}^{\infty}{x^{m - 1} \over \pars{ax + b}^{m + n}}\,\dd x \\[5mm] \stackrel{ax/b\ \mapsto\ x}{=}\,\,\,& {1 \over 2a^{m}b^{n}}\ \underbrace{\int_{0}^{\infty}{x^{m - 1} \over \pars{x + 1}^{m + n}}\,\dd x}_{\ds{\on{B}\pars{m,n}}} \end{align} See this link. Then, \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\pi/2} {\sin^{2m - 1}\pars{\theta}\cos^{2n - 1}\pars{\theta} \over \bracks{a\sin^{2}\pars{\theta} + b\cos^{2}\pars{\theta}}^{m + n}} \,\dd\theta} \\[5mm] = &\ \bbx{{1 \over 2a^{m}b^{n}}\,{\Gamma\pars{m}\Gamma\pars{n} \over \Gamma\pars{m + n}}} \\ & \end{align}