Is there any form of closed form solution for these two integrals? I have two integrals of the from
\begin{equation}
\int_{-\pi/2}^{\pi/2} \mathrm d k e^{-i\omega \sin k} \frac{\lambda}{\alpha+\beta \cos 2k}
\end{equation}
and
\begin{equation}
\int_{-\pi/2}^{\pi/2} \mathrm d k e^{-i\omega \sin k} \frac{\sigma \cos 2k}{\alpha+\beta \cos 2k}
\end{equation}
where $\{\omega,\alpha,\beta,\sigma,\lambda\}\in \mathbb R$.
My attempts were to use $\sin k=p$ and change the integrals to
\begin{equation}
\int^1_{-1} \frac{e^{-i p \omega }\lambda}{\sqrt{1-p^2} \left(\beta  \left(1-2 p^2\right)+\alpha\right)} \mathrm d p
\end{equation}
and
\begin{equation}
\int^1_{-1} \frac{e^{-i p \omega }\sigma (1-2p^2)}{\sqrt{1-p^2} \left(\beta  \left(1-2 p^2\right)+\alpha\right)} \mathrm d p
\end{equation}
But I do not know how to proceed with these integrals.
Any form of closed form solution even in terms of infinite series would be helpful.
 A: We will assume that $\alpha > |\beta|$. Write
\begin{align*}
I
&:= \int_{-\pi/2}^{\pi/2} \mathrm{d}k \, e^{-i\omega \sin k} \frac{\lambda}{\alpha+\beta \cos (2k)} \\
&= \int_{0}^{\pi} \mathrm{d}k \, e^{i\omega \cos(k)} \frac{\lambda}{\alpha - \beta \cos (2k)} \\
&= \int_{0}^{\pi} \mathrm{d}k \, e^{i\omega \cos(k)} \frac{\lambda}{\sqrt{\alpha^2 - \beta^2}} \left( 1 + 2 \sum_{n=1}^{\infty} r^n \cos(2nk) \right),
\end{align*}
where $r$ is given by
$$ r = \frac{\beta}{\alpha + \sqrt{\alpha^2 - \beta^2}} \in (-1, 1). $$
Then by utilizing the integral representation for the Bessel function of the 1st kind,
\begin{align*}
I
&= \frac{\lambda}{\sqrt{\alpha^2 - \beta^2}} \left( \int_{0}^{\pi} \mathrm{d}k \, e^{i\omega \cos(k)} + 2 \sum_{n=1}^{\infty} r^n \int_{0}^{\pi} \mathrm{d}k \, e^{i\omega \cos(k)} \cos(2nk) \right) \\
&= \frac{\pi \lambda}{\sqrt{\alpha^2 - \beta^2}} \left( J_0(\omega) + 2 \sum_{n=1}^{\infty} (-r)^n J_{2n}(\omega) \right).
\end{align*}
Below is a plot for the partial sums
$$ I_m(\omega) := \frac{\pi \lambda}{\sqrt{\alpha^2 - \beta^2}} \left( J_0(\omega) + 2 \sum_{n=1}^{m} (-r)^n J_{2n}(\omega) \right) $$
with $\lambda = 1$, $\alpha = 2$, and $\beta = 1$:

Note that the second integral reduces to the first one:
\begin{align*}
&\int_{-\pi/2}^{\pi/2} \mathrm{d}k \, e^{-i\omega \sin k} \frac{\sigma \cos 2k}{\alpha+\beta \cos 2k} \\
&= \frac{\sigma}{\beta} \left( \int_{-\pi/2}^{\pi/2} \mathrm{d}k \, e^{-i\omega \sin k} - \alpha \int_{-\pi/2}^{\pi/2} \mathrm{d}k \, e^{-i\omega \sin k} \frac{1}{\alpha+\beta \cos 2k} \right)
\end{align*}
