Evaluating $\int_{0}^{2\pi} \frac{\mathrm{e}^{-i k a \cos\phi \sin\theta}}{1+\cos\phi \sin\theta}\,\mathrm d\phi$ How to simplify the following equation: 
$$\int_{0}^{2\pi} \frac{\mathrm{e}^{-i k a \cos\phi  \sin\theta}}{1+\cos\phi \sin\theta}\,\mathrm d\phi$$
where $k$, $a$ and $\theta$ can be regarded as the constants, and $0<\theta<\frac{\pi}{2}$. Call $\sin\theta=b$ for brevity.
Attempt:
As Von Neumann pointed out, which is really helpful, using the residues theorem,
$$\frac{2}{ib}\int_{|z|=1} \frac{\exp\left(-ikab\left(\frac{z^2+1}{2z}\right)\right)}{z^2 + \frac{2}{b}z + 1}\ \text{d}z$$
Where does the pole locate?
$$z_{1,\ 2} = -\frac{1}{b}\pm \sqrt{\frac{1}{b^2}-1}$$ 
noted that:  $\frac{1}{b}=\frac{1}{\sin(\theta)}>1$, then only one pole within the circle is considered,where 
$$z_{1} = -\frac{1}{b}+ \sqrt{\frac{1}{b^2}-1}$$
Using the residues theorem, I got the result 
$$\frac{2\pi}{\cos\theta}\exp(ika)$$ 
Then I programmed the original integral and the derived result, there are discrepancies.  My question is that should we consider the numerator when using residues theorem? That is
 $$\exp\left(-ikab\left(\frac{z^2+1}{2z}\right)\right)$$ because using the pole $z_{1}$ we can derive that 
$$\cos\phi =\frac{z^2+1}{2z}=\frac{\left(-\frac{1}{b}+\sqrt{\frac{1}{b^2}-1}\right)^2+1}{2\left(-\frac{1}{b}+\sqrt{\frac{1}{b^2}-1}\right)} = -\frac{1}{b}=-\frac{1}{\sin\theta}$$
As it is shown that $$\cos\phi=-\frac{1}{\sin\theta}$$ which conflicts with each other. Could someone help? Thank you very much in advance. 

Methinks we can appreciate it carefully with residues theorem. 
First of all we step into the complex plane with the usual rules for sine and cosines, that is:
$$\mathcal{J} = \phi \to e^{i\psi} = z ~~~~~~~ \cos\phi \to \ \frac{1}{2}\left(z + \frac{1}{z}\right) ~~~~~~~ \text{d}\phi = \frac{\text{d}z}{iz}$$
We will also call $\sin\theta = b$ for brevity.
The integral becomes (arrange it a bit)
$$\frac{1}{2i}\int_{|z|=1} \frac{\exp\left(-ikab\left(\frac{z^2+1}{2z}\right)\right)}{1 + \frac{b}{2}\frac{z^2+1}{z}}\ \frac{\text{d}z}{z}$$
That is, 
$$\frac{1}{ib}\int_{|z|=1} \frac{\exp\left(-ikab\left(\frac{z^2+1}{2z}\right)\right)}{z^2 + \frac{2}{b}z + 1}\ \text{d}z$$
Poles occurs at $$z_{1,\ 2} = -\frac{1}{b}\pm \sqrt{\frac{1}{b^2}-1}$$
Now by residues theorem:
$$\mathcal{J} = 2\pi i \frac{1}{ib} \sum_{z_k} \lim_{z\to z_k} (z-z_k)f(z)|_{z = z_k}$$
Where 
$$f(z) = \frac{\exp\left(-ikab\left(\frac{z^2+1}{2z}\right)\right)}{(z-z_1)(z - z_2)}$$
Can you proceed? 
Doing the calculation you shall obtain:
$$\frac{2\pi}{b}\left(\frac{e^{-2 i a \sqrt{\frac{1}{b^2}-1} b k}}{2 \sqrt{\frac{1}{b^2}-1}} -\frac{e^{2 i a \sqrt{\frac{1}{b^2}-1} b k}}{2 \sqrt{\frac{1}{b^2}-1}}\right) = -\frac{2\pi}{b}\frac{i \sin \left(2 a \sqrt{\frac{1}{b^2}-1} b k\right)}{\sqrt{\frac{1}{b^2}-1}} = -\frac{2\pi}{b}\frac{i b \sin \left(2 a \sqrt{1-b^2} k\right)}{\sqrt{1-b^2}}$$
Eventually: 
$$-2\pi i\frac{ \sin \left(2 a \sqrt{1-b^2} k\right)}{\sqrt{1-b^2}}$$
Where $b = \sin\theta$.
I will check this again later, for the sake of correctness.
