Consider the real integration: $\displaystyle{\int_{0}^{2\pi}}\dfrac{dx}{1-2\alpha\sin x+\alpha^2}$ given that $\alpha^2<1$.
In order to perform this calculation, I define a function over the complex plane as follows:
$f(z) = \dfrac{r}{1-2\alpha\sin \theta+\alpha^2} $ ; $z=re^{i\theta}$ where $\theta \in [0, 2\pi), $ $r \in R^{+} \cup \{0\} $.
Now, $1-2\alpha\sin \theta+\alpha^2 = 2\alpha\bigg(\dfrac{1}{2}\bigg(\alpha+\dfrac{1}{\alpha}\bigg)-\sin \theta\bigg)$. Since $\alpha^2<1$, $\dfrac{1}{2}\bigg| \alpha+\dfrac{1}{\alpha}\bigg|>1$. Therefore, $1-2\alpha\sin \theta+\alpha^2$ can never be zero. This is also equivalent to saying that the denominator of the original integrand as well never goes to zero.
Approach 1
I take the contour integration of the function $f(z)$ over the closed path $|z|=1$. Since $1-2\alpha\sin \theta+\alpha^2$ can never be equal to zero, the function $f(z)$ is clearly analytic in the $z$-plane. Therefore, its integration over the closed contour $|z|=1$ will be zero according to the Cauchy integral theorem. But clearly, the contour integration of $f(z)$ over $|z|=1$ is equal to $\displaystyle{\int_{0}^{2\pi}}\dfrac{dx}{1-2\alpha\sin x+\alpha^2}$.
Therefore, $\displaystyle{\int_{0}^{2\pi}}\dfrac{dx}{1-2\alpha\sin x+\alpha^2} = 0$
Approach 2
Since the denominator of the original real integrand can never be zero, it remains either positive or negative everywhere. Therefore, the definite integration over the proposed limits must be non-zero.
There is no paradox. We have $\sin \theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}$, hence by setting $e^{i\theta}=z$
$$\frac{1}{\alpha^2-2\alpha\sin\theta+1} = \frac{1}{\alpha^2+\alpha i\left(z-\frac{1}{z}\right)+1}=\frac{1}{\alpha+iz}\cdot\frac{1}{\alpha-\frac{i}{z}}$$ we get a meromorphic function of the $z$ variable. In particular
$$ \int_{0}^{2\pi}\frac{d\theta}{\alpha^2-2\alpha\sin\theta+1}=\oint_{|z|=1}\frac{-i\,dz}{(\alpha+i z)\left(\alpha z-i\right)} $$ is the integral over the unit circle of a function with a pole at $z=\frac{i}{\alpha}$ and a pole at $z=i\alpha$.
Since $\alpha^2<1$ only the pole at $z=i\alpha$ lies inside the integration contour and
$$\int_{0}^{2\pi}\frac{d\theta}{\alpha^2-2\alpha\sin\theta+1} = 2\pi i\cdot\text{Res}\left(\frac{-i\,dz}{(\alpha+i z)\left(\alpha z-i\right)},z=i\alpha\right)$$ by the residue theorem. I guess you can take it from here (and get $\frac{2\pi}{1-\alpha^2}$ as the actual value of the integral).