The integral presented is:

$\int_{-\pi}^{\pi} \frac{\sin(n\varphi) \, d\varphi}{1 - 2a \sin(\varphi) + a^2}, \quad -1 < a < 1, \quad n = 0, 1, 2, \ldots$

Solution Strategy:

This integral can be approached using contour integration or by recognizing it as a standard integral related to Fourier series. The presence of $\sin(n\varphi)$ and the trigonometric expression in the denominator suggests that we might be dealing with a form of generating function for Chebyshev polynomials or a related function in complex analysis.

Steps to Solve the Integral:

1. Convert the Denominator Using Complex Numbers: Write $\sin(\varphi)$ in terms of complex exponentials using: $\sin(\varphi) = \frac{e^{i\varphi} - e^{-i\varphi}}{2i}$ and then rewrite the integrand.

2. Simplify the Expression: Substituting this into the denominator will allow for a simplification using complex analysis, potentially turning the integral into a contour integral around the unit circle in the complex plane.

3. Use Residue Theorem (if applicable): Apply the residue theorem if the integrand is in a form that allows it, calculating the residues of the poles within the unit circle.

4. Recognize Special Cases: For specific values of $n$ (like $n = 0$), the integral may simplify significantly. For other $n$, standard results from integrals of trigonometric functions over symmetric limits like $-\pi$ to $\pi$ can be useful.

Would you like a detailed walkthrough of the complete calculation using complex analysis, or an approach based on properties of Fourier series?