For $$0 < a < 1$$, the value of the integral $$\int_0^{\pi} \frac{dx}{1 - 2a\cos x + a^2}$$ is :

Use the substitution $$t = \tan(x/2)$$. Then $$dx = \frac{2 dt}{1+t^2}$$ and $$\cos x = \frac{1-t^2}{1+t^2}$$.

Change limits: When $$x=0, t=0$$. When $$x=\pi, t \to \infty$$.

Substitute into the integral:

$$I = \int_{0}^{\infty} \frac{\frac{2 dt}{1+t^2}}{1 - 2a\left(\frac{1-t^2}{1+t^2}\right) + a^2} = \int_{0}^{\infty} \frac{2 dt}{(1+t^2) - 2a(1-t^2) + a^2(1+t^2)}$$

Denominator $$= 1 + t^2 - 2a + 2at^2 + a^2 + a^2t^2 = (1-a)^2 + t^2(1+a)^2$$.

$$I = 2 \int_{0}^{\infty} \frac{dt}{(1-a)^2 + t^2(1+a)^2} = \frac{2}{(1+a)^2} \int_{0}^{\infty} \frac{dt}{t^2 + \left(\frac{1-a}{1+a}\right)^2}$$

Using the formula $$\int \frac{dx}{x^2+k^2} = \frac{1}{k} \tan^{-1}(\frac{x}{k})$$:

$$I = \frac{2}{(1+a)^2} \cdot \frac{1+a}{1-a} \left[ \tan^{-1}\left(t \cdot \frac{1+a}{1-a}\right) \right]_{0}^{\infty} = \frac{2}{1-a^2} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{1-a^2}$$