​​​​∫ dx / 1−2acosx+a^2; x ∈ [0, π] is equal to

Question

\(\int \limits_0^\pi \frac {dx}{1 - 2a\cos x + a^2}\) is equal to

(1) \(\frac{(1 + a^2) \pi}{(1 - a^2) ^2}\)

(2) \(\frac{(1 + a^2) \pi}{(1 - a^2) }\)

(3) \(\frac{(1 - a^2) \pi}{(1 +a^2) }\)

(4) \(\frac{(1 - a^2) \pi}{(1 +a^2)^2 }\)

Answer 1

Correct option is (1) \(\frac{(1 + a^2) \pi}{(1 - a^2) ^2}\)