Verified answer

Rewrite this as a contour integral about |z| = 1 using z = e^(iθ):

∫(0 to 2π) dθ/(1 + a sin θ)

= ∫c (dz/(iz)) / [1 + a * (1/(2i))(z - 1/z)]

= ∫c 2 dz / [2iz + a(z^2 - 1)]

= ∫c 2 dz/(az^2 + 2iz - a)

= (2/a) ∫c dz/(z^2 + (2i/a)z - 1)

This integrand has singularities when z^2 + (2i/a)z - 1 = 0

==> z = [(-2i/a) ± √(-4/a^2 + 4)] / 2

..........= (-i/a) ± i√((1 - a^2)/a^2), since |a| < 1

..........= i(-1 ± √(1 - a^2))/a.

Of these singularities, only z = i(-1 + √(1 - a^2))/a is inside C.

Since it is a simple pole, it has residue

lim(z→ i(-1 + √(1 - a^2))/a) (z - i(-1 + √(1 - a^2))/a) * 1//(z^2 + (2i/a)z - 1)

= lim(z→ i(-1 + √(1 - a^2)/a)) 1/[(z - i(-1 - √(1 - a^2))/a], via factoring

= a / [2i√(1 - a^2)].

So, the integral equals (by the Residue Theorem)

(2/a) * {2πi * a / [2i√(1 - a^2)]} = 2π/√(1 - a^2).

I hope this helps!