DUIS: ∫ log((a+b sin x)/(a−b sin x)) dx/sin x

Let a > |b| > 0 and I(b) = ∫₀^(π/2) log[(a + b sin x)/(a − b sin x)] · dx/sin x. Treat a as constant and differentiate with respect to b:

dI/db = ∫₀^(π/2) [ sin x/(a + b sin x) + sin x/(a − b sin x) ] · dx/sin x
      = ∫₀^(π/2) [ 1/(a + b sin x) + 1/(a − b sin x) ] dx
      = ∫₀^(π/2) 2a dx/(a² − b² sin²x)

Evaluate ∫₀^(π/2) dx/(a² − b² sin²x). Divide by cos²x and put t = tan x, so sin²x = t²/(1 + t²), dx/cos²x = dt:

∫₀^∞ dt / (a²(1 + t²) − b² t²)
  = ∫₀^∞ dt / (a² + (a² − b²)t²)
  = (1/(a√(a² − b²))) [tan⁻¹(t√(a²−b²)/a)]₀^∞
  = π / (2a√(a² − b²))

Therefore

dI/db = 2a · π/(2a√(a² − b²)) = π/√(a² − b²).

Integrate with respect to b:

I(b) = π sin⁻¹(b/a) + C

When b = 0, the integrand is log 1 = 0 (and the limit is 0), so I(0) = 0 and C = 0.

Answer: I = π sin⁻¹(b/a).