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Integral of cos x/((1+sin x)(2+sin x))

Short answer

Let t = sin x, then integral becomes ∫₀¹ dt/((1+t)(2+t)). Partial fractions give ln((1+t)/(2+t)) evaluated from 0 to 1, yielding ln(4/3).

Integrate: ∫₀^{π/2} cos x dx / ((1+sin x)(2+sin x)).

Let I = ∫₀^(π/2) cos x dx / ((1 + sin x)(2 + sin x)).

Put t = sin x, dt = cos x dx. Limits: x = 0 → t = 0; x = π/2 → t = 1.

I = ∫₀¹ dt / ((1 + t)(2 + t))

Partial fractions: 1/((1 + t)(2 + t)) = 1/(1 + t) − 1/(2 + t).

I = [ln(1+t) − ln(2+t)]₀¹
  = [ln((1+t)/(2+t))]₀¹
  = ln(2/3) − ln(1/2)
  = ln(4/3)

Answer: I = ln(4/3) ≈ 0.2877.

✓ Answer checked by Claude Opus 5.5 (high effort)

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