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DUIS: ∫ dx/(a²sin²x+b²cos²x)²

Short answer

Start with I(A,B)=∫₀^{π/2} dx/(A sin²x+B cos²x) = π/(2√(AB)). Differentiate w.r.t. A and B, add to get ∫ dx/(A sin²x+B cos²x)² = (π/4)(A+B)/(AB)^{3/2}. Substitute A=a², B=b² to obtain π(a²+b²)/(4a³b³).

Apply the method of differentiation under integral sign to prove that ∫₀^{π/2} dx/(a²sin²x + b²cos²x)² = π(a²+b²)/(4a³b³).

Method

Differentiation under the integral sign: if I(λ) = ∫ f(x, λ) dx with constant limits, then dI/dλ = ∫ ∂f/∂λ dx.

Step 1: a standard integral with parameters

Let I(A, B) = ∫₀^(π/2) dx / (A sin²x + B cos²x), A, B > 0. Divide by cos²x and put t = tan x:

I = ∫₀^∞ sec²x dx / (A tan²x + B) = ∫₀^∞ dt/(A t² + B) = (1/√(AB)) [tan⁻¹(t√(A/B))]₀^∞ = π / (2√(AB)).

Step 2: differentiate with respect to A and B

∂I/∂A = −∫₀^(π/2) sin²x dx/(A sin²x + B cos²x)² = −(π/4) A^(−3/2) B^(−1/2)
∂I/∂B = −∫₀^(π/2) cos²x dx/(A sin²x + B cos²x)² = −(π/4) A^(−1/2) B^(−3/2)

Add the two results. Since sin²x + cos²x = 1:

∫₀^(π/2) dx/(A sin²x + B cos²x)² = (π/4)[A^(−3/2)B^(−1/2) + A^(−1/2)B^(−3/2)] = (π/4)(A + B)/(AB)^(3/2).

Step 3: put A = a², B = b²

∫₀^(π/2) dx/(a² sin²x + b² cos²x)² = π(a² + b²)/(4a³b³). Proved.

✓ Answer checked by Claude Opus 5.5 (high effort)

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