>>398
F'(a)=∫[0,π/2] {2acos^2(θ)/(a^2cos^2(θ)+sin^2(θ))}dθ
=2a∫[0,π/2] {1/(a^2+tan^2(θ))}dθ
t=tanθ　とおくと　dt=dθ/cos^2(θ)=(1+t^2)dθ
F'(a)=2a∫[0,∞] {1/(1+t^2)(a^2+t^2)}dt
=2a/(a^2-1)∫[0,∞] {1/(1+t^2)-1/(a^2+t^2)}dt
=2a/(a^2-1) {π/2 - (1/a)(π/2)}
=π/(a+1)