10/09/28 20:54:39
>>48
要するに, ∫[0～α] 1/(1+cosθ) dθ = ∫[0～α] 1/{cos(θ/2)^2} d(θ/2) = [tan(θ/2)](θ=0,α) = tan(α/2),
10/09/28 20:54:39
>>48
要するに, ∫[0～α] 1/(1+cosθ) dθ = ∫[0～α] 1/{cos(θ/2)^2} d(θ/2) = [tan(θ/2)](θ=0,α) = tan(α/2),
10/09/28 20:54:39
>>48
要するに, ∫[0～α] 1/(1+cosθ) dθ = ∫[0～α] 1/{cos(θ/2)^2} d(θ/2) = [tan(θ/2)](θ=0,α) = tan(α/2),