- 120 名前：１３２人目の素数さん mailto:sage [2024/04/11(木) 23:04:26.01 ID:xK64JHhj.net]
- ∫[0,π/2] sinx/(1+√sin(2x)) dx
= ∫[0,π/2] cosx/(1+√sin(2x)) dx
= (1/2)∫[0,π/2] (sinx+cosx)/(1+√sin(2x)) dx
= (1/2)∫[0,π/2] (√2)sin(x+π/4)/(1+√sin(2x)) dx
= ∫[0,π/4] (√2)cosx/(1+√cos(2x)) dx
= ∫[0,π/4] √(1+cos(2x))/(1+√cos(2x)) dx
置換 cos(2x)=(cost)^2, sin(2x)dx=cost sint dt
= ∫[0,π/2] √(1+(cost)^2)/(1+cost) cost sint dt/√(1-(cost)^4)
= ∫[0,π/2] cost/(1+cost) dt
= ∫[0,π/2] (1 - 1/(1+cost)) dt
= ∫[0,π/2] (1 - (1/2)/cos(t/2)^2) dt
= t - tan(t/2)|_(t=0,π/2)
= (π/2) - 1
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