■三角関数の積分(その2)
[Q]∫(0,φ)dθ/cosθ=?
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tanθ=tでなく,sinθ=tと変数変換してみたい.
dt=cosθdθ
∫(0,φ)dθ/cosθ=∫(0,φ)cosθdθ/cos^2θ
=∫(0,φ)cosθdθ/(1−sin^2θ)
=∫(0,sinφ)dt(1−t^2)=1/2・∫(0,sinφ){1/(1+t)+1/(1−t)}dt
=1/2・log(1+sinφ)/(1−sinφ)
=1/2・log(1−cos(φ+π/2))/(1+cos(φ+π/2))
cos(φ+π/2)=2cos^2(φ/2+π/4)−1=1−2sin^2(φ/2+π/4)
より
∫(0,φ)dθ/cosθ=1/2・logsin^2(φ/2+π/4)/cos^2(φ/2+π/4)=logtan(φ/2+π/4)
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