■18世紀における微積分(その190)

[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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