■三角関数の積分(その5)

[Q]∫sin x/(sin x + cos x)dxは?

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 t=tan(θ/2)とすると

  tan(θ/2)=sinθ/(1+cosθ)=(1-cosθ)/sinθ,

  cosθ=(1-t^2)/(1+t^2),

  sinθ=2t/(1+t^2),

  dθ/dt=2/(1+t^2)

  ∫sin x/(sin x + cos x)dx=∫4t/(1+2t-t^2)(1+t^2)dt

 ここで,

  4t/(1+2t-t^2)(1+t^2)=1/(1+2t-t^2)+(1/1+t^2)=(t-1)/(1+2t-t^2)+(t+1)/(1+t^2)

=(t-1)/(1+2t-t^2)+t/(1+t^2)+1/(1+t^2)

より

  ∫sin x/(sin x + cos x)dx=-1/2・log|1+2t-t^2|+1/2log|1+t^2|+arctan|t|+C,t=tan(x/2)

 もっと簡単にできるかもしれないが,この辺で止めておく.

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