質問<1609>2004/2/24
∫(1/1-cosx)dxを教えてください
お便り2004/2/25
from=こんにちは
tan(x/2)=tとおくと dt=(1/2)*{1/cos(x/2)}^2dx =(1/2)*{1+{tan(x/2)}^2}dx =(1/2)*(1+t^2)dx よって dx={2/(t^2+1)}dt (1-t^2)/(1+t^2) =(1-{sin(x/2)/cos(x/2)}^2)/(1+{sin(x/2)/cos(x/2)}^2) =({cos(x/2)}^2-{sin(x/2)}^2)/({cos(x/2)}^2+{sin(x/2)}^2) =cosx だから ∫1/(1-cosx)dx =∫{1/(1-{(1-t^2)/(1+t^2)})}*{2/(t^2+1)}dt =∫{(1+t^2)/(1+t^2)-(1-t^2)}*{2/(t^2+1)}dt =∫{(1+t^2)/(2t^2)}*{2/(t^2+1)}dt =∫(1/t^2)dt =-1/t+C =-1/{tan(x/2)}+C Cは積分定数
お便り2004/2/25
from=juin
1/(1-cosx) =(1+cosx)/(1-(cosx)^2) =(1+cosx)/(sinx)^2 ∫dx/(1-cosx) =∫dx/(sinx)^2+∫cosx/(sinx)^2dx =-cotx-1/sinx+C