 | > y=logx(1<=x<=a)
S(x)=∫√{1+(1/x)^2}dx=∫√(1+x^2)/xdx ｛x=sinh(t)とおく(tan(t)と置いても解ける）｝
{sinh^2(t)+1=cosh^2(t) d(sinh(t))/dt=cosh(t) dx=cosh(t)dt}
=∫(cosh(t)/sinh(t))(cosh(t)dt)
=∫(1/sinh(t))dt+∫sinh(t)dt=∫(sinh(t)/(cosh^2(t)-1)dt+cosh(t)+C
=途中略=1/2log{(cosh(t)-1)/(cosh(t)+1)}+cosh(t)+C
{cosh(t)=√(1+sinh^2(t))=√(1+x^2)}
=1/2log{(√(1+x^2)-1)/(√(1+x^2)+1)}+√(1+x^2)+C
s=[S(x)](1〜a)=1/2log{(3+2√2)(√(1+a^2)-1)/(√(1+a^2)+1)}+√(1+a^2)+√2//
> y=log(cosx)(0<=x<=π/4)
y'=-sinx/cosx=-tanx
s=∫[0〜π/4]√(1+(-tanx)^2)dx=∫[0〜π/4]√(1/cos^2x)dx=∫[0〜π/4]cosx/cos^2xdx
=∫[0〜π/4]cosx/(1-sin^2x)dx=[1/2log{(1+sinx)/(1-sinx)}](0〜π/4)=1/2log(3+2√2)//
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