MY MATH NOTE

京都大学(理系) 2021年　問題4

$曲線 \ y=\log (1+\cos x)\ の \ \ 0 \leqq x \leqq \cfrac{\pi}{2}\ \ の部分の長さを求めよ。$

　
$\quad y=\log (1+\cos x) \quad より \quad y'=\cfrac{-\sin x}{1+\cos x}$
\begin{eqnarray*} L &=&\int _0^{\small{\cfrac{\pi}{2}}}\sqrt{1+y'^2}dx\\ &=&\int _0^{\small{\cfrac{\pi}{2}}}\sqrt{1+\big(\cfrac{-\sin x}{1+\cos x}\big)^2}dx\\ &=&\int _0^{\small{\cfrac{\pi}{2}}}\cfrac{\sqrt{(1+\cos x )^2+\sin ^2x}}{1+\cos x}dx\\ &=&\int _0^{\small{\cfrac{\pi}{2}}}\cfrac{\sqrt{2(1+\cos x )}}{1+\cos x}dx\\ &=&\int _0^{\small{\cfrac{\pi}{2}}}\cfrac{\sqrt{2}}{\sqrt{1+\cos x}}dx\\ &=&\int _0^{\small{\cfrac{\pi}{2}}}\cfrac{\sqrt{2}}{\sqrt{1+2\cos ^2\dfrac{x}{2}-1}}dx\\ &=&\int _0^{\small{\cfrac{\pi}{2}}}\cfrac{dx}{\cos \dfrac{x}{2}}\\ \\ &=&\int _0^{\small{\cfrac{\pi}{2}}}\cfrac{\cos \dfrac{x}{2}}{\cos ^2\dfrac{x}{2}}dx\\ \\ &=&\int _0^{\small{\cfrac{\pi}{2}}}\cfrac{\cos \dfrac{x}{2}}{1-\sin ^2\dfrac{x}{2}}dx\\ \end{eqnarray*} \[ \sin \dfrac{x}{2}=t \quad とおくと \quad \cfrac{1}{2}\cos \cfrac{x}{2}dx=dt \hspace{3em} \begin{array}{c|c} x & 0\ \ \rightarrow \cfrac{\pi}{2} \\ \hline t & \ 0 \rightarrow \cfrac{1}{\sqrt{2}} \\ \end{array} \] \begin{eqnarray*} L &=&\int _0^{\small{\cfrac{1}{\sqrt{2}}}}\cfrac{2}{1-t^2}dt\\ &=&\int _0^{\small{\cfrac{1}{\sqrt{2}}}}\cfrac{2}{(1+t)(1-t)}dt\\ &=&\int _0^{\small{\cfrac{1}{\sqrt{2}}}}\big(\cfrac{1}{1+t}+\cfrac{1}{1-t}\big)dt\\ &=&\Big[\log(1+t)-\log(1-t)\Big] _0^{\small{\cfrac{1}{\sqrt{2}}}}\\ \\ &=&\log \cfrac{\sqrt{2}+1}{\sqrt{2}-1}\\ \\ &=&2\log(\sqrt{2}+1)\\ \end{eqnarray*}

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