曲線 \(y = \log ( 1 +\cos x )\) の \(0 \leqq x \leqq \dfrac{\pi}{2}\) の部分の長さを求めよ.
【 解 答 】
\[ y' = -\dfrac{\sin x}{1 +\cos x} \] なので \[\begin{align} \sqrt{1 +\left\{ y' \right\}^2} & = \sqrt{\dfrac{2 +2 \cos x}{( 1 +\cos x )^2}} \\ & = \dfrac{1}{\sqrt{\cos^2 \dfrac{x}{2}}} = \dfrac{1}{\cos \dfrac{x}{2}}\quad \left( \text{∵} \cos \dfrac{x}{2} \gt 0 \right) \end{align}\] よって, 求める長さ \(L\) は \[\begin{align} L & = \displaystyle\int _ {0}^{\frac{\pi}{2}} \dfrac{1}{\cos \dfrac{x}{2}} \, dx = \displaystyle\int _ {0}^{\frac{\pi}{2}} \dfrac{\cos \dfrac{x}{2}}{1 -\sin^2 \dfrac{x}{2}} \, dx \\ & = \displaystyle\int _ {0}^{\frac{\pi}{2}} \left\{ \dfrac{\left( 1 +\sin \dfrac{x}{2} \right)'}{1 +\sin \dfrac{x}{2}} -\dfrac{\left( 1 -\sin \dfrac{x}{2} \right)'}{1 -\sin \dfrac{x}{2}} \right\} \, dx \\ & = \left[ \log \dfrac{1 +\sin \dfrac{x}{2}}{1 -\sin \dfrac{x}{2}} \right]_ {0}^{\frac{\pi}{2}} \\ & = \log \dfrac{\sqrt{2} +1}{\sqrt{2} -1} = \underline{2 \log \left( \sqrt{2} +1 \right)} \end{align}\]