https://spherical-harmonics.hateblo.jp/entry/Kyodai/2021/Rikei

[4] 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．

2021.03.09記

[解答]

$\dfrac{dy}{dx}=\dfrac{-\sin x}{1+\cos x}=-\tan\dfrac{x}{2}$ だから，求める長さ $L$ は
$L=\displaystyle\int_0^{\pi/2} \sqrt{1+\Bigl(\dfrac{dy}{dx}\Bigr)^2}dx$ $=\displaystyle\int_0^{\pi/2} \sqrt{1+\tan^2\dfrac{x}{2}}dx$ $=\displaystyle\int_0^{\pi/2} \dfrac{1}{\cos\dfrac{x}{2}}dx$
（∵ $0\leqq x\leqq\dfrac{\pi}{2}$ で $\cos\dfrac{x}{2}\gt 0$ ）
$=2\displaystyle\int_0^{\pi/4} \dfrac{1}{\cos u}du$ $=2\displaystyle\int_0^{\pi/4} \dfrac{(\sin u)'}{1-\sin^2 u}du$ $=\displaystyle\int_0^{\pi/4} \Bigl(\dfrac{1}{1+\sin u} - \dfrac{1}{1-\sin u}\Bigr) (\sin u)' du$ $=\Bigl[\log\dfrac{1+\sin u}{1-\sin u}\Bigr]_0^{\pi/4}$ $=\log\dfrac{1+1/\sqrt{2}}{1-1/\sqrt{2}}$ $=\log\dfrac{\sqrt{2}+1}{\sqrt{2}-1}$ $=2\log(\sqrt{2}+1)$