#11. Chinookさんの問題
便利メモ/数学/懐石メニュー  Share on Twitter

このページは Chinook さんの問題 を解いた時の解法のメモです。想定解法が綺麗すぎるのでそっちを見るべきですが。
$$ \int _{0}^{\pi }(1-\sin x)^{n}\sqrt {\sin x(1-\cos x)}dx =\frac {\pi \sqrt 2(2n-1)!!}{(2n+2)!!} $$
$$ \begin {aligned} \int _{0}^{\pi }(1-\sin x)^{n}\sqrt {\sin x(1-\cos x)}dx &=\int _{0}^{\frac {\pi }2}(1-\sin x)^{n}\sin x\left (\sqrt {\frac {1-\cos x}{\sin x}}+\sqrt \frac {1+\cos x}{\sin x}\right )dx\\ &=\int _{0}^{\frac {\pi }2}(1-\sin x)^{n}\sin x\left (\sqrt {\tan \frac {x}2}+\sqrt {\cot \frac {x}2}\right )dx\\ &=\int _{0}^{\frac {\pi }2}(1-\cos x)^{n}\cos x\left (\sqrt {\frac {1-\tan \frac {x}2}{1+\tan \frac {x}2}}+\sqrt{\frac {1+\tan \frac {x}2}{1-\tan \frac {x}2}}\right )dx \\ &=\int _{0}^{\frac {\pi }2}(1-\cos x)^{n}\cos x\cdot \frac {2}{\sqrt {1-\tan ^{2}\frac {x}2}}dx\\ &=2\int _{0}^{\frac {\pi }2}(1-\cos x)^{n}\cos x\cdot \frac {\cos \frac {x}2}{\sqrt {\cos x}}dx\\ &=2\int _{0}^{\frac {\pi }2}\left (2\sin ^{2}\frac {x}2\right )^{n}\sqrt {\cos x}\cos \frac {x}2dx\\ &=2\sqrt 2\int _{0}^{1}t^{2n}\sqrt {1-t^{2}}dt\\ &=\sqrt 2B\left (n+\frac {1}2,\frac {3}2\right )\\ &=\textcolor {blue}{\frac {\pi \sqrt 2(2n-1)!!}{(2n+2)!!}}. \end {aligned} $$