積分　√(a^2-x^2)

$3$\displaystyle{ \int \sqrt{ {a}^{2}-{x}^{2} }dx }$ $3$\displaystyle{ $ a> 0 $ }$

部分積分法より． $3$\displaystyle{{f}^{\prime } $x$ =1,g $x$ =\sqrt{ {a}^{2}-{x}^{2} }}$ とおく．

$3$\displaystyle{=\int 1\cdot \sqrt{ {a}^{2}-{x}^{2} }dx }$

$3$\displaystyle{=x\sqrt{ {a}^{2}-{x}^{2} }- \int \frac{\hspace{2} -{x}^{2} \hspace{2}}{\hspace{2} \sqrt{ {a}^{2}-{x}^{2} } \hspace{2}}dx }$

$3$\displaystyle{=x\sqrt{ {a}^{2}-{x}^{2} }- \int \frac{\hspace{2} {a}^{2}-{x}^{2}- {a}^{2} \hspace{2}}{\hspace{2} \sqrt{ {a}^{2}-{x}^{2} } \hspace{2}}dx }$

$3$\displaystyle{=x\sqrt{ {a}^{2}-{x}^{2} }- \int \sqrt{ {a}^{2}-{x}^{2} }dx }$ $3$\displaystyle{+\int \frac{\hspace{2} {a}^{2} \hspace{2}}{\hspace{2} \sqrt{ {a}^{2}-{x}^{2} } \hspace{2}}dx }$

$3$\displaystyle{=x\sqrt{ {a}^{2}- {x}^{2} }}$ $3$\displaystyle{- \int \sqrt{ {a}^{2}- {x}^{2} }dx +{a}^{2}{\sin }^{ - 1 }\frac{\hspace{2}x\hspace{2}}{\hspace{2}a\hspace{2}}}$

$3$\displaystyle{\int \frac{\hspace{2} {a}^{2} \hspace{2}}{\hspace{2} \sqrt{ {a}^{2}-{x}^{2} } \hspace{2}}dx }$ の積分はここを参照のこと

改めて書き直すと

$3$\displaystyle{\int \sqrt{ {a}^{2}- {x}^{2} }dx }$ $3$\displaystyle{=x\sqrt{ {a}^{2}- {x}^{2} }- \int \sqrt{ {a}^{2}- {x}^{2} }dx +{a}^{2}{\sin }^{ - 1 }\frac{\hspace{2}x\hspace{2}}{\hspace{2}a\hspace{2}}}$ ･･････(1)

(1)を $3$\displaystyle{\int \sqrt{ {a}^{2}-{x}^{2} }dx }$ についてとく．

$3$\displaystyle{2\int \sqrt{ {a}^{2}- {x}^{2} }dx }$ $3$\displaystyle{=x\sqrt{ {a}^{2}- {x}^{2} }+{a}^{2}{\sin }^{ - 1 }\frac{\hspace{2}x\hspace{2}}{\hspace{2}a\hspace{2}}}$

$3$\displaystyle{\int \sqrt{ {a}^{2}- {x}^{2} }dx }$ $3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} $ x\sqrt{ {a}^{2}- {x}^{2} }+{a}^{2}{ \sin }^{ - 1 }\frac{\hspace{2}x\hspace{2}}{\hspace{2}a\hspace{2}} $ }$

積分定数を付け加えると

■置換積分による解法

$3$\displaystyle{x=a\sin {\theta}}$ $3$\displaystyle{ $ - \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}}< {\theta}< \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} $ }$ とおくと

$3$\displaystyle{\frac{\hspace{2} dx \hspace{2}}{\hspace{2} d{\theta} \hspace{2}}=a\cos {\theta}\rightarrow dx=a\cos {\theta} d{\theta} }$

となる．よって

与式 $3$\displaystyle{=\int \sqrt{ {a}^{2}-{ $ a\sin {\theta} $ }^{2} }\cdot a \cos {\theta} d{\theta} }$

$3$\displaystyle{=\int \sqrt{ {a}^{2} $ 1-{\sin }^{2}{\theta} $ }\cdot a \cos {\theta} d{\theta} }$

$3$\displaystyle{=\int \sqrt{ {a}^{2} { \cos }^{2}{\theta} }\cdot a \cos {\theta} d{\theta}}$

$3$\displaystyle{ - \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}}< {\theta}< \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} }$ では $3$\displaystyle{\cos {\theta}\geq 0}$ より， $3$\displaystyle{\sqrt{ {a}^{2} { \cos }^{2}{\theta} }=a \cos {\theta} }$

よって

$3$\displaystyle{=\int a \cos {\theta} \cdot a \cos {\theta} d{\theta} }$

$3$\displaystyle{=\int {a}^{2} { \cos }^{2}{\theta} d{\theta} }$

$3$\displaystyle{={a}^{2}\int { \cos }^{2}{\theta} d{\theta}}$

$3$\displaystyle{={a}^{2} $ \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}{\theta}+\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}}\sin 2{\theta} $ +C}$

$3$\displaystyle{={a}^{2} $ \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}{\theta}+\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\sin {\theta}\cos {\theta} $ +C}$

$3$\displaystyle{\int { \cos }^{2}{\theta} d{\theta} }$ の積分はここを参照のこと

$3$\displaystyle{\sin {\theta}=\frac{\hspace{2}x\hspace{2}}{\hspace{2}a\hspace{2}}}$ ， $3$\displaystyle{\cos {\theta}=\sqrt{ 1- { $ \frac{\hspace{2}x\hspace{2}}{\hspace{2}a\hspace{2}} $ }^{2} },{\theta}={\sin }^{ - 1 }\frac{\hspace{2}x\hspace{2}}{\hspace{2}a\hspace{2}}}$ より，変数を $3${\theta}$ から $3$x$ に戻すと

与式 $3$\displaystyle{={a}^{2} $ \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}{ \sin }^{ - 1 }\frac{\hspace{2}x\hspace{2}}{\hspace{2}a\hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\cdot \frac{\hspace{2}x\hspace{2}}{\hspace{2}a\hspace{2}}\sqrt{ 1- { \( \frac{\hspace{2}x\hspace{2}}{\hspace{2}a\hspace{2}} $ }^{2} } \) +C}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} $ x\sqrt{ {a}^{2}- {x}^{2} }+{a}^{2}{ \sin }^{ - 1 }\frac{\hspace{2}x\hspace{2}}{\hspace{2}a\hspace{2}} $ +C}$

ホーム>>カテゴリー分類>>積分>>積分の具体事例>>根号を含む積分>積分　√(a^2-x^2)

最終更新日： 2025年6月30日

積分 √(a^2-x^2)

■置換積分による解法

積分　√(a^2-x^2)