$3$\displaystyle{\int \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{ {x}^{2}- {a}^{2} } \hspace{2}}dx}$ を置換積分で計算（ただし， $3$a> 0$ ）

■ $3$ x+\sqrt{ {x}^{2}- {a}^{2} }=t$ とおく場合

$3$\displaystyle{\frac{\hspace{2} dt \hspace{2}}{\hspace{2} dx \hspace{2}}=1+\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}{\left({ {x}^{2}- {a}^{2} }\right)}^{ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} }\cdot 2x}$ $3$\displaystyle{=1+\frac{\hspace{2}x\hspace{2}}{\hspace{2} \sqrt{ {x}^{2}- {a}^{2} } \hspace{2}}}$ $3$\displaystyle{=\frac{\hspace{2} \sqrt{ {x}^{2}- {a}^{2} }+x \hspace{2}}{\hspace{2} \sqrt{ {x}^{2}- {a}^{2} } \hspace{2}}}$

より

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

$3$\displaystyle{\frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{ {a}^{2}- {x}^{2} } \hspace{2}}dx=\frac{\hspace{2}1\hspace{2}}{\hspace{2}t\hspace{2}}dt}$

以上より，積分変数を $3$x$ から $3$t$ に置換すると

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

となり，積分を進めると

$3$\displaystyle{ =\log \left|{t}\right|+C }$ 　( $3$C$ は積分定数)

$3$\displaystyle{=\log \left|{ x+\sqrt{ {x}^{2}- {a}^{2} } }\right|+C}$

■ $3$\displaystyle{x=\frac{\hspace{2}a\hspace{2}}{\hspace{2} \cos {\theta} \hspace{2}}}$ とおく場合

ただし， $3$\displaystyle{ 0< {\theta}< \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} }$ ， $3$\displaystyle{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}}< {\theta}< {\pi} }$ である．

$3$\displaystyle{ \frac{\hspace{2} dx \hspace{2}}{\hspace{2} d{\theta} \hspace{2}}=\frac{\hspace{2} a\sin {\theta} \hspace{2}}{\hspace{2} { \cos }^{2}{\theta} \hspace{2}}\text{ }\text{ }\rightarrow \text{ }dx=\frac{\hspace{2} a\sin {\theta} \hspace{2}}{\hspace{2} { \cos }^{2}{\theta} \hspace{2}}d{\theta} }$
TeXに変換設定していない数学記号や，特殊文字が含まれています。今後直していきます。

よって，積分変数を $3$x$ から $3$t$ に置換すると

$3$\displaystyle{\int \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{ {x}^{2}- {a}^{2} } \hspace{2}}dx}$ $3$\displaystyle{ =\int \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{ { \left({ \frac{\hspace{2}a\hspace{2}}{\hspace{2} \cos {\theta} \hspace{2}} }\right) }^{2}- {a}^{2} } \hspace{2}}\cdot \frac{\hspace{2} a\sin {\theta} \hspace{2}}{\hspace{2} { \cos }^{2}{\theta} \hspace{2}}d{\theta} }$

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

$3$\displaystyle{{\tan }^{2}{\theta}+1=\frac{\hspace{2}1\hspace{2}}{\hspace{2} { \cos }^{2}{\theta} \hspace{2}}}$ より

$3$\displaystyle{=\int \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{ { \tan }^{2}{\theta} } \hspace{2}}\cdot \frac{\hspace{2} \sin {\theta} \hspace{2}}{\hspace{2} { \cos }^{2}{\theta} \hspace{2}}d{\theta}}$

$3$\displaystyle{ 0< {\theta}< \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} }$ のとき， $3$\displaystyle{\tan {\theta}> 0}$ となり， $3$\displaystyle{\sqrt{ { \tan }^{2}{\theta} }=\tan {\theta}}$ である． $3$\displaystyle{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}}< {\theta}< {\pi} }$ のとき， $3$\displaystyle{\tan {\theta}< 0}$ となり，の範囲では， $3$\displaystyle{\cos {\theta}> 0}$ となり， $3$\displaystyle{\sqrt{ { \tan }^{2}{\theta} }=- \tan {\theta}}$ である．よって，場合分けをして計算をするめる．

● $3$\displaystyle{ 0< {\theta}< \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} }$ 場合

与式 $3$\displaystyle{=\int \frac{\hspace{2}1\hspace{2}}{\hspace{2} \tan {\theta} \hspace{2}}\cdot \frac{\hspace{2} \sin {\theta} \hspace{2}}{\hspace{2} { \cos }^{2}{\theta} \hspace{2}}d{\theta}}$

$3$\displaystyle{=\int \frac{\hspace{2}1\hspace{2}}{\hspace{2} \cos {\theta} \hspace{2}}d{\theta}}$

ここを参照すると

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\log \left({ \frac{\hspace{2} 1+\sin x \hspace{2}}{\hspace{2} 1- \sin x \hspace{2}} }\right)+C}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\log \left({ \frac{\hspace{2} 1+\sin {\theta} \hspace{2}}{\hspace{2} 1- \sin {\theta} \hspace{2}}\cdot \frac{\hspace{2} 1+\sin {\theta} \hspace{2}}{\hspace{2} 1+\sin {\theta} \hspace{2}} }\right)+C}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\log \left({ \frac{\hspace{2} { \left({ 1+\sin {\theta} }\right) }^{2} \hspace{2}}{\hspace{2} 1- { \sin }^{2}{\theta} \hspace{2}} }\right)+C}$

$3$\displaystyle{=\log {\left({ \frac{\hspace{2} { \left({ 1+\sin {\theta} }\right) }^{2} \hspace{2}}{\hspace{2} { \cos }^{2}{\theta} \hspace{2}} }\right)}^{ \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} }+C}$

$3$\displaystyle{\cos {\theta}> 0}$ ， $3$\displaystyle{1+\sin {\theta}> 0}$ より

$3$\displaystyle{=\log \left({ \frac{\hspace{2} 1+\sin {\theta} \hspace{2}}{\hspace{2} \cos {\theta} \hspace{2}} }\right)+C}$

$3$\displaystyle{=\log \left({ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \cos {\theta} \hspace{2}}+\tan {\theta} }\right)+C}$

$3$\displaystyle{=\log \left({ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \cos {\theta} \hspace{2}}+\sqrt{ \frac{\hspace{2}1\hspace{2}}{\hspace{2} { \cos }^{2}{\theta} \hspace{2}}- 1 } }\right)+C}$

$3$\displaystyle{=\log \left({ \frac{\hspace{2}x\hspace{2}}{\hspace{2}a\hspace{2}}+\sqrt{ { \left({ \frac{\hspace{2}x\hspace{2}}{\hspace{2}a\hspace{2}} }\right) }^{2}- 1 } }\right)+C}$

$3$\displaystyle{=\log \left({ \frac{\hspace{2}1\hspace{2}}{\hspace{2}a\hspace{2}}\left({ x+\sqrt{ {x}^{2}- {a}^{2} } }\right) }\right)+C}$

$3$\displaystyle{=\log \left({ x- \sqrt{ {x}^{2}- {a}^{2} } }\right)+\log \left({ \frac{\hspace{2}1\hspace{2}}{\hspace{2}a\hspace{2}} }\right)+C}$

$3$\displaystyle{\log \left({ \frac{\hspace{2}1\hspace{2}}{\hspace{2}a\hspace{2}} }\right)+C}$ を改めて $3$C$ とする．

$3$\displaystyle{=\log \left({ x+\sqrt{ {x}^{2}+1 } }\right)+C}$

備考： $3$\displaystyle{ 0< {\theta}< \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} }$ のとき， $3$x> a$ となり， $3$x+\sqrt{ {x}^{2}- {a}^{2} }> 0$

● $3$\displaystyle{\frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}}< {\theta}< {\pi}}$ の場合

与式 $3$\displaystyle{=\int \frac{\hspace{2}1\hspace{2}}{\hspace{2} - \tan {\theta} \hspace{2}}\cdot \frac{\hspace{2} \sin {\theta} \hspace{2}}{\hspace{2} { \cos }^{2}{\theta} \hspace{2}}d{\theta}}$

$3$\displaystyle{=- \int \frac{\hspace{2}1\hspace{2}}{\hspace{2} \cos {\theta} \hspace{2}}d{\theta}}$

ここを参照すると

$3$\displaystyle{=- \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\log \left({ \frac{\hspace{2} 1+\sin x \hspace{2}}{\hspace{2} 1- \sin x \hspace{2}} }\right)+C}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\log {\left({ \frac{\hspace{2} 1+\sin x \hspace{2}}{\hspace{2} 1- \sin x \hspace{2}} }\right)}^{ - 1 }+C}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\log \left({ \frac{\hspace{2} 1- \sin x \hspace{2}}{\hspace{2} 1+\sin x \hspace{2}} }\right)+C}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\log \left({ \frac{\hspace{2} 1- \sin {\theta} \hspace{2}}{\hspace{2} 1+\sin {\theta} \hspace{2}}\cdot \frac{\hspace{2} 1- \sin {\theta} \hspace{2}}{\hspace{2} 1- \sin {\theta} \hspace{2}} }\right)+C}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\log \left({ \frac{\hspace{2} { \left({ 1- \sin {\theta} }\right) }^{2} \hspace{2}}{\hspace{2} 1- { \sin }^{2}{\theta} \hspace{2}} }\right)+C}$

$3$\displaystyle{=\log {\left({ \frac{\hspace{2} { \left({ 1- \sin {\theta} }\right) }^{2} \hspace{2}}{\hspace{2} { \cos }^{2}{\theta} \hspace{2}} }\right)}^{ \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} }+C}$

$3$\displaystyle{\cos {\theta}< 0}$ ， $3$\displaystyle{1- \sin {\theta}> 0}$ より

$3$\displaystyle{=\log \left({ \frac{\hspace{2} 1- \sin {\theta} \hspace{2}}{\hspace{2} - \cos {\theta} \hspace{2}} }\right)+C}$

$3$\displaystyle{=\log \left({ - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \cos {\theta} \hspace{2}}?\{\tan {\theta} }\right)+C}$
MathMLのtextタグの中に日本語が含まれており、日本語が文字化けしています。今後直していきます。

$3$\displaystyle{\sqrt{ { \tan }^{2}{\theta} }=- \tan {\theta}}$ より

$3$\displaystyle{=\log \left({ - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \cos {\theta} \hspace{2}}- \sqrt{ { \tan }^{2}{\theta} } }\right)+C}$

$3$\displaystyle{=\log \left({ - \frac{\hspace{2}1\hspace{2}}{\hspace{2} \cos {\theta} \hspace{2}}- \sqrt{ \frac{\hspace{2}1\hspace{2}}{\hspace{2} { \cos }^{2}{\theta} \hspace{2}}- 1 } }\right)+C}$

$3$\displaystyle{=\log \left({ - \frac{\hspace{2}x\hspace{2}}{\hspace{2}a\hspace{2}}- \sqrt{ { \left({ \frac{\hspace{2}x\hspace{2}}{\hspace{2}a\hspace{2}} }\right) }^{2}- 1 } }\right)+C}$

$3$\displaystyle{=\log \left({ \frac{\hspace{2}1\hspace{2}}{\hspace{2}a\hspace{2}}\left({ - x- \sqrt{ {x}^{2}- {a}^{2} } }\right) }\right)+C}$

$3$\displaystyle{=\log \left({ - x- \sqrt{ {x}^{2}- {a}^{2} } }\right)+\log \left({ \frac{\hspace{2}1\hspace{2}}{\hspace{2}a\hspace{2}} }\right)+C}$

$3$\displaystyle{\log \left({ \frac{\hspace{2}1\hspace{2}}{\hspace{2}a\hspace{2}} }\right)+C}$ を改めて $3$C$ とする．

$3$\displaystyle{=\log \left({ - x- \sqrt{ {x}^{2}- {a}^{2} } }\right)+C}$

備考： $3$\displaystyle{\frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}}< {\theta}< {\pi}}$ のとき， $3$x< - a$ となり， $3$- x- \sqrt{ {x}^{2}- {a}^{2} }> 0$

● $3$\displaystyle{ 0< {\theta}< \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} }$ ， $3$\displaystyle{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}}< {\theta}< {\pi} }$ の場合

上記の結果を統合すると

$3$\displaystyle{\int \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{ {x}^{2}- {a}^{2} } \hspace{2}}dx}$ $3$\displaystyle{=\log \left|{ x+\sqrt{ {x}^{2}- {a}^{2} } }\right|+C}$

となる．

ホーム>>カテゴリー分類>>積分>>置換積分>> $3$\displaystyle{\int \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{ {x}^{2}- {a}^{2} } \hspace{2}}dx}$ を置換積分で計算（ただし， $3$a> 0$ ）

最終更新日： 2025年9月5日

を置換積分で計算（ただし，）

■とおく場合

■ とおく場合

●場合