積分　√(a^2+x^2)

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

■部分積分法による解法

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

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

$3$\displaystyle{\int \sqrt{ {a}^{2}+{x}^{2} }dx }$ について整理すると

$3$\displaystyle{2\int \sqrt{ {a}^{2}+{x}^{2} }dx }$ $3$\displaystyle{=x\sqrt{ {a}^{2}+{x}^{2} }+{a}^{2}\log \| x+\sqrt{ {a}^{2}+{x}^{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}\log \| x+\sqrt{ {a}^{2}+{x}^{2} } \| \} }$

積分定数を付け加えると

■置換積分法による解法

$3$\displaystyle{x=a\tan {\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{ $ \frac{\hspace{2} \sin {\theta} \hspace{2}}{\hspace{2} \cos {\theta} \hspace{2}} $ }^{\prime }=a\frac{\hspace{2} { \cos }^{2}{\theta}+{ \sin }^{2}{\theta} \hspace{2}}{\hspace{2} { \cos }^{2}{\theta} \hspace{2}}}$ $3$\displaystyle{=a\frac{\hspace{2}1\hspace{2}}{\hspace{2} { \cos }^{2}{\theta} \hspace{2}}}$ となり， $3$\displaystyle{dx=\frac{\hspace{2}a\hspace{2}}{\hspace{2} { \cos }^{2}{\theta} \hspace{2}}d{\theta}}$

よって

与式 $3$\displaystyle{=\int \sqrt{ {a}^{2}+{ $ a\tan {\theta} $ }^{2} }\cdot \frac{\hspace{2}a\hspace{2}}{\hspace{2} { \cos }^{2}{\theta} \hspace{2}}d{\theta} }$

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

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

よって

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

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

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

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

$3$\displaystyle{\int \frac{\hspace{2}1\hspace{2}}{\hspace{2} { \cos }^{3}{\theta} \hspace{2}}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} \sqrt{ {a}^{2}+{x}^{2} } \hspace{2}}\text{\hspace{2}},\text{\hspace{2}}\cos {\theta}=\frac{\hspace{2}a\hspace{2}}{\hspace{2} \sqrt{ {a}^{2}+{x}^{2} } \hspace{2}}}$ より

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

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

これらより，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}}\cdot \frac{\hspace{2}x\hspace{2}}{\hspace{2} {a}^{2} \hspace{2}}\sqrt{ {a}^{2}+{x}^{2} }+\frac{\hspace{2}1\hspace{2}}{\hspace{2}4\hspace{2}}\log { $ \frac{\hspace{2} x+\sqrt{ {a}^{2}+{x}^{2} } \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} }}$ $3$\displaystyle{+\frac{\hspace{2} {a}^{2} \hspace{2}}{\hspace{2}4\hspace{2}} \{ 2\log \| x+\sqrt{ {a}^{2}+{x}^{2} } \| - 2\log a \} +C}$

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

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

$3$\displaystyle{\frac{\hspace{2} {a}^{2} \hspace{2}}{\hspace{2}2\hspace{2}}\log a}$ は定数なので，積分定数を $3$\displaystyle{\frac{\hspace{2} {a}^{2} \hspace{2}}{\hspace{2}2\hspace{2}}\log a+C\rightarrow C}$ におきなおしている．

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　最終更新日： 2023年10月4日

積分 √(a^2+x^2)

■部分積分法による解法

■置換積分法による解法

積分　√(a^2+x^2)