定積分の問題

■問題

次の問題を積分せよ(定積分)．

$3$\displaystyle{\int _{1}^{ \sqrt{2} } \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{ 4- 2{x}^{2} } \hspace{2}}dx }$

■解説動画

◇積分の動画一覧のページへ

■答

$3$\displaystyle{\frac{\hspace{2} \sqrt{2} \hspace{2}}{\hspace{2}8\hspace{2}}{\pi}}$ 　

■ヒント

$3$\displaystyle{ x=\sqrt{2}\sin {\theta} }$ とおく置換積分をする．

あるいは，基本となる関数の積分より

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

を用いる．

■解説

$3$\displaystyle{ x=\sqrt{2}\sin {\theta} }$ とおく置換積分をする．

$3$\displaystyle{\frac{\hspace{2} dx \hspace{2}}{\hspace{2} d{\theta} \hspace{2}}=\sqrt{2}\cos {\theta}}$ より， $3$dx=\sqrt{2}\cos {\theta}d{\theta}$

$3$x$ が $3$\displaystyle{1\rightarrow \sqrt{2}}$ のとき， $3${\theta}$ は $3$\displaystyle{\frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}4\hspace{2}}\rightarrow \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}}}$ 　

よって

$3$\displaystyle{ \int _{1}^{ \sqrt{2} } \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{ 4- 2{x}^{2} } \hspace{2}}dx }$

$3$\displaystyle{=\displaystyle{ \int _{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}4\hspace{2}} }^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} } \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{ 4- 2{ $ \sqrt{2}\sin {\theta} $ }^{2} } \hspace{2}}\sqrt{2}\cos {\theta}d{\theta} }}$

$3$\displaystyle{=\displaystyle{ \int _{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}4\hspace{2}} }^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} } \frac{\hspace{2} \sqrt{2}\cos {\theta} \hspace{2}}{\hspace{2} \sqrt{ 4- 4{ \sin }^{2}{\theta} } \hspace{2}}d{\theta} }}$

$3$\displaystyle{=\frac{\hspace{2} \sqrt{2} \hspace{2}}{\hspace{2}2\hspace{2}}\displaystyle{ \int _{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}4\hspace{2}} }^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} } \frac{\hspace{2} \cos {\theta} \hspace{2}}{\hspace{2} \sqrt{ 1- { \sin }^{2}{\theta} } \hspace{2}}d{\theta} }}$

（ $3$\displaystyle{\frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}4\hspace{2}}\leq {\theta}\leq \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}}}$ では $3$\displaystyle{\cos {\theta}\geq 0}$ なので $3$\displaystyle{ \sqrt{ { \cos }^{2}{\theta} }=\cos {\theta} }$ ）

$3$\displaystyle{=\frac{\hspace{2} \sqrt{2} \hspace{2}}{\hspace{2}2\hspace{2}} \[{\theta}\] _{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}4\hspace{2}} }^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} }}$

$3$\displaystyle{=\frac{\hspace{2} \sqrt{2} \hspace{2}}{\hspace{2}2\hspace{2}} $ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}}- \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}4\hspace{2}} $ }$

$3$\displaystyle{=\frac{\hspace{2} \sqrt{2} \hspace{2}}{\hspace{2}8\hspace{2}}{\pi}}$

●別解

あらかじめ， $3$\displaystyle{\int \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{ 4- 2{x}^{2} } \hspace{2}}dx }$ を求めておく．類題のこれを参照のこと

ヒントの式(1)より

$3$\displaystyle{\int \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{ 4- 2{x}^{2} } \hspace{2}}dx }$

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

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

$3$\displaystyle{ \frac{\hspace{2} \sqrt{2} \hspace{2}}{\hspace{2}2\hspace{2}}=\frac{\hspace{2} \sqrt{2}\cdot \sqrt{2} \hspace{2}}{\hspace{2} 2\sqrt{2} \hspace{2}} }$ $3$\displaystyle{ =\frac{\hspace{2}2\hspace{2}}{\hspace{2} 2\sqrt{2} \hspace{2}} }$ $3$\displaystyle{ =\frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} }$ より

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

（これが　 $3$\displaystyle{ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{ 4- 2{x}^{2} } \hspace{2}} }$ の原始関数である）

よって，定積分の定義より

$3$\displaystyle{\int _{1}^{ \sqrt{2} } \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{ 4- 2{x}^{2} } \hspace{2}}dx }$ $3$\displaystyle{= \[ \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} { \sin }^{ - 1 }\frac{\hspace{2}x\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} \] _{1}^{ \sqrt{2} }}$

となる．

$3$\displaystyle{= \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} {\sin }^{ - 1 }\frac{\hspace{2} \sqrt{2} \hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}}- \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} {\sin }^{ - 1 }\frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}}}$

$3$\displaystyle{= \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} {\sin }^{ - 1 }1- \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} {\sin }^{ - 1 }\frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}}}$

アークサインはここを参照

$3$\displaystyle{= \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} \cdot \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}}- \frac{\hspace{2}1\hspace{2}}{\hspace{2} \sqrt{2} \hspace{2}} \cdot \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}4\hspace{2}}}$

$3$\displaystyle{ =\frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2} 4\sqrt{2} \hspace{2}} }$

$3$\displaystyle{=\frac{\hspace{2} \sqrt{2} \hspace{2}}{\hspace{2}8\hspace{2}}{\pi}}$

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最終更新日： 2025年6月9日