定積分の問題

■問題

次の問題を積分せよ(定積分)．

${\displaystyle {\int }_1^{\sqrt{2}}\frac{1}{\sqrt{4-2x^2}}dx}$

■解説動画

◇積分の動画一覧のページへ

■答

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

■ヒント

${\displaystyle x=\sqrt{2}\sin \theta }$ 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

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

を用いる．

■解説

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

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

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

よって

${\displaystyle {\int }_1^{\sqrt{2}}\frac{1}{\sqrt{4-2x^2}}dx}$

${\displaystyle ={\displaystyle {\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\frac{1}{\sqrt{4-2{\left(\sqrt{2}\sin \theta \right)}^2}}\sqrt{2}\cos \theta d\theta }}$

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

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

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

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

（ ${\displaystyle \frac{\pi }{4}\leqq \theta \leqq \frac{\pi }{2}}$ では ${\displaystyle \cos \theta \geqq 0}$ なので ${\displaystyle \sqrt{{\cos }^2\theta }=\cos \theta }$ ）

${\displaystyle =\frac{\sqrt{2}}{2}{\displaystyle {\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}d\theta }}$

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

${\displaystyle =\frac{\sqrt{2}}{2}\left(\frac{\pi }{2}-\frac{\pi }{4}\right)}$

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

●別解

あらかじめ， ${\displaystyle \int \frac{1}{\sqrt{4-2x^2}}dx}$ 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

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

${\displaystyle =\int \frac{1}{\sqrt{2^2-{\left(\sqrt{2}x\right)}^2}}dx}$

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

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

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

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

よって，定積分の定義より

${\displaystyle {\int }_1^{\sqrt{2}}\frac{1}{\sqrt{4-2x^2}}dx}$ ${\displaystyle ={\left[\frac{1}{\sqrt{2}}{sin}^{-1}\frac{x}{\sqrt{2}}\right]}_1^{\sqrt{2}}}$

となる．

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

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

アークサインはここを参照

${\displaystyle =\frac{1}{\sqrt{2}}·\frac{\pi }{2}-\frac{1}{\sqrt{2}}·\frac{\pi }{4}}$

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

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

　

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最終更新日： 2025年6月9日