定積分の問題

■問題

次の問題を積分せよ(定積分)．

$3$\displaystyle{ \int _{0}^{1} \frac{\hspace{2} {x}^{5} \hspace{2}}{\hspace{2} \sqrt{ 1- {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}8\hspace{2}}{\hspace{2} 15 \hspace{2}}}$

■ヒント

置換積分法

$3$\displaystyle{ \int _{a}^{b} f $x$ dx=\int _{{\alpha}}^{{\beta}} f $ g \(t$ \) {g}^{\prime } $t$ dt }$ ･･････(1)

ウォリス積分

$3$\displaystyle{ \int _{0}^{\frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}}} {\cos }^{n}xdx }$ $3$\displaystyle{ =\int _{0}^{\frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}}} {\sin }^{n}xdx }$ $3$\displaystyle{=\left{{ \begin{array} \frac{\hspace{2} n- 1 \hspace{2}}{\hspace{2}n\hspace{2}}\cdot \frac{\hspace{2} n- 3 \hspace{2}}{\hspace{2} n- 2 \hspace{2}}\cdots \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\cdot \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} & \vspace{6}\\ \frac{\hspace{2} n- 1 \hspace{2}}{\hspace{2}n\hspace{2}}\cdot \frac{\hspace{2} n- 3 \hspace{2}}{\hspace{2} n- 2 \hspace{2}}\cdots \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}\cdot 1 & \vspace{6}\\ \end{array} }$

	$3$n$ ：偶数	･･････(2)
	$3$n$ ：奇数

を用いる．

■解説

あらかじめ， $3$\displaystyle{ \int \frac{\hspace{2} {x}^{5} \hspace{2}}{\hspace{2} \sqrt{ 1- {x}^{2} } \hspace{2}}dx }$ を求めておく．

$3$\displaystyle{ x=\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}} $ }$ ･･････(3)

とおいて置換積分をする．

$3$\displaystyle{ \frac{\hspace{2} dx \hspace{2}}{\hspace{2} d{\theta} \hspace{2}}=\cos {\theta} }$ ∴ $3$\displaystyle{ dx=\cos {\theta}d{\theta} }$ ･･････(4)

（ $3$\displaystyle{ \sin {\theta} }$ を微分すると $3$\displaystyle{ \cos {\theta} }$ になるのは，微分 $3$\displaystyle{ \sin x }$ を参照）

よって

$3$\displaystyle{ \int \frac{\hspace{2} {x}^{5} \hspace{2}}{\hspace{2} \sqrt{ 1- {x}^{2} } \hspace{2}}dx }$ $3$\displaystyle{=\int \frac{\hspace{2} { \sin }^{5}{\theta} \hspace{2}}{\hspace{2} \sqrt{ 1- { \sin }^{2}{\theta} } \hspace{2}}\cos {\theta}d{\theta} }$

ここで， $3$\displaystyle{ 1- { \sin }^{2}{\theta}={ \cos }^{2}{\theta} }$ （三角関数の相互関係の $3$\displaystyle{1}$ 番目の式を参照）より　

$3$\displaystyle{ \sqrt{ 1- { \sin }^{2}{\theta} }=\sqrt{ { \cos }^{2}{\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}} }$ なので， $3$\displaystyle{ \cos {\theta}\geq 0 }$ となり， $3$\displaystyle{ \sqrt{ { \cos }^{2}{\theta} }=\cos {\theta} }$ である．　

したがって

$3$\displaystyle{=\int \frac{\hspace{2} { \sin }^{5}{\theta} \hspace{2}}{\hspace{2} \cos {\theta} \hspace{2}}\cos {\theta}d{\theta} }$

$3$\displaystyle{ =\int { \sin }^{5}{\theta}d{\theta} }$

次に， $3$\displaystyle{ \int _{0}^{1} \frac{\hspace{2} {x}^{5} \hspace{2}}{\hspace{2} \sqrt{ 1- {x}^{2} } \hspace{2}} dx }$ を求める．

はじめに　 $3$\displaystyle{ x=\sin {\theta} }$ と置換していることいり

$3$\displaystyle{x=0}$ のとき $3$\displaystyle{ {\theta}=0 }$ ,　 $3$\displaystyle{ x=1 }$ のとき $3$\displaystyle{{\theta}=\frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}}}$

よって，ヒントの式(1)より

$3$\displaystyle{ \int _{0}^{1} \frac{\hspace{2} {x}^{5} \hspace{2}}{\hspace{2} \sqrt{ 1- {x}^{2} } \hspace{2}}dx=\int _{0}^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} } { \sin }^{5}{\theta}d{\theta} }$

となる．

ヒントの式(2)のを適用する． $3$\displaystyle{n}$ に代入する値が $3$\displaystyle{5}$ （奇数）となる

$3$\displaystyle{=\frac{\hspace{2}4\hspace{2}}{\hspace{2}5\hspace{2}}\cdot \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}\cdot 1}$

$3$\displaystyle{=\frac{\hspace{2}8\hspace{2}}{\hspace{2} 15 \hspace{2}}}$

●別解

$3$\displaystyle{\displaystyle{ \int _{0}^{1} \frac{\hspace{2} {x}^{5} \hspace{2}}{\hspace{2} \sqrt{ 1- {x}^{2} } \hspace{2}}dx }}$ $3$\displaystyle{=\displaystyle{ \int _{0}^{1} \frac{\hspace{2} { \left({ {x}^{2} }\right) }^{2}\cdot x \hspace{2}}{\hspace{2} \sqrt{ 1- {x}^{2} } \hspace{2}}dx }}$

$3$\displaystyle{1- {x}^{2}=t}$ とおく置換積分で計算する．

$3$\displaystyle{\frac{\hspace{2} dt \hspace{2}}{\hspace{2} dx \hspace{2}}=- 2x}$ より $3$\displaystyle{xdx=- \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}dt}$

$3$\displaystyle{x:0\rightarrow 1}$ のとき $3$\displaystyle{t:1\rightarrow 0}$

$3${x}^{2}=1- t$

よって

$3$\displaystyle{=\displaystyle{ \int _{1}^{0} \frac{\hspace{2} { \left({ 1- t }\right) }^{2} \hspace{2}}{\hspace{2} \sqrt{t} \hspace{2}}\left({ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}dt }\right) }}$

$3$\displaystyle{=- \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\displaystyle{ \int _{1}^{0} \frac{\hspace{2} 1- 2t+{t}^{2} \hspace{2}}{\hspace{2} \sqrt{t} \hspace{2}}dt }}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\displaystyle{ \int _{0}^{1} \frac{\hspace{2} 1- 2t+{t}^{2} \hspace{2}}{\hspace{2} \sqrt{t} \hspace{2}}dt }}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\displaystyle{ \int _{0}^{1} \left({ {t}^{ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} }- 2{t}^{ \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} }+{t}^{ \frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}} } }\right)dt }}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\left[{ 2{t}^{ \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}} }- \frac{\hspace{2}4\hspace{2}}{\hspace{2}3\hspace{2}}{t}^{ \frac{\hspace{2}3\hspace{2}}{\hspace{2}2\hspace{2}} }+\frac{\hspace{2}2\hspace{2}}{\hspace{2}5\hspace{2}}{t}^{ \frac{\hspace{2}5\hspace{2}}{\hspace{2}2\hspace{2}} } }\right]_{0}^{1}}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\left({ 2- \frac{\hspace{2}4\hspace{2}}{\hspace{2}3\hspace{2}}+\frac{\hspace{2}2\hspace{2}}{\hspace{2}5\hspace{2}} }\right)}$

$3$\displaystyle{==\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\left({ \frac{\hspace{2}2\hspace{2}}{\hspace{2}3\hspace{2}}+\frac{\hspace{2}2\hspace{2}}{\hspace{2}5\hspace{2}} }\right)}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}3\hspace{2}}+\frac{\hspace{2}1\hspace{2}}{\hspace{2}5\hspace{2}}}$

$3$\displaystyle{=\frac{\hspace{2}8\hspace{2}}{\hspace{2} 15 \hspace{2}}}$

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最終更新日： 2025年6月9日