定積分の問題

■問題

次の問題を積分せよ(定積分)．

$3$\displaystyle{ \int _{0}^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} } { \sin }^{4}x{ \cos }^{2}xdx }$

■解説動画

◇積分の動画一覧のページへ

■答

$3$\displaystyle{ \frac{\hspace{2}1\hspace{2}}{\hspace{2} 32 \hspace{2}}{\pi} }$

■ヒント

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
	$3$n$ ：奇数

を用いる．

■解説

$3$\displaystyle{ { \cos }^{2}x=1- { \sin }^{2}x }$ （三角関数の関係式の $3$\displaystyle{1}$ 番目の式を変形する）より

与式 $3$\displaystyle{ =\int _{0}^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} } { \sin }^{4}x $ 1- { \sin }^{2}x $ dx }$

$3$\displaystyle{ =\int _{0}^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} } $ { \sin }^{4}x- { \sin }^{6}x $ dx }$

$3$\displaystyle{ =\int _{0}^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} } { \sin }^{4}x dx- \int _{0}^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} } { \sin }^{6}x dx }$ （∵ 定積分の基本式(3)）

ヒントの公式の $3$\displaystyle{n}$ に代入する値が $3$\displaystyle{4}$ と $3$\displaystyle{6}$ （偶数）なので，以下のようになる．

$3$\displaystyle{ =\frac{\hspace{2}3\hspace{2}}{\hspace{2}4\hspace{2}}\cdot \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\cdot \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}}- \frac{\hspace{2}5\hspace{2}}{\hspace{2}6\hspace{2}}\cdot \frac{\hspace{2}3\hspace{2}}{\hspace{2}4\hspace{2}}\cdot \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\cdot \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} }$

$3$\displaystyle{ =\frac{\hspace{2}3\hspace{2}}{\hspace{2} 16 \hspace{2}}{\pi}- \frac{\hspace{2}5\hspace{2}}{\hspace{2} 32 \hspace{2}}{\pi} }$

$3$\displaystyle{ =\frac{\hspace{2}1\hspace{2}}{\hspace{2} 32 \hspace{2}}{\pi} }$

●別解

$3$\displaystyle{\displaystyle{ \int _{0}^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} } { \sin }^{4}x{ \cos }^{2}xdx }}$

$3$\displaystyle{=\displaystyle{ \int _{0}^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} } { \left({ \sin x\cos x }\right) }^{2}{ \sin }^{2}xdx }}$

2倍角の公式を用いて式を変形する．

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

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}8\hspace{2}}\displaystyle{ \int _{0}^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} } { \sin }^{2}2x\left({ 1- \cos 2x }\right)dx }}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}8\hspace{2}}\left{{ \displaystyle{ \int _{0}^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} } { \sin }^{2}2xdx }- \displaystyle{ \int _{0}^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} } { \sin }^{2}2x\cos 2xdx } }\right}}$

$3$\displaystyle{\displaystyle{ \int _{0}^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} } { \sin }^{2}2xdx }}$ $3$\displaystyle{=\displaystyle{ \int _{0}^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} } \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\left({ 1- \cos 2x }\right)dx }}$ $3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\left[{ x- \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\sin 2x }\right]_{0}^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} }}$ $3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\left{{ \left({ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}}- \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\sin {\pi} }\right)- \left({ 0- \frac{\hspace{2}1\hspace{2}}{\hspace{2}2\hspace{2}}\sin 0 }\right) }\right}}$ $3$\displaystyle{=\frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}4\hspace{2}}}$
$3$\displaystyle{\displaystyle{ \int _{0}^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} } { \sin }^{2}2x\cos 2xdx }}$
$3$\displaystyle{\sin 2x=t}$ とおく置換積分で解く

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

$3$\displaystyle{x:0\rightarrow \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}}}$ のとき $3$\displaystyle{t:0\rightarrow 0}$

よって

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

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}8\hspace{2}}\left({ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}4\hspace{2}}+0 }\right)}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2} 32 \hspace{2}}{\pi}}$

ホーム>>カテゴリー分類>>積分>>積分の問題>>定積分の問題 >> $3$\displaystyle{ \int _{0}^{ \frac{\hspace{2}{\pi}\hspace{2}}{\hspace{2}2\hspace{2}} } { \sin }^{4}x{ \cos }^{2}xdx }$

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最終更新日： 2025年10月29日