置換積分

■問題

次の計算をせよ（不定積分）．

${\displaystyle {{\displaystyle \int }}_0^1{\left(x\sqrt{1-x^2}\right)}^{3}dx}$

■解説動画

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■答

${\displaystyle \frac{2}{35}}$

■ヒント

${\displaystyle \sqrt{1-x^2}=t}$ とおき， ${\displaystyle t}$ の式に変換する． （置換積分法）

■解説

${\displaystyle {{\displaystyle \int }}_0^1{\left(x\sqrt{1-x^2}\right)}^{3}dx}$ ･･････(1)

${\displaystyle \sqrt{1-x^2}=t}$ とおく．　　･･････(2)

両辺を２乗して， ${\displaystyle 1-x^2=t^2}$

${\displaystyle x^2=1-t^2}$ ･･････(3)

両辺を ${\displaystyle x}$ で微分して，

${\displaystyle 2x=-2t\frac{dt}{dx}}$

よって， ${\displaystyle xdx=-tdt}$ ･･････(4)

${\displaystyle x}$ の積分範囲より，

${\displaystyle x=0}$ のとき， ${\displaystyle t=1}$

${\displaystyle x=1}$ のとき， ${\displaystyle t=0}$

であるから， ${\displaystyle t}$ の積分範囲は， ${\displaystyle 1\to 0}$ ･･････(5)

(1)の式を変形すると，

${\displaystyle {{\displaystyle \int }}_0^1{\left(x\sqrt{1-x^2}\right)}^{3}dx}$

${\displaystyle ={{\displaystyle \int }}_0^1{x}^3\cdot {\left(\sqrt{1-x^2}\right)}^{3}dx}$

${\displaystyle ={{\displaystyle \int }}_0^1{x}^2\cdot {\left(\sqrt{1-x^2}\right)}^3\cdot xdx}$

ここに(2),(3),(4),(5)を代入して，

${\displaystyle {\displaystyle {\int }_1^0\left(1-t^2\right)\cdot t^3}\cdot \left(-tdt\right)}$

${\displaystyle =-{\displaystyle {\int }_1^0\left(1-t^2\right)\cdot t^4}dt}$

${\displaystyle ={\displaystyle {\int }_0^1\left(1-t^2\right)\cdot t^4}dt}$

${\displaystyle ={\displaystyle {\int }_0^1\left(t^4-t^6\right)}dt}$

${\displaystyle ={\left[\frac{1}{5}{t}^5-\frac{1}{7}{t}^7\right]}_0^1}$

${\displaystyle =\left(\frac{1}{5}\cdot 1^5-\frac{1}{7}\cdot 1^7\right)}$ ${\displaystyle -\left(\frac{1}{5}\cdot 0^5-\frac{1}{7}\cdot 0^7\right)}$

${\displaystyle =\frac{1}{5}-\frac{1}{7}}$

${\displaystyle =\frac{2}{35}}$

●別解

$x=\sin \theta$とおく．ただし，${\displaystyle -\frac{1}{2}\pi \leqq \theta \leqq \frac{1}{2}\pi }$ ．

${\displaystyle \frac{dx}{d\theta }=\cos \theta \to dx=\cos \theta d\theta }$

$x:0\to 1$のとき，${\displaystyle \theta :0\to \frac{\pi }{2}}$

よって

与式${\displaystyle ={\displaystyle {\int }_0^{\frac{\pi }{2}}{\left(\sin \theta \sqrt{1-{\sin }^2\theta }\right)}^3\cos \theta d\theta }}$

${\displaystyle ={\displaystyle {\int }_0^{\frac{\pi }{2}}{\left(\sin \theta \sqrt{{\cos }^2\theta }\right)}^3\cos \theta d\theta }}$

${\displaystyle -\frac{1}{2}\pi \leqq \theta \leqq \frac{1}{2}\pi }$ のとき，${\displaystyle \cos \theta \geqq 0}$ ．よって

${\displaystyle ={\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^3\theta {\cos }^4\theta d\theta }}$

${\displaystyle ={\displaystyle {\int }_0^{\frac{\pi }{2}}\left(1-{\cos }^2\theta \right)\sin \theta {\cos }^4\theta d\theta }}$

${\displaystyle ={\displaystyle {\int }_0^{\frac{\pi }{2}}\left({\cos }^4\theta -{\cos }^6\theta \right)\sin \theta d\theta }}$

${\displaystyle {\displaystyle \cos \theta =t}}$とおく．

${\displaystyle {\displaystyle \frac{dt}{d\theta }=-\sin \theta \to \cos \theta d\theta =-dt}}$

${\displaystyle \theta :0\to \frac{\pi }{2}}$ のとき，${\displaystyle t:1\to 0}$

よって

${\displaystyle ={\displaystyle {\int }_1^0\left(t^4-t^6\right)\left(-1\right)dt}}$

${\displaystyle ={{\displaystyle \int }}_0^1\left(t^4-t^6\right)dt}$

${\displaystyle ={\left[\frac{1}{5}{t}^5-\frac{1}{7}{t}^7\right]}_0^1}$

${\displaystyle =\left(\frac{1}{5}\cdot 1^5-\frac{1}{7}\cdot 1^7\right)-\left(\frac{1}{5}\cdot 0^5-\frac{1}{7}\cdot 0^7\right)}$

${\displaystyle =\frac{1}{5}-\frac{1}{7}}$

${\displaystyle =\frac{2}{35}}$

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最終更新日：2025年9月8日