積分　√(a^2-x^2)

${\displaystyle \int \sqrt{a^2-x^2}dx}$     ${\displaystyle \left(a>0\right)}$

部分積分法より． ${\displaystyle f^{\prime }\left(x\right)=1,g\left(x\right)=\sqrt{a^2-x^2}}$ とおく．

${\displaystyle =\int 1\cdot \sqrt{a^2-x^2}dx}$

${\displaystyle =x\sqrt{a^2-x^2}-\int \frac{-x^2}{\sqrt{a^2-x^2}}dx}$

${\displaystyle =x\sqrt{a^2-x^2}-\int \frac{a^2-x^2-a^2}{\sqrt{a^2-x^2}}dx}$

${\displaystyle =x\sqrt{a^2-x^2}-\int \sqrt{a^2-x^2}dx}$ ${\displaystyle +\int \frac{a^2}{\sqrt{a^2-x^2}}dx}$

${\displaystyle =x\sqrt{a^2-x^2}}$ ${\displaystyle -\int \sqrt{a^2-x^2}dx+a^2{sin}^{-1}\frac{x}{a}}$

${\displaystyle \int \frac{a^2}{\sqrt{a^2-x^2}}dx}$  の積分はここを参照のこと

改めて書き直すと

(1)を ${\displaystyle \int \sqrt{a^2-x^2}dx}$  についてとく．

${\displaystyle 2\int \sqrt{a^2-x^2}dx}$ ${\displaystyle =x\sqrt{a^2-x^2}+a^2{sin}^{-1}\frac{x}{a}}$

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

積分定数を付け加えると

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

■置換積分による解法

${\displaystyle x=asin\theta }$      ${\displaystyle \left(-\frac{\pi }{2}<\theta <\frac{\pi }{2}\right)}$  とおくと

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

となる．よって

与式 ${\displaystyle =\int \sqrt{a^2-{\left(asin\theta \right)}^2}\cdot acos\theta d\theta }$

${\displaystyle =\int \sqrt{a^2\left(1-{sin}^2\theta \right)}\cdot acos\theta d\theta }$

${\displaystyle =\int \sqrt{a^2{cos}^2\theta }\cdot acos\theta d\theta }$

${\displaystyle -\frac{\pi }{2}<\theta <\frac{\pi }{2}}$  では ${\displaystyle cos\theta \geqq 0}$  より， ${\displaystyle \sqrt{a^2{cos}^2\theta }=acos\theta }$

よって

${\displaystyle =\int acos\theta \cdot acos\theta d\theta }$

${\displaystyle =\int a^2{cos}^2\theta d\theta }$

${\displaystyle =a^2\int {cos}^2\theta d\theta }$

${\displaystyle =a^2\left(\frac{1}{2}\theta +\frac{1}{4}sin2\theta \right)+C}$

${\displaystyle =a^2\left(\frac{1}{2}\theta +\frac{1}{2}sin\theta cos\theta \right)+C}$

${\displaystyle \int {cos}^2\theta d\theta }$  の積分はここを参照のこと

${\displaystyle sin\theta =\frac{x}{a}}$ ， ${\displaystyle cos\theta =\sqrt{1-{\left(\frac{x}{a}\right)}^2},\theta ={sin}^{-1}\frac{x}{a}}$  より，変数を $\theta$  から x  に戻すと

${\displaystyle =\frac{1}{2}\left(x\sqrt{a^2-x^2}+a^2{sin}^{-1}\frac{x}{a}\right)+C}$

　

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最終更新日： 2025年6月30日