積分　√(a^2+x^2)

$\int \sqrt{a^{2} + x^{2}} d x$ $\int \sqrt{a^{2} + x^{2}} d x$ $(a > 0)$ $(a > 0)$

■部分積分法による解法

$\int \sqrt{a^{2} + x^{2}} d x$ $\int \sqrt{a^{2} + x^{2}} d x$ $= \int 1 \cdot \sqrt{a^{2} + x^{2}} d x$ $= \int 1 \cdot \sqrt{a^{2} + x^{2}} d x$

部分積分法より． $f^{'} (x) = 1, g (x) = \sqrt{a^{2} + x^{2}}$ $f^{'} (x) = 1, g (x) = \sqrt{a^{2} + x^{2}}$ とおく．

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

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

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

$= x \sqrt{a^{2} + x^{2}} - \int \sqrt{a^{2} + x^{2}} d x$ $= x \sqrt{a^{2} + x^{2}} - \int \sqrt{a^{2} + x^{2}} d x$ $+ a^{2} log ∣ ∣ x + \sqrt{a^{2} + x^{2}} ∣ ∣$ $+ a^{2} log | x + \sqrt{a^{2} + x^{2}} |$

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

改めて書き直すと

$\int \sqrt{a^{2} + x^{2}} d x$ $\int \sqrt{a^{2} + x^{2}} d x$ $= x \sqrt{a^{2} + x^{2}}$ $= x \sqrt{a^{2} + x^{2}}$ $- \int \sqrt{a^{2} + x^{2}} d x$ $- \int \sqrt{a^{2} + x^{2}} d x$ $+ a^{2} log ∣ ∣ x + \sqrt{a^{2} + x^{2}} ∣ ∣$ $+ a^{2} log | x + \sqrt{a^{2} + x^{2}} |$

$\int \sqrt{a^{2} + x^{2}} d x$ $\int \sqrt{a^{2} + x^{2}} d x$ について整理すると

$2 \int \sqrt{a^{2} + x^{2}} d x$ $2 \int \sqrt{a^{2} + x^{2}} d x$ $= x \sqrt{a^{2} + x^{2}} + a^{2} log ∣ ∣ x + \sqrt{a^{2} + x^{2}} ∣ ∣$ $= x \sqrt{a^{2} + x^{2}} + a^{2} log | x + \sqrt{a^{2} + x^{2}} |$

$\int \sqrt{a^{2} + x^{2}} d x$ $\int \sqrt{a^{2} + x^{2}} d x$ $= \frac{1}{2} {x \sqrt{a^{2} + x^{2}} + a^{2} log ∣ ∣ x + \sqrt{a^{2} + x^{2}} ∣ ∣}$ $= \frac{1}{2} {x \sqrt{a^{2} + x^{2}} + a^{2} log | x + \sqrt{a^{2} + x^{2}} |}$

積分定数を付け加えると

$\int \sqrt{a^{2} + x^{2}} d x$ $\int \sqrt{a^{2} + x^{2}} d x$ $= \frac{1}{2} {x \sqrt{a^{2} + x^{2}} + a^{2} log ∣ ∣ x + \sqrt{a^{2} + x^{2}} ∣ ∣} + C$ $= \frac{1}{2} {x \sqrt{a^{2} + x^{2}} + a^{2} log | x + \sqrt{a^{2} + x^{2}} |} + C$

■置換積分法による解法

$x = a tan θ$ $x = a tan θ$ $(- \frac{π}{2} < θ < \frac{π}{2})$ $(- \frac{π}{2} < θ < \frac{π}{2})$ とおくと

$\frac{d x}{d θ} = a {(\frac{sin θ}{cos θ})}^{'} = a \frac{{cos}^{2} θ + {sin}^{2} θ}{{cos}^{2} θ}$ $\frac{d x}{d θ} = a {(\frac{sin θ}{cos θ})}^{'} = a \frac{{cos}^{2} θ + {sin}^{2} θ}{{cos}^{2} θ}$ $= a \frac{1}{{cos}^{2} θ}$ $= a \frac{1}{{cos}^{2} θ}$ となり， $d x = \frac{a}{{cos}^{2} θ} d θ$ $d x = \frac{a}{{cos}^{2} θ} d θ$

よって

与式 $= \int \sqrt{a^{2} + {(a tan θ)}^{2}} \cdot \frac{a}{{cos}^{2} θ} d θ$ $= \int \sqrt{a^{2} + {(a tan θ)}^{2}} \cdot \frac{a}{{cos}^{2} θ} d θ$

$= \int \sqrt{a^{2} (1 + tan θ^{2})} \cdot \frac{a}{{cos}^{2} θ} d θ$ $= \int \sqrt{a^{2} (1 + tan θ^{2})} \cdot \frac{a}{{cos}^{2} θ} d θ$

$= \int \sqrt{\frac{a^{2}}{{cos}^{2} θ}} \cdot \frac{a}{{cos}^{2} θ} d θ$ $= \int \sqrt{\frac{a^{2}}{{cos}^{2} θ}} \cdot \frac{a}{{cos}^{2} θ} d θ$

$- \frac{π}{2} < θ < \frac{π}{2}$ $- \frac{π}{2} < θ < \frac{π}{2}$ では $cos θ ≧ 0$ $cos θ ≧ 0$ より， $\sqrt{\frac{a^{2}}{{cos}^{2} θ}} = \frac{a}{cos θ}$ $\sqrt{\frac{a^{2}}{{cos}^{2} θ}} = \frac{a}{cos θ}$

よって

$= \int \frac{a}{cos θ} \cdot \frac{a}{{cos}^{2} θ} d θ$ $= \int \frac{a}{cos θ} \cdot \frac{a}{{cos}^{2} θ} d θ$

$= \int \frac{a^{2}}{{cos}^{3} θ} d θ$ $= \int \frac{a^{2}}{{cos}^{3} θ} d θ$

$= a^{2} \int \frac{1}{{cos}^{3} θ} d θ$ $= a^{2} \int \frac{1}{{cos}^{3} θ} d θ$

$= a^{2} {\frac{1}{2} \cdot \frac{sin θ}{{cos}^{2} θ} + \frac{1}{4} \cdot log (\frac{1 + sin θ}{1 - sin θ})} + C$ $= a^{2} {\frac{1}{2} \cdot \frac{sin θ}{{cos}^{2} θ} + \frac{1}{4} \cdot log (\frac{1 + sin θ}{1 - sin θ})} + C$

$\int \frac{1}{{cos}^{3} θ} d θ$ $\int \frac{1}{{cos}^{3} θ} d θ$ の積分はここを参照のこと

$sin θ = \frac{x}{\sqrt{a^{2} + x^{2}}}, cos θ = \frac{a}{\sqrt{a^{2} + x^{2}}}$ $sin θ = \frac{x}{\sqrt{a^{2} + x^{2}}}, cos θ = \frac{a}{\sqrt{a^{2} + x^{2}}}$ より

$\frac{sin θ}{{cos}^{2} θ} = \frac{x}{\sqrt{a^{2} + x^{2}}} \cdot \frac{1}{{(\frac{a}{\sqrt{a^{2} + x^{2}}})}^{2}}$ $\frac{sin θ}{{cos}^{2} θ} = \frac{x}{\sqrt{a^{2} + x^{2}}} \cdot \frac{1}{{(\frac{a}{\sqrt{a^{2} + x^{2}}})}^{2}}$ $= \frac{x}{a^{2}} \cdot \sqrt{a^{2} + x^{2}}$ $= \frac{x}{a^{2}} \cdot \sqrt{a^{2} + x^{2}}$

$\frac{1 + sin θ}{1 - sin θ} = \frac{1 + \frac{x}{\sqrt{a^{2} + x^{2}}}}{1 - \frac{x}{\sqrt{a^{2} + x^{2}}}}$ $\frac{1 + sin θ}{1 - sin θ} = \frac{1 + \frac{x}{\sqrt{a^{2} + x^{2}}}}{1 - \frac{x}{\sqrt{a^{2} + x^{2}}}}$ $= \frac{{(1 + \frac{x}{\sqrt{a^{2} + x^{2}}})}^{2}}{1^{2} - {(\frac{x}{\sqrt{a^{2} + x^{2}}})}^{2}}$ $= \frac{{(1 + \frac{x}{\sqrt{a^{2} + x^{2}}})}^{2}}{1^{2} - {(\frac{x}{\sqrt{a^{2} + x^{2}}})}^{2}}$ $= \frac{{(\frac{\sqrt{a^{2} + x^{2}} + x}{\sqrt{a^{2} + x^{2}}})}^{2}}{\frac{a^{2} + x^{2} - x^{2}}{a^{2} + x^{2}}}$ $= \frac{{(\frac{\sqrt{a^{2} + x^{2}} + x}{\sqrt{a^{2} + x^{2}}})}^{2}}{\frac{a^{2} + x^{2} - x^{2}}{a^{2} + x^{2}}}$ $= {(\frac{\sqrt{a^{2} + x^{2}} + x}{a})}^{2}$ $= {(\frac{\sqrt{a^{2} + x^{2}} + x}{a})}^{2}$

これらより，変数を $θ$ から $x$ に戻すと

与式 $= a^{2} {\frac{1}{2} \cdot \frac{x}{a^{2}} \sqrt{a^{2} + x^{2}} + \frac{1}{4} log {(\frac{x + \sqrt{a^{2} + x^{2}}}{a})}^{2}} + C$

$= \frac{1}{2} x \sqrt{a^{2} + x^{2}}$ $+ \frac{a^{2}}{4} {2 log | x + \sqrt{a^{2} + x^{2}} | - 2 log a} + C$

$= \frac{1}{2} x \sqrt{a^{2} + x^{2}}$ $+ \frac{a^{2}}{2} log | x + \sqrt{a^{2} + x^{2}} | + \frac{a^{2}}{2} log a + C$

$= \frac{1}{2} {x \sqrt{a^{2} + x^{2}} + a^{2} log | x + \sqrt{a^{2} + x^{2}} |} + C$

$\frac{a^{2}}{2} log a$ は定数なので，積分定数を $\frac{a^{2}}{2} log a + C \to C$ におきなおしている．

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　最終更新日： 2023年10月4日

積分 √(a^2+x^2)

■部分積分法による解法

■置換積分法による解法

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積分　√(a^2+x^2)