1948年(昭和23年)東京大学工学部-数学[3] - [別館]球面倶楽部零八式markIISR

2020.04.03記

[3] 次の積分値を求めよ．
$I=\displaystyle\int_{-a}^{a}\dfrac{|x^3|dx}{\sqrt{a^2-x^2}(\sqrt{2}a-x)}$ 但し $a\gt 0$ とする．

2022.07.19記

[解答]
$I=\displaystyle\int_{-a}^{0}\dfrac{-x^3 dx}{\sqrt{a^2-x^2}(\sqrt{2}a-x)}$ $+\displaystyle\int_{0}^{a}\dfrac{x^3dx}{\sqrt{a^2-x^2}(\sqrt{2}a-x)}$
であり，前の積分を $y=-x$ を置換すると
$\displaystyle\int_{-a}^{0}\dfrac{-x^3 dx}{\sqrt{a^2-x^2}(\sqrt{2}a-x)}$ $=\displaystyle\int_{a}^{0}\dfrac{y^3 (-dy)}{\sqrt{a^2-y^2}(\sqrt{2}a+y)}$ $=\displaystyle\int_{0}^{a}\dfrac{y^3 dy}{\sqrt{a^2-y^2}(\sqrt{2}a+y)}$ $=\displaystyle\int_{0}^{a}\dfrac{x^3 dy}{\sqrt{a^2-x^2}(\sqrt{2}a+x)}$
となるので，
$I=\displaystyle\int_{0}^{a}\dfrac{x^3 dx}{\sqrt{a^2-x^2}(\sqrt{2}a+x)}$ $+\displaystyle\int_{0}^{a}\dfrac{x^3dx}{\sqrt{a^2-x^2}(\sqrt{2}a-x)}$
$=2\sqrt{2}a \displaystyle\int_{0}^{a}\dfrac{x^3 dx}{\sqrt{a^2-x^2}(2a^2-x^2)}$
となる． $\sqrt{a^2-x^2}=t$ とおくと
$a^2-x^2=t^2$ から $-xdx=tdt$ であるから
$I=2\sqrt{2}a \displaystyle\int_{a}^{0}\dfrac{t^2-a^2}{a^2+t^2}dt$
$=2\sqrt{2}a \displaystyle\int_{0}^{a}\dfrac{a^2-t^2}{a^2+t^2}dt$
$=2\sqrt{2}a \displaystyle\int_{0}^{a}\left(\dfrac{2a^2}{a^2+t^2}-1\right)dt$
$=2\sqrt{2}a \Bigl[2a\,\mbox{Arctan}\,\dfrac{t}{a}-t\Bigr]_{0}^{a}$
$=2\sqrt{2}a \left(\dfrac{a\pi}{2}-a\right)$
$=\sqrt{2}(\pi-2) a^2$

[解答]
$x=a\sin\theta$ とおくと
$I=\displaystyle\int_{-\pi/2}^{\pi/2}\dfrac{a^4|\sin^3\theta|\cos\theta d\theta}{a\cos\theta(\sqrt{2}a-a\sin\theta)}$
$=a^2\displaystyle\int_{-\pi/2}^{\pi/2}\dfrac{|\sin^3\theta| d\theta}{\sqrt{2}-\sin\theta}$
$=a^2\displaystyle\int_{-\pi/2}^{0}\dfrac{-\sin^3\theta d\theta}{\sqrt{2}-\sin\theta}$ $+a^2\displaystyle\int_{0}^{\pi/2}\dfrac{\sin^3\theta d\theta}{\sqrt{2}-\sin\theta}$
$=a^2\displaystyle\int_{0}^{\pi/2}\dfrac{\sin^3\theta d\theta}{\sqrt{2}+\sin\theta}$ $+a^2\displaystyle\int_{0}^{\pi/2}\dfrac{\sin^3\theta d\theta}{\sqrt{2}-\sin\theta}$
$=2\sqrt{2} a^2\displaystyle\int_{0}^{\pi/2}\dfrac{\sin^3\theta d\theta}{2-\sin^2\theta}$
$=2\sqrt{2} a^2\displaystyle\int_{0}^{\pi/2}\dfrac{(1-\cos^2\theta)\sin\theta d\theta}{1+\cos^2\theta}$
$=2\sqrt{2} a^2\displaystyle\int_{0}^{1}\dfrac{1-t^2}{1+t^2} dt$ ( $\cos\theta=t$
$=2\sqrt{2}a^2 \displaystyle\int_{0}^{1}\left(\dfrac{2}{1+t^2}-1\right)dt$
$=2\sqrt{2}a^2 \Bigl[2\,\mbox{Arctan}\, t-t\Bigr]_{0}^{1}$
$=2\sqrt{2}a^2 \left(\dfrac{\pi}{2}-1\right)$
$=\sqrt{2}(\pi-2) a^2$