与えられた定積分を計算します。 (1) $\int_{0}^{2} x^4 \sqrt{4-x^2} dx$ (2) $\int_{0}^{2} x^2 \sqrt{2x-x^2} dx$

1. 問題の内容

与えられた定積分を計算します。

(1) $\int_{0}^{2} x^4 \sqrt{4-x^2} dx$

(2) $\int_{0}^{2} x^2 \sqrt{2x-x^2} dx$

2. 解き方の手順

(1) $\int_{0}^{2} x^4 \sqrt{4-x^2} dx$

$x = 2\sin{\theta}$ と置換すると、$dx = 2\cos{\theta} d\theta$ となります。

$x=0$ のとき、$\theta = 0$ であり、$x=2$ のとき、$\theta = \frac{\pi}{2}$ となります。

$\sqrt{4-x^2} = \sqrt{4-4\sin^2{\theta}} = \sqrt{4\cos^2{\theta}} = 2\cos{\theta}$

よって、

$\int_{0}^{2} x^4 \sqrt{4-x^2} dx = \int_{0}^{\frac{\pi}{2}} (2\sin{\theta})^4 (2\cos{\theta}) (2\cos{\theta}) d\theta$

$= \int_{0}^{\frac{\pi}{2}} 16\sin^4{\theta} \cdot 4\cos^2{\theta} d\theta = 64\int_{0}^{\frac{\pi}{2}} \sin^4{\theta} \cos^2{\theta} d\theta$

ここで、$\sin^2{\theta} = \frac{1-\cos{2\theta}}{2}$ 、$\cos^2{\theta} = \frac{1+\cos{2\theta}}{2}$ を用います。

$\sin^4{\theta} = (\sin^2{\theta})^2 = (\frac{1-\cos{2\theta}}{2})^2 = \frac{1}{4}(1-2\cos{2\theta} + \cos^2{2\theta})$

$\cos^2{2\theta} = \frac{1+\cos{4\theta}}{2}$

$\sin^4{\theta} = \frac{1}{4}(1-2\cos{2\theta} + \frac{1+\cos{4\theta}}{2}) = \frac{1}{8}(2-4\cos{2\theta}+1+\cos{4\theta}) = \frac{1}{8}(3-4\cos{2\theta}+\cos{4\theta})$

$\sin^4{\theta} \cos^2{\theta} = \frac{1}{8}(3-4\cos{2\theta}+\cos{4\theta})\frac{1+\cos{2\theta}}{2} = \frac{1}{16}(3-4\cos{2\theta}+\cos{4\theta})(1+\cos{2\theta})$

$= \frac{1}{16}(3+3\cos{2\theta}-4\cos{2\theta}-4\cos^2{2\theta}+\cos{4\theta}+\cos{4\theta}\cos{2\theta})$

$= \frac{1}{16}(3-\cos{2\theta}-4\cos^2{2\theta}+\cos{4\theta}+\cos{4\theta}\cos{2\theta})$

$\cos^2{2\theta} = \frac{1+\cos{4\theta}}{2}$

$\cos{4\theta}\cos{2\theta} = \frac{1}{2}(\cos(4\theta+2\theta) + \cos(4\theta-2\theta)) = \frac{1}{2}(\cos{6\theta}+\cos{2\theta})$

$\sin^4{\theta}\cos^2{\theta} = \frac{1}{16}(3-\cos{2\theta}-4\frac{1+\cos{4\theta}}{2}+\cos{4\theta}+\frac{1}{2}(\cos{6\theta}+\cos{2\theta}))$

$= \frac{1}{16}(3-\cos{2\theta}-2-2\cos{4\theta}+\cos{4\theta}+\frac{1}{2}\cos{6\theta}+\frac{1}{2}\cos{2\theta})$

$= \frac{1}{16}(1-\frac{1}{2}\cos{2\theta}-\cos{4\theta}+\frac{1}{2}\cos{6\theta})$

$64\int_{0}^{\frac{\pi}{2}} \sin^4{\theta} \cos^2{\theta} d\theta = 64\int_{0}^{\frac{\pi}{2}} \frac{1}{16}(1-\frac{1}{2}\cos{2\theta}-\cos{4\theta}+\frac{1}{2}\cos{6\theta})d\theta$

$= 4\int_{0}^{\frac{\pi}{2}}(1-\frac{1}{2}\cos{2\theta}-\cos{4\theta}+\frac{1}{2}\cos{6\theta})d\theta = 4[\theta - \frac{1}{4}\sin{2\theta} - \frac{1}{4}\sin{4\theta} + \frac{1}{12}\sin{6\theta}]_{0}^{\frac{\pi}{2}}$

$= 4(\frac{\pi}{2} - 0 - 0 + 0 - (0 - 0 - 0 + 0)) = 2\pi$

(2) $\int_{0}^{2} x^2 \sqrt{2x-x^2} dx$

$2x - x^2 = -(x^2 - 2x) = -(x^2 - 2x + 1 - 1) = 1 - (x-1)^2$

$x-1 = \sin{\theta}$ と置換すると、$dx = \cos{\theta} d\theta$ となります。

$x = \sin{\theta} + 1$

$x=0$ のとき、$\sin{\theta} = -1$ なので $\theta = -\frac{\pi}{2}$ であり、$x=2$ のとき、$\sin{\theta} = 1$ なので $\theta = \frac{\pi}{2}$ となります。

$\sqrt{2x-x^2} = \sqrt{1-(x-1)^2} = \sqrt{1-\sin^2{\theta}} = \cos{\theta}$

$x^2 = (\sin{\theta} + 1)^2 = \sin^2{\theta} + 2\sin{\theta} + 1$

$\int_{0}^{2} x^2 \sqrt{2x-x^2} dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin^2{\theta} + 2\sin{\theta} + 1)\cos{\theta} \cdot \cos{\theta} d\theta = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin^2{\theta} + 2\sin{\theta} + 1)\cos^2{\theta} d\theta$

$= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin^2{\theta}\cos^2{\theta} + 2\sin{\theta}\cos^2{\theta} + \cos^2{\theta}) d\theta = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin^2{\theta}\cos^2{\theta} + \cos^2{\theta}) d\theta + 2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin{\theta}\cos^2{\theta} d\theta$

$\sin{\theta}\cos^2{\theta}$ は奇関数なので、$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin{\theta}\cos^2{\theta} d\theta = 0$

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2{\theta}\cos^2{\theta} d\theta = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{4}\sin^2{2\theta} d\theta = \frac{1}{4}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1-\cos{4\theta}}{2} d\theta = \frac{1}{8}[\theta - \frac{1}{4}\sin{4\theta}]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{1}{8}(\frac{\pi}{2} - 0 - (-\frac{\pi}{2} - 0)) = \frac{1}{8}\pi$

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2{\theta} d\theta = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\cos{2\theta}}{2} d\theta = \frac{1}{2}[\theta + \frac{1}{2}\sin{2\theta}]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{1}{2}(\frac{\pi}{2} + 0 - (-\frac{\pi}{2} + 0)) = \frac{1}{2}\pi$

$\int_{0}^{2} x^2 \sqrt{2x-x^2} dx = \frac{1}{8}\pi + \frac{1}{2}\pi = \frac{5}{8}\pi$

3. 最終的な答え

(1) $2\pi$

(2) $\frac{5}{8}\pi$