第5章 5.27

(1)

$\int_0^a \frac{x^3}{\sqrt{a^2-x^2}}dx$

$x = a\sin\theta$ とおくと $dx = a\cos\theta d\theta$ $x:0 \to a$ $\theta:0 \to \dfrac{\pi}{2}$ これから

$=\int_0^{\frac{\pi}{2}} \frac{a^3\sin^3\theta}{a\cos\theta}\cdot a\cos\theta d\theta = \int_0^{\frac{\pi}{2}} a^3\sin^3\theta d\theta = a^3\cdot\frac{2}{3} = \frac{2}{3}a^3$

(2)

$\int_0^a \frac{x^4}{\sqrt{a^2-x^2}}dx$

$x = a\sin\theta$ とおくと $dx = a\cos\theta d\theta$ $x:0 \to a$ $\theta:0 \to \dfrac{\pi}{2}$ これから

$=\int_0^{\frac{\pi}{2}} \frac{a^4\sin^4\theta}{a\cos\theta}\cdot a\cos\theta d\theta = \int_0^{\frac{\pi}{2}} a^4\sin^4\theta d\theta = a^4\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2} = \frac{3}{16}a^4\pi$

(3)

$\int_0^a \frac{x^6}{\sqrt{a^2-x^2}}dx$

(1)

と同様に考えて

$=\int_0^{\frac{\pi}{2}} a^6\sin^6\theta d\theta = a^6\cdot\frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2} = \frac{5}{32}\pi a^6$