第9章 9.8

(1)

$I_n = \int_0^{\frac{\pi}{2}} \sin^n\theta\, d\theta$

$\theta = \dfrac{\pi}{2} - t$ とおくと $d\theta = -dt$ $\theta: 0 \to \dfrac{\pi}{2}$ $t: \dfrac{\pi}{2} \to 0$ これから

$I_n = \int_{\frac{\pi}{2}}^{0} -\sin^n\left(\frac{\pi}{2}-t\right) d\theta = \int_0^{\frac{\pi}{2}} \sin^n\left(\frac{\pi}{2}-t\right) d\theta = \int_0^{\frac{\pi}{2}} \cos^n\theta\, d\theta$

(2)

$I_2 = \int_0^{\frac{\pi}{2}} \sin^2\theta\, d\theta = \int_0^{\frac{\pi}{2}} \frac{1-\cos2\theta}{2}\, d\theta = \frac{1}{2}\left[\theta - \frac{1}{2}\sin2\theta\right]_0^{\frac{\pi}{2}} = \frac{1}{2}\left(\frac{\pi}{2}\right) = \frac{\pi}{4}$

(3)

$I_{n+2} = \int_0^{\frac{\pi}{2}} \sin^{n+2}\theta\, d\theta = \int_0^{\frac{\pi}{2}} \sin^n\theta \cdot \sin^2\theta\, d\theta = \int_0^{\frac{\pi}{2}} \sin^n\theta \cdot (1-\cos^2\theta)\, d\theta = I_n - \int_0^{\frac{\pi}{2}} \sin^n\theta \cdot \cos^2\theta\, d\theta$

ここで $\displaystyle\int \sin^n\theta \cdot \cos\theta\, d\theta = \frac{1}{n+1}\sin^{n+1}\theta$ から

$I_{n+2} = I_n - \left[\frac{1}{n+1}\sin^{n+1}\theta \cdot \cos\theta\right]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} \left(\frac{1}{n+1}\sin^{n+1}\theta \cdot (-\sin\theta)\right) d\theta = I_n - \frac{1}{n+1}\int_0^{\frac{\pi}{2}} \sin^{n+2}\theta\, d\theta$

$= I_n - \frac{1}{n+1}I_{n+2}$

$\left(1+\frac{1}{n+1}\right)I_{n+2} = I_n \qquad I_{n+2} = \frac{n+1}{n+2}I_n$

(4)

(3)より

$I_{n+2} = \frac{n+1}{n+2}I_n = \frac{n+1}{n+2}\left(\frac{n-1}{n}I_{n-2}\right) = \frac{n+1}{n+2}\cdot\frac{n-1}{n}\left(\frac{n-3}{n-2}I_{n-4}\right) = \cdots$ となる

$I_0 = \int_0^{\frac{\pi}{2}} d\theta = [\theta]_0^{\frac{\pi}{2}} = \frac{\pi}{2}$

$I_1 = \int_0^{\frac{\pi}{2}} \sin\theta\, d\theta = [-\cos\theta]_0^{\frac{\pi}{2}} = 1$ これより

n が偶数のとき $I_n = \frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdots\cdots\frac{1}{2}\cdot\frac{\pi}{2}$

n が奇数のとき $I_n = \frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdots\cdots\frac{2}{3}$