## 問題の内容

## 問題の内容

以下の4つの定積分を計算する問題です。

(1) $\int_{0}^{\pi/2} \sin^7 x \, dx$

(2) $\int_{0}^{\pi/2} \cos^8 x \, dx$

(3) $\int_{0}^{\pi/2} (1+\cos^2 x)^2 \, dx$

(4) $\int_{0}^{\pi/2} (1+2\sin x)^4 \, dx$

## 解き方の手順

### (1) $\int_{0}^{\pi/2} \sin^7 x \, dx$

ウォリスの公式を利用します。ウォリスの公式は以下の通りです。

$I_n = \int_{0}^{\pi/2} \sin^n x \, dx = \int_{0}^{\pi/2} \cos^n x \, dx$

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

$I_0 = \int_{0}^{\pi/2} 1 \, dx = \frac{\pi}{2}$

$I_1 = \int_{0}^{\pi/2} \sin x \, dx = 1$

よって、

$I_7 = \frac{6}{7}I_5 = \frac{6}{7}\frac{4}{5}I_3 = \frac{6}{7}\frac{4}{5}\frac{2}{3}I_1 = \frac{6}{7}\frac{4}{5}\frac{2}{3} \cdot 1 = \frac{16}{35}$

### (2) $\int_{0}^{\pi/2} \cos^8 x \, dx$

同様にウォリスの公式を利用します。

$I_8 = \frac{7}{8}I_6 = \frac{7}{8}\frac{5}{6}I_4 = \frac{7}{8}\frac{5}{6}\frac{3}{4}I_2 = \frac{7}{8}\frac{5}{6}\frac{3}{4}\frac{1}{2}I_0 = \frac{7}{8}\frac{5}{6}\frac{3}{4}\frac{1}{2} \cdot \frac{\pi}{2} = \frac{35\pi}{256}$

### (3) $\int_{0}^{\pi/2} (1+\cos^2 x)^2 \, dx$

$(1+\cos^2 x)^2 = 1+2\cos^2 x + \cos^4 x$ なので、

$\int_{0}^{\pi/2} (1+\cos^2 x)^2 \, dx = \int_{0}^{\pi/2} (1+2\cos^2 x + \cos^4 x) \, dx = \int_{0}^{\pi/2} 1 \, dx + 2\int_{0}^{\pi/2} \cos^2 x \, dx + \int_{0}^{\pi/2} \cos^4 x \, dx$

$\int_{0}^{\pi/2} 1 \, dx = \frac{\pi}{2}$

$\int_{0}^{\pi/2} \cos^2 x \, dx = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$

$\int_{0}^{\pi/2} \cos^4 x \, dx = \frac{3}{4} \int_{0}^{\pi/2} \cos^2 x \, dx = \frac{3}{4} \cdot \frac{\pi}{4} = \frac{3\pi}{16}$

よって、

$\int_{0}^{\pi/2} (1+\cos^2 x)^2 \, dx = \frac{\pi}{2} + 2\cdot \frac{\pi}{4} + \frac{3\pi}{16} = \frac{8\pi}{16} + \frac{8\pi}{16} + \frac{3\pi}{16} = \frac{19\pi}{16}$

### (4) $\int_{0}^{\pi/2} (1+2\sin x)^4 \, dx$

$(1+2\sin x)^4 = 1 + 8\sin x + 24\sin^2 x + 32\sin^3 x + 16\sin^4 x$

$\int_{0}^{\pi/2} 1 \, dx = \frac{\pi}{2}$

$\int_{0}^{\pi/2} \sin x \, dx = 1$

$\int_{0}^{\pi/2} \sin^2 x \, dx = \frac{1}{2}\cdot \frac{\pi}{2} = \frac{\pi}{4}$

$\int_{0}^{\pi/2} \sin^3 x \, dx = \frac{2}{3}$

$\int_{0}^{\pi/2} \sin^4 x \, dx = \frac{3}{4} \int_{0}^{\pi/2} \sin^2 x \, dx = \frac{3}{4} \cdot \frac{\pi}{4} = \frac{3\pi}{16}$

よって、

$\int_{0}^{\pi/2} (1+2\sin x)^4 \, dx = \int_{0}^{\pi/2} (1 + 8\sin x + 24\sin^2 x + 32\sin^3 x + 16\sin^4 x) \, dx = \frac{\pi}{2} + 8 + 24\cdot \frac{\pi}{4} + 32\cdot \frac{2}{3} + 16\cdot \frac{3\pi}{16} = \frac{\pi}{2} + 8 + 6\pi + \frac{64}{3} + 3\pi = \frac{19\pi}{2} + \frac{24+64}{3} = \frac{19\pi}{2} + \frac{88}{3} = \frac{57\pi+176}{6}$

## 最終的な答え

(1) $\frac{16}{35}$

(2) $\frac{35\pi}{256}$

(3) $\frac{19\pi}{16}$

(4) $\frac{57\pi+176}{6}$