定積分 $\int_{0}^{\pi/2} (1+2\sin x)^4 dx$ を計算します。

1. 問題の内容

定積分 $\int_{0}^{\pi/2} (1+2\sin x)^4 dx$ を計算します。

2. 解き方の手順

まず、二項定理を用いて $(1+2\sin x)^4$ を展開します。

$$(1+2\sin x)^4 = \sum_{k=0}^{4} \binom{4}{k} (2\sin x)^k = \binom{4}{0} + \binom{4}{1} (2\sin x) + \binom{4}{2} (2\sin x)^2 + \binom{4}{3} (2\sin x)^3 + \binom{4}{4} (2\sin x)^4$$

$$= 1 + 4(2\sin x) + 6(4\sin^2 x) + 4(8\sin^3 x) + 16\sin^4 x$$

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

したがって、

$$\int_{0}^{\pi/2} (1+2\sin x)^4 dx = \int_{0}^{\pi/2} (1 + 8\sin x + 24\sin^2 x + 32\sin^3 x + 16\sin^4 x) dx$$

各項を個別に計算します。

$\int_{0}^{\pi/2} 1 dx = [x]_{0}^{\pi/2} = \frac{\pi}{2}$

$\int_{0}^{\pi/2} \sin x dx = [-\cos x]_{0}^{\pi/2} = -\cos(\pi/2) + \cos(0) = 0 + 1 = 1$

$\int_{0}^{\pi/2} \sin^2 x dx = \int_{0}^{\pi/2} \frac{1-\cos(2x)}{2} dx = \frac{1}{2} [x - \frac{1}{2}\sin(2x)]_{0}^{\pi/2} = \frac{1}{2} [\frac{\pi}{2} - 0 - (0 - 0)] = \frac{\pi}{4}$

$\int_{0}^{\pi/2} \sin^3 x dx = \int_{0}^{\pi/2} \sin x (1-\cos^2 x) dx = \int_{0}^{\pi/2} (\sin x - \sin x \cos^2 x) dx = [-\cos x + \frac{\cos^3 x}{3}]_{0}^{\pi/2} = (0+0) - (-1+\frac{1}{3}) = 1 - \frac{1}{3} = \frac{2}{3}$

$\int_{0}^{\pi/2} \sin^4 x dx = \int_{0}^{\pi/2} (\sin^2 x)^2 dx = \int_{0}^{\pi/2} (\frac{1-\cos(2x)}{2})^2 dx = \frac{1}{4} \int_{0}^{\pi/2} (1 - 2\cos(2x) + \cos^2(2x)) dx = \frac{1}{4} \int_{0}^{\pi/2} (1 - 2\cos(2x) + \frac{1+\cos(4x)}{2}) dx = \frac{1}{4} [x - \sin(2x) + \frac{1}{2}x + \frac{1}{8}\sin(4x)]_{0}^{\pi/2} = \frac{1}{4} [\frac{3\pi}{2} - 0 + 0 - 0] = \frac{3\pi}{8 \cdot 2} = \frac{3\pi}{16}$

よって、

$$\int_{0}^{\pi/2} (1+2\sin x)^4 dx = \frac{\pi}{2} + 8(1) + 24(\frac{\pi}{4}) + 32(\frac{2}{3}) + 16(\frac{3\pi}{16}) = \frac{\pi}{2} + 8 + 6\pi + \frac{64}{3} + 3\pi = \frac{\pi}{2} + 9\pi + 8 + \frac{64}{3} = \frac{19\pi}{2} + \frac{24+64}{3} = \frac{19\pi}{2} + \frac{88}{3}$$

3. 最終的な答え

$\frac{19\pi}{2} + \frac{88}{3}$