与えられた定積分を計算する問題です。具体的には、以下の6つの定積分を計算する必要があります。 (1) $\int_{1}^{2} \log x dx$ (2) $\int_{0}^{1} x^{2} \tan^{-1} x dx$ (3) $\int_{0}^{\frac{\pi}{2}} \sin^{7} x dx$ (4) $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{5} x dx$ (5) $\int_{0}^{3\pi} \sin^{4} \frac{x}{3} dx$ (6) $\int_{0}^{1} x^{5} \sqrt{1-x^{2}} dx$

## 解答

1. 問題の内容

与えられた定積分を計算する問題です。具体的には、以下の6つの定積分を計算する必要があります。

(1) $\int_{1}^{2} \log x dx$

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

(3) $\int_{0}^{\frac{\pi}{2}} \sin^{7} x dx$

(4) $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{5} x dx$

(5) $\int_{0}^{3\pi} \sin^{4} \frac{x}{3} dx$

(6) $\int_{0}^{1} x^{5} \sqrt{1-x^{2}} dx$

2. 解き方の手順

**(1) $\int_{1}^{2} \log x dx$**

部分積分を利用します。$\int u v' dx = uv - \int u' v dx$において、$u = \log x$, $v' = 1$とおくと、$u' = \frac{1}{x}$, $v = x$となります。したがって、

$\int_{1}^{2} \log x dx = [x \log x]_{1}^{2} - \int_{1}^{2} x \cdot \frac{1}{x} dx = [x \log x]_{1}^{2} - \int_{1}^{2} dx$

$= (2 \log 2 - 1 \log 1) - [x]_{1}^{2} = 2 \log 2 - (2 - 1) = 2 \log 2 - 1$

**(2) $\int_{0}^{1} x^{2} \tan^{-1} x dx$**

部分積分を利用します。$u = \tan^{-1} x$, $v' = x^{2}$とおくと、$u' = \frac{1}{1+x^{2}}$, $v = \frac{x^{3}}{3}$となります。したがって、

$\int_{0}^{1} x^{2} \tan^{-1} x dx = [\frac{x^{3}}{3} \tan^{-1} x]_{0}^{1} - \int_{0}^{1} \frac{x^{3}}{3} \cdot \frac{1}{1+x^{2}} dx = \frac{1}{3} \tan^{-1} 1 - \frac{1}{3} \int_{0}^{1} \frac{x^{3}}{1+x^{2}} dx$

$= \frac{1}{3} \cdot \frac{\pi}{4} - \frac{1}{3} \int_{0}^{1} \frac{x^{3}}{1+x^{2}} dx = \frac{\pi}{12} - \frac{1}{3} \int_{0}^{1} \frac{x^{3}}{1+x^{2}} dx$

$\int_{0}^{1} \frac{x^{3}}{1+x^{2}} dx$を計算します。$\frac{x^{3}}{1+x^{2}} = x - \frac{x}{1+x^{2}}$なので、

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

したがって、$\int_{0}^{1} x^{2} \tan^{-1} x dx = \frac{\pi}{12} - \frac{1}{3} (\frac{1}{2} - \frac{1}{2} \log 2) = \frac{\pi}{12} - \frac{1}{6} + \frac{1}{6} \log 2$

**(3) $\int_{0}^{\frac{\pi}{2}} \sin^{7} x dx$**

漸化式を用いるか、直接計算します。今回は直接計算します。

$\int_{0}^{\frac{\pi}{2}} \sin^{7} x dx = \int_{0}^{\frac{\pi}{2}} \sin^{6} x \cdot \sin x dx = \int_{0}^{\frac{\pi}{2}} (1 - \cos^{2} x)^{3} \sin x dx$

$t = \cos x$とおくと、$dt = - \sin x dx$。$x: 0 \to \frac{\pi}{2}$のとき、$t: 1 \to 0$なので、

$\int_{0}^{\frac{\pi}{2}} (1 - \cos^{2} x)^{3} \sin x dx = \int_{1}^{0} (1 - t^{2})^{3} (-dt) = \int_{0}^{1} (1 - 3t^{2} + 3t^{4} - t^{6}) dt$

$= [t - t^{3} + \frac{3}{5} t^{5} - \frac{1}{7} t^{7}]_{0}^{1} = 1 - 1 + \frac{3}{5} - \frac{1}{7} = \frac{21 - 5}{35} = \frac{16}{35}$

**(4) $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{5} x dx$**

$\cos x$は偶関数なので、$\cos^{5} x$も偶関数です。したがって、

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{5} x dx = 2 \int_{0}^{\frac{\pi}{2}} \cos^{5} x dx$

$\int_{0}^{\frac{\pi}{2}} \cos^{5} x dx = \int_{0}^{\frac{\pi}{2}} \cos^{4} x \cdot \cos x dx = \int_{0}^{\frac{\pi}{2}} (1 - \sin^{2} x)^{2} \cos x dx$

$t = \sin x$とおくと、$dt = \cos x dx$。$x: 0 \to \frac{\pi}{2}$のとき、$t: 0 \to 1$なので、

$\int_{0}^{\frac{\pi}{2}} (1 - \sin^{2} x)^{2} \cos x dx = \int_{0}^{1} (1 - t^{2})^{2} dt = \int_{0}^{1} (1 - 2t^{2} + t^{4}) dt$

$= [t - \frac{2}{3} t^{3} + \frac{1}{5} t^{5}]_{0}^{1} = 1 - \frac{2}{3} + \frac{1}{5} = \frac{15 - 10 + 3}{15} = \frac{8}{15}$

したがって、$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{5} x dx = 2 \cdot \frac{8}{15} = \frac{16}{15}$

**(5) $\int_{0}^{3\pi} \sin^{4} \frac{x}{3} dx$**

$t = \frac{x}{3}$とおくと、$dx = 3 dt$。$x: 0 \to 3\pi$のとき、$t: 0 \to \pi$なので、

$\int_{0}^{3\pi} \sin^{4} \frac{x}{3} dx = \int_{0}^{\pi} \sin^{4} t \cdot 3 dt = 3 \int_{0}^{\pi} \sin^{4} t dt$

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

$3 \int_{0}^{\pi} \sin^{4} t dt = 3 \int_{0}^{\pi} \frac{1}{8} (3 - 4 \cos 2t + \cos 4t) dt = \frac{3}{8} [3t - 2 \sin 2t + \frac{1}{4} \sin 4t]_{0}^{\pi} = \frac{3}{8} (3\pi - 0 + 0) = \frac{9\pi}{8}$

**(6) $\int_{0}^{1} x^{5} \sqrt{1-x^{2}} dx$**

$t = 1 - x^{2}$とおくと、$x^{2} = 1 - t$。$dt = -2x dx$。$x: 0 \to 1$のとき、$t: 1 \to 0$なので、

$\int_{0}^{1} x^{5} \sqrt{1-x^{2}} dx = \int_{0}^{1} x^{4} \sqrt{1-x^{2}} x dx = \int_{1}^{0} (1-t)^{2} \sqrt{t} (-\frac{1}{2}) dt = \frac{1}{2} \int_{0}^{1} (1 - 2t + t^{2}) t^{\frac{1}{2}} dt = \frac{1}{2} \int_{0}^{1} (t^{\frac{1}{2}} - 2t^{\frac{3}{2}} + t^{\frac{5}{2}}) dt$

$= \frac{1}{2} [\frac{2}{3} t^{\frac{3}{2}} - \frac{4}{5} t^{\frac{5}{2}} + \frac{2}{7} t^{\frac{7}{2}}]_{0}^{1} = \frac{1}{2} (\frac{2}{3} - \frac{4}{5} + \frac{2}{7}) = \frac{1}{2} (\frac{70 - 84 + 30}{105}) = \frac{1}{2} \cdot \frac{16}{105} = \frac{8}{105}$

3. 最終的な答え

(1) $2 \log 2 - 1$

(2) $\frac{\pi}{12} - \frac{1}{6} + \frac{1}{6} \log 2$

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

(4) $\frac{16}{15}$

(5) $\frac{9\pi}{8}$

(6) $\frac{8}{105}$