与えられた8つの定積分を計算する問題です。 (1) $\int_{0}^{1} \sin^{-1} x \, dx$ (2) $\int_{0}^{1} x \cos^{-1} x \, dx$ (3) $\int_{1/4}^{1/2} \frac{1}{\sqrt{1 - 4x^2}} \, dx$ (4) $\int_{0}^{1} \frac{1}{x^2 + 4x + 3} \, dx$ (5) $\int_{0}^{1} \frac{1}{1 + 3x^2} \, dx$ (6) $\int_{0}^{1} \frac{2x + 1}{x^2 + x + 1} \, dx$ (7) $\int_{0}^{\pi/2} \frac{\cos x}{1 + \sin^2 x} \, dx$ (8) $\int_{1}^{2} x^3 \log x \, dx$

1. 問題の内容

与えられた8つの定積分を計算する問題です。

(1) $\int_{0}^{1} \sin^{-1} x \, dx$

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

(3) $\int_{1/4}^{1/2} \frac{1}{\sqrt{1 - 4x^2}} \, dx$

(4) $\int_{0}^{1} \frac{1}{x^2 + 4x + 3} \, dx$

(5) $\int_{0}^{1} \frac{1}{1 + 3x^2} \, dx$

(6) $\int_{0}^{1} \frac{2x + 1}{x^2 + x + 1} \, dx$

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

(8) $\int_{1}^{2} x^3 \log x \, dx$

2. 解き方の手順

各積分を個別に計算します。

(1) $\int_{0}^{1} \sin^{-1} x \, dx$

部分積分を使います。$u = \sin^{-1} x$, $dv = dx$ とすると、$du = \frac{1}{\sqrt{1 - x^2}} \, dx$, $v = x$ です。

$\int \sin^{-1} x \, dx = x \sin^{-1} x - \int \frac{x}{\sqrt{1 - x^2}} \, dx$

$\int \frac{x}{\sqrt{1 - x^2}} \, dx$ は、$t = 1 - x^2$ と置換すると、$dt = -2x \, dx$ となるので、

$\int \frac{x}{\sqrt{1 - x^2}} \, dx = -\frac{1}{2} \int \frac{1}{\sqrt{t}} \, dt = -\sqrt{t} = \sqrt{1 - x^2}$

したがって、$\int \sin^{-1} x \, dx = x \sin^{-1} x + \sqrt{1 - x^2}$

$\int_{0}^{1} \sin^{-1} x \, dx = [x \sin^{-1} x + \sqrt{1 - x^2}]_{0}^{1} = (1 \cdot \frac{\pi}{2} + 0) - (0 + 1) = \frac{\pi}{2} - 1$

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

部分積分を使います。$u = \cos^{-1} x$, $dv = x \, dx$ とすると、$du = -\frac{1}{\sqrt{1 - x^2}} \, dx$, $v = \frac{x^2}{2}$ です。

$\int x \cos^{-1} x \, dx = \frac{x^2}{2} \cos^{-1} x + \frac{1}{2} \int \frac{x^2}{\sqrt{1 - x^2}} \, dx$

$x = \sin \theta$ と置換すると、$dx = \cos \theta \, d\theta$ となり、$\int \frac{x^2}{\sqrt{1 - x^2}} \, dx = \int \frac{\sin^2 \theta}{\cos \theta} \cos \theta \, d\theta = \int \sin^2 \theta \, d\theta = \int \frac{1 - \cos 2\theta}{2} \, d\theta = \frac{\theta}{2} - \frac{\sin 2\theta}{4} = \frac{\theta}{2} - \frac{\sin \theta \cos \theta}{2} = \frac{\sin^{-1} x}{2} - \frac{x\sqrt{1 - x^2}}{2}$

したがって、$\int x \cos^{-1} x \, dx = \frac{x^2}{2} \cos^{-1} x + \frac{1}{2} \left( \frac{\sin^{-1} x}{2} - \frac{x\sqrt{1 - x^2}}{2} \right)$

$\int_{0}^{1} x \cos^{-1} x \, dx = \left[ \frac{x^2}{2} \cos^{-1} x + \frac{\sin^{-1} x}{4} - \frac{x\sqrt{1 - x^2}}{4} \right]_{0}^{1} = \left( \frac{1}{2} \cdot 0 + \frac{\pi}{8} - 0 \right) - (0 + 0 - 0) = \frac{\pi}{8}$

(3) $\int_{1/4}^{1/2} \frac{1}{\sqrt{1 - 4x^2}} \, dx = \int_{1/4}^{1/2} \frac{1}{\sqrt{1 - (2x)^2}} \, dx$

$t = 2x$ と置換すると、$dt = 2 \, dx$ となり、

$\int \frac{1}{\sqrt{1 - (2x)^2}} \, dx = \frac{1}{2} \int \frac{1}{\sqrt{1 - t^2}} \, dt = \frac{1}{2} \sin^{-1} t = \frac{1}{2} \sin^{-1} (2x)$

$\int_{1/4}^{1/2} \frac{1}{\sqrt{1 - 4x^2}} \, dx = \left[ \frac{1}{2} \sin^{-1} (2x) \right]_{1/4}^{1/2} = \frac{1}{2} \left( \sin^{-1} (1) - \sin^{-1} (\frac{1}{2}) \right) = \frac{1}{2} \left( \frac{\pi}{2} - \frac{\pi}{6} \right) = \frac{1}{2} \cdot \frac{\pi}{3} = \frac{\pi}{6}$

(4) $\int_{0}^{1} \frac{1}{x^2 + 4x + 3} \, dx = \int_{0}^{1} \frac{1}{(x + 1)(x + 3)} \, dx = \int_{0}^{1} \frac{1}{2} \left( \frac{1}{x + 1} - \frac{1}{x + 3} \right) \, dx$

$= \frac{1}{2} [\log(x + 1) - \log(x + 3)]_{0}^{1} = \frac{1}{2} \left[ \log(\frac{x + 1}{x + 3}) \right]_{0}^{1} = \frac{1}{2} \left( \log(\frac{2}{4}) - \log(\frac{1}{3}) \right) = \frac{1}{2} \left( \log(\frac{1}{2}) - \log(\frac{1}{3}) \right) = \frac{1}{2} \log(\frac{3}{2})$

(5) $\int_{0}^{1} \frac{1}{1 + 3x^2} \, dx = \int_{0}^{1} \frac{1}{1 + (\sqrt{3}x)^2} \, dx$

$t = \sqrt{3} x$ と置換すると、$dt = \sqrt{3} \, dx$ となり、

$\int \frac{1}{1 + (\sqrt{3}x)^2} \, dx = \frac{1}{\sqrt{3}} \int \frac{1}{1 + t^2} \, dt = \frac{1}{\sqrt{3}} \arctan t = \frac{1}{\sqrt{3}} \arctan(\sqrt{3} x)$

$\int_{0}^{1} \frac{1}{1 + 3x^2} \, dx = \left[ \frac{1}{\sqrt{3}} \arctan(\sqrt{3} x) \right]_{0}^{1} = \frac{1}{\sqrt{3}} \left( \arctan(\sqrt{3}) - \arctan(0) \right) = \frac{1}{\sqrt{3}} \left( \frac{\pi}{3} - 0 \right) = \frac{\pi}{3\sqrt{3}}$

(6) $\int_{0}^{1} \frac{2x + 1}{x^2 + x + 1} \, dx$

$t = x^2 + x + 1$ と置換すると、$dt = (2x + 1) \, dx$ となり、

$\int \frac{2x + 1}{x^2 + x + 1} \, dx = \int \frac{1}{t} \, dt = \log |t| = \log (x^2 + x + 1)$

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

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

$t = \sin x$ と置換すると、$dt = \cos x \, dx$ となり、

$\int \frac{\cos x}{1 + \sin^2 x} \, dx = \int \frac{1}{1 + t^2} \, dt = \arctan t = \arctan(\sin x)$

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

(8) $\int_{1}^{2} x^3 \log x \, dx$

部分積分を使います。$u = \log x$, $dv = x^3 \, dx$ とすると、$du = \frac{1}{x} \, dx$, $v = \frac{x^4}{4}$ です。

$\int x^3 \log x \, dx = \frac{x^4}{4} \log x - \int \frac{x^4}{4} \cdot \frac{1}{x} \, dx = \frac{x^4}{4} \log x - \frac{1}{4} \int x^3 \, dx = \frac{x^4}{4} \log x - \frac{x^4}{16}$

$\int_{1}^{2} x^3 \log x \, dx = \left[ \frac{x^4}{4} \log x - \frac{x^4}{16} \right]_{1}^{2} = \left( \frac{16}{4} \log 2 - \frac{16}{16} \right) - \left( \frac{1}{4} \log 1 - \frac{1}{16} \right) = 4 \log 2 - 1 + \frac{1}{16} = 4 \log 2 - \frac{15}{16}$

3. 最終的な答え

(1) $\frac{\pi}{2} - 1$

(2) $\frac{\pi}{8}$

(3) $\frac{\pi}{6}$

(4) $\frac{1}{2} \log(\frac{3}{2})$

(5) $\frac{\pi}{3\sqrt{3}}$

(6) $\log 3$

(7) $\frac{\pi}{4}$

(8) $4 \log 2 - \frac{15}{16}$