与えられた4つの関数について、不定積分をそれぞれ求める。 (1) $\int \frac{\sqrt{x}}{1+\sqrt{x}} dx$ (2) $\int \frac{1}{1+\sqrt{x^2+1}} dx$ (3) $\int \frac{1}{x^2\sqrt{1-x^2}} dx$ (4) $\int x^2 \sqrt{a^2 - x^2} dx$

## 回答

1. 問題の内容

与えられた4つの関数について、不定積分をそれぞれ求める。

(1) $\int \frac{\sqrt{x}}{1+\sqrt{x}} dx$

(2) $\int \frac{1}{1+\sqrt{x^2+1}} dx$

(3) $\int \frac{1}{x^2\sqrt{1-x^2}} dx$

(4) $\int x^2 \sqrt{a^2 - x^2} dx$

2. 解き方の手順

(1) $\int \frac{\sqrt{x}}{1+\sqrt{x}} dx$

$\sqrt{x} = t$ と置換すると、$x = t^2$、$dx = 2t dt$となる。

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

$\frac{t^2}{1+t} = \frac{t^2-1+1}{1+t} = \frac{(t-1)(t+1)+1}{1+t} = t-1 + \frac{1}{1+t}$

よって、

$2 \int \frac{t^2}{1+t} dt = 2 \int (t-1 + \frac{1}{1+t}) dt = 2(\frac{t^2}{2} - t + \ln|1+t|) + C = t^2 - 2t + 2\ln|1+t| + C = x - 2\sqrt{x} + 2\ln(1+\sqrt{x}) + C$

(2) $\int \frac{1}{1+\sqrt{x^2+1}} dx$

$\sqrt{x^2+1} = t-x$ と置換する。$x^2+1 = t^2-2tx+x^2$ より $1 = t^2 - 2tx$ なので、$x = \frac{t^2-1}{2t}$ となる。

$dx = \frac{2t(2t) - 2(t^2-1)}{4t^2} dt = \frac{4t^2 - 2t^2 + 2}{4t^2} dt = \frac{2t^2 + 2}{4t^2} dt = \frac{t^2+1}{2t^2} dt$

また、$\sqrt{x^2+1} = t-x = t - \frac{t^2-1}{2t} = \frac{2t^2 - t^2 + 1}{2t} = \frac{t^2+1}{2t}$

$1 + \sqrt{x^2+1} = 1 + \frac{t^2+1}{2t} = \frac{2t+t^2+1}{2t} = \frac{(t+1)^2}{2t}$

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

部分分数分解を行う。

$\frac{t^2+1}{t(t+1)^2} = \frac{A}{t} + \frac{B}{t+1} + \frac{C}{(t+1)^2}$

$t^2+1 = A(t+1)^2 + Bt(t+1) + Ct = A(t^2+2t+1) + B(t^2+t) + Ct$

$t^2+1 = (A+B)t^2 + (2A+B+C)t + A$

$A+B = 1, 2A+B+C = 0, A=1$

$B=0, 2+0+C = 0, C=-2$

$\int \frac{t^2+1}{t(t+1)^2} dt = \int (\frac{1}{t} - \frac{2}{(t+1)^2}) dt = \ln|t| + \frac{2}{t+1} + C$

$\ln|\sqrt{x^2+1}+x| + \frac{2}{\sqrt{x^2+1}+x+1} + C$

(3) $\int \frac{1}{x^2\sqrt{1-x^2}} dx$

$x = \sin\theta$ と置換すると、$dx = \cos\theta d\theta$

$\int \frac{1}{x^2\sqrt{1-x^2}} dx = \int \frac{\cos\theta}{\sin^2\theta \sqrt{1-\sin^2\theta}} d\theta = \int \frac{\cos\theta}{\sin^2\theta \cos\theta} d\theta = \int \frac{1}{\sin^2\theta} d\theta = \int \csc^2\theta d\theta = -\cot\theta + C$

$\cot\theta = \frac{\cos\theta}{\sin\theta} = \frac{\sqrt{1-x^2}}{x}$

したがって、

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

(4) $\int x^2 \sqrt{a^2 - x^2} dx$

$x = a\sin\theta$ と置換すると、$dx = a\cos\theta d\theta$

$\int x^2 \sqrt{a^2 - x^2} dx = \int a^2\sin^2\theta \sqrt{a^2 - a^2\sin^2\theta} a\cos\theta d\theta = \int a^2\sin^2\theta \cdot a\cos\theta \cdot a\cos\theta d\theta = a^4 \int \sin^2\theta \cos^2\theta d\theta = a^4 \int (\sin\theta \cos\theta)^2 d\theta = a^4 \int (\frac{1}{2} \sin 2\theta)^2 d\theta = \frac{a^4}{4} \int \sin^2 2\theta d\theta = \frac{a^4}{4} \int \frac{1 - \cos 4\theta}{2} d\theta = \frac{a^4}{8} \int (1-\cos 4\theta) d\theta = \frac{a^4}{8} (\theta - \frac{1}{4}\sin 4\theta) + C$

$\sin 4\theta = 2\sin 2\theta \cos 2\theta = 2(2\sin\theta\cos\theta)(\cos^2\theta - \sin^2\theta) = 4\sin\theta\cos\theta (\cos^2\theta-\sin^2\theta)$

$x = a\sin\theta$ より、$\sin\theta = \frac{x}{a}$

$\cos\theta = \sqrt{1-\sin^2\theta} = \sqrt{1-\frac{x^2}{a^2}} = \frac{\sqrt{a^2-x^2}}{a}$

$\theta = \arcsin(\frac{x}{a})$

$\sin 4\theta = 4\cdot \frac{x}{a} \cdot \frac{\sqrt{a^2-x^2}}{a} \cdot (\frac{a^2-x^2}{a^2} - \frac{x^2}{a^2}) = 4 \frac{x\sqrt{a^2-x^2}}{a^2} \frac{a^2-2x^2}{a^2}$

$\sin 4\theta = \frac{4x(a^2-2x^2)\sqrt{a^2-x^2}}{a^4}$

$\frac{a^4}{8}(\arcsin(\frac{x}{a}) - \frac{1}{4} \frac{4x(a^2-2x^2)\sqrt{a^2-x^2}}{a^4}) + C = \frac{a^4}{8}\arcsin(\frac{x}{a}) - \frac{x(a^2-2x^2)\sqrt{a^2-x^2}}{8} + C$

3. 最終的な答え

(1) $x - 2\sqrt{x} + 2\ln(1+\sqrt{x}) + C$

(2) $\ln|\sqrt{x^2+1}+x| + \frac{2}{\sqrt{x^2+1}+x+1} + C$

(3) $-\frac{\sqrt{1-x^2}}{x} + C$

(4) $\frac{a^4}{8}\arcsin(\frac{x}{a}) - \frac{x(a^2-2x^2)\sqrt{a^2-x^2}}{8} + C$