以下の3つの定積分および広義積分の値を求めます。 (1) $\int_2^3 (x-1)\sqrt{2x-3} \, dx$ (2) $\int_0^{\frac{\pi}{2}} \frac{1}{3+2\cos{x}+3\sin{x}} \, dx$ (3) $\int_2^\infty \frac{1}{x^2+2x-3} \, dx$

## 解答

1. 問題の内容

以下の3つの定積分および広義積分の値を求めます。

(1) $\int_2^3 (x-1)\sqrt{2x-3} \, dx$

(2) $\int_0^{\frac{\pi}{2}} \frac{1}{3+2\cos{x}+3\sin{x}} \, dx$

(3) $\int_2^\infty \frac{1}{x^2+2x-3} \, dx$

2. 解き方の手順

(1) $\int_2^3 (x-1)\sqrt{2x-3} \, dx$

置換積分を行います。$t = 2x - 3$ とおくと、$x = \frac{t+3}{2}$ であり、$dx = \frac{1}{2} dt$ となります。また、$x: 2 \rightarrow 3$ のとき、$t: 1 \rightarrow 3$ となります。

$\int_2^3 (x-1)\sqrt{2x-3} \, dx = \int_1^3 (\frac{t+3}{2}-1)\sqrt{t} \cdot \frac{1}{2} \, dt = \frac{1}{4} \int_1^3 (t+1)\sqrt{t} \, dt$

$= \frac{1}{4} \int_1^3 (t^{\frac{3}{2}} + t^{\frac{1}{2}}) \, dt = \frac{1}{4} [\frac{2}{5}t^{\frac{5}{2}} + \frac{2}{3}t^{\frac{3}{2}}]_1^3$

$= \frac{1}{4} [(\frac{2}{5}3^{\frac{5}{2}} + \frac{2}{3}3^{\frac{3}{2}}) - (\frac{2}{5} + \frac{2}{3})]$

$= \frac{1}{4} [(\frac{2}{5}9\sqrt{3} + \frac{2}{3}3\sqrt{3}) - \frac{16}{15}] = \frac{1}{4} [\frac{18\sqrt{3}}{5} + 2\sqrt{3} - \frac{16}{15}] = \frac{1}{4} [\frac{28\sqrt{3}}{5} - \frac{16}{15}]$

$= \frac{7\sqrt{3}}{5} - \frac{4}{15}$

(2) $\int_0^{\frac{\pi}{2}} \frac{1}{3+2\cos{x}+3\sin{x}} \, dx$

半角の公式を利用します。$t = \tan(\frac{x}{2})$ とおくと、$dx = \frac{2}{1+t^2} dt$、$cos(x) = \frac{1-t^2}{1+t^2}$、$sin(x) = \frac{2t}{1+t^2}$。$x: 0 \rightarrow \frac{\pi}{2}$ のとき、$t: 0 \rightarrow 1$

$\int_0^{\frac{\pi}{2}} \frac{1}{3+2\cos{x}+3\sin{x}} \, dx = \int_0^1 \frac{1}{3+2(\frac{1-t^2}{1+t^2})+3(\frac{2t}{1+t^2})} \cdot \frac{2}{1+t^2} \, dt$

$= \int_0^1 \frac{2}{3(1+t^2) + 2(1-t^2) + 6t} \, dt = \int_0^1 \frac{2}{3+3t^2+2-2t^2+6t} \, dt$

$= \int_0^1 \frac{2}{t^2+6t+5} \, dt = 2 \int_0^1 \frac{1}{(t+1)(t+5)} \, dt$

部分分数分解します。$\frac{1}{(t+1)(t+5)} = \frac{A}{t+1} + \frac{B}{t+5}$ より、$1 = A(t+5) + B(t+1)$

$t=-1$ のとき、$1 = 4A \Rightarrow A = \frac{1}{4}$

$t=-5$ のとき、$1 = -4B \Rightarrow B = -\frac{1}{4}$

$2 \int_0^1 \frac{1}{(t+1)(t+5)} \, dt = 2 \int_0^1 (\frac{1/4}{t+1} - \frac{1/4}{t+5}) \, dt = \frac{1}{2} \int_0^1 (\frac{1}{t+1} - \frac{1}{t+5}) \, dt$

$= \frac{1}{2} [\ln|t+1| - \ln|t+5|]_0^1 = \frac{1}{2} [\ln|\frac{t+1}{t+5}|]_0^1 = \frac{1}{2} [(\ln(\frac{2}{6}) - \ln(\frac{1}{5}))] = \frac{1}{2} [\ln(\frac{1}{3}) - \ln(\frac{1}{5})]$

$= \frac{1}{2} \ln(\frac{5}{3})$

(3) $\int_2^\infty \frac{1}{x^2+2x-3} \, dx$

$\int_2^\infty \frac{1}{x^2+2x-3} \, dx = \int_2^\infty \frac{1}{(x+3)(x-1)} \, dx$

部分分数分解します。$\frac{1}{(x+3)(x-1)} = \frac{A}{x+3} + \frac{B}{x-1}$ より、$1 = A(x-1) + B(x+3)$

$x=1$ のとき、$1 = 4B \Rightarrow B = \frac{1}{4}$

$x=-3$ のとき、$1 = -4A \Rightarrow A = -\frac{1}{4}$

$\int_2^\infty \frac{1}{(x+3)(x-1)} \, dx = \int_2^\infty (\frac{-1/4}{x+3} + \frac{1/4}{x-1}) \, dx = \frac{1}{4} \int_2^\infty (\frac{1}{x-1} - \frac{1}{x+3}) \, dx$

$= \frac{1}{4} [\ln|x-1| - \ln|x+3|]_2^\infty = \frac{1}{4} [\ln|\frac{x-1}{x+3}|]_2^\infty = \frac{1}{4} [\lim_{x\to\infty}\ln|\frac{x-1}{x+3}| - \ln|\frac{2-1}{2+3}|]$

$= \frac{1}{4} [\ln(1) - \ln(\frac{1}{5})] = \frac{1}{4} [0 - (-\ln{5})] = \frac{1}{4} \ln{5}$

3. 最終的な答え

(1) $\frac{7\sqrt{3}}{5} - \frac{4}{15}$

(2) $\frac{1}{2}\ln(\frac{5}{3})$

(3) $\frac{1}{4}\ln{5}$