与えられた積分を計算する問題です。 (1) $I = \int \frac{1}{8 + 7\cos{x} - 4\sin{x}} dx$ の不定積分を計算します。 (2) $J = \int_{e\sqrt{e}}^{e^6} \frac{\sqrt{(\log{x})^2 + 28}}{x} dx$ の定積分を計算します。

1. 問題の内容

与えられた積分を計算する問題です。

(1) $I = \int \frac{1}{8 + 7\cos{x} - 4\sin{x}} dx$ の不定積分を計算します。

(2) $J = \int_{e\sqrt{e}}^{e^6} \frac{\sqrt{(\log{x})^2 + 28}}{x} dx$ の定積分を計算します。

2. 解き方の手順

(1) $I = \int \frac{1}{8 + 7\cos{x} - 4\sin{x}} dx$ の解き方:

半角の公式 $\cos{x} = \frac{1 - t^2}{1 + t^2}$, $\sin{x} = \frac{2t}{1 + t^2}$, $dx = \frac{2}{1 + t^2} dt$ を用いて、積分変数を $t = \tan{\frac{x}{2}}$ に変換します。

\begin{align*}

I &= \int \frac{1}{8 + 7(\frac{1 - t^2}{1 + t^2}) - 4(\frac{2t}{1 + t^2})} \frac{2}{1 + t^2} dt \\

&= \int \frac{2}{8(1 + t^2) + 7(1 - t^2) - 8t} dt \\

&= \int \frac{2}{8 + 8t^2 + 7 - 7t^2 - 8t} dt \\

&= \int \frac{2}{t^2 - 8t + 15} dt \\

&= 2 \int \frac{1}{(t - 3)(t - 5)} dt \\

&= 2 \int \frac{-1/2}{t - 3} + \frac{1/2}{t - 5} dt \\

&= -\int \frac{1}{t - 3} dt + \int \frac{1}{t - 5} dt \\

&= -\log{|t - 3|} + \log{|t - 5|} + C \\

&= \log{\left|\frac{t - 5}{t - 3}\right|} + C \\

&= \log{\left|\frac{\tan{\frac{x}{2}} - 5}{\tan{\frac{x}{2}} - 3}\right|} + C

\end{align*}

(2) $J = \int_{e\sqrt{e}}^{e^6} \frac{\sqrt{(\log{x})^2 + 28}}{x} dx$ の解き方:

$u = \log{x}$ とおくと、$\frac{du}{dx} = \frac{1}{x}$ より $dx = x du$。

積分範囲は $x = e\sqrt{e} = e^{3/2}$ のとき $u = \log{e^{3/2}} = \frac{3}{2}$、

$x = e^6$ のとき $u = \log{e^6} = 6$。

\begin{align*}

J &= \int_{3/2}^{6} \sqrt{u^2 + 28} du \\

\end{align*}

ここで $u = 2\sqrt{7}\sinh{v}$ とおくと、$du = 2\sqrt{7}\cosh{v} dv$。

$u^2 + 28 = 28\sinh^2{v} + 28 = 28(\sinh^2{v} + 1) = 28\cosh^2{v}$

$\sqrt{u^2 + 28} = 2\sqrt{7}\cosh{v}$

$u = \frac{3}{2}$ のとき $\sinh{v_1} = \frac{3}{4\sqrt{7}}$

$u = 6$ のとき $\sinh{v_2} = \frac{6}{2\sqrt{7}} = \frac{3}{\sqrt{7}}$

\begin{align*}

J &= \int_{v_1}^{v_2} 2\sqrt{7}\cosh{v} \cdot 2\sqrt{7}\cosh{v} dv \\

&= 28 \int_{v_1}^{v_2} \cosh^2{v} dv \\

&= 28 \int_{v_1}^{v_2} \frac{1 + \cosh{2v}}{2} dv \\

&= 14 \int_{v_1}^{v_2} 1 + \cosh{2v} dv \\

&= 14 \left[ v + \frac{1}{2}\sinh{2v} \right]_{v_1}^{v_2} \\

&= 14 \left[ v + \sinh{v}\cosh{v} \right]_{v_1}^{v_2}

\end{align*}

$\sinh{v_1} = \frac{3}{4\sqrt{7}}$, $\cosh{v_1} = \sqrt{1 + \sinh^2{v_1}} = \sqrt{1 + \frac{9}{16 \cdot 7}} = \sqrt{\frac{112 + 9}{112}} = \sqrt{\frac{121}{112}} = \frac{11}{4\sqrt{7}}$

$\sinh{v_2} = \frac{3}{\sqrt{7}}$, $\cosh{v_2} = \sqrt{1 + \frac{9}{7}} = \sqrt{\frac{16}{7}} = \frac{4}{\sqrt{7}}$

$v_1 = \text{arcsinh}(\frac{3}{4\sqrt{7}})$, $v_2 = \text{arcsinh}(\frac{3}{\sqrt{7}})$

$J = 14 \left[ \text{arcsinh}(\frac{3}{\sqrt{7}}) + \frac{3}{\sqrt{7}}\cdot \frac{4}{\sqrt{7}} - \text{arcsinh}(\frac{3}{4\sqrt{7}}) - \frac{3}{4\sqrt{7}}\cdot \frac{11}{4\sqrt{7}} \right]$

$= 14 \left[ \text{arcsinh}(\frac{3}{\sqrt{7}}) + \frac{12}{7} - \text{arcsinh}(\frac{3}{4\sqrt{7}}) - \frac{33}{16 \cdot 7} \right]$

$= 14 \left[ \text{arcsinh}(\frac{3}{\sqrt{7}}) - \text{arcsinh}(\frac{3}{4\sqrt{7}}) + \frac{12}{7} - \frac{33}{112} \right]$

$= 14 \left[ \text{arcsinh}(\frac{3}{\sqrt{7}}) - \text{arcsinh}(\frac{3}{4\sqrt{7}}) + \frac{192 - 33}{112} \right]$

$= 14 \left[ \text{arcsinh}(\frac{3}{\sqrt{7}}) - \text{arcsinh}(\frac{3}{4\sqrt{7}}) + \frac{159}{112} \right]$

$= 14 \left[ \text{arcsinh}(\frac{3}{\sqrt{7}}) - \text{arcsinh}(\frac{3}{4\sqrt{7}}) + \frac{159}{112} \right]$

$= 14 \text{arcsinh}(\frac{3}{\sqrt{7}}) - 14 \text{arcsinh}(\frac{3}{4\sqrt{7}}) + \frac{159}{8}$

$= 14 \log{(\frac{3}{\sqrt{7}} + \sqrt{\frac{9}{7} + 1})} - 14 \log{(\frac{3}{4\sqrt{7}} + \sqrt{\frac{9}{16\cdot 7} + 1})} + \frac{159}{8}$

$= 14 \log{(\frac{3}{\sqrt{7}} + \frac{4}{\sqrt{7}})} - 14 \log{(\frac{3}{4\sqrt{7}} + \frac{11}{4\sqrt{7}})} + \frac{159}{8}$

$= 14 \log{(\frac{7}{\sqrt{7}})} - 14 \log{(\frac{14}{4\sqrt{7}})} + \frac{159}{8}$

$= 14 \log{(\sqrt{7})} - 14 \log{(\frac{7}{2\sqrt{7}})} + \frac{159}{8}$

$= 14 \log{(\sqrt{7})} - 14 \log{(\frac{\sqrt{7}}{2})} + \frac{159}{8}$

$= 14 (\log{\sqrt{7}} - \log{\sqrt{7}} + \log{2}) + \frac{159}{8}$

$= 14 \log{2} + \frac{159}{8}$

3. 最終的な答え

(1) $I = \log{\left|\frac{\tan{\frac{x}{2}} - 5}{\tan{\frac{x}{2}} - 3}\right|} + C$

(2) $J = 14\log{2} + \frac{159}{8}$