与えられた2つの積分 $I$ と $J$ を計算します。 $I = \int \frac{1}{8 + 7\cos{x} - 4\sin{x}} dx$ $J = \int_{e\sqrt{e}}^{e^6} \frac{\sqrt{(\log{x})^2 + 28}}{x} dx$

1. 問題の内容

与えられた2つの積分 $I$ と $J$ を計算します。

$I = \int \frac{1}{8 + 7\cos{x} - 4\sin{x}} dx$

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

2. 解き方の手順

(1) 積分 $I$ の計算

$t = \tan{\frac{x}{2}}$ と置換すると、

$\cos{x} = \frac{1 - t^2}{1 + t^2}$, $\sin{x} = \frac{2t}{1 + t^2}$, $dx = \frac{2}{1 + t^2} dt$

これらの置換を $I$ に代入すると、

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

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

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

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

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

部分分数分解を用いて、

$\frac{1}{(t - 3)(t - 5)} = \frac{A}{t - 3} + \frac{B}{t - 5}$

$1 = A(t - 5) + B(t - 3)$

$t = 3$ のとき、 $1 = -2A \implies A = -\frac{1}{2}$

$t = 5$ のとき、 $1 = 2B \implies B = \frac{1}{2}$

$I = 2 \int \left( -\frac{1}{2(t - 3)} + \frac{1}{2(t - 5)} \right) dt$

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

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

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

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

(2) 積分 $J$ の計算

$u = \log{x}$ と置換すると、 $du = \frac{1}{x} dx$

$x = e\sqrt{e} = e^{3/2}$ のとき、 $u = \log{(e^{3/2})} = \frac{3}{2}$

$x = e^6$ のとき、 $u = \log{(e^6)} = 6$

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

$u = 2\sqrt{7} \sinh{v}$ と置換すると、 $du = 2\sqrt{7} \cosh{v} dv$

$v = \sinh^{-1}{\left(\frac{u}{2\sqrt{7}}\right)}$

$u = \frac{3}{2}$ のとき、 $v_1 = \sinh^{-1}{\left(\frac{3}{4\sqrt{7}}\right)}$

$u = 6$ のとき、 $v_2 = \sinh^{-1}{\left(\frac{6}{2\sqrt{7}}\right)} = \sinh^{-1}{\left(\frac{3}{\sqrt{7}}\right)}$

$J = \int_{v_1}^{v_2} \sqrt{(2\sqrt{7}\sinh{v})^2 + 28} (2\sqrt{7}\cosh{v}) dv$

$J = \int_{v_1}^{v_2} \sqrt{28\sinh^2{v} + 28} (2\sqrt{7}\cosh{v}) dv$

$J = \int_{v_1}^{v_2} \sqrt{28(\sinh^2{v} + 1)} (2\sqrt{7}\cosh{v}) dv$

$J = \int_{v_1}^{v_2} \sqrt{28\cosh^2{v}} (2\sqrt{7}\cosh{v}) dv$

$J = \int_{v_1}^{v_2} 2\sqrt{7}\cosh{v} (2\sqrt{7}\cosh{v}) dv$

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

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

$J = 14 \int_{v_1}^{v_2} (1 + \cosh{2v}) dv$

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

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

$J = 14 \left[ \sinh^{-1}{\left(\frac{u}{2\sqrt{7}}\right)} + \frac{u}{2\sqrt{7}} \sqrt{1 + \left(\frac{u}{2\sqrt{7}}\right)^2} \right]_{3/2}^{6}$

$J = 14 \left[ \sinh^{-1}{\left(\frac{u}{2\sqrt{7}}\right)} + \frac{u}{2\sqrt{7}} \sqrt{1 + \frac{u^2}{28}} \right]_{3/2}^{6}$

$J = 14 \left[ \sinh^{-1}{\left(\frac{6}{2\sqrt{7}}\right)} + \frac{6}{2\sqrt{7}} \sqrt{1 + \frac{36}{28}} - \sinh^{-1}{\left(\frac{3}{4\sqrt{7}}\right)} - \frac{3}{4\sqrt{7}} \sqrt{1 + \frac{9}{112}} \right]$

$J = 14 \left[ \sinh^{-1}{\left(\frac{3}{\sqrt{7}}\right)} + \frac{3}{\sqrt{7}} \sqrt{\frac{64}{28}} - \sinh^{-1}{\left(\frac{3}{4\sqrt{7}}\right)} - \frac{3}{4\sqrt{7}} \sqrt{\frac{121}{112}} \right]$

$J = 14 \left[ \sinh^{-1}{\left(\frac{3}{\sqrt{7}}\right)} + \frac{3}{\sqrt{7}} \frac{8}{2\sqrt{7}} - \sinh^{-1}{\left(\frac{3}{4\sqrt{7}}\right)} - \frac{3}{4\sqrt{7}} \frac{11}{4\sqrt{7}} \right]$

$J = 14 \left[ \sinh^{-1}{\left(\frac{3}{\sqrt{7}}\right)} + \frac{12}{7} - \sinh^{-1}{\left(\frac{3}{4\sqrt{7}}\right)} - \frac{33}{112} \right]$

$J = 14 \sinh^{-1}{\left(\frac{3}{\sqrt{7}}\right)} - 14 \sinh^{-1}{\left(\frac{3}{4\sqrt{7}}\right)} + 24 - \frac{33}{8}$

$J = 14 \sinh^{-1}{\left(\frac{3}{\sqrt{7}}\right)} - 14 \sinh^{-1}{\left(\frac{3}{4\sqrt{7}}\right)} + \frac{192 - 33}{8}$

$J = 14 \sinh^{-1}{\left(\frac{3}{\sqrt{7}}\right)} - 14 \sinh^{-1}{\left(\frac{3}{4\sqrt{7}}\right)} + \frac{159}{8}$

3. 最終的な答え

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

$J = 14 \sinh^{-1}{\left(\frac{3}{\sqrt{7}}\right)} - 14 \sinh^{-1}{\left(\frac{3}{4\sqrt{7}}\right)} + \frac{159}{8}$