与えられた5つの定積分を計算し、解答欄に適切な値を入力する問題です。

はい、承知しました。以下の形式で回答します。

1. 問題の内容

与えられた5つの定積分を計算し、解答欄に適切な値を入力する問題です。

2. 解き方の手順

(1) $\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} (\tan x + \frac{1}{\tan x})^2 dx$

$(\tan x + \frac{1}{\tan x})^2 = \tan^2 x + 2 + \frac{1}{\tan^2 x} = \tan^2 x + 2 + \cot^2 x = (\tan^2 x + 1) + (\cot^2 x + 1) = \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} = \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} = \frac{1}{\sin^2 x \cos^2 x} = \frac{4}{4 \sin^2 x \cos^2 x} = \frac{4}{\sin^2 2x} = 4 \csc^2 2x$

$\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} 4 \csc^2 2x dx = 4 \left[ -\frac{1}{2} \cot 2x \right]_{\frac{\pi}{6}}^{\frac{\pi}{4}} = -2 \left[ \cot \frac{\pi}{2} - \cot \frac{\pi}{3} \right] = -2 (0 - \frac{1}{\sqrt{3}}) = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$

(2) $\int_{0}^{1} (e^x + \frac{1}{e^x})^2 dx = \int_{0}^{1} (e^x + e^{-x})^2 dx = \int_{0}^{1} (e^{2x} + 2 + e^{-2x}) dx = \left[ \frac{1}{2} e^{2x} + 2x - \frac{1}{2} e^{-2x} \right]_{0}^{1} = (\frac{1}{2} e^2 + 2 - \frac{1}{2} e^{-2}) - (\frac{1}{2} + 0 - \frac{1}{2}) = \frac{1}{2}e^2 + 2 - \frac{1}{2e^2}$

$= \frac{e^2}{2} + 2 - \frac{1}{2e^2}$

なので、$e^{2}/2 + 2 - e^{-2}/2$

(3) $\int_{0}^{\frac{\pi}{8}} \cos^2 4x dx = \int_{0}^{\frac{\pi}{8}} \frac{1 + \cos 8x}{2} dx = \frac{1}{2} \int_{0}^{\frac{\pi}{8}} (1 + \cos 8x) dx = \frac{1}{2} \left[ x + \frac{1}{8} \sin 8x \right]_{0}^{\frac{\pi}{8}} = \frac{1}{2} \left[ \frac{\pi}{8} + \frac{1}{8} \sin \pi - (0 + \frac{1}{8} \sin 0) \right] = \frac{1}{2} \left[ \frac{\pi}{8} + 0 - 0 \right] = \frac{\pi}{16}$

(4) $\int_{1}^{2} \frac{(x^2 + 1)^2}{x} dx = \int_{1}^{2} \frac{x^4 + 2x^2 + 1}{x} dx = \int_{1}^{2} (x^3 + 2x + \frac{1}{x}) dx = \left[ \frac{1}{4} x^4 + x^2 + \log |x| \right]_{1}^{2} = (\frac{1}{4} (16) + 4 + \log 2) - (\frac{1}{4} + 1 + 0) = 4 + 4 + \log 2 - \frac{1}{4} - 1 = 7 - \frac{1}{4} + \log 2 = \frac{27}{4} + \log 2$

(5) $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{\tan x} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cot x dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\cos x}{\sin x} dx = [\log|\sin x|]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \log \sin \frac{\pi}{3} - \log \sin \frac{\pi}{6} = \log \frac{\sqrt{3}}{2} - \log \frac{1}{2} = \log \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \log \sqrt{3} = \log 3^{\frac{1}{2}} = \frac{1}{2} \log 3$

3. 最終的な答え

(1) $\frac{2\sqrt{3}}{3}$

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

(3) $\frac{\pi}{16}$

(4) $\frac{27}{4} + \log 2$

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