与えられた5つの定積分を計算し、空欄を埋める問題です。 (1) $\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} (\tan x + \frac{1}{\tan x})^2 dx$ (2) $\int_0^1 (e^x + \frac{1}{e^x})^2 dx$ (3) $\int_0^{\pi} \cos^2 4x dx$ (4) $\int_1^2 \frac{(x^2+1)^2}{x} dx$ (5) $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{\tan x} dx$

1. 問題の内容

与えられた5つの定積分を計算し、空欄を埋める問題です。

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

(2) $\int_0^1 (e^x + \frac{1}{e^x})^2 dx$

(3) $\int_0^{\pi} \cos^2 4x dx$

(4) $\int_1^2 \frac{(x^2+1)^2}{x} dx$

(5) $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{\tan x} dx$

2. 解き方の手順

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

$(\tan x + \frac{1}{\tan x})^2 = \tan^2 x + 2 + \frac{1}{\tan^2 x} = \tan^2 x + 2 + \cot^2 x$

$\tan^2 x = \frac{1}{\cos^2 x} - 1$

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

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

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

(2) $\int_0^1 (e^x + \frac{1}{e^x})^2 dx = \int_0^1 (e^{2x} + 2 + e^{-2x}) dx = [\frac{1}{2} e^{2x} + 2x - \frac{1}{2} e^{-2x}]_0^1 = (\frac{1}{2}e^2 + 2 - \frac{1}{2}e^{-2}) - (\frac{1}{2} + 0 - \frac{1}{2}) = \frac{1}{2}e^2 + 2 - \frac{1}{2e^2} = \frac{e^4 + 4e^2 - 1}{2e^2}$

(3) $\int_0^{\pi} \cos^2 4x dx = \int_0^{\pi} \frac{1 + \cos 8x}{2} dx = [\frac{1}{2}x + \frac{1}{16}\sin 8x]_0^{\pi} = \frac{\pi}{2} + \frac{1}{16} \sin 8\pi - (0 + 0) = \frac{\pi}{2}$

(4) $\int_1^2 \frac{(x^2+1)^2}{x} dx = \int_1^2 \frac{x^4 + 2x^2 + 1}{x} dx = \int_1^2 (x^3 + 2x + \frac{1}{x}) dx = [\frac{1}{4}x^4 + x^2 + \ln |x|]_1^2 = (\frac{1}{4} \cdot 16 + 4 + \ln 2) - (\frac{1}{4} + 1 + 0) = 4 + 4 + \ln 2 - \frac{1}{4} - 1 = 7 - \frac{1}{4} + \ln 2 = \frac{27}{4} + \ln 2$

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

3. 最終的な答え

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

(2) $\frac{e^4 + 4e^2 - 1}{2e^2}$

(3) $\frac{\pi}{2}$

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

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

空欄を埋めると以下のようになります。

(1) 1: 2, 2: 3

(2) 3: 1, 4: 4, 5: 4, 6: 1, 7: 2, 8: 2

(3) 9: 2

(4) 10: 2, 11: 7, 12: 2

(5) 13: 1, 14: 3