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

はい、承知いたしました。画像にある定積分の問題を解きます。

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

被積分関数を展開します。

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

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

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

したがって、

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

(2) $\int_{0}^{1} (e^x + \frac{1}{e^x})^2 dx = \int_{0}^{1} (e^x + e^{-x})^2 dx = \int_{0}^{1} (e^{2x} + 2 + e^{-2x}) dx$

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

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

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

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

(4) $\int_{1}^{2} \frac{(x^2+1)^2}{x} dx = \int_{1}^{2} \frac{x^4 + 2x^2 + 1}{x} dx = \int_{1}^{2} (x^3 + 2x + \frac{1}{x}) dx$

$= \left[\frac{1}{4} x^4 + x^2 + \ln |x| \right]_{1}^{2} = \left(\frac{1}{4} (16) + 4 + \ln 2 \right) - \left(\frac{1}{4} + 1 + 0\right) = (4 + 4 + \ln 2) - (\frac{5}{4}) = 8 + \ln 2 - \frac{5}{4} = \frac{27}{4} + \ln 2$

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

3. 最終的な答え

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

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

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

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

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