次の定積分を計算する問題です。 (1) $\int_{1}^{3} \frac{(x^2-1)^2}{x^4} dx$ (2) $\int_{0}^{1} (x+1-\sqrt{x})^2 dx$ (3) $\int_{0}^{\pi} (2\sin x + \cos x)^2 dx$ (4) $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (3-4\sin^2 x) dx$ (5) $\int_{0}^{\log 7} \frac{e^x}{1+e^x} dx$

1. 問題の内容

次の定積分を計算する問題です。

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

(2) $\int_{0}^{1} (x+1-\sqrt{x})^2 dx$

(3) $\int_{0}^{\pi} (2\sin x + \cos x)^2 dx$

(4) $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (3-4\sin^2 x) dx$

(5) $\int_{0}^{\log 7} \frac{e^x}{1+e^x} dx$

2. 解き方の手順

(1)

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

$= [x + \frac{2}{x} - \frac{1}{3x^3}]_{1}^{3} = (3 + \frac{2}{3} - \frac{1}{81}) - (1+2-\frac{1}{3}) = 3 + \frac{2}{3} - \frac{1}{81} - 3 + \frac{1}{3} = 1 - \frac{1}{81} = \frac{80}{81}$

(2)

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

$= \int_{0}^{1} (x^2 + 3x + 1 - 2x^{3/2} - 2x^{1/2}) dx = [\frac{x^3}{3} + \frac{3x^2}{2} + x - \frac{4}{5}x^{5/2} - \frac{4}{3}x^{3/2}]_{0}^{1}$

$= \frac{1}{3} + \frac{3}{2} + 1 - \frac{4}{5} - \frac{4}{3} = \frac{10 + 45 + 30 - 24 - 40}{30} = \frac{21}{30} = \frac{7}{10}$

(3)

$\int_{0}^{\pi} (2\sin x + \cos x)^2 dx = \int_{0}^{\pi} (4\sin^2 x + 4\sin x \cos x + \cos^2 x) dx = \int_{0}^{\pi} (3\sin^2 x + \sin^2 x + \cos^2 x + 4\sin x \cos x) dx$

$= \int_{0}^{\pi} (3\sin^2 x + 1 + 2\sin 2x) dx = \int_{0}^{\pi} (3 \cdot \frac{1-\cos 2x}{2} + 1 + 2\sin 2x) dx$

$= \int_{0}^{\pi} (\frac{3}{2} - \frac{3}{2}\cos 2x + 1 + 2\sin 2x) dx = [\frac{5}{2}x - \frac{3}{4}\sin 2x - \cos 2x]_{0}^{\pi} = \frac{5}{2}\pi - \frac{3}{4}\sin 2\pi - \cos 2\pi - (0 - \frac{3}{4}\sin 0 - \cos 0) = \frac{5}{2}\pi - 0 - 1 + 1 = \frac{5}{2}\pi$

(4)

$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (3-4\sin^2 x) dx = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (3-4\cdot \frac{1-\cos 2x}{2}) dx = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (3-2+2\cos 2x) dx = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (1+2\cos 2x) dx$

$= [x + \sin 2x]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = (\frac{\pi}{2} + \sin \pi) - (\frac{\pi}{4} + \sin \frac{\pi}{2}) = \frac{\pi}{2} + 0 - \frac{\pi}{4} - 1 = \frac{\pi}{4} - 1$

(5)

$\int_{0}^{\log 7} \frac{e^x}{1+e^x} dx = [\log(1+e^x)]_{0}^{\log 7} = \log(1+e^{\log 7}) - \log(1+e^0) = \log(1+7) - \log(1+1) = \log 8 - \log 2 = \log \frac{8}{2} = \log 4$

3. 最終的な答え

(1) $\frac{80}{81}$

(2) $\frac{7}{10}$

(3) $\frac{5}{2}\pi$

(4) $\frac{\pi}{4} - 1$

(5) $\log 4$