以下の3つの定積分を計算します。 (1) $\int_{0}^{\frac{\pi}{2}} \cos^2 x dx$ (2) $\int_{0}^{\frac{\pi}{4}} \sin 3x \sin x dx$ (3) $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{\tan^2 x}$

1. 問題の内容

以下の3つの定積分を計算します。

(1) $\int_{0}^{\frac{\pi}{2}} \cos^2 x dx$

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

(3) $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{\tan^2 x}$

2. 解き方の手順

(1) $\int_{0}^{\frac{\pi}{2}} \cos^2 x dx$ の計算

$\cos^2 x = \frac{1 + \cos 2x}{2}$ を利用します。

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

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

(2) $\int_{0}^{\frac{\pi}{4}} \sin 3x \sin x dx$ の計算

$\sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)]$ を利用します。

$\sin 3x \sin x = \frac{1}{2} [\cos(3x - x) - \cos(3x + x)] = \frac{1}{2} (\cos 2x - \cos 4x)$

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

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

$= \frac{1}{2} \left[ \frac{1}{2} (1) - \frac{1}{4} (0) - (0 - 0) \right] = \frac{1}{2} \left[ \frac{1}{2} \right] = \frac{1}{4}$

(3) $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{\tan^2 x}$ の計算

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

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

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{\tan^2 x} = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cot^2 x dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} (\frac{1}{\sin^2 x} - 1) dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} (\csc^2 x - 1) dx$

$= \left[ - \cot x - x \right]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \left( - \cot \frac{\pi}{3} - \frac{\pi}{3} \right) - \left( - \cot \frac{\pi}{6} - \frac{\pi}{6} \right)$

$= \left( - \frac{1}{\sqrt{3}} - \frac{\pi}{3} \right) - \left( - \sqrt{3} - \frac{\pi}{6} \right) = - \frac{1}{\sqrt{3}} - \frac{\pi}{3} + \sqrt{3} + \frac{\pi}{6} = \sqrt{3} - \frac{\sqrt{3}}{3} - \frac{\pi}{6} = \frac{2\sqrt{3}}{3} - \frac{\pi}{6}$

3. 最終的な答え

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

(2) $\frac{1}{4}$

(3) $\frac{2\sqrt{3}}{3} - \frac{\pi}{6}$