次の定積分を求めよ。 (1) $\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \sin 2x \cos x dx$ (2) $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \cos 4x \sin x dx$ (3) $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos 2x \cos 3x dx$ (4) $\int_{\frac{\pi}{4}}^{\pi} \sin 4x \sin 2x dx$

1. 問題の内容

次の定積分を求めよ。

(1) $\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \sin 2x \cos x dx$

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

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

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

2. 解き方の手順

(1) $\sin 2x \cos x = 2 \sin x \cos x \cos x = 2 \sin x \cos^2 x$

$\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} 2 \sin x \cos^2 x dx$

$t = \cos x$ とすると $dt = - \sin x dx$

$x = \frac{\pi}{6}$ のとき $t = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$

$x = \frac{\pi}{2}$ のとき $t = \cos \frac{\pi}{2} = 0$

$\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} 2 \sin x \cos^2 x dx = \int_{\frac{\sqrt{3}}{2}}^{0} 2 t^2 (-dt) = -2 \int_{\frac{\sqrt{3}}{2}}^{0} t^2 dt = 2 \int_{0}^{\frac{\sqrt{3}}{2}} t^2 dt = 2 \left[ \frac{1}{3} t^3 \right]_{0}^{\frac{\sqrt{3}}{2}} = \frac{2}{3} \left( \frac{\sqrt{3}}{2} \right)^3 = \frac{2}{3} \frac{3\sqrt{3}}{8} = \frac{\sqrt{3}}{4}$

(2) $\cos 4x \sin x = \frac{1}{2} \left( \sin (x+4x) - \sin (4x-x) \right) = \frac{1}{2} \left( \sin 5x - \sin 3x \right)$

$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \cos 4x \sin x dx = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{1}{2} \left( \sin 5x - \sin 3x \right) dx = \frac{1}{2} \left[ - \frac{1}{5} \cos 5x + \frac{1}{3} \cos 3x \right]_{\frac{\pi}{4}}^{\frac{\pi}{3}} = \frac{1}{2} \left[ - \frac{1}{5} \cos \frac{5\pi}{3} + \frac{1}{3} \cos \pi - \left( - \frac{1}{5} \cos \frac{5\pi}{4} + \frac{1}{3} \cos \frac{3\pi}{4} \right) \right] = \frac{1}{2} \left[ - \frac{1}{5} \cdot \frac{1}{2} - \frac{1}{3} - \left( - \frac{1}{5} \cdot \left( - \frac{\sqrt{2}}{2} \right) + \frac{1}{3} \cdot \left( - \frac{\sqrt{2}}{2} \right) \right) \right] = \frac{1}{2} \left[ - \frac{1}{10} - \frac{1}{3} - \frac{\sqrt{2}}{10} + \frac{\sqrt{2}}{6} \right] = \frac{1}{2} \left[ - \frac{13}{30} + \frac{5\sqrt{2}-3\sqrt{2}}{30} \right] = \frac{1}{2} \left( - \frac{13}{30} + \frac{2\sqrt{2}}{30} \right) = \frac{-13+2\sqrt{2}}{60}$

(3) $\cos 2x \cos 3x = \frac{1}{2} \left( \cos (2x+3x) + \cos (3x-2x) \right) = \frac{1}{2} \left( \cos 5x + \cos x \right)$

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos 2x \cos 3x dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{2} \left( \cos 5x + \cos x \right) dx = \frac{1}{2} \left[ \frac{1}{5} \sin 5x + \sin x \right]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{1}{2} \left[ \frac{1}{5} \sin \frac{5\pi}{3} + \sin \frac{\pi}{3} - \left( \frac{1}{5} \sin \frac{5\pi}{6} + \sin \frac{\pi}{6} \right) \right] = \frac{1}{2} \left[ \frac{1}{5} \left( - \frac{\sqrt{3}}{2} \right) + \frac{\sqrt{3}}{2} - \frac{1}{5} \cdot \frac{1}{2} - \frac{1}{2} \right] = \frac{1}{2} \left[ - \frac{\sqrt{3}}{10} + \frac{\sqrt{3}}{2} - \frac{1}{10} - \frac{1}{2} \right] = \frac{1}{2} \left[ \frac{- \sqrt{3} + 5 \sqrt{3}}{10} - \frac{1+5}{10} \right] = \frac{1}{2} \left( \frac{4\sqrt{3}}{10} - \frac{6}{10} \right) = \frac{2\sqrt{3} - 3}{10}$

(4) $\sin 4x \sin 2x = \frac{1}{2} \left( \cos (4x-2x) - \cos (4x+2x) \right) = \frac{1}{2} \left( \cos 2x - \cos 6x \right)$

$\int_{\frac{\pi}{4}}^{\pi} \sin 4x \sin 2x dx = \int_{\frac{\pi}{4}}^{\pi} \frac{1}{2} \left( \cos 2x - \cos 6x \right) dx = \frac{1}{2} \left[ \frac{1}{2} \sin 2x - \frac{1}{6} \sin 6x \right]_{\frac{\pi}{4}}^{\pi} = \frac{1}{2} \left[ \frac{1}{2} \sin 2\pi - \frac{1}{6} \sin 6\pi - \left( \frac{1}{2} \sin \frac{\pi}{2} - \frac{1}{6} \sin \frac{3\pi}{2} \right) \right] = \frac{1}{2} \left[ 0 - 0 - \frac{1}{2} + \frac{1}{6} (-1) (-1) \right] = \frac{1}{2} \left( - \frac{1}{2} + \frac{1}{6} \right) = \frac{1}{2} \left( \frac{-3+1}{6} \right) = \frac{1}{2} \cdot \frac{-2}{6} = - \frac{1}{6}$

3. 最終的な答え

(1) $\frac{\sqrt{3}}{4}$

(2) $\frac{-13+2\sqrt{2}}{60}$

(3) $\frac{2\sqrt{3} - 3}{10}$

(4) $-\frac{1}{6}$