$\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \cos{t} \cos{3t} dt$ を計算する問題です。

1. 問題の内容

$\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \cos{t} \cos{3t} dt$ を計算する問題です。

2. 解き方の手順

積和の公式 $\cos{A}\cos{B} = \frac{1}{2}(\cos{(A+B)} + \cos{(A-B)})$ を用いて積分を計算します。

$\cos{t}\cos{3t} = \frac{1}{2}(\cos{(t+3t)} + \cos{(t-3t)}) = \frac{1}{2}(\cos{4t} + \cos{(-2t)}) = \frac{1}{2}(\cos{4t} + \cos{2t})$

したがって、

$\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \cos{t} \cos{3t} dt = \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \frac{1}{2}(\cos{4t} + \cos{2t}) dt = \frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} (\cos{4t} + \cos{2t}) dt$

$\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} (\cos{4t} + \cos{2t}) dt = \frac{1}{2}[\frac{1}{4}\sin{4t} + \frac{1}{2}\sin{2t}]_{\frac{\pi}{6}}^{\frac{5\pi}{6}}$

$= \frac{1}{2}[\frac{1}{4}\sin{\frac{20\pi}{6}} + \frac{1}{2}\sin{\frac{10\pi}{6}} - (\frac{1}{4}\sin{\frac{4\pi}{6}} + \frac{1}{2}\sin{\frac{2\pi}{6}})]$

$= \frac{1}{2}[\frac{1}{4}\sin{\frac{10\pi}{3}} + \frac{1}{2}\sin{\frac{5\pi}{3}} - (\frac{1}{4}\sin{\frac{2\pi}{3}} + \frac{1}{2}\sin{\frac{\pi}{3}})]$

$= \frac{1}{2}[\frac{1}{4}\sin{(\frac{4\pi}{3})} + \frac{1}{2}\sin{(\frac{5\pi}{3})} - (\frac{1}{4}\sin{(\frac{2\pi}{3})} + \frac{1}{2}\sin{(\frac{\pi}{3})})]$

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

$= \frac{1}{2}[-\frac{\sqrt{3}}{8} - \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{8} - \frac{\sqrt{3}}{4}]$

$= \frac{1}{2}[-\frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{2}]$

$= \frac{1}{2}[-\frac{3\sqrt{3}}{4}] = -\frac{3\sqrt{3}}{8}$

3. 最終的な答え

$-\frac{3\sqrt{3}}{8}$