선분 길이의 정적분

[출제의도] 삼각함수의 적분법과 부분적분법을 이용하여 정적분의 값을 구할 수 있는가?

$\displaystyle \overline{QB} = 2 + 2 \cos \theta = 2 \left( 1 + \cos \theta \right)$이고 직각삼각형 $\mathrm{QRB}$에서 $\displaystyle \angle {\mathrm{Q}} \mathit{\mathrm{B}} \mathit{\mathrm{R}} \mathit{=} \frac{\pi}{2} - \theta$이므로 $\displaystyle \overline{{\mathrm{B}} \mathit{\mathrm{R}}} = \overline{{\mathrm{Q}} \mathit{\mathrm{B}}} \times \cos \left( \frac{\pi}{2} - \theta \right) = 2 \left( 1 + \cos \theta \right) \sin \theta$ 삼각형 $\mathrm{\mathrm{A}} \mathit{\mathrm{P}} \mathit{\mathrm{B}}$의 외접원의 반지름의 길이가 $2$이므로 사인법칙에 의해 $\displaystyle \frac{\overline{{\mathrm{B}} \mathit{\mathrm{P}}}}{\sin \theta} = 2 \times 2$이므로 $\displaystyle \overline{{\mathrm{B}} \mathit{\mathrm{P}}} = 4 \sin \theta$ 따라서 $f \left( \theta \right)$$\displaystyle = \overline{{\mathrm{B}} \mathit{\mathrm{P}}} - \overline{{\mathrm{B}} \mathit{\mathrm{R}}}$ $= 4 \sin \theta - 2 \left( 1 + \cos \theta \right) \sin \theta$ $= 2 \sin \theta - 2 \cos \theta \sin \theta$이므로 $\displaystyle \int _{\frac{\pi}{6}} ^{\frac{\pi}{3}} {f \left( \theta \right) d \theta}$$\displaystyle = \int _{\frac{\pi}{6}} ^{\frac{\pi}{3}} {\left( 2 - 2 \cos \theta \right) \sin \theta d \theta}$ $\displaystyle = \left[ \begin{array}{l} - 2 \cos \theta - \sin ^{2} \theta \begin{matrix} \\ \end{matrix} \end{array} \right] _{\frac{\pi}{6}} ^{\frac{\pi}{3}}$ $\displaystyle = \left( - 2 \cos \frac{\pi}{3} - \sin ^{2} \frac{\pi}{3} \right) - \left( - 2 \cos \frac{\pi}{6} - \sin ^{2} \frac{\pi}{6} \right)$ $\displaystyle = \left( - 1 - \frac{3}{4} \right) - \left( - \sqrt{3} - \frac{1}{4} \right)$ $\displaystyle = \frac{2 \sqrt{3} - 3}{2}$