치환적분

[출제의도] 치환적분법을 활용하여 삼각함수의 정적분의 값을 구한다. $\displaystyle \mathrm{\overline{OP}} \mathit{=} r$라 하면 $\displaystyle \overline{\mathrm{O}} H = \mathit{r} \cos \theta$, $\displaystyle \overline{\mathrm{P}} H = \mathit{r} \sin \theta$이므로 $\displaystyle f ( \theta ) = \frac{{\overline{\mathrm{OH}}}}{{\overline{\mathrm{PH}}}}$$\displaystyle = \frac{r \cos \theta}{r \sin \theta} = \frac{\cos \theta}{\sin \theta}$ $\displaystyle \int _{\frac{\pi}{6}} ^{\frac{\pi}{3}} {f ( \theta ) d \theta} = \int _{\frac{\pi}{6}} ^{\frac{\pi}{3}} {\frac{\cos \theta}{\sin \theta}} d \theta$에서 $\sin \theta = t$로 놓으면 $\displaystyle \frac{dt}{d \theta} = \cos \theta$ $\displaystyle \theta = \frac{\pi}{6}$일 때 $\displaystyle t = \frac{1}{2}$, $\displaystyle \theta = \frac{\pi}{3}$일 때 $\displaystyle t = \frac{\sqrt{3}}{2}$이므로 $\displaystyle \int _{\frac{\pi}{6}} ^{\frac{\pi}{3}} {\frac{\cos \theta}{\sin \theta} d \theta} = \int _{\frac{1}{2}} ^{\frac{\sqrt{3}}{2}} {\frac{1}{t} dt}$ $\displaystyle = \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} \ln t \begin{matrix} \\ \end{matrix} \end{array} \right] _{\frac{1}{2}} ^{\frac{\sqrt{3}}{2}}$ $\displaystyle = \ln \frac{\sqrt{3}}{2} - \ln \frac{1}{2}$ $\displaystyle = \ln \sqrt{3} = \frac{1}{2} \ln 3$