대칭성 이용 정적분

[출제의도] 치환적분법을 이해하고 정적분 문제를 해결한다. $\displaystyle f ( x ) = \frac{e ^{\cos x}}{1 + e ^{\cos x}}$에서 $f ( \pi - x )$$\displaystyle = \frac{e ^{\cos ( \pi - x )}}{1 + e ^{\cos ( \pi - x )}}$$\displaystyle = \frac{e ^{- \cos x}}{1 + e ^{- \cos x}}$$\displaystyle = \frac{e ^{- \cos x} \times e ^{\cos x}}{\left( 1 + e ^{- \cos x} \right) \times e ^{\cos x}}$ $\displaystyle = \frac{1}{e ^{\cos x} + 1}$ 그러므로 $a = f ( \pi - x ) + f ( x )$ $\displaystyle = \frac{1}{e ^{\cos x} + 1} + \frac{e ^{\cos x}}{1 + e ^{\cos x}}$ $\displaystyle = \frac{1 + e ^{\cos x}}{1 + e ^{\cos x}} = 1$ $\displaystyle b = \int _{0} ^{\pi} {} f ( x ) dx$ $\displaystyle = \int _{0} ^{\pi} {} \left\{ 1 - f ( \pi - x ) \right\} dx$ $\displaystyle = \int _{0} ^{\pi} {1 dx - \int _{0} ^{\pi} {f ( \pi - x ) dx}}$ $\pi - x = t$로 놓으면 $x = \pi - t$이므로 $\displaystyle \frac{dx}{dt} = - 1$ $x = 0$일 때, $t = {\pi}$이고 $x = \pi$일 때, $t = 0$이므로 $\displaystyle b = \pi + \int _{\pi} ^{0} {f ( t ) dt}$ $\displaystyle = \pi - \int _{0} ^{\pi} {f ( t ) dt}$$= \pi - b$ $b = \pi - b$이므로 $\displaystyle b = \frac{\pi}{2}$ 따라서 $\displaystyle a + \frac{100}{\pi} b = 1 + \frac{100}{\pi} \times \frac{\pi}{2}$ $= 1 + 50 = 51$ [다른풀이] $\displaystyle b = \int _{0} ^{\frac{\pi}{2}} {} f ( x ) dx + \int _{\frac{\pi}{2}} ^{\pi} {} f ( x ) dx$ $\displaystyle = \int _{0} ^{\frac{\pi}{2}} {} f ( x ) dx + \int _{0} ^{\frac{\pi}{2}} {} f ( \pi - x ) dx$ $\displaystyle = \int _{0} ^{\frac{\pi}{2}} {\left\{ f ( x ) + f ( \pi - x ) \right\} dx}$ $\displaystyle = \int _{0} ^{\frac{\pi}{2}} {1 dx}$$\displaystyle = \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} x \end{array} \right] _{0} ^{\frac{\pi}{2}}$$\displaystyle = \frac{\pi}{2} - 0$$\displaystyle = \frac{\pi}{2}$