절댓값이 있는 x^2 cos x 의 정적분

$\displaystyle \int x ^{2} \cos x dx = x ^{2} \sin x - \int 2 x \sin x dx = x ^{2} \sin x + 2 x \cos x - 2 \sin x + C$이므로 $G ( x ) = x ^{2} \sin x + 2 x \cos x - 2 \sin x$라 하면 $G' ( x ) = x ^{2} \cos x$ $\displaystyle 0 \leq x \leq \frac{\pi}{2}$에서 $\cos x \geq 0$, $\displaystyle \frac{\pi}{2} \leq x \leq \pi$에서 $\cos x \leq 0$이므로 $\displaystyle \int _{0} ^{\pi} x ^{2} \left| \cos x \right| dx = \int _{0} ^{\frac{\pi}{2}} x ^{2} \cos x dx - \int _{\frac{\pi}{2}} ^{\pi} x ^{2} \cos x dx$ $\displaystyle = \left\{ G \left( \frac{\pi}{2} \right) - G ( 0 ) \right\} - \left\{ G ( \pi ) - G \left( \frac{\pi}{2} \right) \right\} = 2 G \left( \frac{\pi}{2} \right) - G ( 0 ) - G ( \pi )$ $\displaystyle G \left( \frac{\pi}{2} \right) = \frac{\pi ^{2}}{4} - 2 , G ( 0 ) = 0 , G ( \pi ) = - 2 \pi$이므로 $\displaystyle \int _{0} ^{\pi} x ^{2} \left| \cos x \right| dx = 2 \left( \frac{\pi ^{2}}{4} - 2 \right) - 0 + 2 \pi = \frac{\pi ^{2}}{2} + 2 \pi - 4$ 따라서 구하는 값은 $\displaystyle \frac{\pi ^{2}}{2} + 2 \pi - 4$이다. 답 ②