integral from 0 to pi of (cos(x)+4)^2

解

\int_{0}^{π} (cos (x) + 4)^{2} d x

16 π + \frac{π}{2}

十進法表記

51.83627 \dots

解答ステップ

\int_{0}^{π} (cos (x) + 4)^{2} d x

拡張

(cos (x) + 4)^{2} : cos^{2} (x) + 8 cos (x) + 16

= \int_{0}^{π} cos^{2} (x) + 8 cos (x) + 16 d x

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{π} cos^{2} (x) d x + \int_{0}^{π} 8 cos (x) d x + \int_{0}^{π} 16 d x

\int_{0}^{π} cos^{2} (x) d x = \frac{π}{2}

\int_{0}^{π} 8 cos (x) d x = 0

\int_{0}^{π} 16 d x = 16 π

= \frac{π}{2} + 0 + 16 π

簡素化

= 16 π + \frac{π}{2}