integral from 0 to l of cos^2((pix)/l)

解

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

\frac{l}{2}

解答ステップ

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

u置換積分法を適用する

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

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{l}{π} \cdot \int_{0}^{π} cos^{2} (u) d u

三角関数の公式を使用して書き換える

= \frac{l}{π} \cdot \int_{0}^{π} \frac{1 + cos ( 2 u )}{2} d u

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{l}{π} \cdot \frac{1}{2} \cdot \int_{0}^{π} 1 + cos (2 u) d u

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

= \frac{l}{π} \cdot \frac{1}{2} (\int_{0}^{π} 1 d u + \int_{0}^{π} cos (2 u) d u)

\int_{0}^{π} 1 d u = π

\int_{0}^{π} cos (2 u) d u = 0

= \frac{l}{π} \cdot \frac{1}{2} (π + 0)

簡素化

\frac{l}{π} \cdot \frac{1}{2} (π + 0) : \frac{l}{2}

= \frac{l}{2}