integral from 0 to l of sin^2((npix)/l)

解

\int_{0}^{l} sin^{2} (\frac{nπ x}{l}) d x

\frac{1}{2} (l - \frac{l sin ( 2 πn )}{2 πn})

解答ステップ

\int_{0}^{l} sin^{2} (\frac{nπ x}{l}) d x

三角関数の公式を使用して書き換える

= \int_{0}^{l} \frac{1}{2} (1 - cos (\frac{2 πn x}{l})) d x

定数を除く:

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

= \frac{1}{2} \cdot \int_{0}^{l} 1 - cos (\frac{2 πn x}{l}) 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

= \frac{1}{2} (\int_{0}^{l} 1 d x - \int_{0}^{l} cos (\frac{2 πn x}{l}) d x)

\int_{0}^{l} 1 d x = l

\int_{0}^{l} cos (\frac{2 πn x}{l}) d x = \frac{l sin ( 2 πn )}{2 πn}

= \frac{1}{2} (l - \frac{l sin ( 2 πn )}{2 πn})