integral from 0 to L of sin^2(kx)

解

\int_{0}^{L} sin^{2} (k x) d x

\frac{1}{2} (L - \frac{sin ( 2 L k )}{2 k})

解答ステップ

\int_{0}^{L} sin^{2} (k x) d x

三角関数の公式を使用して書き換える

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

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

\int_{0}^{L} cos (2 k x) d x = \frac{sin ( 2 L k )}{2 k}

= \frac{1}{2} (L - \frac{sin ( 2 L k )}{2 k})