integral from 0 to pi of (sin(nx))^2

解

\int_{0}^{π} (sin (n x))^{2} d x

\frac{1}{2} (π - \frac{sin ( 2 πn )}{2 n})

解答ステップ

\int_{0}^{π} (sin (n x))^{2} d x

簡素化

= \int_{0}^{π} sin^{2} (n x) d x

三角関数の公式を使用して書き換える

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

\int_{0}^{π} 1 d x = π

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

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