integral from 0 to pi of sin^2(9x)

解

\int_{0}^{π} sin^{2} (9 x) d x

\frac{π}{2}

十進法表記

1.57079 \dots

解答ステップ

\int_{0}^{π} sin^{2} (9 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}{2} (1 - cos (18 x)) 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 (18 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 (18 x) d x)

\int_{0}^{π} 1 d x = π

\int_{0}^{π} cos (18 x) d x = 0

= \frac{1}{2} (π - 0)

簡素化

\frac{1}{2} (π - 0) : \frac{π}{2}

= \frac{π}{2}