integral from 0 to pi/(12) of sin^2(3x)

解

\int_{0}^{\frac{π}{12}} sin^{2} (3 x) d x

\frac{π - 2}{24}

十進法表記

0.04756 \dots

解答ステップ

\int_{0}^{\frac{π}{12}} sin^{2} (3 x) d x

三角関数の公式を使用して書き換える

= \int_{0}^{\frac{π}{12}} \frac{1}{2} (1 - cos (6 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}^{\frac{π}{12}} 1 - cos (6 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}^{\frac{π}{12}} 1 d x - \int_{0}^{\frac{π}{12}} cos (6 x) d x)

\int_{0}^{\frac{π}{12}} 1 d x = \frac{π}{12}

\int_{0}^{\frac{π}{12}} cos (6 x) d x = \frac{1}{6}

= \frac{1}{2} (\frac{π}{12} - \frac{1}{6})

簡素化

\frac{1}{2} (\frac{π}{12} - \frac{1}{6}) : \frac{π - 2}{24}

= \frac{π - 2}{24}